电力电子建模与仿真实验
(四)
题目:1-Phase, Bipolar-Voltage Switching Inverter
Three-Phase PWM Inverter
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EXAMPLE 12
1-Phase, Bipolar-Voltage Switching Inverter
Problems :
1. Obtain the following waveforms using 1Phbsinv:
(a) vo and io.
(b) vo and id.
(c) vo, io and po
Fig1-1(io and vo)
Fig1-2(id and vo)
Fig1-3(vo, io and po)
2. Obtain v 01 by means of Fourier analysis of the v o waveform. Compare v 01 with its
precalculated nominal value.
The Fourier analysis of the vo :
DC COMPONENT = -1.367907E+00
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 4.000E+01 2.152E+02 1.000E+00 1.444E+02 0.000E+00
2 8.000E+01 8.203E-01 3.811E-03 6.700E+00 -2.821E+02
3 1.200E+02 1.088E+00 5.054E-03 1.356E+02 -2.976E+02
4 1.600E+02 6.913E-01 3.212E-03 -1.541E+02 -7.317E+02
5 2.000E+02 5.708E-01 2.652E-03 -9.746E+01 -8.195E+02
6 2.400E+02 6.413E-01 2.980E-03 1.581E+02 -7.083E+02
7 2.800E+02 9.704E-01 4.509E-03 1.783E+02 -8.325E+02
8 3.200E+02 1.027E+00 4.771E-03 -4.752E+01 -1.203E+03
9 3.600E+02 1.233E+00 5.730E-03 8.358E+01 -1.216E+03
10 4.000E+02 1.536E+00 7.138E-03 3.865E+01 -1.405E+03
11 4.400E+02 6.210E-01 2.885E-03 -1.606E+02 -1.749E+03
12 4.800E+02 3.179E-01 1.477E-03 1.087E+02 -1.624E+03
13 5.200E+02 1.612E+00 7.491E-03 5.246E+01 -1.825E+03
14 5.600E+02 1.071E+00 4.974E-03 -8.479E+01 -2.106E+03
15 6.000E+02 2.222E-01 1.033E-03 7.063E+01 -2.095E+03
16 6.400E+02 1.067E+00 4.956E-03 2.316E+01 -2.287E+03
17 6.800E+02 6.656E-01 3.092E-03 -1.299E+01 -2.468E+03
18 7.200E+02 1.167E+00 5.424E-03 1.339E+02 -2.465E+03
19 7.600E+02 8.236E-01 3.827E-03 -1.529E+02 -2.897E+03
20 8.000E+02 1.060E+00 4.923E-03 1.143E+02 -2.774E+03
21 8.400E+02 2.719E+00 1.263E-02 -1.453E+02 -3.178E+03
22 8.800E+02 3.026E-01 1.406E-03 -1.157E+02 -3.292E+03
23 9.200E+02 5.888E+01 2.735E-01 1.622E+02 -3.159E+03
TOTAL HARMONIC DISTORTION = 2.745433E+01 PERCENT
From the Fourier analysis, we know
152.17V
Because Vo1(rms) = 153.33 V ,so Vo1 approximately equals to the nominal value of Vo1(rms).
3. Using the results of Problem 2, obtain the ripple component v ripple waveform in the
output voltage.
the ripple component vripple = vo- vo1=[ v(L1:1)-v(V5:-)]-Vo1
Fig3-1(vo- vo1)
4. Obtain i01 by means of Fourier analysis of the i o waveform. Compare i 01 with its
precalculated nominal value.
FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1)
DC COMPONENT = -4.341032E-01
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 4.000E+01 1.359E+01 1.000E+00 -2.799E+01 0.000E+00
2 8.000E+01 1.007E-02 7.413E-04 7.046E+01 1.264E+02
3 1.200E+02 6.720E-02 4.946E-03 4.604E+01 1.300E+02
4 1.600E+02 9.693E-03 7.135E-04 -6.629E+01 4.565E+01
5 2.000E+02 2.342E-02 1.724E-03 -7.893E+01 6.100E+01
6 2.400E+02 6.280E-03 4.623E-04 2.867E+01 1.966E+02
7 2.800E+02 7.838E-03 5.770E-04 -1.259E+02 7.001E+01
8 3.200E+02 5.757E-03 4.238E-04 -1.743E+02 4.955E+01
9 3.600E+02 2.774E-03 2.042E-04 1.070E+02 3.589E+02
10 4.000E+02 8.593E-03 6.326E-04 -7.327E+01 2.066E+02
11 4.400E+02 7.374E-03 5.428E-04 5.145E+01 3.593E+02
12 4.800E+02 3.456E-03 2.544E-04 -3.619E+01 2.996E+02
13 5.200E+02 1.061E-02 7.814E-04 -1.133E+02 2.505E+02
14 5.600E+02 5.217E-03 3.840E-04 -4.503E+01 3.468E+02
15 6.000E+02 3.839E-03 2.826E-04 -8.663E+01 3.331E+02
16 6.400E+02 2.546E-03 1.874E-04 -1.618E+02 2.860E+02
17 6.800E+02 1.683E-03 1.239E-04 -5.285E+01 4.229E+02
18 7.200E+02 4.983E-03 3.668E-04 -7.910E+01 4.246E+02
19 7.600E+02 1.790E-03 1.318E-04 -8.766E+01 4.441E+02
20 8.000E+02 2.168E-03 1.596E-04 5.595E+00 5.653E+02
21 8.400E+02 3.123E-02 2.299E-03 1.066E+00 5.888E+02
22 8.800E+02 2.108E-03 1.551E-04 1.617E+02 7.774E+02
23 9.200E+02 1.028E+00 7.570E-02 2.208E+00 6.459E+02
TOTAL HARMONIC DISTORTION = 7.593596E+00 PERCENT
According to the Fourier analysis,
We know that its precalculated nominal value Io1(rms) = 10 A
So Io1 approximately equals to the nominal value of Io1(rms)
5. Using the results of Problem 4, obtain the ripple component i ripple in the output current.
From the problem 4,we know THD=7.59% ,I01=13.59
So
6. Obtain Id(avg) and id2 (the component at the 2nd harmonic frequency) by means of the Fourier analysis of the id waveform. Compare them with their precalculated nominal values
FOURIER COMPONENTS OF TRANSIENT RESPONSE I(V_V1)
DC COMPONENT = -4.908958E+00
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 4.000E+01 4.289E-01 1.000E+00 -1.946E+00 0.000E+00
2 8.000E+01 5.451E+00 1.271E+01 6.056E+01 6.445E+01
3 1.200E+02 4.544E-02 1.060E-01 -1.795E+02 -1.736E+02
4 1.600E+02 7.639E-02 1.781E-01 1.319E+02 1.396E+02
5 2.000E+02 3.974E-02 9.267E-02 -1.334E+01 -3.614E+00
6 2.400E+02 3.045E-02 7.100E-02 -8.175E+01 -7.007E+01
7 2.800E+02 4.784E-02 1.116E-01 9.565E+01 1.093E+02
8 3.200E+02 4.988E-02 1.163E-01 -1.385E+02 -1.229E+02
9 3.600E+02 3.620E-02 8.441E-02 -9.852E+01 -8.100E+01
10 4.000E+02 5.078E-02 1.184E-01 -8.064E+01 -6.118E+01
11 4.400E+02 4.159E-02 9.697E-02 1.219E+02 1.433E+02
12 4.800E+02 1.063E-02 2.479E-02 -6.397E+01 -4.062E+01
13 5.200E+02 4.512E-02 1.052E-01 -2.976E+01 -4.464E+00
14 5.600E+02 2.213E-02 5.160E-02 -3.340E+01 -6.157E+00
15 6.000E+02 3.584E-02 8.358E-02 1.738E+02 2.030E+02
16 6.400E+02 4.414E-02 1.029E-01 -3.200E+01 -8.660E-01
17 6.800E+02 4.462E-02 1.040E-01 7.998E-01 3.388E+01
18 7.200E+02 5.566E-02 1.298E-01 5.333E+01 8.835E+01
19 7.600E+02 3.212E-02 7.489E-02 1.077E+02 1.447E+02
20 8.000E+02 1.147E-01 2.674E-01 7.242E+00 4.616E+01
21 8.400E+02 2.905E-02 6.773E-02 -8.042E+01 -3.956E+01
22 8.800E+02 1.266E+00 2.952E+00 7.332E+00 5.014E+01
23 9.200E+02 1.661E-01 3.873E-01 9.427E+01 1.390E+02
TOTAL HARMONIC DISTORTION = 1.306384E+03 PERCENT
From the Fourier analysis, we know : Id=-DC= 4.91A①
②
Vd =
,cos
So:
③
④
①≈③, ②≈④
So Id and Id2 approximately equal to their precalculated nominal values
7. Using the results of Problem 6, obtain the high frequency ripple component id,ripple in the input dc current. Calculate its rms value.
From Problem 6,we know THD=13.06
So
=13.06*0.43=5.62A
RMS
5.62/1.414=3.97
EXAMPLE 16
Three-Phase PWM Inverter
Problems
1. Obtain the following waveforms using :
(a) vAN and iA.
(b) vAn and iA.
(c) vAN and id.
Fig 1-1 ( vAN and iA )
Fig 1-2 ( vAn and iA )
Fig1-3(vAN and id )
2. Obtain vAn1 by means of Fourier analysis of the vAn waveform. Compare vAn1 with its precalculated nominal value.
The Fourier analysis of the vAn is:
DC COMPONENT = -2.239930E-02
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 4.762E+01 1.491E+02 1.000E+00 1.372E+02 0.000E+00
2 9.524E+01 4.832E-01 3.240E-03 -8.232E+01 -3.568E+02
3 1.429E+02 4.600E-02 3.085E-04 -1.106E+02 -5.222E+02
4 1.905E+02 4.246E-01 2.848E-03 1.929E+01 -5.296E+02
5 2.381E+02 6.052E-01 4.059E-03 3.124E+00 -6.830E+02
6 2.857E+02 4.407E-02 2.955E-04 -1.323E+02 -9.557E+02
7 3.333E+02 7.786E-01 5.222E-03 1.476E+02 -8.130E+02
8 3.810E+02 6.795E-01 4.557E-03 1.786E+02 -9.192E+02
9 4.286E+02 4.730E-02 3.172E-04 -1.491E+02 -1.384E+03
10 4.762E+02 9.098E-01 6.101E-03 1.662E+02 -1.206E+03
11 5.238E+02 1.551E-01 1.040E-03 -7.099E+01 -1.580E+03
12 5.714E+02 4.816E-02 3.230E-04 -1.729E+02 -1.820E+03
13 6.190E+02 1.455E+00 9.758E-03 1.256E+02 -1.658E+03
14 6.667E+02 8.385E-01 5.623E-03 -9.956E+01 -2.021E+03
15 7.143E+02 4.384E-02 2.940E-04 1.718E+02 -1.887E+03
16 7.619E+02 2.030E-01 1.361E-03 -9.822E+01 -2.294E+03
17 8.095E+02 1.799E+00 1.207E-02 -9.840E+01 -2.431E+03
18 8.571E+02 4.842E-02 3.247E-04 1.518E+02 -2.318E+03
19 9.048E+02 4.585E+01 3.075E-01 1.751E+02 -2.432E+03
20 9.524E+02 7.544E-01 5.059E-03 7.145E+01 -2.673E+03
21 1.000E+03 4.487E-02 3.009E-04 1.273E+02 -2.754E+03
22 1.048E+03 6.855E-01 4.597E-03 1.205E+02 -2.898E+03
23 1.095E+03 4.598E+01 3.084E-01 4.126E+00 -3.152E+03
TOTAL HARMONIC DISTORTION = 4.359678E+01 PERCENT
From Fourier analysis ,we know:
precalculated nominal value :VAn1=105.39V
the VAn1 equals to its precalculated nominal value.
3. Using the results of Problem 2, obtain the ripple component vripple waveform in the output voltage.
vripple = VAn- VAn1= [v(L1:1)-v(V5:-)]- 105.6*
4.Obtain iA1 by means of Fourier analysis of iA waveform. Compare iA1 with its
precalculated nominal value.
FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1)
DC COMPONENT = -1.633990E-02
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 4.762E+01 1.417E+01 1.000E+00 1.076E+02 0.000E+00
2 9.524E+01 1.690E-02 1.193E-03 -1.546E+02 -3.698E+02
3 1.429E+02 8.724E-03 6.157E-04 1.718E+02 -1.510E+02
4 1.905E+02 7.686E-03 5.424E-04 1.213E+02 -3.090E+02
5 2.381E+02 1.073E-02 7.575E-04 -1.011E+02 -6.390E+02
6 2.857E+02 3.529E-03 2.491E-04 1.441E+02 -5.014E+02
7 3.333E+02 4.672E-03 3.297E-04 3.532E+01 -7.177E+02
8 3.810E+02 5.291E-03 3.734E-04 1.213E+02 -7.393E+02
9 4.286E+02 2.874E-03 2.028E-04 1.227E+02 -8.455E+02
10 4.762E+02 6.673E-03 4.709E-04 1.030E+02 -9.728E+02
11 5.238E+02 7.769E-03 5.483E-04 9.967E+01 -1.084E+03
12 5.714E+02 1.698E-03 1.198E-04 1.012E+02 -1.190E+03
13 6.190E+02 6.125E-03 4.322E-04 7.720E+01 -1.321E+03
14 6.667E+02 1.612E-03 1.137E-04 1.249E+02 -1.381E+03
15 7.143E+02 1.685E-03 1.189E-04 8.254E+01 -1.531E+03
16 7.619E+02 2.819E-03 1.990E-04 1.944E+00 -1.719E+03
17 8.095E+02 4.433E-02 3.129E-03 1.731E+02 -1.656E+03
18 8.571E+02 1.128E-03 7.959E-05 6.326E+01 -1.873E+03
19 9.048E+02 8.088E-01 5.708E-02 8.755E+01 -1.957E+03
20 9.524E+02 9.749E-03 6.880E-04 -3.756E+01 -2.189E+03
21 1.000E+03 1.190E-03 8.397E-05 4.404E+01 -2.215E+03
22 1.048E+03 1.136E-02 8.015E-04 4.363E+01 -2.323E+03
23 1.095E+03 6.660E-01 4.700E-02 -8.389E+01 -2.558E+03
TOTAL HARMONIC DISTORTION = 7.403818E+00 PERCENT
We can learn that
We know that precalculated nominal value of IA1=10A
So IA1 equals to its precalculated nominal value
5. Using the results of Problem 4, obtain the ripple component iripple in the output current.
From the problem 4,we know THD=7.4%,
so we can get that:
6.Obtain Id(avg) by means of Fourier analysis and obtain the high frequency ripple
id,ripple = id - Id(avg) in the input current.
The Fourier analysis of the iv1 is as follows:
DC COMPONENT = -8.787830E+00
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 4.762E+01 1.441E-02 1.000E+00 1.634E+02 0.000E+00
2 9.524E+01 1.172E-02 8.135E-01 2.644E+01 -3.004E+02
3 1.429E+02 5.315E-02 3.689E+00 1.406E+02 -3.497E+02
4 1.905E+02 6.418E-03 4.454E-01 7.955E+01 -5.742E+02
5 2.381E+02 7.029E-03 4.879E-01 -1.301E+01 -8.302E+02
6 2.857E+02 4.560E-02 3.165E+00 -1.116E+02 -1.092E+03
7 3.333E+02 5.715E-03 3.967E-01 7.054E+01 -1.073E+03
8 3.810E+02 6.181E-03 4.290E-01 -2.357E+01 -1.331E+03
9 4.286E+02 9.775E-02 6.784E+00 -1.723E+01 -1.488E+03
10 4.762E+02 5.177E-03 3.593E-01 5.656E+01 -1.578E+03
11 5.238E+02 5.693E-03 3.951E-01 -4.294E+01 -1.841E+03
12 5.714E+02 9.631E-02 6.684E+00 -7.650E+01 -2.038E+03
13 6.190E+02 5.416E-03 3.759E-01 3.786E+01 -2.087E+03
14 6.667E+02 5.940E-03 4.123E-01 -6.016E+01 -2.348E+03
15 7.143E+02 6.415E-02 4.452E+00 8.983E+01 -2.362E+03
16 7.619E+02 5.151E-03 3.575E-01 1.546E+01 -2.599E+03
17 8.095E+02 5.751E-03 3.992E-01 -9.197E+01 -2.870E+03
18 8.571E+02 2.787E+00 1.934E+02 -3.141E+01 -2.973E+03
19 9.048E+02 2.586E-03 1.794E-01 -7.696E+01 -3.182E+03
20 9.524E+02 2.128E-03 1.477E-01 -9.392E+01 -3.363E+03
21 1.000E+03 9.176E-02 6.369E+00 -9.376E+01 -3.526E+03
22 1.048E+03 5.374E-03 3.730E-01 2.560E+01 -3.570E+03
23 1.095E+03 8.048E-03 5.586E-01 -7.191E+01 -3.831E+03
TOTAL HARMONIC DISTORTION = 1.939046E+04 PERCENT
Id(avg)=-DC=8.79A
7. Obtain the load neutral voltage with respect to the mid-point of the dc input voltage.
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