Hart Chapter 8 solutions

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CHAPTER 8 SOLUTIONS
4/24/10
8-1) 
	
	 
8-2)
	
c) PSpice results are consistent with parts (a) and (b). The current waveform reaches steady state after approximately 100 ms, corresponding to 5 time constants.
	
8-3)
 b) 
	
8-4)
​
8-5)
	
	n
	Vn
	Zn
	In,rms
	1
	331
	29.3
	8.0
	3
	110
	77
	1.02
	5
	66
	127
	0.37
	
8-6)
.37ms________
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	n
	Vn
	Zn
	In,rms
	1
	88.6
	31.3
	2.0
	3
	29.5
	61.8
	0.34
	5
	17.7
	97.5
	0.13
Using PSpice,
FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L)
 DC COMPONENT = -3.668708E-06
 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
 NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
 1 1.200E+02 2.830E+00 1.000E+00 -3.716E+01 0.000E+00
 2 2.400E+02 5.377E-06 1.900E-06 -1.203E+02 -4.594E+01
 3 3.600E+02 4.778E-01 1.688E-01 -6.658E+01 4.490E+01
 4 4.800E+02 3.589E-06 1.268E-06 -1.223E+02 2.629E+01
 5 6.000E+02 1.818E-01 6.422E-02 -7.587E+01 1.099E+02
 6 7.200E+02 2.858E-06 1.010E-06 -1.162E+02 1.068E+02
 7 8.400E+02 9.427E-02 3.331E-02 -8.028E+01 1.798E+02
 8 9.600E+02 2.523E-06 8.913E-07 -1.095E+02 1.878E+02
 9 1.080E+03 5.743E-02 2.029E-02 -8.292E+01 2.515E+02
 TOTAL HARMONIC DISTORTION = 1.847695E+01 PERCENT
8-7)
Using a restricted time interval of 33.33 ms to 50 ms to analyze steady-state current, the peak value is 8.26 A and the rms value is 4.77 A. The THD from the output file is 32%.
8-8)
	
	n
	|Vn|
	Zn
	In,rms
	1
	90
	12.5
	5.08
	3
	51.6
	24.7
	1.5
	5
	4.43
	39
	0.08
8-9)
	
8-10)
	α = 30° 
Using the FFT function in Probe shows that voltages at frequencies at multiples of n = 3 are absent.
	 b) α = 15° 
Using the FFT function in Probe shows that voltages at frequencies at multiples of n = 5 are absent.
8-11)
	From Eq. (8-22),
	
	Using the FFT function in Probe, the n = 7 harmonic is absent.
8-12) 
 Letting T = 360 seconds and taking advantage of half-wave symmetry,
	
8-13) 
The VPWL_FILE source is convenient for this simulation. A period of 360 seconds is used, making each second equal to one degree. A transient simulation with a run time of 360 second and a maximum step size of 1m gives good results. The FFT of the Probe output confirms that the 3rd and 5th harmonics and their multiples are eliminated.
	
	0 		0 
30 		0 
30.01 		1 
54 		1 
54.01 		0 
66 		0 
66.01 		1
114 		1 
114.01 		0 
126 		0 
126.01 		1 
150 		1 
150.01 		0 
210 		0 
210.01 		-1 
234 		-1 
234.01 		0 
246 		0 
246.01 		-1 
294 		-1 
294.01 		0 
306 		0 
306.01		 -1 
330 		-1 
330 		0 
360 		0
8-14)
a) 
	
	n
	1
	3
	5
	7
	9
	Vn
	149.5
	0
	-2.79
	-3.04
	-14.4
	
8-15)
	
	α1
	α2
	α3
	Mi
	15
	25
	55
	0.815
	20
	30
	40
	0.857
	10
	30
	50
	0.831
	10
	30
	70
	0.731
8-16)
This inverter is designed to eliminate harmonics n = 5, 7, 11, and 13. The normalized coefficients through n = 17 are
 
	n
	Vn/Vdc
	1
	4.4593
	3
	-0.8137
	5
	0.0057 ≈ 0
	7
	-0.0077 ≈ 0
	9
	-0.3810
	11
	0.0043 ≈ 0
	13
	-0.0078 ≈ 0
	15
	-0.0370
	17
	0.1725
The coefficients are not exactly zero for those harmonics because of rounding of the angle values.
8-17)
8-18)
	
From Table 8-3,
	
	n
	Vn/Vdc
	Vn
	Zn
	In=Vn/Zn
	
	1
	0.8
	76.8
	33.3
	2.30
	mf
	17
	0.82
	78.7
	157
	0.50
	mf - 2
	15
	0.22
	21.1
	139
	0.151
	mf + 2
	19
	0.22
	21.1
	175
	0.121
	
8-19)
	
From Table 8-3,
	
	n
	Vn/Vdc
	Vn
	Zn
	In=Vn/Zn
	
	1
	0.9
	225
	27.5
	8.18
	mf
	31
	0.71
	178
	585
	0.305
	mf - 2
	29
	0.27
	67
	547
	0.122
	mf + 2
	33
	0.27
	67
	622
	0.108
	
�
8-20)
The circuit “Inverter Bipolar PWM Function” is suitable to verify the design results. The parameters are modified to match the problem values.
Transient Analysis and Fourier Analysis are establish in the Simulation Setup menu:
The output file contains the THD of the load current, verifying that the THD is less than 10%.
TOTAL HARMONIC DISTORTION = 9.387011E+00 PERCENT
8-21)
Example solution:
	
8-22)
Example solution:
	
8-23)
Use the bipolar PWM function circuit of Fig. 8-23a, and use the unipolar PWM function circuit of Fig.8-26 with mf = 10. Use ma = 0.8 for V1 = 120 V from the 150-V dc source.
The THD for bipolar, mf = 21, is 10.2 %, for bipolar mf = 41 is 5.2%, and for unipolar mf = 10 is 5.9%.
Bipolar mf = 21:
Bipolar mf = 41:
Unipolar, mf = 10:
8-24)
	
8-25)
	
		For f = 25 Hz:
	n
	VnL-N
	Zn
	In
	In,rms
	1
	255
	11.1
	23.0
	16.3
	5
	50.9
	25.6
	2.0
	1.41
	7
	36.4
	34.5
	1.06
	0.75
	11
	23.1
	52.8
	0.44
	0.31
	13
	19.6
	62.0
	0.32
	0.22
	
		For f = 100 Hz,
	n
	VnL-N
	Zn
	In
	In,rms
	1
	255
	21.3
	11.9
	8.43
	5
	50.9
	94.8
	0.54
	0.38
	7
	36.4
	132
	0.27
	0.19
	11
	23.1
	208
	0.12
	0.08
	13
	19.6
	245
	0.08
	0.06
The THD for current is reduced from 10% to 5.19% as f is increased from 25 Hz to 100 Hz. The THD of the line-to-neutral voltage remains at 27.3%.
These results can also be determined from a PSpice simulation for the six-step inverter.
�
8-26)
	
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