PROJECT #2 SPACE VECTOR PWM INVERTER
JIN-WOO JUNG, PH.D STUDENT
E-mail: [e-mail]
Tel.: [phone]
ADVISOR: PROF. ALI KEYHANI
DATE: FEBRUARY 20, 2005
MECHATRONIC SYSTEMS LABORATORY
DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING
THE OHIO STATE UNIVERSITY
1. Problem Description
In this simulation, we will study Space Vector Pulse Width Modulation (SVPWM) technique.
We will use the SEMIKRON� IGBT Flexible Power Converter for this purpose. The system
configuration is given below:
Fig. 1 Circuit model of three-phase PWM inverter with a center-taped grounded DC bus.
The system parameters for this converter are as follows:
IGBTs: SEMIKRON SKM 50 GB 123D, Max ratings: VCES = 600 V, IC = 80 A
DC- link voltage: Vdc = 400 V
Fundamental frequency: f = 60 Hz
PWM (carrier) frequency: fz = 3 kHz
Modulation index: a = 0.6
Output filter: Lf = 800 �H and Cf = 400 �F
Load: Lload = 2 mH and Rload = 5 &!
Using Matlab/Simulink, simulate the circuit model described in Fig. 1 and plot the
waveforms of Vi (= [ViAB ViBC ViCA]), Ii (= [iiA iiB iiC]), VL (= [VLAB VLBC VLCA]), and IL (= [iLA
iLB iLC]).
2
2. Space Vector PWM
2.1 Principle of Pulse Width Modulation (PWM)
Fig. 2 shows circuit model of a single-phase inverter with a center-taped grounded DC bus,
and Fig 3 illustrates principle of pulse width modulation.
Fig. 2 Circuit model of a single-phase inverter.
Fig. 3 Pulse width modulation.
As depicted in Fig. 3, the inverter output voltage is determined in the following:
When Vcontrol > Vtri, VA0 = Vdc/2
When Vcontrol < Vtri, VA0 = -Vdc/2
3
Also, the inverter output voltage has the following features:
PWM frequency is the same as the frequency of Vtri
Amplitude is controlled by the peak value of Vcontrol
Fundamental frequency is controlled by the frequency of Vcontrol
Modulation index (m) is defined as:
vcontrol peak of (VA0 )1
4" m = = ,
vtri Vdc / 2
where, (VA0 )1 : fundamental frequecny component of VA0
2.2 Principle of Space Vector PWM
The circuit model of a typical three-phase voltage source PWM inverter is shown in Fig. 4.
S1 to S6 are the six power switches that shape the output, which are controlled by the switching
variables a, a2 , b, b2 , c and c2 . When an upper transistor is switched on, i.e., when a, b or c is 1,
the corresponding lower transistor is switched off, i.e., the corresponding a2 , b2 or c2 is 0.
Therefore, the on and off states of the upper transistors S1, S3 and S5 can be used to determine the
output voltage.
Fig. 4 Three-phase voltage source PWM Inverter.
4
The relationship between the switching variable vector [a, b, c]t and the line-to-line voltage
vector [Vab Vbc Vca]t is given by (2.1) in the following:
Vab 1
Ą# ń# Ą# -1 0 a
ń#Ą# ń#
ó#V Ą#
= Vdc ó# 0 1 -1Ą#ó#bĄ# . (2.1)
bc
ó# Ą# ó# Ą#ó# Ą#
ó#Vca Ą# ó#-1 0 1
Ą#ó#
Ś#
Ł# Ś# Ł# Ś#Ł#cĄ#
Also, the relationship between the switching variable vector [a, b, c]t and the phase voltage
vector [Va Vb Vc]t can be expressed below.
Van 2
Ą# ń# Ą# -1 -1 a
ń#Ą# ń#
Vdc ó#
ó#V Ą#
= . (2.2)
bn
ó# Ą# ó#-1 2 -1Ą#ó#bĄ#
Ą#ó# Ą#
3
ó#Vcn Ą# ó#-1 -1 2
Ą#ó#
Ś#
Ł# Ś# Ł# Ś#Ł#cĄ#
As illustrated in Fig. 4, there are eight possible combinations of on and off patterns for the
three upper power switches. The on and off states of the lower power devices are opposite to the
upper one and so are easily determined once the states of the upper power transistors are
determined. According to equations (2.1) and (2.2), the eight switching vectors, output line to
neutral voltage (phase voltage), and output line-to-line voltages in terms of DC-link Vdc, are
given in Table1 and Fig. 5 shows the eight inverter voltage vectors (V0 to V7).
Table 1. Switching vectors, phase voltages and output line to line voltages
5
Fig. 5 The eight inverter voltage vectors (V0 to V7).
Space Vector PWM (SVPWM) refers to a special switching sequence of the upper three
power transistors of a three-phase power inverter. It has been shown to generate less harmonic
distortion in the output voltages and or currents applied to the phases of an AC motor and to
6
provide more efficient use of supply voltage compared with sinusoidal modulation technique as
shown in Fig. 6.
Fig. 6 Locus comparison of maximum linear control voltage in Sine PWM and SVPWM.
To implement the space vector PWM, the voltage equations in the abc reference frame
can be transformed into the stationary dq reference frame that consists of the horizontal (d) and
vertical (q) axes as depicted in Fig. 7.
7
Fig. 7 The relationship of abc reference frame and stationary dq reference frame.
From this figure, the relation between these two reference frames is below
fdq0 = Ksfabc
(2.3)
1
Ą# -1/ 2 -1/ 2
ń#
2
ó#
where, , fdq0=[fd fq f0]T, fabc=[fa fb fc]T, and f denotes either a voltage
Ks = 0 3 2 - 3 2Ą#
ó# Ą#
3
ó#
Ł#1/ 2 1/ 2 1/ 2 Ą#
Ś#
or a current variable.
As described in Fig. 7, this transformation is equivalent to an orthogonal projection of [a, b,
c]t onto the two-dimensional perpendicular to the vector [1, 1, 1]t (the equivalent d-q plane) in a
three-dimensional coordinate system. As a result, six non-zero vectors and two zero vectors are
possible. Six nonzero vectors (V1 - V6) shape the axes of a hexagonal as depicted in Fig. 8, and
feed electric power to the load. The angle between any adjacent two non-zero vectors is 60
degrees. Meanwhile, two zero vectors (V0 and V7) are at the origin and apply zero voltage to the
load. The eight vectors are called the basic space vectors and are denoted by V0, V1, V2, V3, V4,
8
V5, V6, and V7. The same transformation can be applied to the desired output voltage to get the
desired reference voltage vector Vref in the d-q plane.
The objective of space vector PWM technique is to approximate the reference voltage vector
Vref using the eight switching patterns. One simple method of approximation is to generate the
average output of the inverter in a small period, T to be the same as that of Vref in the same
period.
Fig. 8 Basic switching vectors and sectors.
Therefore, space vector PWM can be implemented by the following steps:
Step 1. Determine Vd, Vq, Vref, and angle (ą)
Step 2. Determine time duration T1, T2, T0
Step 3. Determine the switching time of each transistor (S1 to S6)
2.2.1 Step 1: Determine Vd, Vq, Vref, and angle (ą)
From Fig. 9, the Vd, Vq, Vref, and angle (ą) can be determined as follows:
9
Vd = Van - Vbn �" cos60 - Vcn �" cos60
1 1
= Van - Vbn - Vcn
2 2
Vq = 0 + Vbn �" cos30 - Vcn �" cos30
3 3
= Van + Vbn - Vcn
2 2
Ą#1 - 1 1
ń#Ą#Van ń#
-
Vd 2 ó# Ą#ó# Ą#
Ą# ń#
2 2
4" =
ó# Ą#ó#V Ą#
ó#V Ą# bn
3
q
Ł# Ś# ó#0 3 3 Ą#ó#V Ą#
-
ó# Ą#Ł# cn Ś#
Ł# 2 2 Ś#
4" Vref = Vd 2 + Vq 2
Vq
�# ś#
ś# ź#
4"ą = tan-1ś# ź# = �t = 2Ąft, where f = fundamental frequency
d
�#V #
Fig. 9 Voltage Space Vector and its components in (d, q).
2.2.2 Step 2: Determine time duration T1, T2, T0
From Fig. 10, the switching time duration can be calculated as follows:
10
 Switching time duration at Sector 1
Tz T1 T1+T2 Tz
ref 1 2 0
+"V = +"V dt + +"V dt + +"V
0 0 T1 T1+T2
4"Tz �" V = (T1 �" V + T2 �" V )
ref 1 2
cos(ą) 1 cos(Ą / 3)
Ą# ń# 2 Ą# ń# 2 Ą# ń#
�! Tz �" V �" = T1 �" �" Vdc �" + T2 �" �" Vdc �"
ref
ó#sin (ą) Ą# ó#0Ą# ó#sin (Ą / 3) Ą#
3 3
Ł# Ś# Ł# Ś# Ł# Ś#
(where, 0 d" ą d" 60�)
sin (Ą / 3 - ą )
4"T1 = Tz �" a �"
sin (Ą / 3)
sin (ą )
4"T2 = Tz �" a �"
sin (Ą / 3)
�# ś#
ś# ź#
V
ref
1
ś# ź#
4"T0 = Tz - (T1 + T2 ), where, Tz = and a =
2
ś# fz
Vdc ź#
ś# ź#
�# 3 #
Switching time duration at any Sector
3 �"Tz �" Vref
�# Ą n -1 ś#
ś#ź#
4"T1 = ś#sin�# - ą + Ą
ś# ź#ź#
Vdc ś# �# 3 3
#
�# #
3 �"Tz �" Vref
n
�#sin Ą - ą ś#
=
ś# ź#
Vdc �# 3
#
3 �"Tz �" Vref
n n
�#sin Ą cosą - cos Ą siną ś#
=
ś# ź#
Vdc �# 3 3
#
3 �"Tz �" Vref
�# n -1 ś#
ś#ź#
4"T2 = ś#sin�#ą - Ą
ś# ź#ź#
Vdc ś# �# 3
#
�# #
3 �"Tz Vref
n -1 n -1
�# ś#
=
ś#- cosą �" sin Ą + siną �" cos Ą
ź#
Vdc �# 3 3
#
where, n = 1 through 6(that is,Sector1 to 6)
�# ś#
4"T0 = Tz - T1 - T2 , ś# ź#
ś# ź#
0 d" ą d" 60�
�# #
11
Fig. 10 Reference vector as a combination of adjacent vectors at sector 1.
2.2.3 Step 3: Determine the switching time of each transistor (S1 to S6)
Fig. 11 shows space vector PWM switching patterns at each sector.
(a) Sector 1. (b) Sector 2.
12
(c) Sector 3. (d) Sector 4.
(e) Sector 5. (f) Sector 6.
Fig. 11 Space Vector PWM switching patterns at each sector.
Based on Fig. 11, the switching time at each sector is summarized in Table 2, and it will be
built in Simulink model to implement SVPWM.
13
Table 2. Switching Time Calculation at Each Sector
14
3. State-Space Model
Fig. 12 shows L-C output filter to obtain current and voltage equations.
Fig. 12 L-C output filter for current/voltage equations.
By applying Kirchoff s current law to nodes a, b, and c, respectively, the following current
equations are derived:
node  a :
dVLCA dVLAB
. (3.1)
iiA + ica = iab + iLA �! iiA + C = C + iLA
f f
dt dt
node  b :
dVLAB dVLBC
. (3.2)
iiB + iab = ibc + iLB �! iiB + C = C + iLB
f f
dt dt
node  c :
15
dVLBC dVLCA
. (3.2)
iiC + ibc = ica + iLC �! iiC + C = C + iLC
f f
dt dt
dVLAB dVLBC dVLCA
where, iab = C , ibc = C , ica = C .
f f f
dt dt dt
Also, (3.1) to (3.3) can be rewritten as the following equations, respectively:
subtracting (3.2) from (3.1):
dVLCA dVLAB dVLAB dVLBC
�# ś# �# ś#
iiA - iiB + C ś# - ź# ś# - ź# - iLB
= C + iLA
f f
dt dt dt dt
�# # �# #
. (3.4)
dVLCA dVLBC dVLAB
�# ś#
�! C ś# + - 2 �" ź# -iiA + iiB + iLA - iLB
=
f
dt dt dt
�# #
subtracting (3.3) from (3.2):
dVLAB dVLBC dVLBC dVLCA
�# ś# �# ś#
iiB - iiC + C ś# - ź# ś# - ź# - iLC
= C + iLB
f f
dt dt dt dt
�# # �# #
. (3.5)
dVLAB dVLCA dVLBC
�# ś#
�! C ś# + - 2 �" ź# -iiB + iiC + iLB - iLC
=
f
dt dt dt
�# #
subtracting (3.1) from (3.3):
dVLBC dVLCA dVLCA dVLAB
�# ś# �# ś#
iiC - iiA + C ś# - ź# ś# - ź# - iLA
= C + iLC
f f
dt dt dt dt
�# # �# #
. (3.6)
dVLAB dVLBC dVLCA
�# ś#
�! C ś# + - 2 �" ź# -iiC + iiA + iLC - iLA
=
f
dt dt dt
�# #
To simplify (3.4) to (3.6), we use the following relationship that an algebraic sum of line to line
load voltages is equal to zero:
VLAB + VLBC + VLCA = 0. (3.7)
16
Based on (3.7), the (3.4) to (3.6) can be modified to a first-order differential equation,
respectively:
ż#
dVLAB 1 1
= iiAB - (iLAB )
�#
dt 3C 3C
�# f f
�#
dVLBC
1 1
(3.8)
= iiBC - (iLB C ),
�#
dt 3C 3C
f f
�#
�# dVLCA
1 1
= iiCA - (iLCA )
�#
dt 3C 3C
f f
�#
where, iiAB = iiA % iiB, iiBC = iiB % iiC, iiCA = iiC % iiA and iLAB = iLA % iLB, iLBC = iLB % iLC,
iLCA = iLC % iLA.
By applying Kirchoff s voltage law on the side of inverter output, the following voltage
equations can be derived:
ż#
diiAB 1 1
= - VLAB + ViAB
�#
dt Lf Lf
�#
�#
diiBC 1 1
. (3.9)
= - VLBC + ViBC
�#
dt Lf Lf
�#
�# diiCA 1 1
= - VLCA + ViCA
�#
dt Lf Lf
�#
By applying Kirchoff s voltage law on the load side, the following voltage equations can be
derived:
diLA diLB
ż#
�#VLAB = Lload dt + Rload iLA - Lload dt - Rload iLB
�#
diLB diLC
�#V = Lload + Rload iLB - Lload - Rload iLC
. (3.10)
�#
LBC
dt dt
�#
�# diLC diLA
LCA
�#V = Lload dt + Rload iLC - Lload dt - Rload iLA
�#
17
Equation (3.10) can be rewritten as:
ż#
diLAB Rload
1
= - iLAB + VLAB
�#
dt Lload Lload
�#
�#diLBC Rload
1
�#
. (3.11)
= - iLBC + VLBC
�#
dt Lload Lload
�#
�#diLCA Rload
1
= - iLCA + VLCA
�#
�# dt Lload Lload
�#
Therefore, we can rewrite (3.8), (3.9) and (3.11) into a matrix form, respectively:
dVL 1 1
= Ii - IL
dt 3C 3C
f f
dIi 1 1
= - VL + Vi , (3.12)
dt L L
f f
dIL 1 Rload
= VL - I
dt Lload Lload L
where, VL = [VLAB VLBC VLCA]T , Ii = [iiAB iiBC iiCA]T = [iiA-iiB iiB-iiC iiC-iiA]T , Vi = [ViAB ViBC ViCA]T ,
IL = = [iLAB iLBC iLCA]T = [iLA-iLB iLB-iLC iLC-iLA]T.
Finally, the given plant model (3.12) can be expressed as the following continuous-time state
space equation
&
X(t) = AX(t) + Bu(t) , (3.13)
Ą# ń#
1 1
03�3 I3�3 - I3�3 Ą#
ó#
03�3
Ą# ń#
3C 3C
f f
ó# Ą#
VL
Ą# ń#
ó# Ą#
1
Ą#
ó# Ą#
Ii A = ó# 1 I3�3 03�3
where, X = , , , u = [Vi ]3�1.
03�3 Ą# B = ó# L I3�3 Ą#
ó#-
ó# Ą#
f
ó# Ą#
L
f
I ó# Ą#
ó# Ą#9�1
Ł# L Ś#
ó#0 Ą#
1 Rload
ó#
Ł# 3�3 Ś#9�3
I3�3 03�3 - I3�3 Ą#
ó#
Lload Lload Ą#
Ł# Ś#9�9
18
Note that load line to line voltage VL, inverter output current Ii, and the load current IL are the
state variables of the system, and the inverter output line-to-line voltage Vi is the control input
(u).
4. Simulation Steps
1). Initialize system parameters using Matlab
2). Build Simulink Model
Determine sector
Determine time duration T1, T2, T0
Determine the switching time (Ta, Tb, and Tc) of each transistor (S1 to S6)
Generate the inverter output voltages (ViAB, ViBC, ViCA,) for control input (u)
Send data to Workspace
3). Plot simulation results using Matlab
19
5. Simulation results
Inverter output line to line voltages (ViAB, ViBC, ViCA)
500
0
-500
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
500
0
-500
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
500
0
-500
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
Time [Sec]
Fig. 13 Simulation results of inverter output line to line voltages (ViAB, ViBC, ViCA)
20
iAB
V
[V]
iBC
V
[V]
iCA
V
[V]
Inverter output currents (iiA, iiB, iiC)
100
50
0
-50
-100
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
100
50
0
-50
-100
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
100
50
0
-50
-100
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
Time [Sec]
Fig. 14 Simulation results of inverter output currents (iiA, iiB, iiC)
21
iA
i
[A]
iB
i [A]
iC
i [A]
Load line to line voltages (VLAB, VLBC, VLCA)
400
200
0
-200
-400
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
400
200
0
-200
-400
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
400
200
0
-200
-400
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
Time [Sec]
Fig. 15 Simulation results of load line to line voltages (VLAB, VLBC, VLCA)
22
LA B
V
[V]
LBC
V
[V]
LCA
V
[V]
Load phase currents (iLA, iLB, iLC)
50
0
-50
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
50
0
-50
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
50
0
-50
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
Time [Sec]
Fig. 16 Simulation results of load phase currents (iLA, iLB, iLC)
23
LA
i
[A]
LB
i
[A]
LC
i
[A]
500
0
-500
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
100
iiA
iiB
0
iiC
-100
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
400
VLAB
200
VLBC
0
VLCA
-200
-400
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
50
iLA
iLB
0
iLC
-50
0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1
Time [Sec]
Fig. 17 Simulation waveforms.
(a) Inverter output line to line voltage (ViAB)
(b) Inverter output current (iiA)
(c) Load line to line voltage (VLAB)
(d) Load phase current (iLA)
24
iAB
V
[V]
LA B
LBC
LCA
iA
iB
iC
V
, V
, V
[V]
i
, i , i [A]
LA
LB
LC
i
, i
, i
[A]
Appendix
Matlab/Simulink Codes
25
A.1 Matlab Code for System Parameters
% Written by Jin Woo Jung
% Date: 02/20/05
% ECE743, Simulation Project #2 (Space Vector PWM Inverter)
% Matlab program for Parameter Initialization
clear all % clear workspace
% Input data
Vdc= 400; % DC-link voltage
Lf= 800e-6;% Inductance for output filter
Cf= 400e-6; % Capacitance for output filter
Lload = 2e-3; %Load inductance
Rload= 5; % Load resistance
f= 60; % Fundamental frequency
fz = 3e3; % Switching frequency
a= 0.6;% Modulation index
w= 2*pi*60; %angular frequency
Tz= 1/fz; % Sampling time
V_ref= (2/3)*a*Vdc; % Reference voltage
% Coefficients for State-Space Model
A=[zeros(3,3) eye(3)/(3*Cf) -eye(3)/(3*Cf)
-eye(3)/Lf zeros(3,3) zeros(3,3)
eye(3,3)/Lload zeros(3,3) -eye(3)*Rload/Lload]; % system matrix
B=[zeros(3,3)
eye(3)/Lf
zeros(3,3)]; % coefficient for the control variable u
26
C=[eye(9)]; % coefficient for the output y
D=[zeros(9,3)]; % coefficient for the output y
Ks = 1/3*[-1 0 1; 1 -1 0; 0 1 -1]; % Conversion matrix to transform [iiAB iiBC iiCA] to [iiA iiB
iiC]
27
A.2 Matlab Code for Plotting the Simulation Results
% Written by Jin Woo Jung
% Date: 02/20/05
% ECE743, Simulation Project #2 (Space Vector PWM)
% Matlab program for plotting Simulation Results
% using Simulink
ViAB = Vi(:,1);
ViBC = Vi(:,2);
ViCA = Vi(:,3);
VLAB= VL(:,1);
VLBC= VL(:,2);
VLCA= VL(:,3);
iiA= IiABC(:,1);
iiB= IiABC(:,2);
iiC= IiABC(:,3);
iLA= ILABC(:,1);
iLB= ILABC(:,2);
iLC= ILABC(:,3);
figure(1)
subplot(3,1,1)
plot(t,ViAB)
axis([0.9 1 -500 500])
ylabel('V_i_A_B [V]')
title('Inverter output line to line voltages (V_i_A_B, V_i_B_C, V_i_C_A)')
28
grid
subplot(3,1,2)
plot(t,ViBC)
axis([0.9 1 -500 500])
ylabel('V_i_B_C [V]')
grid
subplot(3,1,3)
plot(t,ViCA)
axis([0.9 1 -500 500])
ylabel('V_i_C_A [V]')
xlabel('Time [Sec]')
grid
figure(2)
subplot(3,1,1)
plot(t,iiA)
axis([0.9 1 -100 100])
ylabel('i_i_A [A]')
title('Inverter output currents (i_i_A, i_i_B, i_i_C)')
grid
subplot(3,1,2)
plot(t,iiB)
axis([0.9 1 -100 100])
ylabel('i_i_B [A]')
grid
subplot(3,1,3)
29
plot(t,iiC)
axis([0.9 1 -100 100])
ylabel('i_i_C [A]')
xlabel('Time [Sec]')
grid
figure(3)
subplot(3,1,1)
plot(t,VLAB)
axis([0.9 1 -400 400])
ylabel('V_L_A_B [V]')
title('Load line to line voltages (V_L_A_B, V_L_B_C, V_L_C_A)')
grid
subplot(3,1,2)
plot(t,VLBC)
axis([0.9 1 -400 400])
ylabel('V_L_B_C [V]')
grid
subplot(3,1,3)
plot(t,VLCA)
axis([0.9 1 -400 400])
ylabel('V_L_C_A [V]')
xlabel('Time [Sec]')
grid
figure(4)
subplot(3,1,1)
plot(t,iLA)
axis([0.9 1 -50 50])
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ylabel('i_L_A [A]')
title('Load phase currents (i_L_A, i_L_B, i_L_C)')
grid
subplot(3,1,2)
plot(t,iLB)
axis([0.9 1 -50 50])
ylabel('i_L_B [A]')
grid
subplot(3,1,3)
plot(t,iLC)
axis([0.9 1 -50 50])
ylabel('i_L_C [A]')
xlabel('Time [Sec]')
grid
figure(5)
subplot(4,1,1)
plot(t,ViAB)
axis([0.9 1 -500 500])
ylabel('V_i_A_B [V]')
grid
subplot(4,1,2)
plot(t,iiA,'-', t,iiB,'-.',t,iiC,':')
axis([0.9 1 -100 100])
ylabel('i_i_A, i_i_B, i_i_C [A]')
legend('i_i_A', 'i_i_B', 'i_i_C')
grid
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subplot(4,1,3)
plot(t,VLAB,'-', t,VLBC,'-.',t,VLCA,':')
axis([0.9 1 -400 400])
ylabel('V_L_A_B, V_L_B_C, V_L_C_A [V]')
legend('V_L_A_B', 'V_L_B_C', 'V_L_C_A')
grid
subplot(4,1,4)
plot(t,iLA,'-', t,iLB,'-.',t,iLC,':')
axis([0.9 1 -50 50])
ylabel('i_L_A, i_L_B, i_L_C [A]')
legend('i_L_A', 'i_L_B', 'i_L_C')
xlabel('Time [Sec]')
grid
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A.3 Simulink Code
Simulink Model for Overall System
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Subsystem Simulink Model for  Space Vector PWM Generator
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Subsystem Simulink Model for  Making Switching Time
1). T1 = u[1]*(sin(u[3]*pi/3)*cos(u[2])-cos(u[3]*pi/3)*sin(u[2]))
2). T2 = u[1]*(cos((u[3]-1)*(pi/3))*sin(u[2])-sin((u[3]-1)*(pi/3))*cos(u[2]))
3). Ta = (u[4]==1)*(u[1]+u[2]+u[3])+(u[4]==2)*(u[1]+u[2]+u[3]) + (u[4]==3)*(u[1]+u[3]) +
(u[4]==4)*(u[1])+ (u[4]==5)*(u[1])+ (u[4]==6)*(u[1]+u[2])
4). Tb = (u[4]==1)*(u[1])+(u[4]==2)*(u[1]+u[2]) + (u[4]==3)*(u[1]+u[2]+u[3]) +
(u[4]==4)*(u[1]+u[2]+u[3])+ (u[4]==5)*(u[1]+u[3])+ (u[4]==6)*(u[1])
5). Tc = (u[4]==1)*(u[1]+u[3])+(u[4]==2)*(u[1]) + (u[4]==3)*(u[1]) + (u[4]==4)*(u[1]+u[2])+
(u[4]==5)*(u[1]+u[2]+u[3])+ (u[4]==6)*(u[1]+u[2]+u[3])
35