Attached please find:
(a) 1) the questions for the final exam
(b) 2) the Simulink file for Problem 2
(c) 3) the FFT example for Problem 2
Please follow the exam instruction for submission.
metadata/coreProperties.xml
model 2016-03-01T15:19:44Z Tqvo hatle 2020-05-06T21:30:42Z 1.98 R2019a
metadata/mwcoreProperties.xml
application/vnd.mathworks.simulink.model Simulink Model R2019a
metadata/mwcorePropertiesExtension.xml
9.6.0.1063848
metadata/thumbnail
simulink/ScheduleCore.xml
HighNumberLast
false
Default
-2147483648
true
false
1
Cont
40
Cont
255
true
0
0
FiM
-2139062017
true
1
0
1
D1
-436207361
true
1.9999999999999999e-06
2
D2
13697279
true
5.1440000000000002e-06
3
simulink/ScheduleEditor.xml
0,0
0,0,0,0
Cont
base
0,0
Cont
graph.Graph
Default
graph.Graph
40
#000000
true
0
Cont
#808080
true
[0, 1]
FiM
#e60000
true
2e-06
D1
#00d100
true
5.144e-06
D2
Cont
base
Cont
Default
simulink/bddefaults.xml
landscape
auto
usletter
inches
[0.500000, 0.500000, 0.500000, 0.500000]
1
off
off
200
white
100
off
off
black
white
off
normal
Helvetica
10
normal
normal
on
on
0
off
center
middle
black
white
off
Helvetica
10
normal
normal
model
off
note_annotation
off
off
off
Helvetica
9
normal
normal
1
on
Sample based
[]
[]
Inherit: Inherit from ‘Constant value’
off
inf
inf
off
4
none
off
1
A
Tag
local
1
Element-wise(K.*u)
[]
[]
Inherit: Same as input
[]
[]
Inherit: Same as input
off
Floor
on
-1
A
Tag
local
1
off
[]
[]
Inherit: auto
off
off
inherit
-1
Inherit
-1
auto
auto
off
off
on
AND
2
rectangular
on
Inherit: Logical (see Configuration Parameters: Optimization)
-1
4
none
off
BusObject
off
1
[]
[]
Inherit: auto
off
off
inherit
-1
Inherit
-1
auto
auto
off
Dialog
held
[]
off
off
0
off
1
2
Element-wise(.*)
All dimensions
1
on
[]
[]
Inherit: Same as first input
off
Zero
on
-1
>=
on
Inherit: Logical (see Configuration Parameters: Optimization)
on
-1
Nearest
Simulink.scopes.TimeScopeBlockCfg
FromPortIcon
ReadWrite
All
off
off
off
Sample time
-1
Auto
Auto
Auto
void_void
off
Inherit from model
Inherit from model
Inherit from model
Inherit from model
Inherit from model
off
UseLocalSettings
AllNumericTypes
UseLocalSettings
off
off
NONE
Expression
off
off
off
off
on
off
off
off
0
off
off
rectangular
++
All dimensions
1
on
Inherit: Inherit via internal rule
[]
[]
Inherit: Same as first input
off
Floor
on
-1
u2 >= Threshold
0
on
[]
[]
Inherit: Inherit via internal rule
off
Floor
on
on
-1
off
sin
None
11
auto
-1
simulink/blockdiagram.xml
windows-1252
on
Normal
0.035000
on
off
UseLocalSettings
AllNumericTypes
UseLocalSettings
Overwrite
Run 1
120
win64
1
[-7.0, -7.0, 1295.0, 704.0]
0
Left
50
50
9
Unset
1
1
SimulinkTopLevel
0
[1864.0, 809.0]
0.60000000000000009
[712.39612691660568, 542.65927267813152]
GLUE2:PropertyInspector
Property Inspector
0
0
Right
426
320
Unset
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
on
4175268304
2197470448
Power_System_Blocks
UpdateHistoryNever
%
%
510676228
1.%
off
off
disabled
off
off
off
off
AliasTypeOnly
on
on
off
off
off
on
off
off
on
on
on
off
off
off
off
on
on
on
off
off
off
off
on
on
off
off
off
on
normal
5
1
10
10
0
off
1
none
off
MATLABWorkspace
accel.tlc
accel_default_tmf
make_rtw
off
$bdroot
0U
$bdroot
[]
off
on
manual
normal
1
any
1000
auto
0
0
rising
0
off
off
off
off
off
on
off
on
on
off
off
on
Ensure deterministic transfer (maximum delay)
Ensure data integrity only
Ensure deterministic transfer (minimum delay)
None
off
[-7, -7, 1288, 697]
on
off
Deduce
60
simulink-default.rpt
836
Simulink
[0, 0, 0, 0, 0, 1, 1]
[1689, 965, 1741, 1020]
579
270
on
alternate
\n
1.2814
powerlib/Elements/Parallel RLC Load
Parallel RLC Load
Simscape Electrical
PS
a
__new0
240
60
1e3
0
0
off
0
off
0
None
constant Z
[0, 0, 0, 0, 0, 1, 1]
[1749, 965, 1801, 1020]
543
270
on
alternate
\n
1.2814
powerlib/Elements/Parallel RLC Load
Parallel RLC Load
Simscape Electrical
PS
a
__new0
240
60
1e3
0
0
off
0
off
0
None
constant Z
[0, 0, 0, 0, 0, 1, 1]
[1799, 965, 1851, 1020]
580
270
on
alternate
\n
1.2814
powerlib/Elements/Parallel RLC Load
Parallel RLC Load
Simscape Electrical
PS
a
__new0
240
60
1e3
0
0
off
0
off
0
None
constant Z
[1, 0, 0, 0, 0, 3, 2]
[1305, 951, 1375, 1054]
325
on
[0.435294, 0.768627, 0.921569]
off
[-8, -8, 1928, 1058]
off
Deduce
Simulink
[155, 183, 185, 197]
-1
Port number
[1, 6]
[220, 50, 225, 325]
-2
off
6
bar
[760, 53, 795, 67]
-3
off
tagdialog Close
s5
[735, 208, 775, 222]
-4
off
tagdialog Close
s6
[485, 53, 525, 67]
-5
off
tagdialog Close
s1
[500, 208, 535, 222]
-6
off
tagdialog Close
s2
[625, 53, 660, 67]
-7
off
tagdialog Close
s3
[625, 207, 665, 223]
-8
off
tagdialog Close
s4
[260, 65, 300, 85]
-9
s1
[260, 110, 300, 130]
-10
s2
[260, 155, 300, 175]
-11
s3
[260, 200, 300, 220]
-12
s4
[260, 245, 300, 265]
-13
s5
[260, 290, 300, 310]
-14
s6
[1, 0, 0, 0, 0, 1, 1]
[533, 80, 577, 135]
-15
270
on
alternate
1.2814
powerlib/Power
Electronics/Mosfet
Mosfet
Simscape Electrical
PS
off
0.1
0
0.01
0
0
1e5
inf
off
[1, 0, 0, 0, 0, 1, 1]
[523, 240, 567, 295]
-16
270
on
alternate
1.2814
powerlib/Power
Electronics/Mosfet
Mosfet
Simscape Electrical
PS
off
0.1
0
0.01
0
0
1e5
inf
off
[1, 0, 0, 0, 0, 1, 1]
[673, 75, 717, 130]
-17
270
on
alternate
1.2814
powerlib/Power
Electronics/Mosfet
Mosfet
Simscape Electrical
PS
off
0.1
0
0.01
0
0
1e5
inf
off
[1, 0, 0, 0, 0, 1, 1]
[663, 240, 707, 295]
-18
270
on
alternate
1.2814
powerlib/Power
Electronics/Mosfet
Mosfet
Simscape Electrical
PS
off
0.1
0
0.01
0
0
1e5
inf
off
[1, 0, 0, 0, 0, 1, 1]
[793, 75, 837, 130]
-19
270
on
alternate
1.2814
powerlib/Power
Electronics/Mosfet
Mosfet
Simscape Electrical
PS
off
0.1
0
0.01
0
0
1e5
inf
off
[1, 0, 0, 0, 0, 1, 1]
[783, 240, 827, 295]
-20
270
on
alternate
1.2814
powerlib/Power
Electronics/Mosfet
Mosfet
Simscape Electrical
PS
off
0.1
0
0.01
0
0
1e5
inf
off
[935, 33, 965, 47]
-21
on
Right
[395, 123, 425, 137]
-22
2
Left
[395, 168, 425, 182]
-23
3
Left
[395, 218, 425, 232]
-24
4
Left
[995, 305, 1010, 335]
-25
on
5
Right
1
27#lconn:1
[0, -20]
[-120, 0]
[-140, 0]
23#lconn:1
25#lconn:1
29#rconn:1
[-95, 0]
6
28#rconn:1
[0, 10]
[685, 320; 120, 0]
24#rconn:1
[0, 10; 140, 0]
26#rconn:1
[0, 10]
33#rconn:1
[-175, 0]
11
[555, 150; 0, 75]
24#lconn:1
23#rconn:1
[0, 0]
30#rconn:1
[39, 0; 0, 20; 76, 0]
14
25#rconn:1
[0, 30]
26#lconn:1
31#rconn:1
[255, 0]
17
[815, 195; 0, -50]
27#rconn:1
28#lconn:1
[0, -30]
32#rconn:1
[23, 0; 0, -30; 352, 0]
20
13#out:1
[0, 5]
23#in:1
21
14#out:1
[-5, 0]
24#in:1
22
15#out:1
25#in:1
23
16#out:1
[5, 0]
26#in:1
24
11#out:1
27#in:1
25
12#out:1
[15, 0]
28#in:1
26
10#out:1
17#in:1
27
10#out:2
18#in:1
28
10#out:3
19#in:1
29
10#out:4
20#in:1
30
10#out:5
21#in:1
31
10#out:6
22#in:1
32
9#out:1
10#in:1
[2]
[1990, 951, 2030, 1029]
329
lightBlue
Simulink.scopes.TimeScopeBlockCfg(‘CurrentConfiguration’, extmgr.ConfigurationSet(extmgr.Configuration(‘Core’,’General UI’,true),extmgr.Configuration(‘Core’,’Source UI’,true),extmgr.Configuration(‘Sources’,’WiredSimulink’,true,’DataLoggingVariableName’,’Vabc_B1′,’DataLoggingSaveFormat’,’StructureWithTime’,’DataLoggingMaxPoints’,’1000000′,’DataLoggingDecimation’,’1′,’DataLoggingDecimateData’,true),extmgr.Configuration(‘Visuals’,’Time Domain’,true,’SerializedDisplays’,{struct(‘MinYLimReal’,’-1000′,’MaxYLimReal’,’800′,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’V – 1 ‘,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,3,’LineNames’,{{‘Vabc_B1:1′,’Vabc_B1:2′,’Vabc_B1:3′}},’ShowContent’,true,’Placement’,1),struct(‘MinYLimReal’,’-5′,’MaxYLimReal’,’5′,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’LegendVisibility’,’off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’I – 1 ‘,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,3,’LineNames’,{{‘Iabc_B1:1′,’Iabc_B1:2′,’Iabc_B1:3′}},’ShowContent’,true,’Placement’,2)},’DisplayPropertyDefaults’,struct(‘MinYLimReal’,’-1000′,’MaxYLimReal’,’800′,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’V – 1 ‘,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,0,’LineNames’,{{[]}},’ShowContent’,true,’Placement’,1),’DisplayLayoutDimensions’,[2 1]),extmgr.Configuration(‘Tools’,’Plot Navigation’,true,’OnceAtStop’,false),extmgr.Configuration(‘Tools’,’Measurements’,true,’Version’,’2019a’)),’Version’,’2019a’,’Location’,[154.333333333333 80.3333333333334 1098.33333333333 730.333333333333])
2
off
[0, 0, 0, 0, 0, 3, 3]
[1520, 979, 1555, 1051]
361
1.2814
powerlib/Measurements/Three-Phase
V-I Measurement
Three-Phase VI Measurement
Simscape Electrical
PS
off
phase-to-ground
on
Vabc_B1
off
off
yes
on
Iabc_B1
off
1e3
380
Magnitude
off
0
[0, 1, 0, 0, 0, 2]
[1095, 1009, 1165, 1076]
534
on
1.1623
electricdrivelib/Extra Sources/Battery
Battery
Simscape Electrical
PS
off
Lithium-Ion
off
no
off
400
251.6
100
30
on
251.6
300
465.5949
109.3913
0.015898
227.5339
[432.1543 12.36122]
[32.5 32.5 32.5]
Time
20
20
-30
4.8
7.1
5.655
[6.58 1]
0.6
2000
0
0
1e6
25
5.4*0.9
0.013333*1.2
[2.3478, 3]
[2.3478, 10]
1500
10500
1000
1400
45
950
[1, 3]
[865, 947, 965, 1013]
316
off
[-8, -8, 1928, 1058]
off
Deduce
Simulink
[50, 83, 80, 97]
6
on
Port number
[1, 3]
[210, 22, 215, 118]
3
off
Voltage (V),SOC (%),Current (A)
off
1
2
3
[410, 73, 465, 107]
5
1/100
Inherit: Inherit via internal rule
Inherit: Inherit via internal rule
off
1
SOC
[500, 33, 530, 47]
7
Port number
[545, 83, 575, 97]
8
2
Port number
[500, 128, 530, 142]
9
3
Port number
1
[-1, 1]
65#out:1
67#in:1
SOC
2
[0, 0; -1, 1]
66#out:1
68#in:1
3
[2, 1]
65#out:2
[175, 0]
66#in:1
4
64#out:1
[-9, 0; 0, -20]
65#in:1
5
[0, 0]
65#out:3
[70, 0; 0, 36; 195, 0]
69#in:1
[1, 4]
[1200, 797, 1205, 873]
323
black
alternate
off
bar
[1, 2]
[2100, 801, 2105, 839]
561
off
2
bar
[2, 1]
[2175, 737, 2205, 768]
558
*/
off
Inherit: Inherit via internal rule
Floor
off
[1, 1]
[2225, 740, 2255, 770]
557
atan
[1875, 998, 1940, 1022]
332
Iabc_B1
global
1
Iabc_B1
signal
[1875, 961, 1930, 979]
330
off
Vabc_B1
global
1
Vabc_B1
signal
[2065, 1003, 2130, 1027]
368
Iabc_B3
global
1
Iabc_Load
signal
[2070, 966, 2125, 984]
367
off
Vabc_B3
global
1
Vabc_Load
signal
[1, 1]
[2275, 740, 2305, 770]
559
cos
[1245, 794, 1290, 816]
322
off
G1_2
[1245, 814, 1290, 836]
321
off
G2_2
[1245, 834, 1290, 856]
320
off
G3_2
[1245, 854, 1290, 876]
319
off
G4_2
[0, 0, 0, 0, 0, 1]
[1859, 880, 1881, 905]
553
270
on
off
1.2814
powerlib/Elements/Ground
Ground
Simscape Electrical
PS
a
[0, 0, 0, 0, 0, 1]
[1704, 1040, 1726, 1065]
539
270
on
off
1.2814
powerlib/Elements/Ground
Ground
Simscape Electrical
PS
a
[0, 0, 0, 0, 0, 1]
[1764, 1040, 1786, 1065]
540
270
on
off
1.2814
powerlib/Elements/Ground
Ground
Simscape Electrical
PS
a
[0, 0, 0, 0, 0, 1]
[1814, 1040, 1836, 1065]
541
270
on
off
1.2814
powerlib/Elements/Ground
Ground
Simscape Electrical
PS
a
PoWeRsYsTeMmEaSuReMeNt
[0, 1, 0, 0, 0, 1, 1]
[1705, 790, 1730, 825]
551
\n
1.2814
powerlib/Measurements/Current Measurement
Current Measurement
Simscape Electrical
PS
off
off
Magnitude
0
[1]
[1880, 741, 1925, 799]
555
green
alternate
Simulink.scopes.TimeScopeBlockCfg(‘CurrentConfiguration’, extmgr.ConfigurationSet(extmgr.Configuration(‘Core’,’General UI’,true,’FigureColor’,[0.501960784313725 0.501960784313725 0.501960784313725]),extmgr.Configuration(‘Core’,’Source UI’,true),extmgr.Configuration(‘Sources’,’WiredSimulink’,true,’DataLoggingVariableName’,’Ic_Load’,’DataLoggingSaveFormat’,’Array’,’DataLoggingDecimation’,’1′,’DataLoggingDecimateData’,true),extmgr.Configuration(‘Visuals’,’Time Domain’,true,’TimeAxisLabels’,’All’,’SerializedDisplays’,{struct(‘MinYLimReal’,’-70′,’MaxYLimReal’,’70’,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[1 1 1],’AxesTickColor’,[0 0 0],’ColorOrder’,[0 0.447058823529412 0.741176470588235;0.850980392156863 0.325490196078431 0.0980392156862745;0.929411764705882 0.694117647058824 0.125490196078431;0.494117647058824 0.184313725490196 0.556862745098039;0.466666666666667 0.674509803921569 0.188235294117647;0.301960784313725 0.745098039215686 0.933333333333333;0.635294117647059 0.0784313725490196 0.184313725490196],’Title’,’Load_Current’,’LinePropertiesCache’,{{struct(‘Color’,[1 0 0],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[1 0 1],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[0 1 1],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[1 0 0],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[0 1 0],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[0 0 1],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’)}},’UserDefinedChannelNames’,{{}},’NumLines’,1,’LineNames’,{{‘IC’}},’ShowContent’,true,’Placement’,1)},’DisplayPropertyDefaults’,struct(‘MinYLimReal’,’-70′,’MaxYLimReal’,’70’,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[1 1 1],’AxesTickColor’,[0 0 0],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’Load_Current’,’LinePropertiesCache’,{{struct(‘Color’,[1 0 0],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[1 0 1],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[0 1 1],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[1 0 0],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[0 1 0],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’),struct(‘Color’,[0 0 1],’LineStyle’,’-‘,’LineWidth’,0.5,’Marker’,’none’,’Visible’,’on’)}},’UserDefinedChannelNames’,{{}},’NumLines’,0,’LineNames’,{{[]}},’ShowContent’,true,’Placement’,1)),extmgr.Configuration(‘Tools’,’Plot Navigation’,true,’OnceAtStop’,false),extmgr.Configuration(‘Tools’,’Measurements’,true,’Version’,’2019a’)),’Version’,’2019a’,’Location’,[468.333333333333 116.333333333333 1163 585.666666666667])
1
off
[0, 0, 0, 0, 0, 1, 1]
[1200, 1013, 1240, 1037]
318
alternate
\n
1.2814
powerlib/Elements/Series RLC Branch
Series RLC Branch
Simscape Electrical
PS
__new0
__new0
L
1.0
200e-6
off
0
1.0
off
0
None
[2]
[2185, 955, 2225, 1035]
366
lightBlue
Simulink.scopes.TimeScopeBlockCfg(‘CurrentConfiguration’, extmgr.ConfigurationSet(extmgr.Configuration(‘Core’,’General UI’,true),extmgr.Configuration(‘Core’,’Source UI’,true),extmgr.Configuration(‘Sources’,’WiredSimulink’,true,’DataLoggingVariableName’,’Vabc_Load’,’DataLoggingSaveFormat’,’StructureWithTime’,’DataLoggingMaxPoints’,’1000000′,’DataLoggingDecimation’,’1′,’DataLoggingDecimateData’,true),extmgr.Configuration(‘Visuals’,’Time Domain’,true,’SerializedDisplays’,{struct(‘MinYLimReal’,’-300′,’MaxYLimReal’,’300′,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’V – 3 ‘,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,3,’LineNames’,{{‘Vabc_Load:1′,’Vabc_Load:2′,’Vabc_Load:3′}},’ShowContent’,true,’Placement’,1),struct(‘MinYLimReal’,’-60′,’MaxYLimReal’,’60’,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’LegendVisibility’,’off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’I – 3′,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,3,’LineNames’,{{‘Iabc_Load:1′,’Iabc_Load:2′,’Iabc_Load:3′}},’ShowContent’,true,’Placement’,2)},’DisplayPropertyDefaults’,struct(‘MinYLimReal’,’-300′,’MaxYLimReal’,’300′,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’V – 3 ‘,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,0,’LineNames’,{{[]}},’ShowContent’,true,’Placement’,1),’DisplayLayoutDimensions’,[2 1]),extmgr.Configuration(‘Tools’,’Plot Navigation’,true,’OnceAtStop’,false),extmgr.Configuration(‘Tools’,’Measurements’,true,’Version’,’2019a’)),’Version’,’2019a’,’Location’,[-70.3333333333333 157.666666666667 1212.33333333333 807.666666666667])
2
off
[0, 0, 0, 0, 0, 3, 3]
[1729, 885, 1801, 920]
400
270
1.2814
powerlib/Measurements/Three-Phase
V-I Measurement
Three-Phase VI Measurement
Simscape Electrical
PS
off
phase-to-ground
on
Vabc_B3
off
off
yes
on
Iabc_B3
off
1e3
380
Magnitude
off
0
[1]
[2335, 726, 2370, 784]
560
green
alternate
Simulink.scopes.TimeScopeBlockCfg(‘CurrentConfiguration’, extmgr.ConfigurationSet(extmgr.Configuration(‘Core’,’General UI’,true),extmgr.Configuration(‘Core’,’Source UI’,true),extmgr.Configuration(‘Sources’,’WiredSimulink’,true,’DataLoggingVariableName’,’ScopeData22′,’DataLoggingSaveFormat’,’Array’,’DataLoggingDecimation’,’1′,’DataLoggingDecimateData’,true),extmgr.Configuration(‘Visuals’,’Time Domain’,true,’TimeAxisLabels’,’All’,’SerializedDisplays’,{struct(‘MinYLimReal’,’0.9′,’MaxYLimReal’,’1.1′,’YLabelReal’,”,’MinYLimMag’,’0.9′,’MaxYLimMag’,’1.1′,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’PF’,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,1,’LineNames’,{{‘Func 2′}},’ShowContent’,true,’Placement’,1)},’DisplayPropertyDefaults’,struct(‘MinYLimReal’,’0.9′,’MaxYLimReal’,’1.1′,’YLabelReal’,”,’MinYLimMag’,’0.9′,’MaxYLimMag’,’1.1′,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’PF’,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,0,’LineNames’,{{[]}},’ShowContent’,true,’Placement’,1)),extmgr.Configuration(‘Tools’,’Plot Navigation’,true,’OnceAtStop’,false,’AutoscaleMode’,’Auto’),extmgr.Configuration(‘Tools’,’Measurements’,true,’Version’,’2019a’)),’Version’,’2019a’,’Location’,[5 108.333333333333 1287.66666666667 749.666666666667])
1
off
[2, 1]
[2025, 800, 2065, 840]
556
alternate
1.676
powerlib_extras/Measurements/Active & Reactive
Power
Active & Reactive Power
Simscape Electrical
PS
off
60
[2]
[2180, 801, 2215, 874]
562
lightBlue
Simulink.scopes.TimeScopeBlockCfg(‘CurrentConfiguration’, extmgr.ConfigurationSet(extmgr.Configuration(‘Core’,’General UI’,true),extmgr.Configuration(‘Core’,’Source UI’,true),extmgr.Configuration(‘Sources’,’WiredSimulink’,true,’DataLogging’,true,’DataLoggingVariableName’,’ScopeData23′,’DataLoggingSaveFormat’,’StructureWithTime’,’DataLoggingMaxPoints’,’1000000′,’DataLoggingDecimation’,’1′,’DataLoggingDecimateData’,true),extmgr.Configuration(‘Visuals’,’Time Domain’,true,’SerializedDisplays’,{struct(‘MinYLimReal’,’-59.85892′,’MaxYLimReal’,’538.73029′,’YLabelReal’,”,’MinYLimMag’,’0.00000′,’MaxYLimMag’,’538.73029′,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’P’,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,1,’LineNames’,{{‘Demux2/1′}},’ShowContent’,true,’Placement’,1),struct(‘MinYLimReal’,’-0.00000000017′,’MaxYLimReal’,’0.0000000001′,’YLabelReal’,”,’MinYLimMag’,’0′,’MaxYLimMag’,’10’,’LegendVisibility’,’off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’Q ‘,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,1,’LineNames’,{{‘Demux2/2′}},’ShowContent’,true,’Placement’,2)},’DisplayPropertyDefaults’,struct(‘YLabelReal’,”,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’P’,’LinePropertiesCache’,{{}},’UserDefinedChannelNames’,{{}},’NumLines’,0,’LineNames’,{{[]}},’ShowContent’,true,’Placement’,1),’DisplayLayoutDimensions’,[2 1]),extmgr.Configuration(‘Tools’,’Plot Navigation’,true,’OnceAtStop’,false,’AutoscaleMode’,’Auto’),extmgr.Configuration(‘Tools’,’Measurements’,true,’Version’,’2019a’)),’Version’,’2019a’,’Location’,[443.666666666667 80.3333333333334 1047 730.333333333333])
2
off
[0, 1]
[1090, 823, 1155, 847]
324
alternate
1.168
Arial
powerlib_meascontrol/Pulse & Signal
Generators/PWM Generator
(2-Level)
PWM Generator (2-Level)
Simscape Electrical
PS
off
Single-phase full-bridge (4 pulses)
Unsynchronized
1080
90
27
[ -1 1 ]
Natural
on
0.8
60
0
Ts
off
[0, 1]
[1395, 916, 1475, 944]
326
on
[0.988235, 0.905882, 0.254902]
off
Control signal freq(Fc)
60
Swicting frequency
1900
Modulation Index
1
3rd harmonic Injection
off
Phase shift(Degrees)
0
[960, 0, 1920, 1050]
off
Deduce
Simulink
[0, 1]
[85, 340, 125, 400]
-1
off
[960, 0, 1920, 1050]
off
Deduce
125
Simulink
[885, 740, 1005, 770]
-1
off
0
Ts
[950, 614, 1015, 656]
-2
off
2*pi*fc*3
[1190, 680, 1220, 710]
-3
1/6
Inherit: Inherit via internal rule
Inherit: Inherit via internal rule
off
[960, 676, 1010, 714]
-4
off
pi/180
[3, 1]
[1050, 603, 1085, 787]
-5
off
+++
Ts
[1, 1]
[1105, 682, 1135, 708]
-6
off
Ts
[885, 685, 925, 705]
-7
off
shift
Ts
[880, 624, 915, 646]
-8
Ts
[1250, 688, 1280, 702]
-9
Port number
1
200#out:1
206#in:1
2
203#out:1
200#in:1
3
202#out:1
203#in:1
4
198#out:1
202#in:3
5
204#out:1
201#in:1
6
201#out:1
202#in:2
7
199#out:1
202#in:1
8
205#out:1
199#in:1
3 Harmonic Ref
[884, 784, 956, 798]
[0, 0, 0, 0]
-1
[200, 449, 245, 471]
-2
off
inj
Ts
[1, 6]
[790, 497, 800, 623]
-3
off
6
bar
[0, 1]
[100, 510, 140, 570]
-4
off
[960, 0, 1920, 1050]
off
Deduce
175
Simulink
[660, 530, 780, 560]
-1
off
[0 -2*pi/3 2*pi/3]
Ts
[725, 404, 790, 446]
-2
off
2*pi*fc
[735, 466, 785, 504]
-3
off
pi/180
[3, 1]
[825, 393, 860, 577]
-4
off
+++
Ts
[1, 1]
[880, 472, 910, 498]
-5
off
Ts
[660, 475, 700, 495]
-6
off
shift
Ts
[655, 414, 690, 436]
-7
Ts
[1015, 478, 1045, 492]
-8
Port number
1
214#out:1
215#in:1
2
211#out:1
214#in:3
3
216#out:1
213#in:1
4
213#out:1
214#in:2
5
212#out:1
214#in:1
6
217#out:1
212#in:1
7
215#out:1
218#in:1
[1, 1]
[560, 419, 645, 501]
-5
Ts=2e-6
off
[960, 0, 1920, 1050]
off
Deduce
Simulink
[20, 113, 50, 127]
-1
Port number
[1, 1]
[655, 139, 685, 161]
-2
alternate
off
NOT
1
Ts
[1, 1]
[655, 209, 685, 231]
-3
alternate
off
NOT
1
Ts
[1, 1]
[655, 69, 685, 91]
-4
alternate
off
NOT
1
Ts
[1, 3]
[130, 73, 135, 167]
-5
off
3
bar
[80, 105, 110, 135]
-6
m
Inherit: Inherit via internal rule
Inherit: Inherit via internal rule
off
[6, 1]
[755, 29, 760, 236]
-7
off
6
bar
[2, 1]
[450, 97, 480, 128]
-8
off
<=
off
[2, 1]
[450, 27, 480, 58]
-9
off
<=
off
[2, 1]
[450, 167, 480, 198]
-10
off
<=
off
[0, 1]
[315, 20, 345, 50]
-11
1.453
simulink/Sources/Repeating
Sequence
Repeating table
Simulink
SL
off
[0 1/fs/4 3/fs/4 1/fs]
[0 -1 1 0]
[845, 128, 875, 142]
-12
Port number
1
225#out:1
224#in:1
2
220#out:1
225#in:1
3
224#out:2
227#in:2
4
224#out:3
[138, 0; 0, 40]
229#in:2
5
222#out:1
226#in:6
6
229#out:1
[101, 0]
7
226#in:5
8
[0, 35]
222#in:1
9
221#out:1
226#in:4
10
230#out:1
[71, 0]
11
228#in:1
12
[0, 70]
13
227#in:1
14
[0, 70]
229#in:1
15
224#out:1
[244, 0; 0, -40]
228#in:2
16
227#out:1
[101, 0]
17
226#in:3
18
[0, 35]
221#in:1
19
223#out:1
226#in:2
20
228#out:1
[103, 0]
21
[0, 35]
223#in:1
22
226#in:1
23
226#out:1
231#in:1
[3]
[410, 555, 480, 655]
-6
C++SS(StrPVP('Location','[5, 53, 1925, 1031]'),StrPVP('Open','off'),MxPVP('AxesTitles',53,'struct(''axes1'',''%
3
off
[6]
[915, 502, 975, 618]
-7
C++SS(StrPVP(‘Location’,'[5, 48, 1925, 1049]’),StrPVP(‘Open’,’off’),MxPVP(‘AxesTitles’,53,’struct(”axes1”,”%
6
off
[2, 1]
[340, 420, 360, 440]
-8
off
round
|++
off
Inherit: Inherit via internal rule
off
[400, 414, 435, 506]
-9
off
0.5
Ts
[795, 373, 825, 387]
-10
alternate
Port number
1
210#out:1
[75, 0]
2
[50, 0; 0, -50; 80, 0]
3
235#in:3
4
234#in:2
5
[0, 65]
232#in:2
6
197#out:1
[105, 0; 0, 60; 68, 0]
7
234#in:1
8
[0, 140]
232#in:1
9
234#out:1
[14, 0]
10
235#in:1
11
[0, 210]
232#in:3
12
235#out:1
219#in:1
13
208#out:1
235#in:2
14
219#out:1
[92, 0]
15
[0, 100]
209#in:1
16
[0, -80]
236#in:1
17
209#out:6
233#in:6
18
209#out:5
233#in:5
19
209#out:4
233#in:4
20
209#out:3
233#in:3
21
209#out:2
233#in:2
22
209#out:1
233#in:1
Note:To enable the 3rd Harmonic injection, tick the check box of the block properties
[117, 165, 643, 183]
[0, 0, 0, 0]
-1
14
[3]
[1015, 951, 1065, 1009]
317
C++SS(StrPVP(‘Location’,'[9, 56, 953, 1041]’),StrPVP(‘Open’,’off’),MxPVP(‘AxesTitles’,53,’struct(”axes1”,”Battery Voltage”,”axes2”,”Battery SOC”,”axes3”,”Battery Current Output”)’),MxPVP(‘ScopeGraphics’,56,’struct(”FigureColor”,”[0.501960784313725 0.501960784313725 0.501960784313725]”,”AxesColor”,”[0 0 0]”,”AxesTickColor”,”[1 1 1]”,”LineColors”,”[1 1 0;1 0 1;0 1 1;1 0 0;0 1 0;0 0 1]”,”LineStyles”,”-|-|-|-|-|-”,”LineWidths”,”[0.5 0.5 0.5 0.5 0.5 0.5]”,”MarkerStyles”,”none|none|none|none|none|none”)’),StrPVP(‘ShowLegends’,’on’),StrPVP(‘YMin’,’258.35~0.999998~-225′),StrPVP(‘YMax’,’259.5~0.999999~100′),StrPVP(‘LimitDataPoints’,’off’),StrPVP(‘DataFormat’,’Array’),StrPVP(‘Decimation’,’1′),StrPVP(‘BlockParamSampleInput’,’off’))
3
off
PoWeRsYsTeMmEaSuReMeNt
[0, 1, 0, 0, 0, 2]
[1900, 843, 1925, 867]
552
\n
1.2814
powerlib/Measurements/Voltage Measurement
Voltage Measurement
Simscape Electrical
PS
off
off
Magnitude
0
[1]
[1960, 741, 2005, 799]
554
green
alternate
Simulink.scopes.TimeScopeBlockCfg(‘CurrentConfiguration’, extmgr.ConfigurationSet(extmgr.Configuration(‘Core’,’General UI’,true),extmgr.Configuration(‘Core’,’Source UI’,true),extmgr.Configuration(‘Sources’,’WiredSimulink’,true,’DataLoggingVariableName’,’ScopeData20′,’DataLoggingSaveFormat’,’Array’,’DataLoggingDecimation’,’1′,’DataLoggingDecimateData’,true),extmgr.Configuration(‘Visuals’,’Time Domain’,true,’TimeAxisLabels’,’All’,’SerializedDisplays’,{struct(‘MinYLimReal’,’-795.57042′,’MaxYLimReal’,’795.56837′,’YLabelReal’,”,’MinYLimMag’,’ 0.00000′,’MaxYLimMag’,’795.57042′,’LegendVisibility’,’Off’,’XGrid’,true,’YGrid’,true,’PlotMagPhase’,false,’AxesColor’,[0 0 0],’AxesTickColor’,[0.686274509803922 0.686274509803922 0.686274509803922],’ColorOrder’,[1 1 0.0666666666666667;0.0745098039215686 0.623529411764706 1;1 0.411764705882353 0.16078431372549;0.392156862745098 0.831372549019608 0.0745098039215686;0.717647058823529 0.274509803921569 1;0.0588235294117647 1 1;1 0.0745098039215686 0.650980392156863],’Title’,’%
1
off
[]
[895, 795, 1011, 844]
333
1
*1.2814
powerlib/powergui
PSB option menu block
Simscape Electrical
PS
off
Discrete
5.144e-06
60
50
0
100e6
1e-4
kV
MW
0
on
off
off
off
off
off
off
off
off
Tustin
off
off
off
0
blocks
off
0
off
0
15
off
warning
off
[0:2:500]
off
on
off
off
ZData
Vabc_Load
off
0.0
1
1
60
off
10000
1
1
off
328
180#rconn:1
[18, 0; 0, -45]
8#rconn:1
28
63#out:3
70#in:3
29
63#out:2
70#in:2
30
63#out:1
70#in:1
942
761#out:1
[-239, 0; 0, -65]
63#in:1
32
170#out:4
178#in:1
33
170#out:1
175#in:1
34
170#out:2
176#in:1
35
170#out:3
177#in:1
36
238#out:1
170#in:1
37
196#out:1
[-2, 0; 0, 35]
8#in:1
Vabc_B1
38
[0, 1]
172#out:1
60#in:1
Iabc_B1
39
[0, 0]
171#out:1
60#in:2
941
761#lconn:1
180#lconn:1
940
761#lconn:2
[104, 0; 0, -30]
8#rconn:2
1702
8#lconn:2
336#lconn:2
1703
8#lconn:3
336#lconn:3
Vabc_Load
196
[0, 1]
343#out:1
341#in:1
Iabc_Load
197
[0, 0]
342#out:1
341#in:2
1623
778#lconn:1
[-23, 0; 0, 230]
336#rconn:3
1617
375#rconn:1
[0, -13; -124, 0; 0, 133]
336#rconn:1
1701
8#lconn:1
336#lconn:1
1618
375#rconn:2
[-125, 0; 0, 145]
336#rconn:2
1123
767#lconn:1
[-10, 0]
375#lconn:2
1489
806#lconn:1
[25, 0]
375#lconn:1
1490
806#rconn:1
769#lconn:1
1121
767#rconn:1
770#lconn:1
1491
807#rconn:1
771#lconn:1
1492
807#lconn:1
[-35, 0]
375#lconn:3
1146
778#rconn:1
[45, 0; 0, 35]
779#lconn:1
[-95, 0]
375#rconn:3
1147
778#out:1
[73, 0]
1700
[0, -30]
782#in:1
1625
[0, 30]
783#in:2
1150
779#lconn:2
[-15, 0]
780#lconn:1
1151
779#out:1
[9, 0; 0, -45]
1602
[0, -40]
781#in:1
1580
783#in:1
1157
784#out:1
786#in:1
1158
786#out:1
787#in:1
1159
787#out:1
785#in:1
1160
783#out:1
788#in:1
1161
788#out:1
[45, 0]
1604
[0, 10]
789#in:1
1583
[0, -50]
784#in:2
1595
788#out:2
[18, 0]
1605
[0, 25]
789#in:2
1600
[0, -85]
784#in:1
simulink/configSet0.xml
[]
[]
[]
0.0
0.1
auto
on
auto
auto
5
auto
10*128*eps
1000
4
1
auto
auto
1
1e-3
on
off
ode45
ode45
auto
DisableAll
UseLocalSettings
Nonadaptive
TrustRegion
off
off
Fast
off
off
Unconstrained
Whenever possible
[]
off
off
3
[]
[]
1
[t, u]
xFinal
xInitial
on
1000
off
off
off
off
Array
Dataset
on
off
on
on
off
on
off
streamout
on
off
xout
tout
yout
logsout
dsmout
RefineOutputTimes
[]
out
1
off
timeseries
out.mat
[-inf, inf]
BooleansAsBitfields
PassReuseOutputArgsAs
PassReuseOutputArgsThreshold
ZeroExternalMemoryAtStartup
ZeroInternalMemoryAtStartup
OptimizeModelRefInitCode
NoFixptDivByZeroProtection
UseSpecifiedMinMax
[]
on
on
on
Tunable
off
off
off
double
off
off
on
on
off
off
on
off
on
off
uint_T
on
64
Structure reference
12
128
on
5
off
off
Native Integer
on
on
off
off
off
on
inf
Inherit from target
on
off
off
off
on
on
off
level2
Balanced
on
off
off
off
GradualUnderflow
UseOnlyExistingSharedCode
[]
error
none
none
none
error
none
UseLocalSettings
UseLocalSettings
UseLocalSettings
warning
warning
warning
warning
on
Simplified
error
off
off
UseLocalSettings
warning
warning
none
error
warning
warning
warning
warning
error
error
error
none
warning
none
warning
none
warning
warning
error
error
none
warning
warning
none
none
none
none
none
none
error
error
none
warning
warning
warning
error
none
error
none
warning
warning
UseLocalSettings
on
off
none
error
none
none
warning
warning
warning
error
none
warning
error
none
warning
none
warning
ErrorLevel1
WarnAndRepair
none
warning
warning
error
error
none
warning
warning
warning
warning
warning
warning
warning
warning
error
warning
warning
none
warning
warning
all
warning
on
warning
warning
off
none
off
warning
[]
[]
8
16
32
32
64
32
64
32
32
32
Char
None
Undefined
Unspecified
32
on
off
32-bit Generic
8
16
32
32
64
32
64
32
32
32
Char
None
on
off
Undefined
Unspecified
32
32
32
Specified
off
on
on
on
EmbeddedCoderHSP
[]
[]
IfOutOfDateOrStructuralChange
on
error
off
on
None
Multi
Infer from blocks in model
on
off
off
off
[]
[]
[]
on
on
on
off
on
off
sf_incremental_build
off
on
50
on
on
65536
[]
NotSpecified
IncludeHyperlinkInReport
GenerateTraceInfo
GenerateTraceReport
GenerateTraceReportSl
GenerateTraceReportSf
GenerateTraceReportEml
PortableWordSizes
GenerateWebview
GenerateCodeMetricsReport
GenerateCodeReplacementReport
GenerateErtSFunction
CreateSILPILBlock
CodeExecutionProfiling
CodeProfilingSaveOptions
CodeProfilingInstrumentation
GenerateMissedCodeReplacementReport
grt.tlc
None
off
off
off
make_rtw
on
off
grt_default_tmf
off
on
off
[]
off
off
off
off
off
off
Automatically locate an installed toolchain
Faster Builds
[]
off
off
off
None
off
executionProfile
SummaryOnly
off
off
C
off
off
off
off
off
off
off
off
off
off
off
[]
Off
1024
IgnoreCustomStorageClasses
IgnoreTestpoints
InsertBlockDesc
InsertPolySpaceComments
SFDataObjDesc
MATLABFcnDesc
SimulinkDataObjDesc
DefineNamingRule
SignalNamingRule
ParamNamingRule
InternalIdentifier
InlinedPrmAccess
CustomSymbolStr
CustomSymbolStrGlobalVar
CustomSymbolStrType
CustomSymbolStrField
CustomSymbolStrFcn
CustomSymbolStrFcnArg
CustomSymbolStrBlkIO
CustomSymbolStrTmpVar
CustomSymbolStrMacro
CustomSymbolStrUtil
ReqsInCode
BlockCommentType
CustomSymbolStrModelFcn
CustomSymbolStrEmxType
CustomSymbolStrEmxFcn
CustomUserTokenString
[]
off
on
Auto
on
off
31
off
off
off
off
off
off
off
1
8
$R$N$M
$N$R$M_T
$N$M
$R$N$M$F
$R$N
rt$I$N$M
rtb_$N$M
$N$M
$R$N$M
$N$C
emxArray_$M$N
emx$M$N
None
None
None
off
off
on
BlockPathComment
on
off
off
Shortened
Literals
off
off
[]
GeneratePreprocessorConditionals
IncludeMdlTerminateFcn
PreserveStateflowLocalDataDimensions
SuppressErrorStatus
ERTCustomFileBanners
GenerateSampleERTMain
GenerateTestInterfaces
ModelStepFunctionPrototypeControlCompliant
GenerateAllocFcn
PurelyIntegerCode
SupportComplex
SupportAbsoluteTime
SupportContinuousTime
SupportNonInlinedSFcns
ExistingSharedCode
RemoveDisableFunc
RemoveResetFunc
[]
ansi_tfl_table_tmw.mat
NOT IN USE
C89/C90 (ANSI)
None
Auto
System defined
2048
256
on
off
off
off
on
on
on
on
on
Disable all
on
off
off
off
Auto
off
on
rt_
on
off
Nonreusable function
off
on
on
on
on
off
off
off
Nominal
Nominal
off
Simulink.SoftwareTarget.GRTCustomization
off
on
off
off
off
off
on
on
[]
[]
1,2,3,4,…
Size,Breakpoints,Table
Size,Breakpoints,Table
Column-major
error
$R$E
$R$E
$R_data
off
off
off
off
1000000
0
ext_comm
Level1
off
off
off
off
off
Error
[]
Simulink Coverage Configuration Component
[]
Simulink Coverage
off
EntireSystem
on
off
/
covdata
dw
off
on
on
on
on
on
covCumulativeData
off
on
slcov_output/$ModelName$
$ModelName$_cvdata
on
on
off
off
off
1e-05
0.01
off
0
0
Masking
Name
[]
Simscape
Full
warning
warning
off
[{“value”:”1″,”unit”:”A”},{“value”:”1″,”unit”:”bar”},{“value”:”1″,”unit”:”cm^2″},{“value”:”1″,”unit”:”cm^3/s”},{“value”:”1″,”unit”:”kJ/kg”},{“value”:”1″,”unit”:”kW”},{“value”:”1″,”unit”:”l”},{“value”:”1″,”unit”:”N”},{“value”:”1″,”unit”:”N*m”},{“value”:”1″,”unit”:”V”}]
none
off
off
off
simlog
1
on
5000
off
Configuration
Solver
[ 420, 220, 1500, 860 ]
simulink/configSetInfo.xml
Configuration
simulink/graphicalInterface.xml
0
0
0
0
0
0
0
0
0
Unset
0
0
0
23
powerlib/Elements/Parallel RLC Load
$bdroot/ A
806
LIBRARY_BLOCK
powerlib/Elements/Parallel RLC Load
$bdroot/ B
767
LIBRARY_BLOCK
powerlib/Elements/Parallel RLC Load
$bdroot/ C
807
LIBRARY_BLOCK
powerlib/Power
Electronics/Mosfet
$bdroot/3P Converter/Mosfet1
23
LIBRARY_BLOCK
powerlib/Power
Electronics/Mosfet
$bdroot/3P Converter/Mosfet2
24
LIBRARY_BLOCK
powerlib/Power
Electronics/Mosfet
$bdroot/3P Converter/Mosfet3
25
LIBRARY_BLOCK
powerlib/Power
Electronics/Mosfet
$bdroot/3P Converter/Mosfet4
26
LIBRARY_BLOCK
powerlib/Power
Electronics/Mosfet
$bdroot/3P Converter/Mosfet5
27
LIBRARY_BLOCK
powerlib/Power
Electronics/Mosfet
$bdroot/3P Converter/Mosfet6
28
LIBRARY_BLOCK
powerlib/Measurements/Three-Phase
V-I Measurement
$bdroot/B1 Meas
336
LIBRARY_BLOCK
electricdrivelib/Extra Sources/Battery
$bdroot/Battery (90 kWh)
761
LIBRARY_BLOCK
powerlib/Elements/Ground
$bdroot/Ground11
780
LIBRARY_BLOCK
powerlib/Elements/Ground
$bdroot/Ground7
769
LIBRARY_BLOCK
powerlib/Elements/Ground
$bdroot/Ground8
770
LIBRARY_BLOCK
powerlib/Elements/Ground
$bdroot/Ground9
771
LIBRARY_BLOCK
powerlib/Measurements/Current Measurement
$bdroot/IC
778
LIBRARY_BLOCK
powerlib/Elements/Series RLC Branch
$bdroot/L1
180
LIBRARY_BLOCK
powerlib/Measurements/Three-Phase
V-I Measurement
$bdroot/Load Meas
375
LIBRARY_BLOCK
powerlib_extras/Measurements/Active & Reactive
Power
$bdroot/PQ Load
783
LIBRARY_BLOCK
powerlib_meascontrol/Pulse & Signal
Generators/PWM Generator
(2-Level)
$bdroot/PWM Generator (4-pulses)
238
LIBRARY_BLOCK
simulink/Sources/Repeating
Sequence
$bdroot/PWM Pulse/PWM Generator/Triangle2
230
LIBRARY_BLOCK
powerlib/Measurements/Voltage Measurement
$bdroot/VC
779
LIBRARY_BLOCK
powerlib/powergui
$bdroot/powergui
240
LIBRARY_BLOCK
1
simulink/modelDictionary.xml
simulink/plugins/DiagnosticSuppressor.xml
on
simulink/plugins/LogicAnalyzerPlugin.xml
on
simulink/plugins/NotesPlugin.xml
on
simulink/plugins/SLCCPlugin.xml
on
simulink/plugins/WebScopes_FoundationPlugin.xml
on
[Content_Types].xml
_rels/.rels
simulink/_rels/blockdiagram.xml.rels
simulink/_rels/configSetInfo.xml.rels
% % ECE 5750 DS & PQ
% % Example on how to use FFT to obtain a waveform frequency spectrum
% % Based on a Matlab example
% Initialization
clear all
clc;
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sample time
L = 1000; % Length of signal
t = (0:L-1)*T; % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
y = x + 2*randn(size(t)); % Two sinusoids plus noise
figure(1)
plot(Fs*t(1:50),y(1:50))
title(‘Signal Corrupted with Zero-Mean Random Noise’)
xlabel(‘Time, milliseconds’)
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
% Plot single-sided amplitude spectrum
% It is visible from Fig. 2 that the two significant components in y(t)
% are 50-Hz and 120-Hz. Other components are noises.
figure(2)
plot(f,2*abs(Y(1:NFFT/2+1)))
title(‘Single-Sided Amplitude Spectrum of y(t)’)
xlabel(‘Frequency (Hz)’)
ylabel(‘|Y(f)|’)
Cal Poly Pomona ECE 5750 Distribution System and Power Quality – Instructor Dr. Ha Thu Le
1
California State Polytechnic University Pomona
Department of Electrical and Computer Engineering
ECE 5750 Distribution system and power quality – Spring 2020
Final exam
Exam policy:
1. The exam is open-book and take-home. Students can use any materials such as lecture PPT, books,
and notes. Any computer program may be used.
2. Exam duration: From 9am May 15 to midnight May 17, 2020.
3. NO discussion with anyone is allowed during exam.
4. Submission: Submit a soft copy of the solution to the instructor via email. Also submit a code and/or
Simulink file as specified in the exam question(s) to the instructor via email.
Problem 1
Consider the feeder in the following figure.
The substation transformer is connected to an infinite bus. The infinite bus voltages (i.e. the substation
primary voltages) are balanced and being held at 138 kV for all power-flow problems. The substation
transformer ratings are:
5000 kVA, 138 kV delta – 25 kV grounded Y, Z = 1.7 + j8.5 %
The phase impedance matrix for a four-wire wye line between Node 2-3 is
The four-wire wye feeder between Node 2-3 is 9.5 miles long. An unbalanced wye-connected load is
located at Node 3 and has the following values:
Phase a: 700 kVA at 0.85 lagging power factor
Phase b: 600 kVA at 0.90 lagging power factor
Phase c: 1020 kVA at 0.95 lagging power factor
Cal Poly Pomona ECE 5750 Distribution System and Power Quality – Instructor Dr. Ha Thu Le
2
Two single-phase loads are also located at Node 3 and have the following values:
Load 1: 120 kVA at 0.82 lagging power factor, connected to Phase a
Load 2: 100 kVA at 0.92 lagging power factor, connected to Phase c
Assuming that the regulators are in the neutral position:
1) Determine the forward and backward sweep matrices for the substation transformer and the
line segment.
2) Use the modified ladder technique to determine the line-to-ground voltages at all nodes. Use
a tolerance of 0.0001 per unit.
3) Calculate all node voltages in volts, in per unit, and on a 120-V base.
Problem 2
Consider the following system where a battery is connected to a load via a 3-phase converter. The
system diagram is shown in Fig. 1 below and its Simulink file is provided via email.
Fig. 1 Battery and converter system generating harmonics
The battery DC power is converted to AC 60-Hz power by the 3-phase converter to supply a 240-V 3-
phase AC resistive load. The converter generates a number of harmonics. Therefore, the load voltages
and currents are severely distorted, as seen by the scopes (IC Load, VC Load, Converter, and Load).
Design a filter to reduce and/or eliminate the harmonic content of the converter output. Perform the
following task:
1) Use a Fourier waveform analyzer (e.g. FFT function, a spectrum analyzer, or other waveform
analyzing functions) to determine significant harmonic voltages and currents that the converter
generates. Provide a list of the frequencies and magnitudes (as percentage of the fundamental)
of the obtained voltage and current harmonics (up to harmonic order 55 or 3300Hz). Also
include a graph of the frequency spectrum to show the voltage harmonics and a graph to show
current harmonics.
Hint: Run the provided m-file (an example by Matlab) to see how to use FFT function to analyze
a waveform and obtain its frequency spectrum.
Cal Poly Pomona ECE 5750 Distribution System and Power Quality – Instructor Dr. Ha Thu Le
3
2) Based on the result in Part 1, design a filter to eliminate the most significant harmonic
component(s). The design MUST include calculation of the filter parameters. If you design a
single-tuned LC filter, calculate its capacitance and inductance values. NO need to evaluate
the filter duty requirements.
Note: You can have more than one filter to reduce the harmonics. The filter must be 3-phase.
3) Connect the LC filter to the output terminals of the converter (as shown in Fig. 1) and verify by
simulation that the filter does eliminate most of the harmonics. If this is true, the load voltages
and currents must be SINUSOIDAL with little or no noises. Include a graph of the load IC and
VC to show that the waveforms are clean and sinusoidal after the filter is connected.
4) Submit the *.slx file with the designed filter via email to the instructor to prove that your
filter is actually effective. Make sure to replace the word “Student” in the *.slx file name
with your last name before submission.
Problem 3
(Optional, Extra Credit of 15 points)
A current i(t) through a load is non-sinusoidal, as shown in the figure below.
Determine:
1) The Trigonometric Fourier series of i(t).
2) The RMS value of the fundamental component of the load current. Consider the first order
harmonic component as the fundamental component for this circuit.
3) The RMS value of the harmonic load current (i.e. including all harmonics except the
fundamental component)
4) The RMS value of the load current (i.e. including the DC, the fundamental, and all harmonics)
5) The THD of the load current.
Note: Include the first 5000 harmonics (zero and non-zero values) for calculating the RMS values in
part 2 through 4.
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
1
California State Polytechnic University Pomona
Department of Electrical and Computer Engineering
ECE 5750 Distribution system and power quality – Spring 2020
Solution for Homework 4
Distribution feeder analysis
Problem 1
Consider the feeder in the following figure.
The substation transformer is connected to an infinite bus. The infinite bus voltages (i.e. the
substation primary voltages) are balanced and being held at 69 kV for all power-flow problems. The
substation transformer ratings are:
5000 kVA, 69 kV delta – 13.8 kV grounded Y, Z = 1.7 + j8.5 %
The phase impedance matrix for a four-wire wye line between Node 2-3 is
The four-wire wye feeder between Node 2-3 is 0.75 miles long. An unbalanced wye-connected load
is located at node 3 and has the following values:
Phase a: 650 kVA at 0.85 lagging power factor
Phase b: 500 kVA at 0.90 lagging power factor
Phase c: 950 kVA at 0.95 lagging power factor
Assuming that the regulators are in the neutral position:
1) Determine the forward and backward sweep matrices for the substation transformer and
the line segment.
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
2
2) Use the modified ladder technique to determine the line-to-ground voltages at all nodes.
Use a tolerance of 0.0001 per unit.
3) Calculate all node voltages in actual values in volts and on a 120-V base.
Solution:
tol = 0.0001; % tolerance level
number of iterations: 4
V1 is the voltage at the primary of the substation
sweep matrices for line segment
a1 = 1 0 0
0 1 0
0 0 1
b1 =
0.3432 + 0.8085i 0.1170 + 0.3763i 0.1151 + 0.2887i
0.1170 + 0.3763i 0.3499 + 0.7862i 0.1185 + 0.3177i
0.1151 + 0.2887i 0.1185 + 0.3177i 0.3461 + 0.7988i
d1 =
1 0 0
0 1 0
0 0 1
A1 =
1 0 0
0 1 0
0 0 1
B1 =
0.3432 + 0.8085i 0.1170 + 0.3763i 0.1151 + 0.2887i
0.1170 + 0.3763i 0.3499 + 0.7862i 0.1185 + 0.3177i
0.1151 + 0.2887i 0.1185 + 0.3177i 0.3461 + 0.7988i
Q1 TRF forward and backward sweep matrices
Ztabc =
0.6475 + 3.2375i 0.0000 + 0.0000i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.6475 + 3.2375i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.6475 + 3.2375i
at =
0 -5.7735 -2.8868
-2.8868 0 -5.7735
-5.7735 -2.8868 0
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
3
bt =
0.0000 + 0.0000i -3.7383 -18.6916i -1.8692 – 9.3458i
-1.8692 – 9.3458i 0.0000 + 0.0000i -3.7383 -18.6916i
-3.7383 -18.6916i -1.8692 – 9.3458i 0.0000 + 0.0000i
dt =
0.1155 -0.1155 0
0 0.1155 -0.1155
-0.1155 0 0.1155
At =
0.1155 0 -0.1155
-0.1155 0.1155 0
0 -0.1155 0.1155
Bt =
0.6475 + 3.2375i 0.0000 + 0.0000i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.6475 + 3.2375i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.6475 + 3.2375i
Specified line-neutral source voltage and load
VLN_ABC = 1.0e+04 *
3.4500 – 1.9919i
-3.4500 – 1.9919i
0.0000 + 3.9837i
VLN_ABC_mag = 1.0e+04 *
3.9837
3.9837
3.9837
VLN_ABC_ang =
-30.0000
-150.0000
90.0000
S3 = 1.0e+05 *
5.5250 + 3.4241i
4.5000 + 2.1794i
9.0250 + 2.9664i
V3 = 1.0e+03 *
7.9674 + 0.0000i
-3.9837 – 6.9000i
-3.9837 + 6.9000i
V3_mag = 1.0e+03 *
7.9674
7.9674
7.9674
V3_ang =
0
-120.0000
120.0000
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
4
End of first iteration
V1_mag = 1.0e+04 *
4.1254
4.0597
4.0785
V1_ang =
30.8211
-87.0387
151.8043
V2_mag = 1.0e+03 *
7.9967
7.9930
8.0190
V2_ang =
0.1723
-119.9450
120.4171
V3_mag = 1.0e+03 *
7.9674
7.9674
7.9674
V3_ang =
0
-120.0000
120.0000
I3_mag =
81.5821
62.7555
119.2354
I3_ang =
-31.7883
-145.8419
101.8051
End of second iteration
V1_mag = 1.0e+04 *
4.1021
4.2111
4.1460
V1_ang =
-31.3711
-149.5510
89.7250
V2_mag = 1.0e+03 *
8.0873
8.0690
8.2524
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
5
V2_ang =
-61.9272
178.5134
57.5030
V3_mag = 1.0e+03 *
8.0574
8.0447
8.2025
V3_ang =
-62.0986
178.4547
57.1146
I3_mag =
80.6716
62.1530
115.8188
I3_ang =
-93.8869
152.6127
38.9197
Final result when system converged
V1_mag = 1.0e+04 *
3.9837
3.9847
3.9840
V1_ang =
-30.0081
-149.9951
90.0010
V2_mag = 1.0e+03 *
7.7754
7.8380
7.7583
V2_ang =
-61.4511
178.7891
57.4620
V3_mag = 1.0e+03 *
7.7452
7.8130
7.7042
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
6
V3_ang =
-61.6420
178.7369
57.0198
Final voltages in pu & volt on a 120-V base
V1_pu =
1.0000
1.0002
1.0001
V2_pu =
0.9759
0.9838
0.9737
V3_pu =
0.9721
0.9806
0.9670
V3_120V =
116.6528
117.6745
116.0355
Error history
er =
0.0356
0.0191
0.0238
er =
0.0297
0.0571
0.0407
er =
0.0019
0.0011
0.0002
er = 1.0e-03 *
0.0062
0.2419
0.0591
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
7
Problem 2
Three type B step-voltage regulators are installed in a wye connection at the substation in order to
hold the load voltages (node 3) at a voltage level of 121 V and a bandwidth of 2 V.
1) Compute the actual equivalent line impedance between nodes 2 and 3.
2) Determine a potential transformer ratio and current transformer ratio given that the
compensator circuit ratings are 120V and 5A. Determine the R and X compensator settings
calibrated in volts and Ohms. The settings must be the same for all three regulators.
3) For the load conditions of Problem 1 with the regulators in the neutral position, compute the
voltages across the voltage relays in the compensator circuits.
4) Determine the appropriate tap settings for the three regulators to hold the node 3 voltages
at 121 V in a bandwidth of 2 V.
5) With the regulators taps set, compute the actual load voltages in volts and in per unit.
Solution:
Q 1 n 2 Line impedance – Compensator R and X setting
Zline_ohm =
0.1425 + 0.4516i
0.3022 + 0.2705i
0.2632 + 0.5976i
Zline_aver = 0.2360 + 0.4399i ohm
Npt = 66.3953 CTp = 209.1849 CTs = 5
Zcomp_volt = 0.4489 + 1.4227i
0.9520 + 0.8521i
0.8293 + 1.8829i
Zcomp_ohm = 0.0898 + 0.2845i
0.1904 + 0.1704i
0.1659 + 0.3766i
Q3 Compensator voltage and current input – Load new
Icomp_mag = 2.0060
1.5296
2.9474 A
Icomp_ang = -93.4303
152.8949
38.8249 deg.
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
8
Vrelay_mag = 116.6528
117.6745
116.0355 V
Vrelay_ang = -61.6420
178.7369
57.0198
Q4 Final tap position, V=121V, bandwidth=2V
tap_req =
4.4630
3.1007
5.2860
Tap = [5; 4; 6]
Q5 With the regulators taps set, compute the actual load voltages in volts and in per unit.
tol = 0.0001; % tolerance level
number of iterations: 4
V2r is the input voltage to regulator
V2 is the output voltage from regulator
V1 is the substation primary voltage
Regulator matrices
a_reg = 0.9688 0 0
0 0.9750 0
0 0 0.9625
d_reg = 1.0323 0 0
0 1.0256 0
0 0 1.0390
Areg = 1.0323 0 0
0 1.0256 0
0 0 1.0390
First iteration
I3_mag =
81.5821
62.7555
119.2354
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
9
I3_ang =
-31.7883
-145.8419
101.8051
V1_mag = 1.0e+04 *
3.9837
3.9837
3.9837
V1_ang =
-30.0000
-150.0000
90.0000
V2r_mag = 1.0e+03 *
8.0527
8.0523
8.1826
V2r_ang =
-61.8932
178.6064
57.5315
V2_mag = 1.0e+03 *
8.3124
8.2588
8.5014
V2_ang = -61.8932
178.6064
57.5315
V3_mag = 1.0e+03 *
8.3200
8.2533
8.5297
V3_ang = -62.1487
178.4249
57.0442
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le
10
Final result when system converged
V1_mag = 1.0e+04 *
3.9837
3.9847
3.9840
V1_ang = -30.0079
-149.9956
90.0009
V2_mag = 1.0e+03 *
8.0263
8.0390
8.0609
V2_ang = -61.4512
178.7894
57.4622
V2r_mag = 1.0e+03 *
7.7754
7.8380
7.7586
V2r_ang = -61.4512
178.7894
57.4622
V3_mag = 1.0e+03 *
7.9968
8.0146
8.0091
V3_ang = -61.6288
178.7365
57.0535
V3_pu = 1.0037
1.0059
1.0052
V3_120V = 120.4417
120.7107
120.6276
ECE
5
750 Supporting document
Part
9
: Fourier series and applications
Instructor: Dr. Ha Le
Department of Electrical and Computer Engineering
California State Polytechnic University, Pomona
2
What will be presented?
1. Trigonometric Fourier Series
2. Symmetry Considerations
3. Exponential Fourier Series
4. Applications:
Steady-state response of circuits
Average power calculations with periodic functions
The RMS value of periodic functions
Spectrum analyzers
Filters
(Alexander and Sadiku “Fundamentals of Electric Circuits”, Chapter
17
)
3
Why need Fourier series?
Fourier series is a means for analyzing circuits
with periodic, nonsinusoidal excitations.
Most of the functions of a circuit are periodic.
They can be decomposed into infinite number of sine
and cosine functions that are harmonically related
(harmonics).
A complete response of a forcing function =
∑ Partial response to each harmonics
Practical applications:
Steady-state response of circuits
Spectrum analyzers
Filters
Average power calculations etc.
4
Trigonometric Fourier series (1)
Definition: The Fourier series of a periodic function f(t) is a
representation that resolves f(t) into a DC component and an AC
component comprising an infinite series of harmonic sinusoids.
Given a periodic function f(t)=f(t+nT) where n is an integer and T
is the period of the function. The Fourier series of f(t) has the form
where 0=2/T is the fundamental frequency in rad/s, a0 is the DC
component or the average value of f(t), an and bn are Fourier
coefficients.
Harmonics have frequencies: 0 20 30 40 …
How to calculate a0, an, bn?
0 0
0
1
( ) ( cos sin )n
n
n
dc
ac
f t a a n t b n t
Sufficient conditions on f(t) to yield a convergent
Fourier series:
1. f(t) is single-valued everywhere.
2. f(t) has a finite number of discontinuities in any
one period. The discontinuities must be of finite
size.
3. f(t) has a finite number of maxima and minima in
any one period.
4. The integral
Fourier series (2)
Dirichlet conditions
0
0
0( ) for any
t
T
t
f t dt t
5
Example: Convergence of Fourier series
Fourier series (3)
Dirichlet conditions
6
0sin
0 0
dttnw
T
0cos
0 0
dttnw
T
(b)
0 0 0 00 0
1sin cos sin sin 0
2
T T
kw t nw tdt k n w t k n w t dt
nkif
nkifT
dttwnktwnk
dttnwtkw
T
T
,0
,2
coscos
2
1
sinsin
0
00
0 00
nkif
nkifT
dttwnktwnk
dttnwtkw
T
T
,0
,2
coscos
2
1
coscos
0 00
0 00
Trigonometric Fourier series (4)
Useful trigonometric integrals
7
(a)
(c)
(d) (e)
T
n
nn
TT
dttnwbtnwadtadttf
0
1
000 00
sincos
Based on (a) (b)
0sincos
0
1
00
T
n
nn dttnwbtnw
a
T
dttf
T
a
00
1
0a is called the DC component or the average value of tf
Trigonometric Fourier series (5)
Evaluation of Fourier coefficient a0
8
Integrating both sides of the Fourier series definition of f(t)
Based on (b)
T
n
n
T
n
n
TT
dttkwtnw
b
dttkwtnwadttkwatdtkwtf
0
1
00
0
1
000 000 0
cossin
coscoscoscos
0 00
1
cos cos
2
T
n n
n
Ta nw t kw tdt a
0cossin
0
1
00
T
n
n dttkwtnwb
0cos
0 00
dttkwa
T
Based on (c)
Based on (e)
When k=n
Trigonometric Fourier series (6)
Evaluation of Fourier coefficient an
00
2 cos
T
na f t nw tdtT
9
Based on (a)
T
n
n
T
n
n
TT
dttkwtnwb
dttkwtnwadttkwatdtkwtf
0
1
00
0
1
000 000 0
sinsin
sincossinsin
0 00
1
sin sin
2
T
n n
n
Tb nw t kw tdt b
0sincos
0
1
00
T
n
n dttkwtnwa
0sin
0 00
dttkwa
T
Based on (c)
Based on (d)
When k=n
Trigonometric Fourier series (7)
Evaluation of Fourier coefficient bn
00
2 sin
T
nb f t nw tdtT
10
11
Summary of trigonometric Fourier series (8)
FS and evaluation of coefficients a0, an, bn, sine and
cosine form
Alternative form of f(t): amplitude-phase form
T
on dttntfT
a
0
)cos()(2
T
on dttntfT
b
0
)sin()(2
ac
n
nn
dc
tnAatf
1
00 )cos(()(
2 2 1
n , tan nn n n
n
bA a
b
a
T
dttf
T
a
00
1
0 0 0
1
( ) ( cos sin )n n
ndc
ac
f t a a n t b n t
0 0 0cos sin cosn n n na nw t b nw t A nw t
2 2
n n nA a b
Phase spectrum
Trig. Fourier series: Frequency spectrum (9)
12
Harmonic amplitude
1tan nn
n
b
a
The frequency
spectrum of a signal
consists of the plots of
the amplitudes
and phases of the
harmonics versus
frequency.
Example 1 (Prob. 17.3): Calculate the Fourier coefficients a0, an,
and bn of the following waveform. Plot the amplitude and phase
spectra.
Examples: Trig. Fourier series (1)
13
Solution:
Solution (cont.):
Examples: Trig. Fourier series (2)
0 00 0
00
2 1( )cos( )
2 ( )sin( )
T T
n
T
n
a f t n t dt a f t dt
T T
b f t n t dt
T
14
Solution (cont.):
Examples: Trig. Fourier series (3)
15
0 00 0
00
2 1( ) cos( )
2 ( )sin( )
T T
n
T
n
a f t n t dt a f t dt
T T
b f t n t dt
T
Solution (end):
Examples: Trig. Fourier series (4)
2 2
n n nA a b
16
1tan nn
n
b
a
Amplitude and phase spectrum
Example 2: Determine the Fourier series of the pulse
waveform shown in the following figure.
Examples: Trig. Fourier series (5)
17
Solution for Example 2: Matlab plot of the Fourier series of the
pulse waveform, n=500
Examples: Trig. Fourier series (4)
18
Three types of symmetry: even, odd, half-wave
1. Even Symmetry: a function f(t) if its plot is symmetrical about
the vertical axis (even function)
In this case,
Symmetry considerations (1)
)()( tftf
0
)cos()(4
)(2
2/
0 0
2/
00
n
T
n
T
b
dttntf
T
a
dttf
T
a
Examples of even periodic function
19
Note: FS is a cosine series
(cosine function is even)
2. Odd Symmetry : a function f(t) if its plot is anti-symmetrical
about the vertical axis (odd function)
In this case,
)()( tftf
0
/ 2
00
0, 0
4 ( )sin( )
n
T
n
a a
b f t n t dt
T
Examples of odd periodic function
20
Symmetry considerations (2)
Note: FS is a sine series
(sine function is odd)
3. Half-wave Symmetry: a function f(t)
Symmetry considerations (3)
0
/ 2
00
/ 2
00
0
4 ( )cos( ) , for n=odd
0 , for n=even
4 ( )sin( ) , for n=odd
0 , for n=even
T
n
T
n
a
f t n t dt
a T
f t n t dt
b T
Typical examples of half-wave odd periodic functions 21
( )
2
Tf t f t
Symmetry considerations (4)
Summary of Fourier coefficients
22
Example 17.3: Find the Fourier series expansion of f(t) given below.
Examples: Symmetry (1)
23
Solution:
Determine the fundamental frequency and specify the type of
symmetry present in the following functions.
Examples: Symmetry (2)
24
Fourier series of selected waveforms (1)
25
Fourier series of selected waveforms (2)
26
Application to circuit analysis
Steps for applying Fourier series (FS)
to find the steady-state response of a circuit
forced by a nonsinusoidal periodic excitation
(function)
1. Express the excitation signal as a Fourier series.
2. Transform the circuit from the time domain to the
frequency domain.
3. Find the response of the DC and AC components
in the Fourier series.
4. Add the individual DC and AC responses using
the superposition principle Circuit response.
27
Example 1: Find the steady-state response v0(t) of the
following RC circuit, given that the input vs(t) consists of first
3 terms of the Fourier series of a square wave (n = 0, 1, 3,
n=0 is the DC term)
Examples: Circuit analysis using FS (1)
28
Example 2: Find i(t) in the following circuit given that
Examples: Circuit analysis using FS (2)
29
Average power:
Voltage and current as FS (amplitude-phase form):
The average power is
Average Power and RMS Values (1)
dc n 0 dc n 0
1 1
( ) V V cos( ) and ( ) I I cos( )n n
n n
v t n t i t m t
dc dc n n
1
1P V I V I cos( )
2 n nn
30
Total average power is the sum of the average powers
resulting from each harmonically related voltage and
current.
After substituting v(t) and i(t) and evaluating, we obtain:
RMS value: The RMS value, or the effective value, of a periodic
function is given by
Average Power and RMS Values (2)
31
► RMS value of a periodic function is the square root of the
sum that involves the amplitudes of all the harmonic
components in its FS.
After substituting f(t) and evaluating:
2 2 2
0
1
1 ( )
2rms n nn
F a a b
ac
n
nn
dc
tnAatf
1
00 )cos(()(
Example (Prob. 17.48): For the following circuit
Average Power and RMS Values (3)
32
Solution:
(a) find v(t), and
(b) calculate the average power dissipated in the resistor.
Average Power and RMS Values (4)
33
Solution (end):
Average Power and RMS Values (5)
34
Example 17.9: Find an estimate for the RMS value of the voltage
Solution:
Exponential Fourier series (1)
35
0 0 0
1
( ) ( cos sin )n n
ndc
ac
f t a a n t b n t
A compact way of expressing the Fourier series
Complex or exponential Fourier
series representation of f(t)
The exponential Fourier series of a periodic function f(t) describes
the spectrum of f(t) in terms of the amplitude and phase angle of AC
components at positive and negative harmonic frequencies.
Exponential Fourier series (2)
0 0 0
1
( ) ( cos sin )n n
ndc
ac
f t a a n t b n t
2 2 1
n , tan nn n n
n
bA a b
a
36
Exponential Fourier series versus Trigonometric Fourier series
2 2 2
0
1
1 ( )
2rms n nn
F a a b
ac
n
nn
dc
tnAatf
1
00 )cos(()(
0 0 0
1 Tc a f t dt
T
Exponential Fourier series (3)
Example 17.10: Find the exponential Fourier series expansion of the
periodic function f(t)=et, 0 37
Solution: Exponential Fourier series (4) 38(a) Complex amplitude spectrum (b) Complex phase spectrum Practice: Exp. Fourier series
39
Determine the coefficients c0 and cn and the Exponential Fourier series (5)
40
Symmetry that impacts Cn
0 )()( tftf Even f(t)
Cn = Cn
Odd f(t) )()( tftf
0 Exponential Fourier series (6) 41
Solution: Verify ?
A is amplitude and T is period of v(t) Exp. Fourier series of some waveforms (1)
42 Exp. Fourier series of some waveforms (2)
43
Reference: The Fourier series provides Spectrum analyzer: can be used to conduct noise Application: Spectrum analyzer
44 Average power Example: Spectrum analyzer
45
Solution
dc dc n n After analyzing the signal of a voltage v(t) and a current i(t), a For v(t), this signal is made up of three AC components: 120 For i(t), this signal is made up of a DC and three AC 1) Write v(t) and i(t) as Fourier series using the 2) Calculate the RMS value of v(t) and i(t) resistor if v(t) is applied across a 5-Ω resistor.
Practice: Spectrum analyzer
46 Band-limited periodic function is the one whose Sampling theorem
47
Sampling theorem: A band-limited periodic function Filters are an important components of electronics and communications Example:
Application: filters (1)
(a) Input and output spectra of a low-pass filter 48
► : ωc large, Application: filters (2)
(a) Input and output spectra of a band-pass filter
(b) The band-pass filter passes only the 49
► Highly-selective filter:
► Passed band of frequencies
where B is its bandwidth Application: filters (3)
50
Example 17.14: If the sawtooth waveform in Fig. 17.45(a) is applied to an Solution: Only the DC and fundamental 51
References Circuits”, 4th Ed. (2009), 5th Ed. (2013), 6th Ed. (2017). McGraw Hill.
2. H. W. Jackson, D. Temple, B. Kelly, Introduction to Electric Circuits, 3. H. Saadat, Power System Analysis, 3rd and previous editions, 4. R. C. Dorf, J. A. Svoboda, Introduction to Electric Circuits, 9th Ed. 5. J.D. Irvin, R.M. Nelms, Basic Engineering Circuit Analysis, 10th Ed. 6. J. W. Nilsson, S. A. Riedel, Electric Circuits, 10th Ed. (2014) and 7. Other sources. ECE 5750 n System & Power Quality
Part 8: Basic principles, analysis & control h armonic
s
Instructor: Dr. Ha Le California State Polytechnic University, Pomona
2
What will be presented? 2. Fourier series of distorted waveforms
3. Sources of harmonics
4. Effects of harmonic distortion on power systems
5. Harmonic voltage and distortion limits
6. Principles for controlling harmonics
Reading: PQ textbook, Chapter 6, 7 3
Fundamentals 4
What is harmonics? devices
V(t )
I(t)
V
I
Nonlinear Resistor 5
Linear loads drawing currents in linear For a resistive load, the For an inductive load, the 90 °. It is 6
Non-linear loads draw currents in nonlinear Current waveshapes are Current waveforms are Current is discontinuous and periodic 7
Idealized six-pulse rectifier 8
Adjustable speed drive diode transistor Total harmonic distortion (THD) Peak current, approx. 300 A
Double hump due to rapid
capacitor charging
100 HP, 6 pulse AC drive
Source: www.schneider-electric.us
ASD with 9
Fourier series ꞏ
+
+ ꞏꞏ
+ 60 Hz 300 Hz 420 Hz 540 Hz 660 Hz 780 Hz 180 Hz )sincos()(
1
1
10 tnbtnaati n
n
n
T
n tdtnti b
tdtnti a
dtti a 0 1
0 1 0 0
sin )(
2
cos)( )(
i(t) = i(t + T) 10
Example: Fourier series (1) 11
Example: Fourier series (2) ,…7,5,3,1
6 cos
6
7 6 cos cos sin sin 00
1 6/11
6/7 01
6/5
6/ 0
10
nwhere
n n I nn
n tdtnI tdtnI b n
The calculation of bn can be simplified since the waveform has both odd and ,
cos sin sin)( 0 2/
0 nn tdtnI 12
Example: Fourier series (3) The Fourier series of the current is as follows. Harmonic currents of the
tt
tt tI
tnbti n 1 25sin 5 23sin 23
1
19sin 19
1
17sin 17
1
13sin 13
1
11sin 11 7sin 7 5sin sin 32
sin)( 13 Example: Fourier series (4) 6.2 is decomposed into a series Harmonic content can be 14
Harmonic spectrum of a charger
Harmonic Spectrum for EVSE DC Parameter Value
Rated Voltage (V) 208
Rated Power (kW) 50
Rated PF 1.0
Specifications
Harmonic Spectrum
Courtesy of A. Maitra, DC Fast Chargers for PEVs – Real and Harmonic Magnitude 3 5. 93
5 2.13 52
9 0. 83 11 0. 45 13 1.07 37
17 0. 40
19 0.57 15
Harmonics: Causes & effects on PS (1) because power electronic devices Non-sinusoidal currents The harmonic currents passing )(sin 1 ,…7,5,3
n nsn tnIZV )(sin)sin( 1 111 n nsssload tnIZtZIVV
Fig. 6.6. Harmonic current following through the 16
Harmonics: Causes & effects on PS (2) magnitude of harmonic currents
2) If the impedance Zs at the load bus is very large, the voltage distortion 3) The worst case is experienced when large harmonic currents are According to IEEE Std. 519-19 92 The control over the amount of harmonic current injected into the Assuming that the harmonic current injection is within reasonable limits, Impact of harmonics on RMS current (1) )(sin)sin()( 1 11 tnItIti n Let (6.15) 18
Impact of harmonics on RMS current (2) The fundamental and harmonic terms can be separated and written as,
where Irms and IrmsH are the RMS of fundamental and harmonic frequency of i(t) 19 Impact of harmonics on RMS voltage (1) represented as follows:
The RMS value of the distorted voltage waveform:
where Vrms1 and VrmsH are the RMS of the fundamental and )(sin)sin()( 1 11 tnVtVtv n 22
222
1 5 31
…
H rmsrms
rmsrmsrmsrms
VV
VVVV
20
Example: RMS value with harmonics Solution: Based on Fourier series of the current i(t) obtained in Example 6.1
00 22222220
8165.0 19 17 13 11 2 II
IIrms
21
Power in sinusoidal conditions (1) The active power is the rate at which energy is expended, dissipated or rmsrms IVS
)cos(
)cos()cos( )()(
1
11 1 111
11
0
S
IV dttitv P
rmsrms )2sin()]2cos(1[)()()( tQtPtitvtp
Active power can also be computed from the instantaneous power:
(6.23)
(6.24) 22 Power in sinusoidal conditions (2) The relationship between P,Q and S can be summarized in this )sin( )sin(
2
)sin( 11 rmsrms IV SQ 23 Power in non-sinusoidal conditions (1)
The instantaneous voltage and current under non-sinusoidal Apparent power (S) is given by
)(sin)sin()( 1 ]][[ 2222
222 11 HH rmsrmsrmsrms
rmsrms IIVV
IVS
24
Power in non-sinusoidal conditions (2) 25
Power in non-sinusoidal conditions (3) power SDP and the harmonic apparent power SHA. voltage and the RMS of the harmonic currents.
Voltage distortion power SVDP is given as the product of the fundamental RMS Distortion power SDP is
2 n 11 2 n 22 26
Power in non-sinusoidal conditions (4) current harmonics. The term can be further resolved into total harmonic Note: There is no such thing as reactive harmonic power. Under )sin( 22222
nnrms rmsH
nnrms HHrmsrmsH
nn HH
IV N
IVP
NPIVS
27
Power factor: Displacement & Total The power factor in sinusoidal condition is also known as displacement The power factor that takes into account contribution from all active 1 S PF
)cos( 11 1 P F
S PF H 1 28
Practice: I, V, P, THD under harmonics
Fig. 6.21 29
Solution: I, V, P, THD under harmonics (1)
The Fourier series coefficients are: 30
Solution: I, V, P, THD under harmonics (2)
For n = 3, 5, 7 … an = 0;
For n = 2, 4, 6 …
For n = 1 31 Solution: I, V, P, THD under harmonics (3)
For n = 1 32 Solution: I, V, P, THD under harmonics (4) 33
Solution: I, V, P, THD under harmonics (5) 34
Solution: I, V, P, THD under harmonics (6) 35
Solution: I, V, P, THD under harmonics (7)
Note: Irms1 is the magnitude of fundamental component of load current; 36
Solution: I, V, P, THD under harmonics (8)
Part (e):
By inspection of the voltage V(t) is an ideal source with 37 Harmonic phase sequences (1) assumed to be balanced. Let phase ‘A’ current, ia(t) be represented as,
Phase ‘B’ current lags phase ‘A’ current by 120°. The Fourier series of Phase ‘C’ current leads phase ‘A’ current by 120°. The Fourier series of 1 )sin()( nna tnIti
,…3,2,1 ,…3,2,1 1 ) 2 )) (sin[)(
n n ntnI
tnIti
38
Harmonic phase sequences (2)
The first three components of ia(t), ib(t) and ic(t) are: 39
Harmonic phase sequences (3) 40 Triplen harmonics issue for grounded-wye system Two typical problems which For the system with perfectly Summing currents at node ‘N’, 41
Overloading of the neutral balanced fundamental currents sum to 0, neutral current contains no 300% of phase current
A B
C
N Neutral has no fundamental
Neutral has harmonic currents In = 3 I3 + 3 I9 + 3 I15 + … 42
Triplen harmonics The neutral current is simply the sum of the line currents, thus, 43
Impact of triplen harmonics on TRF single-phase loads on the wye-side.
2. Measuring the current on the delta side of a transformer will NOT 3. The flow of triplen currents can be interrupted by the appropriate Distribution Transmission 44
Harmonic indices THD: A measure of the effective value where M1 is fundamental component THD is related to the RMS value of the TDD: Ratio between RMS value of h1 is fundamental and hmax is highest Example: Calc. of harmonic indices Solution:
45 46
Harmonic sources (1) 1) Switch-mode power supplies: A very high third harmonic 2) Fluorescent lighting: Discharge lamps requiring a ballast to 3) Three phase power converters: They do not generate third 4) Three phase power converters – DC drives: The two largest 47
Harmonic sources (2)
Harmonic sources from industrial loads:
1) Arcing devices: Arc furnaces, arc welders etc. Electric arc is 2) Saturable devices:
Transformers: Harmonic voltage distortion from TRF Motors: Exhibit some distortion in the current when 48
Effect of harmonics on capacitors (1) voltage distortion during resonance
The capacitor current is also significantly The harmonic current shows up IEEE Std. 18-2002 specifies continuous 1) 135% of nameplate kVA R
2) 110% of rated RMS voltage
3) 135% of rated RMS current based 4) 1 20% of peak voltage (including Under presence of 49
Effect of harmonics on capacitors (2) 50 Effect of harmonics on TRF: Losses (1) 1) I2R losses L
The transformer total load loss in watts is,
When the transformer is supplying a sinusoidal current at power The total load loss in per unit of PI2R-R is as follows,
][
][ WPP
WP PPP
ECRI
OSL ECRILL
][2 WPPP RECRRIRLL
][1 puPP RECRLL 51
Effect of harmonics on TRF: Losses (2) consists of the fundamental I1 and harmonic components I3, I5…, I2R losses is
The winding eddy-current losses is given by,
Therefore, the transformer total load loss when supplying non-sinusoidal ][ 22 W I PP h
RRIRI
][2 Wh PP h ][2 2 P P h R h ECRILL 52 Effect of harmonics on TRF: Losses (3) current IR, total load loss PLL becomes
][2
][ ][1][
][][][][
2 22 222 puI puPpuI
puIhpuPpuIpuP
h h hREChLL TRF supply Non-sinusoidal current TRF supply sinusoidal current 53
Derating TRF for non-linear loads (1)
K-factor has been used by transformer manufacturers to design 54
Derating TRF for non-linear loads (2) Solution: K-9 transformer would be an appropriate transformer for the given nonlinear 55
Analysis & Control 56
Root causes: Current & Impedance 57
Harmonics control: IEEE 519 approach Limit the level of harmonic current injection at the point of Current injections: Users can control them (think: nonlinear Utility: Recall: For the same amount of harmonic current injections
Strong system (low impedance system) voltage distortions Weak system (high impedance system) voltage distortions will 58
Concept of common coupling point (1)
Utility System
Customer Under Study
Other Utility PCC
IL
PCC at the transformer primary 59
Concept of common coupling point (2)
Utility System PCC at the transformer secondary 60
IEEE 519 voltage distortion limit in % of nominal fundamental frequency voltage. 61
Short-circuit ratio 2. Find the load average kilowatt demand Pd over the most recent 3. Convert the average kilowatt demand to the average demand 4. Short-circuit ratio is
A 1000 I SC
kVPF 3 L SCR 62
Current distortion limits (1) 63
Current distortion limits (2) 64
Current distortion limits (3) 65
Harmonic current flow
NORMAL PATH 66
System impedance under harmonics capacity (SCC, measured in MVA), which is calculate as
SCSC
SCSCSC
I MVA jXR Z
3 1,, shs hXX
ZSC is assume to be Xs, h is inductive reactance at hth harmonic
Xs, 1 is inductive reactance at fundamental frequency
XC, h is capacitive reactance at hth harmonic
Xc is capacitive reactance at fundamental h X
X chc ,
(7.1)
(7.8) 67
SCC at substation
tSC XX
(%) t LL kV Equivalent system impedance is often dominated by the service 68
Example: Equiv. impedance under harmonics (1) 69
Example: Equiv. impedance under harmonics (2) 70
Example: Capacitor impedance under harmonics
h X Q X X chcrated p
LL ,
2 2 71
Parallel resonance (1)
p p p p p p p XQXQ
R R R XXjR jXjXRZ
22 )()(
)( )( )( Simplified distribution system
Subscript “p” means “parallel” resonance. 72
Parallel resonance (2) the harmonic frequencies being produced by the nonlinear load C L L CL s s 1 1 Rs = Resistance of equiv. Ls = Inductance of equiv. C = Capacitance of capacitor (7.9) 73
Parallel resonance (3) p s
pp s p R Q
p pp p p During parallel resonance, a small harmonic current can cause a large (7.11) 74
Parallel resonance: Current behavior The currents are magnified Qp times.
This may lead to capacitor failure, fuse blowing, transformer The extent of voltage and current magnification depends on p p p p p pp ap
IQ X IQ
75
Impact of capacitor size on parallel resonance 76
Estimating resonant frequency rated sc
s c MVA h 3 (7.13) 77
Example: Estimating resonant frequency 78
Effect of resistive load on parallel resonance
h 0% Resistive Load
10%
20% R 33 2 ap B KEs
Sys tem voltage = 69 kV Series res is tive dam ping = 1 ohm
Cap bank is rated 60 Mvar at 69 kV
Z(f)
0.0 – 79
Frequency – Impedance scan
0 200 400 600 800 1000 1200 1400 1600 1800 2000 50 100
150
200
250
300
frequency |Z Frequency Scan
With Capacitor 80
Series resonance The voltage at the PF capacitor is magnified and highly distorted:
(7.14)
Vh and Vs,pf is harmonic voltage corresponding to harmonic current Ih and svct
c X , 81
Series and Parallel resonance
In many systems with potential series resonance, parallel 82
Practice: Capacitor under harmonics & resonance (1) Practice: Capacitor under harmonics & resonance (2)
83 Practice: Capacitor under harmonics & resonance (3)
84
System Solution: Capacitor under harmonics… (1)
85 Solution: Capacitor under harmonics… (2)
86 Solution: Capacitor under harmonics… (3)
87 Solution: Capacitor under harmonics… (4)
88 Solution: Capacitor under harmonics… (5)
89 Solution: Capacitor under harmonics… (6)
90 Solution: Capacitor under harmonics… (7)
91 Solution: Capacitor under harmonics… (8)
92 93 Three common causes of harmonics
1. The source of harmonic currents is too great: Large 2. The path in which the currents flow is too long 3. The response of the system magnifies one or more 94
Principles of controlling harmonics (1) Nonlinear loads: Producers of harmonic currents. End-users Utility systems: Utility system impedance may interact adversely Strong system: Voltage distortions are likely low and benign.
Weak system (large reactance): Voltage distortions are likely Harmonic resonance: One of system impedance resonant 95
Principles of controlling harmonics (2) currents. Change transformer configurations: Use a pair of delta-delta, and delta- 2) Large HARMONIC CURRENTS: Add filters to siphon currents or block of distortion as practical.
3) HARMONIC RESONANCE: Modify frequency response by filters, with a different short-circuit impedance. 96
Harmonics control: Utility feeders harmonic evaluations.
2) When harmonic resonance occurs, change capacitor bank 3) Triplen harmonics on wye-ground capacitor banks, change 4) Widespread harmonic sources along the feeder: Distribute 97
Harmonic filter installation on Oil-insulated iron-core reactorsCapacitor banks and switches 98
Harmonic filters on a substation 99
Harmonics control: End-user facilities Is it due to resonance with power factor capacitors YES Change capacitor size.
NO Use filters. 100 Harmonics control: Load side
Most common method:
Use a pair of transformers configured as delta- Goal: To reduce characteristic harmonics from 101
Reducing load harmonic current using TRF
A simple delta-connected TRF can block the flow of triplen By using TRF to shift the phase of half of the 6-pulse power Zig-zag and grounding TRFs can eliminate triplen harmonics off 102
In-line reactors or chokes
A typical 3% input Line input choke (% on Drive Base)
In t ur nt is rt n ) 0.0%
10.0%
20.0%
30.0%
40.0%
50.0%
60.0%
70.0%
80.0%
0% 1% 2% 3% 4% 5%
Choke (on ASD kVA)
The inductance slows the rate at which the capacitor on the DC bus can The net effect is a lower-magnitude current with much less harmonic 103
Harmonic control: Filters
Passive filters
These are inductance, capacitance, and resistance Tuned to control harmonics.
Advantage: Economical and straightforward.
Disadvantage: May interact adversely with the power 104
Harmonic filters: Shunt filters
Known as single-tuned notch filters. harmonic current. SINGLE-TUNED FIRST ORDER 2ND ORDER 3RD ORDER 105
Response of a 5th harmonic filter
Harmonic Number h
0 1 3 5 7 9 11 13 15 17
(a) Typical low voltage (b) Equivalent circuit of B X C
F XF SC (c) frequency hZ capacitor capacitor to hNotch
F C X 106
Case study: Shunt harmonic filter design 1. A capacitor bank to be installed at 480V bus to improve PF to 98%
2. Single-tuned notch filter to be designed based on capacitor bank size
HARMONIC EQUIVALENT CIRCUIT
h=2, 3, 4, …C
R L I h
with no reactor 107
Step 1: Select a tuned frequency for the filter (1)
The tuned frequency is selected based on the harmonic Start at the lowest harmonic frequency generated by the Tune the filter slightly below the harmonic frequency of The above situation is worse than without the filter as 108
Step 1: Select a tuned frequency for the filter (2) 109
Step 2: Compute capacitor bank size The filter size is based on Assume that no capacitor is The desired power factor is Thus, the net reactive power Real power of the load at 0.75 PF P = 1600 x 0.75 = 1200 kW
Required reactive compensation Qfilt = 1200 * [ tan(acos(0.75)) – = 814.28 kVAR 110
Step 2: Compute capacitor bank size For a nominal 480V system, the net Note that XFilt above is for Y- For tuning at the 4.7th harmonic,
Thus, the desired capacitive At this point, it is not known whether Try a capacitor rated at 480 V
Standard capacitor 750 kVAR (note LCapFilt XXX
2828.0 48.01000 22
filt
LL Q 2962.0 )7.4(2828.0 2 h X Filt Cap
kVAR kV Cap LL 75.777 )1000(48.0)1000( 22 XCap = 0.3072
XCap = h2XL = 4.72 XL 111 Step 3: Compute filter reactor size
The filter reactor size can now From Step 1, the desired The filter reactor size is computed Alternatively, the reactor size can 0.0139 0.3072 )( X wyeCapfundL
mH 0.0369 )( fundLXL
(wye) L C2 Hz 282607.4 hf 112
Step 4: Evaluate filter duty requirements
Evaluation of filter duty requirements typically involves capacitor peak voltage, current, kVAR produced, and RMS voltage.
IEEE Std 18-2002 is used as the limiting standard to evaluate these Computation of the duties are fairly lengthy, therefore, they are computation for fundamental duties,
harmonic duties, and
RMS current and peak voltage duties. 113
Step 5: Compute fundamental duty requirements
Determine a fundamental 1) The apparent reactance of the 2) The fundamental frequency filter 3) The fundamental frequency 4) The actual reactive power Evaluation:
Filter draws more current than
2933.03072.00139.0
)(wyeCapL A 86.449 3 3 fund
kV fundduty X actual
V 76.5023 )(,, wyeCap
filt
fundduty fundduty XIV
kVAR 822.8 3 ,, fundduty fundduty 114
Step 6: Compute harmonic duty requirements (1)
1) Since the nonlinear load 2) Harmonic current contributed to It is assumed that the 2% fifth harmonic The utility impedance is neglected
Fundamental frequency impedance of The fifth harmonic impedance of the A 9.483 1600 3 actual kV upII
0.0058 48.0 22 )( actual kV 0.02880.00585)(5, fundTT hXX 115
Step 6: Compute harmonic duty requirements (2)
The harmonic impedance of the Harmonic impedance of the reactor is:
Given that the voltage distortion on 3) The maximum harmonic 4) The harmonic voltage across 0614.0 3072.0)( X 06951.00139.05)(5, fundLL hXX
5,5),(5, )( )( LwyeCapT
actualutilityh XXX I 06951.00614.00288.03 A 13.535.23501 384.9)( totalhI
h IV wyeCaptotalh cap )( V 95.56 0.3072 116
Step 7: Evaluate total RMS current 1) Total RMS current passing through This is the total RMS current rating 2) Assuming the harmonic and 3) The RMS voltage across the 4) The total kVAR seen by the A 89.108513.53586.944 22
2 2
utilityhfundtotalRMS III
cap cap cap peakV 55.791295.562502.76
2 2 cap cap V 505.9795.56502.76 22
cap filt cap kVAR 65.51997.50589.10853 117
Step 8: Evaluate capacitor rating limits
Table for evaluating filter duty limit
This would be a very marginal application Use a capacitor bank Required kVAR =
Choose 1200 kVAR
Repeat all the kVAR1172 750 2 118
Step 9: Evaluate filter frequency response
The harmonic at which the parallel resonance below the notch This assumes the service transformer reactance dominates the Hence, the parallel resonance frequency (3.95) is safely away Beware of transformer energization events.
95.3 3072.0 )()( )(‘ wyeCap XX h 119
References
1. S. Santoso, Fundamentals of Electric Power Quality, 2012.
2. R. C. Dugan, M. F. McGranaghan, S. Santoso, W. Beaty, 3. T. A. Short, Electric Power Distribution Handbook, 2003.
4. J. D. Glover, M. S. Sarma, T. J Overbye, Power System Analysis 5. T. Gonen, Electric Power Distribution Engineering, 3rd ed., 2014, 6. W. H. Kersting, Distribution System Modeling and Analysis, 3rd 7. IEEE Std. 519-1992.
8. Other sources Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le 1
California State Polytechnic University Pomona ECE 5750 Distribution system and power quality – Spring 2020 Homework 4 Problem 1 The substation transformer is connected to an infinite bus. The infinite bus voltages (i.e. the 5000 kVA, 69 kV delta – 13.8 kV grounded Y, Z = 1.7 + j8.5 %
The phase impedance matrix for a four-wire wye line between Node 2-3 is
The four-wire wye feeder between Node 2-3 is 0.75 miles long. An unbalanced wye-connected Phase a: 650 kVA at 0.85 lagging power factor Assuming that the regulators are in the neutral position:
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le 2
1) Determine the forward and backward sweep matrices for the substation transformer 2) Use the modified ladder technique to determine the line-to-ground voltages at all 3) Calculate all node voltages in actual values in volts and on a 120-V base. Problem 2 1) Compute the actual equivalent line impedance between nodes 2 and 3.
2) Determine a potential transformer ratio and current transformer ratio given that the 3) For the load conditions of Problem 1 with the regulators in the neutral position, 4) Determine the appropriate tap settings for the three regulators to hold the node 3 5) With the regulators taps set, compute the actual load voltages in volts and in per unit.
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le 1
California State Polytechnic University Pomona ECE 5750 Distribution system and power quality – Spring 2020
Short Answer for Homework 4 Problem 1
tol = 0.0001; % tolerance level
number of iterations: 4
V1 is the voltage at the primary of the substation
Q1 TRF forward and backward sweep matrices Ztabc = Specified line-neutral source voltage and load VLN_ABC = 1.0e+04 * S3 = 1.0e+05 *
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le 2
Final result when system converged
V1_mag = 1.0e+04 * 3.9837
3.9847
3.9840
V1_ang =
V2_mag = 1.0e+03 * 7.7754
7.8380
7.7583 V2_ang = V3_mag = 1.0e+03 *
7.7452
V3_ang = 178.7369
57.0198
Final voltages in pu & volt on a 120-V base V1_pu = V2_pu = Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le 3
V3_pu = 0.9721 V3_120V = 116.6528 Problem 2
Q 1 n 2 Line impedance – Compensator R and X setting
Zline_aver = 0.2360 + 0.4399i ohm
Npt = 66.3953 CTp = 209.1849 CTs = 5
Zcomp_volt = 0.4489 + 1.4227i
0.9520 + 0.8521i
0.8293 + 1.8829i
Zcomp_ohm = 0.0898 + 0.2845i
0.1904 + 0.1704i
0.1659 + 0.3766i
Q3 Compensator voltage and current input – Load new
Icomp_mag = 2.0060
1.5296
2.9474 A
Icomp_ang = -93.4303
152.8949
38.8249 deg.
Vrelay_mag = 116.6528
117.6745
116.0355 V
Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le 4
Vrelay_ang = -61.6420
178.7369 57.0198
Q4 Final tap position, V=121V, bandwidth=2V
Tap = [5; 4; 6] Q5 With the regulators taps set, compute the actual load voltages in volts and in per tol = 0.0001; % tolerance level V2r is the input voltage to regulator
V2 is the output voltage from regulator
V1 is the substation primary voltage Regulator matrices (Self)
Final result when system converged V1_mag = 1.0e+04 *
3.9837 V1_ang = -30.0079
-149.9956
90.0009
V2_mag = 1.0e+03 *
8.0263
8.0390
8.0609
V2_ang = -61.4512
178.7894
57.4622
V2r_mag = 1.0e+03 *
7.7754 7.7586 Cal Poly Pomona ECE 5750 Distribution system and power quality – Instructor Dr. Ha Thu Le 5
V2r_ang = -61.4512
178.7894 7.9968
8.0146
8.0091
V3_ang = -61.6288
178.7365
57.0535
V3_pu = 1.0037
1.0059
1.0052
V3_120V = 120.4417
120.7107
120.6276 ECE 5750 Part 5: Distribution Feeder Analysis & Overview of distribution automation, substation design,
role of capacitors, renewable generators, and energy storage systems
Instructor: Dr. Ha Le California State Polytechnic University, Pomona 2
What will be presented? Power-flow analysis
Ladder iterative technique
2. Overview (some basic concepts)
Overview of feeder automation and substation n
Role of capacitors, renewable generators, Reading: DS textbook Chapter 10 and Gonen book 3
Distribution Feeder Analysis 4
Analysis of distribution feeder
Study of the feeder under normal steady-state Study of the feeder under short-circuit conditions In this course, we focus on power flow analysis
Technique for power flow: An iterative technique 5
What does power flow determine?
1) Voltage magnitudes and angles at all nodes
2) Power flow in each line section (e.g. real and reactive 3) Power loss in each section
4) Total feeder input kW and kVAR
5) Total feeder power loss
6) Load kW and kVAR 6
Ladder network iterative method (1)
54321 Z45Z34Z23Z 12
ZL2 ZL3 ZL4 ZL5
II2 I 3I23I12 I34 I45 I4 5VS +
–
Solution:
Assumed that all of the line impedances and load impedances are 1. Forward sweep: Calculating the voltage at node 5 (V5) under a 7
Ladder network iterative method (2) ZL2 ZL3 ZL4 ZL5 Solution (cont.):
5 5L
VI
4 5 45 45V V Z I
45 5I I
34 45 4 …I I I
This procedure continues V1 is compared with specified VS
1
sVRatio ( )final solution currents or voltages Ratio
2. Backward sweep: 3. Compare and scale to obtain solution
Scaling is possible as the network is linear
I4 = V4 / ZL4 8
Non-linear network
– SV 12Z 23Z 34Z 45Z1 2 3 4 5
S2 3S 4S 5S
Solution: Applying the procedure for linear network with modification
*
n n SI 1. Step 1
Use backward sweep to 2. Step 2
Use (a) specified source Second forward sweep: 3. Step 3: The forward and backward 1SV V specified
Use voltages from the 9
Practice: Iterative method for non-linear network Line segment 2-3 impedance:
Loads: 10
Solution: Iterative method … (1) The first forward sweep:
Vold = V3 11
Solution: Iterative method … (2) The current flowing in the line segment 2–3 is
The load current at node 2 is 12 Solution: Iterative method … (3) The second forward sweep: 13
Solution: Iterative method … (4) compute the new line currents. This is followed by the third After 4 iterations, the voltages have converged to an error of 14
Typical distribution feeder (1)
Solid line: Overhead lines 15
Typical distribution feeder (2)
All series elements The line between nodes 3 One way to model the 16
Model for series components 17
Model for shunt components 1) Spot static loads: Induction machines, capacitor banks. An 2) Spot static loads can be modeled as constant complex power, 3) Line: Most of the time, the shunt capacitance of the line For all the loads, the “node” currents may be 3-phase, 2- 18
Ladder iterative technique 2) 1st forward sweep: Assuming no-load condition, calculate 3) Compare computed source voltage to specified source voltage 4) 1st backward sweep: Calculate all line currents using the 5) 2nd forward sweep: Using specified VS and the line currents 6) Repeat forward and backward sweeps until computed and 19
Example: Feeder analysis Phase impedance Load
Infinite [ZeqS] [ZeqL]
[Iabc][IABC]
21 3 4 3-P TRF consists of three 3‐phase wye‐connected 20
Solution for Example: Feeder analysis (1) for each series element. The modified ladder method use only [A], Specified tolerance: 0.001 pu
Source line segment with shunt admittance neglected: 21
Solution for Example: Feeder analysis (2) 22
Solution for Example: Feeder analysis (3)
Load line 23
Solution for Example: Feeder analysis (4) 24
Solution for Example: Feeder analysis (5) 25
Solution for Example: Feeder analysis (6) 26
Solution for Example: Feeder analysis (7)
Node 4 loads: complex power
Infinite bus line-to-line and line-to-neutral voltages: 27
Solution for Example: Feeder analysis (8) 1ST ITERATION: Backward sweep Voltages and currents at Node 3: 28
Solution for Example: Feeder analysis (9)
Voltages and currents at Node 2: 29
Solution for Example: Feeder analysis (10) L-L voltages at Node 1: 30
Solution for Example: Feeder analysis (11)
Magnitude of the line-to-line voltage errors
These errors are greater than the specified tolerance of 0.001 Forward sweep: It uses the equivalent L-N voltage at the Load 31
Solution for Example: Feeder analysis (12)
This completes the first iteration. 32
Solution for Example: Feeder analysis (13)
2nd ITERATION begins by computing the new currents at the The forward and backward sweeps continue until the error at The Mathcad program is used to analyze the system, after 8 33
Solution for Example: Feeder analysis (14)
The flowchart of a Mathcad® program
Load voltages V4 on a 120 V base are:
The voltages at node 4 are below Hence, three step-voltage 34
Solution for Example: Feeder analysis (15) Potential transformer ratio = 2400 / 120 V i.e. Npt = 20
Rated current of the transformer bank:
The CT ratio is selected to be 1000 / 5 = CT = 200 35
Solution for Example: Feeder analysis (16) The three regulators are to have the same R and X compensator settings. 36
Solution for Example: Feeder analysis (17) Compensator current:
Voltages across the voltage relays: 37
Solution for Example: Feeder analysis (18)
Assume that the voltage level has been set at 121 V with a The taps setting are: 38
Solution for Example: Feeder analysis (19) All other matrices are zero. 39
Solution for Example: Feeder analysis (20) taps.
The same modified ladder iterative technique is repeated as The system converges after 4 iterations. The load voltages at The load voltages at Node 4 are within the desired limits. 40
Sweep equations including regulator Forward sweep:
Backward sweep: 41
Tap changing routine 42
Computational flowchart
The computational 43
Load allocation computed source voltage to determine if the tolerance is However, usually the input complex power (kW and kVAR, total 3- 1) Compute Ratio = metered input / computed input
2) Scale phase loads by multiplying the loads by Ratio
3) Use ladder iterative method to determine a new computed 4) Repeat Step 1-3 until the computed input is within a 44
Overview of substation design, capacitors role, & energy storage systems 45
Distribution automation
2) The supervisory control and data acquisition (SCADA) system involves 3) The distribution automation and control (DAC) system oversees the The power grid 1) The EMS exercises 46
Motivation for DS automation 1) Increased reporting requirements of reliability councils 2) Operation of the electric system closer to design limits.
3) Increased efficiency requirements because of much 4) The tendency of utilities to monitor lower voltages than Overall: Development of Smart Grid DAC automation & control functions 47 48
Vision of future grid
Network of microgrids.
Distributed generation (conventional, renewable such as Distributed energy storage systems (ESS): pumped- Sensors, IT, communication systems everywhere.
Real-time monitoring, pervasive control and automation. 49
Substation design 1) Substation costs: Site costs, transmission cost, transformer 2) Substation diagram: The electrical and physical arrangements 3) Substation location: Dictated by the voltage levels, voltage 4) Rating: Substation capacity determined based on pre-determined 50
Examples of substation diagram
A typical double bus–double breaker scheme A typical main-and-transfer bus scheme 51
Role & benefits of capacitors 1) Power factor correction
2) Reactive power compensation and voltage regulation
3) Application in power quality: Harmonic filters
Question: Name an application of series capacitors?
Economic benefits from capacitor installation:
1) Released generation and transmission capacity
2) Released distribution substation capacity
3) Reduced energy (copper) losses
4) Reduced voltage drop and improved voltage regulation
5) Saving on capital expenditure by delaying system expansions
6) Revenue increase due to voltage improvements etc. 52
Role of renewable generators They are called Dispersed Storage and Generation Examples of DSG technologies: Hydroelectric and diesel If properly planned and operated, DSG may provide benefits to But, integrating them also creates lots of problems to power 53
Role of renewable generators Some technical, economic, and environmental benefits:
1) Reduction of environmental pollution and global warming 2) Diversify power supply sources: Renewable sources in addition 3) Overall grid power quality and reliability improvement.
4) Making use of local power supply, which may reduce 5) Enabling the development and usage of Microgrids 54
Role of renewable generators Some technical problems:
1) Islanding: Islanding is an undesirable condition where a 2) Power quality: Harmonics from converter systems used by PV 3) Power fluctuation: Intermittent flow, causing power deficit or 4) Reverse power flow: Many utility substations cannot handle Overall, grid operation is much harder. 55
References
1. S. Santoso, Fundamentals of Electric Power Quality, 2012.
2. R. C. Dugan, M. F. McGranaghan, S. Santoso, W. Beaty, 3. T. A. Short, Electric Power Distribution Handbook, 2003.
4. J. D. Glover, M. S. Sarma, T. J Overbye, Power System Analysis 5. T. Gonen, Electric Power Distribution Engineering, 3rd ed., 2014, 6. W. H. Kersting, Distribution System Modeling and Analysis, 3rd 7. IEEE Std. 519-1992.
8. Other sources
Solution (end): Plotting the complex frequency spectrum of f(t)
exponential Fourier series of f(t) in the figure below.
)cos()(4
)(2
2/
0 0
2/
00
n
T
n
T
b
dttntf
T
a
dttf
T
a
/ 2
00
0, 0
4 ( )sin( )
n
T
n
a a
b f t n t dt
T
Cn = – Cn
Example: Find the exponential Fourier series expansion of the square wave v(t) below.
Table 15.5-1, p.759, Introduction to Electric Circuits,
Richard C.Dorf, James A. Svoboda, 9th Edition, Wiley, 2013.
amplitudes and phases of the
harmonics versus frequency, showing
which frequencies are playing an
important role in the shape of the
output and which ones are not.
is an instrument that displays the
amplitude of the components of
a signal versus frequency.
and spurious signal analysis,
phase checks, electromagnetic
interference and filter
examinations, vibration
measurements etc.
A spectrum analyzer indicates that a signal is made up of three
components only: 640 kHz at 2 V, 644 kHz at 1 V, 636 kHz at 1
V. If the signal is applied across a 10-Ω resistor, what is the
average power absorbed by the resistor?
1
1P V I V I cos( )
2 n nn
spectrum analyzer indicates the following result:
rad/s and phase 450 at 7 V, 240 rad/s at 5 V, and 300 rad/s at 3
V.
components: The DC component is 3.6741 A; the three AC
components are 120 rad/s at 6A, 240 rad/s at 4A, 300 rad/s at
2A.
amplitude-phase form.
3) Determine the average power absorbed by the
amplitude spectrum contains only a finite number of
coefficients An or cn
In this case, the Fourier series is
whose Fourier series contains N harmonics is uniquely
specified by its values at 2N+1 instants in one period.
system. This filtering process cannot be accomplished without the Fourier
series expansion of the input signal.
Filter design: To select the fundamental component (or any desired
harmonics) of the input signal and reject other harmonics.
(b) The low-pass filter passes only the dc component
when c << 0
a large number of
harmonics can be
passed.
► : ωc small,
a large number of the
ac components are
blocked, only DC
passed.
fundamental component when << 0
(B = ω2 – ω1). The filter
passes only the
fundamental component
ω0 (n=1)
ideal low-pass filter with the transfer function shown in Fig. 17.45(b), determine
the filter output.
components are passed. Filter output:
1. C. K. Alexander and M. N. O. Sadiku, “Fundamentals of Electric
9th Ed. (2012), Oxford University Press.
McGraw-Hill.
(2014) and previous editions, Wiley.
(2010) and previous editions, Wiley.
previous editions, Prentice Hall.
Distributio
of power system
Department of Electrical and Computer Engineering
1. Linear and nonlinear loads
of Harmonics
Fundamental power system (PS) frequency = f1 = 60 Hz
Harmonic frequency = n f1 where n is integer
Harmonic distortion: typically caused by nonlinear
Linear loads: Those
proportion to the applied
voltage.
impedance is constant over
time as the voltage and
current waveforms are
sinusoidal but they are also
perfectly in phase.
current lags the voltage
waveform by
linearly proportional to the
applied voltage.
Nonlinear loads: They
proportion to the applied
voltage.
NOT sinusoidal, i.e. contain
fundamental + harmonic
frequencies
periodic. Hence, they can be
decomposed into a weighted
sum of sinusoids whose
frequencies is an integer
multiple of fundamental
frequency of the periodic
waveform.
v(t)
rectifier
inverter
current = 80%
six-pulse
PWM VSI
Any periodic waveforms can be expressed as a weighted sum of
sinusoids, known as the Fourier series.
+
+
+
+
(h = 1)
(h = 5)
(h = 7)
(h = 9)
(h = 11)
(h = 13)
(h = 3)
T
n
T
T
T
T
2
1
Example 6.1:
Solution for Example 6.1: From Fig. 6.4, it is clear that the current
waveform has odd symmetry. Therefore, a0=0 and an=0. Coefficients bn are:
11
cos
5
6
2
2
sin)(
2
1
1
1
1
n
I
T
T
tdtnti
T
T
half-wave symmetry. Therefore
6
5
cos
6
2
4
4
1
6/5
6/ 0
1
1
1
n
I
T
tdtnti
T
b
T
n
Solution for Example 6.1 (Cont.):
rectangular current are simply the individual sinusoids of the Fourier series.
tt
tt
n
11
11
11
11
10
1
2
1
1
1
5
1
The rectangular current of Fig.
of sinusoids.
quantified.
Fast Charger is based on lab testing
carried out by EPRI.
Simulated Results, EPRI PQ and Smart Distribution Conference
and Exhibition, San Antonio, Texas, June 5, 2012
(% Fund)
7 1.
15 0.
Load current becomes distorted
inside the nonlinear load are
switched on and off to achieve
desired functions.
interact with PS impedance
giving rise to voltage distortion
through the impedance of the
system cause a voltage drop Vn
for each harmonic, given by
n
,…7,5,3
n
system impedance results in harmonic voltage at
the load
1) The amount of voltage distortion depends on the impedance and
can be significant even though the harmonic currents may be small
injected into a weak bus. While the load current ultimately causes the
voltage distortion, the load has no control over the voltage distortion.
system takes place at the end-use application.
the control over the voltage distortion is exercised by entity having
control over the system impedance, which often is the utility.
Under the presence of harmonics, the RMS value of i(t) is computed
as follows:
,…7,5,3
n
17
The individual terms is given as,
For a distorted voltage waveform, its instantaneous voltage is
harmonic frequency of v(t).
,…7,5,3
n
Example 6.2: Calculate the RMS value for the current waveform in Fig. 6.2.
3
2
1
1
1
1
7
1
5
1
1
1
132
Apparent power, S, is given as
consumed by the load and is measured in watts (W). ‘P’ is computed by
averaging the product of instantaneous voltage and current.
2
11
IV
T
T
In a sinusoidal case, the reactive power is simply defined as,
diagram:
1111
11
11
IV
condition is give as,
,…7,5,3
11 tnVtVtv
n
n
)(sin)sin()( 1
,…7,5,3
11 tnItIti
n
n
Carrying out multiplication for each term, we obtain
Non-fundamental apparent power SN consists of two components, the distortion
Current distortion power SCDP is defined as the product of the fundamental RMS
current and the RMS of the harmonic voltages.
,…7,5,3
11
rmsrmsrmsrmsCDP nH
IVIVS
,…7,5,3
rms
rmsrmsrmsVDP IVIVS nH
VDPCDP
SSSDP
Harmonic Apparent Power SH is the product of the RMS of voltage and
active power and total harmonic nonactive power. The harmonic
apparent power is given by
non-sinusoidal condition, the part that is in quadrature to the total
harmonic power is called as total harmonic nonactive power.
)cos(
,…7,5,3
,…7,5,3
n
n
rmsH
nn
In a sinusoid condition, power factor is simply given as,
power factor and is written as,
power both fundamental and harmonic frequencies is known as true
or total power factor. It is simply the ratio of all the total active power
for all frequencies to the apparent power delivered by the utility:
1
P
1
S
P
PP
Hence, the current expression is
Irms2 is the magnitude of the second harmonic component etc.
waveform (top figure), the
load voltage waveform Er is
not distorted because there
is no system impedance.
infinite short-circuit capacity
In determining harmonic phase sequences, the 3-phase power system is
phase ‘B’ current is,
phase ‘C’ current is
,…3,2,1
n
1
1
3
sin(
3
2
nn
nnb
Triplens become an important
with current flowing in the neutral.
arise are overloading the neutral
and telephone interference.
balanced 1-phase loads, an
assumption is made that
fundamental and third
harmonic components are
present.
fundamental current components
in the neutral are zero but the 3rd
harmonic currents are three
times the phase currents because
they naturally coincide in phase
and time.
Condition: Perfectly balanced single-phase loads in grounded-wye
system.
but balanced third harmonic currents coincide
fundamental, but third harmonic is
where its values are:
Consider the following balanced currents, ia(t), ib(t) and ic(t).
1. Transformers with neutral connections: Overheating when serving
show the triplens.
isolation transformer connection.
side
side
Two indices: Total Harmonic Distortion
(THD) and Total Demand Distortion
(TDD)
of the harmonic components Mh of a
distorted waveform.
(50Hz or 60Hz).
waveform as follows:
harmonic components and maximum
load (demand) current.
harmonic order to be considered.
Example 6.5:
Harmonic sources from commercial loads:
content in the current
provide high initial voltage to initiate discharge between the two
electrodes. The current distortion is at a moderate 15%.
harmonic currents, but they can still be significant sources of
harmonics at their characteristic frequencies.
harmonic currents for the 6-pulse drive are the 5-th and the
7-th.
best represented as a source of voltage harmonics.
over-excitation is generally only apparent under light load
conditions.
overexcited, but it has little consequence.
A capacitor bank experiences high
large and rich in a monotonic harmonic.
distinctly, resulting in a waveform that is
essentially riding on top of the
fundamental frequency (e.g. 11th order)
capacitor ratings as follows:
on rated kVAR and rated voltage
harmonics but excluding transients)
harmonics, care must be taken to
ensure that capacitors operate
within the rating limits.
There are three effects that result in increased transformer heating when
the load current includes harmonic components:
2) Winding eddy-current losses, PEC
3) Other stray losses, POS
frequency, the total load loss is,
2
2
When the transformer is supplying a non-sinusoidal current which
currents is
2
I
R
2
I
I
R
RECEC
22
2
Wh
I
I
I
I
PPP
R
REC
RRI
With a per-unit basis of transformer rated I2R loss and rated full load
22
2
2
Wh
I
I
P
I
I
P
PPP
R
h
REC
R
h
RRI
ECRILL
][
2
puIh
RECh
][1 puPP RECRLL
transformers for nonlinear load applications.
Example 6.8
loads. Although K-4 transformer may be acceptable at present, when new
nonlinear loads are added, K-4 transformer may no longer be sufficient.
of Harmonics
End users:
common coupling (PCC).
loads).
Limit the level of harmonic voltage distortions.
Voltage distortions utility can control this (think: impedance)
will be low
be high.
Customers
where multiple customers are served
Customer Under Study
Other Utility
Customers
PCC
IL
where multiple customers are served
Harmonic voltage distortion limits
(from IEEE Std. 519-1992, Table 11.1)
1. Determine the three-phase short-circuit duty, Isc, at the PCC.
12 months. This can be found from billing information.
current in amperes:
3
kV
MVA
kWIL
SC
I
I
ALTERED PATH
The system Thevenin equivalent impedance is called Short-circuit
kV
kV
2
purely reactive
frequency
2
t
t ZMVA
X
TRF impedance:
X
kV
fC
ca
cc
1
Parallel circuit from
the nonlinear load perspective
c
s
s
p
c
s
p
s
s
ss
c
p
c
p
ss
p
ss
p
c
p
c
ssp
X
X
jXRjX
jXRjX
)(
QP is quality factor that determines the
sharpness of the frequency response.
When the system resonant frequencies corresponds to one of
(characteristic frequencies), harmonic resonance can occur.
f
R
f
p
s
s
p
2
1
2
1
2
2
sys. impedance (not shown)
sys. impedance
bank
Apparent impedance Zp is maximum; Denominator is minimum, limited to Rs.
p
cap IXQV
p
c
s
sp
X
R
X
c
s
s
p
c
s
p
s
p
c
p
ssp
XQXQ
R
X
R
X
jXjXRZ
22 )()(
)(
ss XR
voltage drop across Zp. Hence, the voltage near the capacitor bank
will be heavily distorted:
The currents flowing in the capacitor bank and in the power
system are:
overheating.
the shunt capacitor size.
p
p
s
p
s
p
s
p
p
p
p
c
p
c
p
c
c
X
IXQ
V
X
IXQ
X
V
I
Parallel resonance frequency is computed as follows:
cap
p Q
X
X
Example 7.4:
Z
LC
.4
Ic
R
Sys tem s hort-circuit capacity is 1500 MVA
2000 [Hz]
0
|
Without Capacitor
During resonance,
the PF capacitor
forms a series
circuit with the TRF
and harmonic
sources. Resonance
frequency is
voltage at the PF capacitor, respectively. Reactances are shown in Fig. 7.7
s X
h
resonance also possible due to circuit topology (Fig. 7.7).
impedance seen
by non-linear
load for 0-1500
Hz when the
capacitor is ON
and OFF line
Peak Z =252.4
at 300 Hz
HARMONIC CURRENTS.
(electrically), resulting in either high voltage distortion or
telephone interference HIGH IMPEDANCE
(mostly X) WEAK SYSTEM.
harmonics to a greater degree than can be tolerated
HARMONIC RESONANCE
Recall these key concepts in harmonic analysis:
own these loads.
with harmonic currents resulting in voltage distortions.
higher.
frequencies matches one of harmonic frequencies produced
by nonlinear loads (large reactance, small R).
1) Large HARMONIC CURRENTS: Make it smaller, i.e. reducing harmonic
PWM drives that charge the DC bus capacitor directly from the line
Add inductors, i.e., line reactors.
wye.
them from entering the system.
Shunt filters Short-circuiting harmonic currents as close to the source
inductors, capacitors.
Change the capacitor size, move a capacitor to a point of the system
Remove the capacitor and live with it.
1) Capacitor bank placed on utility feeders usually done without
size or move to other locations.
the neutral connection make it floats or better add a
reactor in the neutral to convert the bank into a tuned
resonant shunt for a zero-sequence harmonic.
a few filters toward the ends of the feeder.
an overhead distribution-feeder
Harmonic problems at end-user facilities
in the facility?
delta and delta-wye, or wye-wye and wye-delta.
six-pulse converters (5, 7, 11,13, 17,19, etc.).
harmonics (3rd order).
converters in a load by 30 deg. can significantly reduce 5th and 7th
harmonics.
the line.
choke can reduce
the harmonic
current distortion
for a PWM-based
drives from
approximately 80%
to 40%.
pu
C
re
D
to
io
(%
(0 to 5) % Z
be charged and forces the drive to draw current over a longer time
period.
content while still delivering the same energy.
elements.
system.
Provide a low impedance path to a particular
Connected in shunt.
HIGH-PASS
HIGH-PASS
HIGH-PASS
10
20
30
40
50
filter configuation.
system with filter.
C
A
X
X
X C
3
Harmonic
Source
System
response
(Z = 1.0).
1
only
converted
filter
notch X
h
3
A single-tuned notch filter will be designed for an industrial facility and applied at a
480-volt bus. The load where the filter will be installed is approximately 1600 kVA
with a relatively poor displacement PF of 0.75 lagging.
The total harmonic current produced by this load is dominated by fifth harmonic
(20% of fundamental). The facility is supplied by a 2000 kVA transformer with
5.0% of impedance. The fifth harmonic background voltage distortion on the utility
side of the transformer is 2.0% of the fundamental when there is no load.
SOURCE
characteristics of the loads involved.
load.
concern. If the filter is tuned exactly to the harmonic,
impedances or other changes would shift the parallel
frequency higher into the harmonic being filtered.
resonance is very sharp, which require higher duty from
filter components.
and the resonant frequency (1)
the load reactive power
requirement for power factor
correction.
installed.
0.98.
from the filter required to
correct from 75% to 98%
power factor can be computed
as follows:
lagging is
Qfilt from the filter:
tan(acos(0.98)) ]
and the resonant frequency (2)
wye-equivalent filter reactance
(capacitive), XFilt, is determined by:
connected capacitor bank. Also, XFilt
is
reactance can be determined by
the filter capacitor can be rated the
same as the system, 480V, or would
have to be rated one step higher at
600V.
that this value is slightly lower than
reactive power that Xcap calculated
above produces)
28.814
Filt
kV
X
17.4
1 2
2
2
hX
X
kVAR
2962.0
be selected to tune the capacitor
to the desired frequency.
frequency is at 4.7th harmonic or
282 Hz.
from the equivalent wye capacitive
reactance above as follows:
be computed by solving L for in
the following equation
7.4
22
)( h
X
602
h
f
1
bank duties. These duties include
duties.
divided into three steps, i.e.,
frequency operating voltage
across the capacitor bank.
combined capacitor and reactor at
the fundamental frequency is:
current is:
operating voltage across the capacitor
bank is
produced
capacitor alone, therefore actual
reactive power produced by
capacitor is greater than its 750
kVAR rating.
filt
fund XXX
0.2933
480
,
filt
I
cap
cap
filt
cap
duty,fund VIQ
produces 20% harmonic of the
fundamental current at the fifth
harmonic, the harmonic current in
amperes produced by the load would
be:
the filter from the source side is
estimated as follows:
voltage distortion present on the utility
system will be limited only by the
impedances of the service TRF and the
filter;
the service transformer:
service transformer:
48.03
20.0
.).()(
hampsh
kVA
0.2
05.0(%)
Xfmr
TfundT MVA
ZX
capacitor bank is:
the utility system is 0.02 pu, the
estimated amount of fifth harmonic
current contributed to the filter from
the source side would be:
current is the sum of the harmonic
current produced by the load and
the harmonic current contributed
from the utility side:
the capacitor can be computed as
follows:
5
5),( h
X wyeCapwyeCap
)(
3
utilityh
kVpuV
A 23.150
48002.0
X
duty
)(5, 3
5
13.3553
and peak voltage requirements
the filter:
required for the filter reactor.
fundamental components add together,
the maximum peak voltage across the
capacitor is:
capacitor is:
capacitor:
)(
,
duty
fundduty
peakLLduty VVV 5,,, 22
5,
,, )()(
duty
cap
fundduty
rmsLLduty VVV
rmsLLduty
totalduty
totalduty VIQ ,,, 3
because the capacitor duties are
essentially at the maximum limits.
rated at 600 V
above.
480
600
2
frequency will occur is computed as follows:
source impedance. Including the utility system impedance will
lower the frequency.
from the harmonic (5th).
0139.00058.0
0
fundLfundT
X
Electrical Power Systems Quality, McGraw Hill 2012.
and Design, 5th Ed., CENGAGE Learning, 2012.
CRC Press, ISBN 9781482207002.
ed., CRC Press, 2012.
Department of Electrical and Computer Engineering
Distribution feeder analysis
Consider the feeder in the following figure.
substation primary voltages) are balanced and being held at 69 kV for all power-flow problems.
The substation transformer ratings are:
load is located at node 3 and has the following values:
Phase b: 500 kVA at 0.90 lagging power factor
Phase c: 950 kVA at 0.95 lagging power factor
and the line segment.
nodes. Use a tolerance of 0.0001 per unit.
Three type B step-voltage regulators are installed in a wye connection at the substation in order
to hold the load voltages (node 3) at a voltage level of 121 V and a bandwidth of 2 V.
compensator circuit ratings are 120V and 5A. Determine the R and X compensator
settings calibrated in volts and Ohms. The settings must be the same for all three
regulators.
compute the voltages across the voltage relays in the compensator circuits.
voltages at 121 V in a bandwidth of 2 V.
Department of Electrical and Computer Engineering
Distribution feeder analysis
0.6475 + 3.2375i 0.0000 + 0.0000i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.6475 + 3.2375i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.0000 + 0.0000i 0.6475 + 3.2375i
3.4500 – 1.9919i
-3.4500 – 1.9919i
0.0000 + 3.9837i
5.5250 + 3.4241i
4.5000 + 2.1794i
9.0250 + 2.9664i
V3 = 1.0e+03 *
7.9674 + 0.0000i
-3.9837 – 6.9000i
-3.9837 + 6.9000i
-30.0081
-149.9951
90.0010
-61.4511
178.7891
57.4620
7.8130
7.7042
-61.6420
1.0000
1.0002
1.0001
0.9759
0.9838
0.9737
0.9806
0.9670
117.6745
116.0355
unit.
number of iterations: 4
3.9847
3.9840
7.8380
57.4622
V3_mag = 1.0e+03 *
Distribution System & Power Quality
Department of Electrical and Computer Engineering
1. Distribution Feeder Analysis (focus)
desig
and energy storage systems in distribution system
operating conditions (power-flow analysis).
(short-circuit analysis).
based on modification of the “ladder” network theory
of linear systems.
power, current magnitude and angle).
known along with the voltage (VS) at the source.
no-load condition. With no load currents, there are no line
currents and voltage drops, so V5 = VS.
54321 Z45Z34Z23Z12
II2 I 3I23I12 I34 I45 I4 5VS
+
–
5
Z
until V1 is found.
V
+
54 I45I34I12I 23I 3I2I I
n
V
compute V1 . If V1 differs
from VS more than
specified tolerance
voltage VS; (b) the line
currents from the previous
sweep, then
Compute V2, V3, V4, V5
Using the new voltage V5,
perform 2nd backward sweep
to find new V1 .
sweep process continues until
1st forward sweep to
calculate
Kersting Example 10.1: Use the modified ladder method to compute the load
voltage for the single‐phase lateral shown below. The source voltage at node 1 is
7200 V.
Line segment 1–2 impedance:
Set initial conditions:
The first backward sweep:
The current in line segment 1–2 is
At this point, the second backward sweep is used to
forward sweep.
0.000017 with the final voltages and currents of
Dashed lines: Underground lines
(lines, transformers, and
regulators) can be
represented by their
models.
and 4 and between 4 and 5
have “distributed” loads.
distributed loads is to
combined them into a
lump load which is then
connected at the center.
The shunt components of a distribution feeder are spot static loads.
induction machine is modeled using the shunt admittance matrix
(not covered). Capacitor banks are modeled as constant
admittances.
constant current, constant impedance, or a combination of
the three.
segment can be ignored. However, for long underground line
segments, the shunt capacitance should be included.
phase, or 1-phase.
1) Set initial conditions and desired tolerance.
voltage at all nodes except for source node (VS) whose voltage
is specified.
VS. If specified tolerance is not met, continue to Step 4.
specified load power and node voltages obtained in the 1st
forward sweep.
obtained from the 1st backward sweep, calculate new node
voltages. Compare computed source voltage to specified source
voltage VS (i.e. go to Step 3).
specified source voltages are within the desired tolerance.
Kersting Example 10.2: Consider a simple distribution feeder shown below. The “source”
line segment from node 1 to node 2 is a 3-wire delta 2000-feet long line. The “load” line
segment from node 3 to node 4 is 2500 ft long and is a 4-wire wye with a neutral.
matrices for two
line segments:
Bus
1-phase TRF each rated:
constant PQ load:
The forward and backward sweep matrices must be computed
[B], and [d] matrices.
segment
with shunt
admittance
neglected:
Transformer: Converting TRF per unit impedance to Ohms
referenced to the low-voltage windings:
Transformer: Sweep matrices
Defining L-N voltages and load at Node 4: V4 lags ELNs by 30 deg.
Load currents at Node 4
Equivalent LN voltages and line currents at Node 1:
per unit.
source as the Node 1 voltage and calculates V2, V3, V4 using
the line currents from the previous sweep.
Infinite
Bus
[ZeqS] [ZeqL]
[Iabc][IABC]
21 3 4
Node 4 load using the new values of the Node 4 voltages.
the source is less than the specified tolerance of 0.001 pu.
iterations, the load voltages at Node 4 are:
the desired 120 V.
regulators connected in wye on the
secondary bus (node 3) of the
substation to raise the load
voltages.
The new configuration of the feeder after the step-voltage regulator
installation: The regulator is treated as an additional component between
Node 3r and 3.
R and X settings of the compensator: The equivalent phase impedance
between node 3 and node 4 is computed using the converged voltages at the
two nodes:
With the regulator in the neutral position, the input voltage to the
compensator is based on the converged voltage from the previous iteration:
bandwidth of 2 V. The regulators will change taps until the
phase voltage is at least 120V.
Using the taps 9, 12, 13, the sweep matrices for the regulators are:
Check the load voltages with the regulator set at the obtained
before. The only change is that the regulators are included as
an added element between Nodes 3r and 3.
Node 4 on a 120-V base are:
The regulators are included as an added element
between Nodes 3r and 3:
sequence for the
determination of
the final tap
settings and
convergence of the
system.
So far we compare the specified source voltage VS to the
acceptable.
phase or per phase) to a feeder is known via metering at the
substation. Then, it is desirable to force the computed input
complex power to the feeder matching the metered input. The
steps are:
input to the feeder.
specified tolerance of the metered input.
distribution automation,
renewable generators
generation and transmission systems using significant automatic monitoring
and control features.
distribution system, including connected load.
operations involve
geographically
dispersed and
functionally complex
monitoring and control
systems (shown in the
diagram).
overall control over
the total system.
The need for gathering substation and power plant data and
motivation for DS automation are increasing because of:
and government agencies.
higher fuel prices.
previously.
Utilities generally do not agree on types of functions that should be handled by
a DAC. Some of the automated distribution functions correlated with
locations are shown in the table below.
wind, solar, geothermal etc.).
storage, compressed-air ES, battery, electric vehicles etc.
Some important issues are:
cost, feeder bus-work/getaway costs:
of the switching and busing which are selected based on safety,
reliability, economy, simplicity, and other considerations.
regulation, subtransmission costs, substation costs, and the costs
of primary feeders, mains, and distribution TRF, and other factors.
substation service area, system load changes in short-term and
long-term planning.
Role / applications:
& energy storage systems (1)
(DSG) devices.
generators, wind turbines, photovoltaics (PVs), fuel cells, batteries,
electric vehicles, hydroelectric pumped storage, compressed-air
energy storage (CAES) etc.
distribution systems by reducing capacity requirements,
improving reliability, and reducing losses.
grid operation.
& energy storage systems (2)
concerns.
to conventional (coal, oil, gas).
transmission losses.
can increase grid reliability and customer energy independence.
& energy storage systems (3)
Distributed Energy Resource (DER) system continues to supply
power to the utility grid during a utility outage.
and wind turbines impact power quality.
power surplus many times per day in distribution networks,
leading to increased losses, overload, voltage instability etc.
reverse power flow. Protection setting is more difficult.
Electrical Power Systems Quality, McGraw Hill 2012.
and Design, 5th Ed., CENGAGE Learning, 2012.
CRC Press, ISBN 9781482207002.
ed., CRC Press, 2012.