22-Elec-B8 Power Electronics and Drives · December 2013
Question 1 of 6: Multiple-Pulse versus Sinusoidal PWM, and a Two-Pulse Uniform-PWM Inverter
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2013 — 07-Elec-B8 Power Electronics and Drives. Open-book, three hours, any non-communicating calculator permitted. Six problems are printed and any five constitute a complete paper; all are of equal value (20 points each). Because this set is a study resource, all six problems are solved in full below, every lettered sub-part included.
Reference texts for this subject
M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed. — inverters and PWM schemes (Ch. 6), dc–dc converters (Ch. 5), controlled rectifiers (Ch. 10), drives (Ch. 15).
N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics: Converters, Applications and Design, 3rd ed. — switch-mode inverters and harmonic analysis (Ch. 8).
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — synchronous machines (Ch. 5), induction machines (Ch. 7), dc machines (Ch. 8–9).
B. K. Bose, Modern Power Electronics and AC Drives — constant volts-per-hertz induction-motor drives and PWM side effects.
R. Krishnan, Electric Motor Drives: Modeling, Analysis and Control — converter-fed dc drives.
C. W. Lander, Power Electronics, 3rd ed. — choppers and controlled rectifiers.
J. J. Grainger and W. D. Stevenson, Power System Analysis — salient-pole power-angle equations and capability curves.
IEEE Std 519 and CSA C22.1 (Canadian Electrical Code, Part I) — harmonic limits and harmonic-related conductor sizing in Canadian practice.
Check — two anomalies carried over from the printed paper. (i) Problem 6 is headed “December 2010” on the December 2013 question sheet and is solved here as printed. (ii) Problem 5 supplies a total leakage inductance of 1.25 mH but its accompanying formula describes $L_T$ as the total leakage reactance. The formula is dimensionally consistent only if $L_T$ is an inductance, so it is read as an inductance throughout, which is also what the numbers require.
Question 1: Multiple-Pulse versus Sinusoidal PWM, and a Two-Pulse Uniform-PWM Inverter (20 marks)
Part (a) — Multiple-pulse modulation compared with sinusoidal PWM
Both schemes generate the gate pattern by comparing a reference (modulating) signal of amplitude $A_r$ against a triangular carrier of amplitude $A_c$, and in both the modulation index is $M = A_r/A_c$. The difference lies entirely in the shape of the reference, and that one difference propagates all the way into the harmonic spectrum.
In multiple-pulse modulation — also called uniform PWM — the reference is a constant-amplitude square wave at the output frequency. Every intersection with the carrier therefore occurs at the same relative position within its carrier period, so all $p$ pulses in a half cycle come out with the same width $\delta$. Voltage control is obtained by widening or narrowing all pulses together. In sinusoidal PWM the reference is a sinusoid at the output frequency, so the pulse widths vary sinusoidally across the half cycle: wide near the crest at $\omega t = \pi/2$ and narrow near the zero crossings. The pulse pattern therefore already carries the shape of the wanted sine wave, not just its average value.
The consequence for lower-order harmonics is decisive. A uniform pulse train is a rectangular waveform whose spectrum contains substantial third, fifth and seventh harmonics whenever $p$ is small; those components only begin to fall away once the harmonic order exceeds roughly $2p$, so with two or three pulses per half cycle the low-order content is barely improved on a plain square wave. Sinusoidal PWM instead pushes the first significant harmonic cluster up to the neighbourhood of the frequency-modulation ratio $m_f = f_c/f_o$ and its sidebands, leaving the low-order odd harmonics strongly attenuated. This is why practical drives use SPWM with $m_f$ of 20 or more.
The distortion factor quantifies exactly that difference. Rashid defines it as
the $1/n^{2}$ weighting representing the second-order attenuation a load inductance provides. Because the weighting punishes low-order harmonics most heavily, a scheme that leaves large third and fifth harmonics is penalised severely. Uniform PWM with a small $p$ therefore shows a distortion factor of the order of several percent, whereas sinusoidal PWM at the same fundamental and the same switching frequency typically shows a distortion factor an order of magnitude smaller, because the harmonics that survive sit at $n \approx m_f$ where the $1/n^{2}$ weight has already reduced them to insignificance. The practical pay-offs are a smaller output filter, less harmonic heating in the machine, and lower torque pulsation.
Parts (b) and (c) — the two-pulse scheme
Given.
Quantity
Symbol
Value
Inverter topology
—
single-phase full bridge, dc source $V_s$
Pulses per half cycle
$p$
2
Modulation index
$M = A_r/A_c$
0.4 (part c)
Carrier
—
triangular, $p$ periods per half cycle
Reference
—
square wave of amplitude $A_r$
Find. The redrawn gating and output waveforms for $p = 2$; closed-form expressions for $\alpha_1$, $\alpha_2$ and $\delta$ in terms of $M$; and the fundamental and third-harmonic components of $v_o(t)$, expressed as multiples of $V_s$, when $M = 0.4$.
[Figure not reproduced: Figure 1 — Uniform (multiple-pulse) PWM redrawn for p = 2 at M = 0.4: the triangular carrier completes two periods per half cycle, and each pulse is centred on its slot centre with width δ = πM/p = 36°, giving α₁ = 27° and α₂ = 117°. See the official exam paper.]
Approach. Read the pulse edges off the carrier–reference intersection geometry to get $\delta$ and $\alpha_m$ as functions of $M$, then substitute those angles into the supplied Fourier sums for $n = 1$ and $n = 3$.
Part (b) — fix the carrier geometry for $p = 2$. With $p$ pulses required in the half cycle $0 \le \omega t \le \pi$, the carrier must complete $p$ periods in that interval, so one carrier period spans an angle $\pi/p$. For $p = 2$ each carrier period occupies $90^\circ$, and the half cycle divides into two equal slots $[0,\,\pi/2]$ and $[\pi/2,\,\pi]$, one pulse per slot. The carrier peaks at the slot boundaries and passes through zero at the slot centres, which places each pulse symmetrically about its own slot centre at
$$\omega t = \frac{(2m-1)\pi}{2p}, \qquad m = 1, 2 \quad \Rightarrow \quad 45^\circ \text{ and } 135^\circ .$$
Obtain the pulse width from the comparison level. Inside one slot the carrier magnitude rises linearly from $0$ at the slot centre to $A_c$ at each slot edge. Writing $\xi$ for the angle measured from the slot centre, the carrier magnitude is $A_c\,\xi/(\pi/2p)$, and the gate signal is present while this is below the reference level $A_r$. The switching angle is therefore $\xi = (\pi/2p)(A_r/A_c)$ on each side of the centre, and the total pulse width is twice that:
$$\boxed{\;\delta = \frac{\pi M}{p} = \frac{\pi M}{2}\;}$$
for $p = 2$. As required, $\delta$ sweeps from $0$ to the full slot width $\pi/p$ as $M$ goes from 0 to 1.
Convert the centres and width into leading edges. Each pulse starts half a width before its slot centre, so $\alpha_m = (2m-1)\pi/(2p) - \delta/2$. Substituting $p = 2$ and $\delta = \pi M/2$,
$$\boxed{\;\alpha_1 = \frac{\pi}{4}\left(1 - M\right), \qquad \alpha_2 = \frac{\pi}{4}\left(3 - M\right)\;}$$
with the trailing edges at $\alpha_1 + \delta$ and $\alpha_2 + \delta$. The negative half cycle repeats the pattern displaced by $\pi$, which is the half-wave symmetry the Fourier expressions assume.
Evaluate the angles at $M = 0.4$. Substituting $M = 0.4$ into the three closed forms gives
$$\delta = \frac{180^\circ \times 0.4}{2} = 36^\circ, \qquad \alpha_1 = 45^\circ(1-0.4) = 27^\circ, \qquad \alpha_2 = 45^\circ(3-0.4) = 117^\circ .$$
The pulses therefore occupy $27^\circ$ to $63^\circ$ and $117^\circ$ to $153^\circ$, each still centred on $45^\circ$ and $135^\circ$ respectively, and the figure above is drawn to these angles.
Part (c) — show that the cosine coefficients vanish. The pulse centres are $\alpha_m + \delta/2 = 45^\circ$ and $135^\circ$, so for the fundamental
$$A_1 = \frac{4V_s}{\pi}\sin 18^\circ\left[\cos 45^\circ + \cos 135^\circ\right] = \frac{4V_s}{\pi}(0.3090)(0.7071 - 0.7071) = 0 .$$
The same cancellation occurs for every odd $n$, because the two centres are symmetric about $\omega t = \pi/2$. The waveform is an odd function about the origin once the half-wave symmetry is accounted for, so only sine terms survive and $v_o(t) = \sum B_n \sin n\omega t$.
Compute the fundamental sine coefficient. With $n = 1$, $\sin(n\delta/2) = \sin 18^\circ = 0.30902$ and the bracket becomes $\sin 45^\circ + \sin 135^\circ = 1.41421$:
$$B_1 = \frac{4V_s}{\pi}(0.30902)(1.41421) = 1.27324\,V_s \times 0.43700 = \boxed{\;0.5564\,V_s\;}$$
so the fundamental output is $v_1(t) = 0.556\,V_s\sin\omega t$, an rms value of $0.5564/\sqrt{2} = 0.3935\,V_s$.
Compute the third-harmonic coefficient. For $n = 3$ the width factor becomes $\sin(3\delta/2) = \sin 54^\circ = 0.80902$, while the angle bracket is $\sin 135^\circ + \sin 405^\circ = 1.41421$ again:
$$B_3 = \frac{4V_s}{3\pi}(0.80902)(1.41421) = 0.42441\,V_s \times 1.14412 = \boxed{\;0.4856\,V_s\;}$$
giving $v_3(t) = 0.486\,V_s\sin 3\omega t$, an rms value of $0.3434\,V_s$.
Check the result against the total rms and comment. For uniform PWM the total output rms is $V_o = V_s\sqrt{p\delta/\pi} = V_s\sqrt{2(0.2\pi)/\pi} = 0.6325\,V_s$, and the fundamental accounts for $0.3935/0.6325 = 62.2$ percent of it — consistent with a spectrum in which the third harmonic is nearly as large as the fundamental. Indeed $B_3/B_1 = 0.873$: with only two pulses per half cycle the scheme suppresses almost nothing, which is precisely the weakness identified in part (a).
