Solved Problems · Set 27

PWM Techniques and Harmonic Control

Part 4 · Inverters — Set 26 ended on a constraint: plain SPWM cannot make rated voltage. Six problems on every scheme that can, and what each one costs.

Prof. Mithun Mondal 6 solved problems GATE · ESE · University

Set 27 — PWM Techniques and Harmonic Control

Set 26 ended on a constraint rather than a result. A 415 V drive fed from a 415 V supply has 587 V of DC link, and plain sinusoidal PWM converts only 61% of it — 359 V where 415 is needed. Third-harmonic injection just closes the gap. That makes the modulation scheme a load-bearing design decision, not a preference.

There are more schemes than two, and they compete on three axes that pull against each other: how much of the DC link they convert, how much switching loss they cost, and what they leave in the spectrum. Some buy the last 9.3% by deliberately distorting; some place switching angles to null chosen harmonics exactly; some cut switching loss by a third by clamping each leg for part of every cycle.

Part 4 · Chapter 18 · 6 solved problems

i Method Recap
  • Two ratios define a carrier-based scheme:

    \[ m_a = \frac{\hat V_{ref}}{\hat V_{tri}}, \qquad m_f = \frac{f_{tri}}{f_{out}} \]
  • The linear range ends at \(m_a = 1\); beyond it the output saturates towards the square wave:

    \[ \hat V_1: \ m_aV_{dc} \ \longrightarrow\ \frac{4V_{dc}}{\pi} \quad\text{— a further }27.3\% \]
  • Choose \(m_f\) odd and a multiple of three for a three-phase inverter:

    \[ \text{odd} \Rightarrow \text{no even harmonics}; \qquad \times3 \Rightarrow \text{carrier is zero sequence} \]
  • Selective harmonic elimination places \(k\) angles per quarter cycle to satisfy \(k\) conditions:

    \[ V_n = \frac{4V_{dc}}{n\pi}\left[1-2\cos n\alpha_1+2\cos n\alpha_2-\cdots\right] \]
  • Discontinuous modulation clamps each leg for 120° of every cycle:

    \[ \text{switchings reduced by }\tfrac13 \;\Longrightarrow\; P_{sw}\ \text{reduced by }\tfrac13 \]
  • Synchronous modulation locks the carrier to the output; asynchronous does not:

    \[ m_f > 21 \Rightarrow \text{asynchronous acceptable}; \quad m_f \le 21 \Rightarrow \text{must synchronise} \]
Problem 1CoreTwo Ratios That Define Everything

For a single-phase full bridge on a 400 V link, tabulate the fundamental against modulation index from 0 to the square-wave limit. Identify the linear range, the total gain available beyond it, and explain what is lost in that region.

509 V 400 V ma = 1 linear overmodulation square wave ma →
Linear to the carrier peak, then a slow 27% climb bought with low-order harmonics
Solution

The linear region. While the reference peak stays below the carrier peak, every carrier period produces a pulse whose width is proportional to the reference:

\[ \hat V_1 = m_aV_{dc}, \qquad 0 \le m_a \le 1 \]
\[ m_a = 1 \;\Longrightarrow\; \hat V_1 = 400\ \text{V peak} = 283\ \text{V rms} \]

Exactly proportional, with no trigonometry — which is the whole reason PWM displaced quasi-square control. A regulator that wants 5% more output simply asks for 5% more \(m_a\).

What lies beyond. Raising the reference above the carrier means some carrier intersections are missed, and pulses begin to merge:

\[ m_a \to \infty:\quad \hat V_1 \to \frac{4V_{dc}}{\pi} = 509.3\ \text{V} \]
\[ \text{gain available} = \frac{509.3}{400}-1 = 27.3\% \]

More than a quarter again — a substantial amount to leave on the table when Set 26's drive was 13% short. But it is not free.

The whole range:

\(m_a\)Region\(\hat V_1\)Low-order harmonics
0.5linear200 Vnone
0.8linear320 Vnone
1.0limit400 Vnone
1.2overmodulation I~440 Vappearing
2.0overmodulation II~490 Vsignificant
\(\to\infty\)square wave509 V48.3% THD

The fourth column is what is being bought and sold. Every volt gained past \(m_a = 1\) comes with low-order harmonics that PWM existed to remove — and by the square-wave limit the spectrum is exactly the one Set 25, Problem 1 started with.

Why the relationship goes nonlinear. In the linear range each carrier period contributes a pulse; past the limit some do not:

\[ m_a > 1 \;\Longrightarrow\; \text{reference exceeds carrier for part of the cycle} \;\Longrightarrow\; \text{pulses drop} \]

Near the peaks of the reference the output simply stays clamped at the rail for several carrier periods together. Those missing transitions are what produce the low-order harmonics — and they also mean the switching loss falls in overmodulation, which is occasionally useful.

The three-phase version uses the same idea against a different baseline:

SchemeLinear limitUtilisationRemaining gain to six-step
SPWM\(m_a = 1\)0.61227.3%
Third-harmonic / SVM\(m_a = 1.155\)0.70710.3%
Six-step0.780

Injection has already collected two-thirds of the available gain within the linear range, leaving only 10.3% for overmodulation to fight over. That is why a modern drive uses injection first and overmodulates only in field weakening.

The linear range is where PWM earns its name. Output proportional to reference, no low-order harmonics, and a controller that needs no correction table. Past \(m_a = 1\) the converter is being asked for something the link cannot supply cleanly, and it responds by degrading the spectrum — gradually at first, then all the way back to a square wave.
Answera linear for \(m_a \le 1\), \(\hat V_1 = m_aV_{dc}\)   b square-wave limit \(509.3\ \text{V}\)   c 27.3% more, paid for in low-order harmonics
Problem 2CoreChoosing the Carrier

A three-phase drive must operate from 5 Hz to 60 Hz output. Discuss the choice of carrier ratio: why it should be odd and a multiple of three, when synchronous operation is required, and how the choice interacts with switching loss and filter size.

Solution

Why odd. An odd carrier ratio makes the modulated waveform half-wave symmetric:

\[ m_f\ \text{odd} \;\Longrightarrow\; v(\theta+\pi) = -v(\theta) \;\Longrightarrow\; \text{even harmonics vanish} \]

And with them the DC component. An even carrier ratio leaves a DC offset in the output — small, but enough to saturate an output transformer or produce a standing torque in a machine, and entirely avoidable at zero cost.

Why a multiple of three. If \(m_f\) is divisible by 3, the carrier harmonic appears identically in all three phases:

\[ m_f = 3k \;\Longrightarrow\; \text{carrier component is zero sequence} \;\Longrightarrow\; \text{cancels in }v_{L} \]

The largest single harmonic in the spectrum — 81.8 V of the 200 V link in Set 25's example — disappears from every line voltage. Only the sidebands remain, and they are smaller. So the standard choices are 15, 21, 27, 33, 39: odd and divisible by three.

Synchronous versus asynchronous. A synchronous modulator locks the carrier to a fixed multiple of the output frequency; an asynchronous one runs the carrier at a fixed frequency regardless:

SynchronousAsynchronous
\(f_{carrier}\)\(m_ff_{out}\)fixed
Subharmonicsnonepossible
Spectrumclean, predictablesmeared
Switching lossvaries with speedconstant
Implementationneeds a phase-locked carriertrivial

The critical row is subharmonics. If the carrier is not an integer multiple of the output, the waveform does not repeat every output cycle — and the difference appears as components below the fundamental. In a motor those produce low-frequency torque pulsations and a slow mechanical oscillation, which is far worse than any high-frequency harmonic.

The rule. Subharmonic amplitude scales roughly as \(1/m_f\), so it becomes negligible when the carrier ratio is large:

\[ m_f > 21:\ \text{asynchronous acceptable}; \qquad m_f \le 21:\ \text{synchronise} \]

Applied to this drive at a fixed 4 kHz carrier:

\(f_{out}\)\(m_f\) at 4 kHzMode
5 Hz800asynchronous — easily
50 Hz80asynchronous
60 Hz67asynchronous
200 Hz (field weakening)20synchronise

A general-purpose drive is asynchronous throughout its normal range and switches to synchronous modulation only at high output frequency — which is precisely where it is also overmodulating. Traction drives, which spend their lives there, are synchronous throughout.

The interaction with loss and filtering:

\[ P_{sw} \propto f_{carrier}, \qquad \text{filter reactance} \propto \frac{1}{f_{carrier}^2} \]
CarrierSwitching lossFilter / rippleAcoustic
2 kHzlowestworstaudible whine
4 kHztypicalacceptableaudible
8 kHzdouble4× betterless intrusive
16 kHz16× betterinaudible

Set 26's drive found switching loss already at 63% of the total at 4 kHz, so 16 kHz would take the inverter from 98.1% to about 94%. Silence is expensive, and large drives choose noise.

Two integer properties of the carrier ratio remove two whole families of harmonics. Odd kills the evens and the DC; divisible by three makes the dominant carrier component zero-sequence so it cancels in the line voltages. Neither costs anything, and both are lost the moment the modulator goes asynchronous — which is why the low-frequency end of a drive's range is where the spectrum is worst.
Answera odd → no evens or DC; \(\times3\) → carrier cancels in the line   b synchronise when \(m_f \le 21\)   c loss \(\propto f_c\), filter \(\propto 1/f_c^2\)
Problem 3Exam levelEliminating Harmonics Exactly

A three-phase inverter uses two switching angles per quarter cycle to eliminate the fifth and seventh harmonics exactly. Set up the equations, solve for the angles, and evaluate the resulting fundamental and residual spectrum. Compare with carrier-based PWM at the same switching count.

Solution

The waveform. Instead of comparing against a carrier, the switching instants are computed offline. With two notches per quarter cycle at \(\alpha_1\) and \(\alpha_2\), the Fourier coefficients are:

\[ \hat V_n = \frac{4V_{dc}}{n\pi}\left[1-2\cos n\alpha_1+2\cos n\alpha_2\right] \]

Two unknowns, so two conditions can be imposed. Choosing to null the fifth and seventh uses both — leaving the fundamental to be whatever it turns out to be, with no voltage control. Set 25, Problem 2 met the same arithmetic with one angle.

The equations:

\[ 1-2\cos5\alpha_1+2\cos5\alpha_2 = 0 \]
\[ 1-2\cos7\alpha_1+2\cos7\alpha_2 = 0 \]

Transcendental and coupled — no closed form exists. They are solved numerically once, offline, and the angles stored in a lookup table. That offline solution is the defining feature of the method: nothing is computed in real time.

The solution:

\[ \alpha_1 = 16.25^\circ, \qquad \alpha_2 = 22.07^\circ \]
\[ \text{check: } V_5 = V_7 = 0\ \text{to machine precision}\ \checkmark \]

The resulting spectrum, normalised to \(V_{dc}\):

Order\(\hat V_n/V_{dc}\)% of fundamentalNote
11.188100%above the 1.0 of SPWM
30.20717.4%cancels in three-wire
50eliminated
70eliminated
90.1099.1%cancels in three-wire
110.24120.3%the new dominant
130.32227.1%larger still

Two observations. The fundamental is 1.188 times the link — well above the 1.0 that SPWM manages, because the waveform is closer to a square wave. And the eliminated energy has not disappeared: the 11th and 13th are now larger than the 5th and 7th were.

Why that redistribution is still a win. In a motor the harmonic current is limited by the leakage reactance, which rises with frequency:

\[ I_n = \frac{V_n}{n\omega L} \;\Longrightarrow\; \frac{I_{13}}{I_5} = \frac{0.322/13}{0.200/5} = \frac{0.0248}{0.0400} = 0.62 \]

Comparing against the six-step spectrum's 5th harmonic: a 27% voltage harmonic at the 13th produces only 62% of the current that a 20% harmonic at the 5th did. Pushing distortion up in frequency reduces the current it causes even when the voltage grows — and current is what heats the machine and pulsates the torque.

Where SHE belongs:

PropertySHECarrier PWM
Switchings per cyclevery fewmany
Chosen harmonicsexactly zerosmall but nonzero
Fundamental utilisationhighestlower
Real-time computationnone — lookup tabletrivial
Variable frequencya table per operating pointcontinuous
Dynamic responsepoorfast

The last two rows confine it. SHE angles are valid for one modulation index, so a variable-speed drive needs a table across the whole range and interpolation between entries — and the waveform cannot be changed mid-cycle without breaking the elimination. It is therefore the method of choice for high-power, low-switching-frequency, slowly-varying applications: traction inverters, large industrial drives, grid-connected converters where each switching transition costs kilowatts.

Selective elimination does not remove harmonic energy — it moves it upward. Nulling the 5th and 7th made the 11th and 13th larger than they were. That is a good trade because harmonic current falls as \(1/n\) against the load's reactance, so the same voltage distortion at a higher order does less damage. Every PWM scheme is a variation on this one idea.
Answera\(\alpha_1 = 16.25^\circ,\ \alpha_2 = 22.07^\circ\)   b\(\hat V_1 = 1.188V_{dc}\)   c 5th and 7th exactly zero; 11th and 13th grow to 20% and 27%
Problem 4Exam levelPast the Linear Limit

A 415 V drive from Set 26 has a 587 V link and needs 415 V of line voltage, rising to 458 V during field weakening. Determine what modulation is required in each case, describe the two overmodulation regions, and quantify the harmonic penalty.

Solution

Locate each requirement on the utilisation scale:

RequirementNeeded utilisationScheme
359 V (SPWM limit)0.612SPWM at \(m_a = 1\)
415 V (rated)0.707THI/SVM at the linear limit
458 V (field weakening)0.780six-step — full overmodulation

So rated operation sits exactly at the linear ceiling of third-harmonic injection, with no margin at all, and anything above it requires overmodulation. That is the normal condition of every mains-fed drive.

The two overmodulation regions. They differ in what the modulator is doing:

RegionRangeBehaviour
Linear\(m \le 1\)reference tracked exactly
Overmodulation I\(1 < m \lesssim 1.05\)reference clipped near its peaks; still continuous
Overmodulation II\(1.05 \lesssim m < 1.103\)output holds at a vertex for part of each sector
Six-step\(m = 1.103\)one state per 60° — no modulation

In region I the modulator is still switching every carrier period but cannot reproduce the reference near its peaks. In region II it abandons whole intervals, dwelling on a single state — which is where the low-order harmonics grow rapidly.

The gain-versus-distortion trade:

\[ \text{available above the linear limit} = \frac{0.780}{0.707}-1 = 10.3\% \]
Operating pointUtilisationLow-order THD
Linear limit0.707≈ 0
Mid overmodulation~0.75~10%
Six-step0.78031.1%

Ten per cent more voltage for thirty-one per cent distortion. Whether that is a good trade depends entirely on what the extra voltage is for — and in field weakening, it is for speed.

Why field weakening makes it acceptable. Above base speed the machine's flux is reduced and the frequency raised:

\[ I_n = \frac{V_n}{n\omega L} \;\Longrightarrow\; \text{harmonic currents fall as } \omega\ \text{rises} \]

A 31% voltage distortion at 100 Hz produces half the harmonic current that the same distortion would at 50. So the operating region where overmodulation is needed is also the region where its consequences are mildest — a genuinely fortunate alignment, and the reason six-step operation at high speed is standard rather than exceptional.

What it costs elsewhere:

ConsequenceDetail
Torque ripple at \(6f\)from the 5th and 7th beating with the fundamental
Extra rotor heatingharmonic currents do no useful work
Nonlinear control gain\(V_1\) no longer proportional to \(m\)
Loss of current controlthe modulator cannot follow the reference
Lower switching lossfewer transitions — the one benefit

The third row is what makes overmodulation awkward for a controller. Inside the linear range the loop sees a constant gain; in region II the gain falls towards zero as six-step is approached, so a current regulator tuned for the linear range becomes sluggish and can lose control entirely. Modern drives handle it with an explicit inverse mapping from desired to commanded modulation index.

Overmodulation is a deliberate surrender of waveform quality for voltage. The last 10.3% of a DC link is only available by degrading the spectrum all the way back to six-step — and it is worth taking, because the operating region that needs it is field weakening, where high frequency suppresses the harmonic currents that the distortion would otherwise cause.
Answera rated 415 V needs THI/SVM at the linear limit   b 458 V needs full six-step   c 10.3% gain for 31.1% THD, tolerable in field weakening
Problem 5ChallengeNot Switching At All

Discontinuous PWM clamps each inverter leg to a rail for 120° of every output cycle. Explain how a zero-sequence injection achieves this, quantify the switching-loss saving, and identify when the saving is larger and smaller than the nominal one-third.

Solution

The insight from Set 26, Problem 4 was that any signal added to all three references cancels in the line voltages. Third-harmonic injection used that freedom to flatten the peak. It can be used for something else entirely:

\[ v_{zs}(\theta)\ \text{chosen so that one reference always equals} \pm1 \]

If a reference is pinned at the carrier peak, that leg never crosses the carrier — so it never switches. It is clamped to the rail, dissipating only conduction loss, while the other two legs carry the modulation.

How much can be clamped. By symmetry, each of the three legs takes a turn:

\[ \frac{360^\circ}{3} = 120^\circ\ \text{of clamping per leg per cycle} \]
\[ \frac{120^\circ}{360^\circ} = \frac13 \;\Longrightarrow\; P_{sw}\ \text{reduced by }33\% \]

A third off the switching loss, from a scheme that produces exactly the same line voltages. Applied to Set 26's drive, that is 31 W of the 92 W switching budget — taking the inverter from 98.1% to 98.5%, or allowing the carrier to rise by half for the same loss.

But the saving is not exactly a third, and this is the interesting part. Switching loss depends on the current at the switching instant:

\[ E_{sw} \propto \left|i\right| \;\Longrightarrow\; \text{clamping matters most where }\left|i\right|\text{ is largest} \]
Clamping placed atCurrent thereActual saving
Current peakmaximumup to 50%
Voltage peak (fixed DPWM)depends on power factor33% at unity PF
Current zero crossingminimumas little as 15%

So the placement of the clamped interval should follow the current, not the voltage — and since the phase angle between them depends on the load, the optimum shifts with power factor. Generalised DPWM schemes take the power factor angle as an input and position the clamping accordingly.

What it costs:

PropertyContinuous (SVM)Discontinuous
Switching loss0.67×
Current ripple at low \(m\)lowerhigher
Current ripple at high \(m\)higherlower
Common-mode swing\(V_{dc}\)reduced
Linear rangesamesame
Complexityone extra zero-sequence term

Rows two and three cross over at around \(m = 0.6\). Above it discontinuous modulation is better on both counts — less loss and less ripple — because fewer, larger voltage pulses produce less current ripple than many small ones once the modulation depth is high. Below it the continuous scheme wins on ripple.

The natural strategy is therefore to switch schemes with operating point:

\[ m < 0.6:\ \text{continuous SVM}; \qquad m > 0.6:\ \text{discontinuous} \]

Which most modern drives do, changing over on the fly. The transition is seamless because both schemes produce identical line voltages — only the common-mode component differs, and nothing downstream can detect it.

The fourth row of the table is a bonus. Clamping a leg to a rail means it is at a known potential rather than switching:

\[ \text{fewer transitions} \;\Longrightarrow\; \text{fewer common-mode steps} \;\Longrightarrow\; \text{less bearing current} \]

The 4 A spikes of Set 26, Problem 5 occur at every transition, so removing a third of the transitions removes a third of them. Discontinuous modulation therefore improves efficiency, ripple at high modulation, and bearing life simultaneously.

The zero-sequence freedom can buy voltage or it can buy efficiency. Third-harmonic injection spends it flattening the reference for 15.5% more output; discontinuous modulation spends it pinning one leg to a rail for 33% less switching loss. Both produce identical line voltages, and a drive can choose between them — or move between them — without the load ever knowing.
Answera zero-sequence injection pins one reference to a rail   b 120° per leg → 33% less switching loss   c up to 50% if clamped at the current peak, as little as 15% at the zero crossing
Problem 6ChallengeA Strategy for a Drive

Specify a complete modulation strategy for the 7.5 kW drive of Set 26 across its full operating range — 0 to 200 Hz output at a 4 kHz nominal carrier — selecting the scheme, carrier ratio and synchronisation for each region, and justifying each transition.

Solution

Map the operating range against the constraints established so far:

Region\(f_{out}\)\(m\)Binding constraint
Standstill / creep0–5 Hz< 0.1dead-time distortion
Low speed5–25 Hz0.1–0.5current ripple, torque quality
Mid speed25–45 Hz0.5–0.9efficiency
Base speed50 Hz1.0DC link utilisation
Field weakening50–200 Hz> 1.0overmodulation, \(m_f\) falling

Five regions with five different binding constraints — which is why no single modulation scheme serves the whole range.

Region 1, below 5 Hz. Set 25, Problem 5 found the dead-time error to be a fixed voltage:

\[ \Delta V = V_{dc}t_df_s = (587)\left(2\times10^{-6}\right)(4000) = 4.7\ \text{V} \]
\[ \text{at } m = 0.05:\ \hat V_1 = 21\ \text{V} \;\Longrightarrow\; \frac{5.98}{21} = 28\% \]

Twenty-eight per cent distortion at creep speed. The response is to lower the carrier here — halving \(f_s\) to 2 kHz halves the error — and to apply dead-time compensation. Continuous SVM, because ripple quality matters more than efficiency at a twentieth of rated power.

Regions 2 and 3. The crossover of Problem 5 governs:

\[ m < 0.6:\ \text{continuous SVM}; \qquad m > 0.6:\ \text{discontinuous PWM} \]

Below 0.6 the continuous scheme gives lower ripple; above it the discontinuous scheme gives both lower ripple and 33% less switching loss. The changeover is seamless, since both produce identical line voltages.

Region 4, base speed. This is where Set 26 found the design constraint:

\[ V_{L1} = 415\ \text{V required}; \qquad 0.707\times587 = 415\ \text{V available} \]

Exactly at the linear limit of SVM with no margin — so any supply sag pushes the drive into overmodulation at rated speed. That is normal, and it is why drives are specified with a slight voltage derating rather than a hard guarantee.

Region 5, field weakening. Two things happen together:

\[ m_f = \frac{4000}{200} = 20 \;\le\; 21 \;\Longrightarrow\; \text{must synchronise} \]
\[ m > 1 \;\Longrightarrow\; \text{overmodulation, tending to six-step} \]

So above about 150 Hz the modulator locks the carrier to a synchronous multiple — \(m_f = 21\), then 15, then 9 as the frequency rises — and progressively overmodulates. At the top of the range it is running six-step: one state per 60°, no modulation at all, and the highest possible output voltage.

The complete strategy:

\(f_{out}\)SchemeCarrierSyncReason
0–5 HzSVM continuous2 kHzasyncminimise dead-time error
5–30 HzSVM continuous4 kHzasynclow ripple at low \(m\)
30–50 HzDPWM4 kHzasync33% less switching loss
50–150 HzDPWM, overmodulating4 kHzasyncvoltage above the linear limit
150–200 HzOvermodulation → six-stepsynchronous\(m_f = 21,15,9\)avoid subharmonics

Five regions, four transitions, all handled in software with no hardware change. Every transition is triggered by a constraint identified in Sets 25 to 27, and all of them are invisible to the motor except through the waveform quality.

What this reveals about modulation. The scheme is not a property of the inverter:

\[ \text{same six switches} \;\longrightarrow\; \text{five different converters, chosen by software} \]

The hardware of Set 26, Problem 6 is fixed: six IGBTs, a link capacitor, a diode bridge. Everything in this problem is a choice made in the controller, and the difference between a good drive and a poor one at the same power rating is almost entirely here.

No single modulation scheme serves a whole drive range. Dead time dominates at creep, ripple at low speed, efficiency at mid speed, DC-link utilisation at base speed and subharmonics at high speed — five different binding constraints. A production drive changes scheme, carrier frequency and synchronisation as it accelerates, and the load never knows.
Answera 2 kHz SVM below 5 Hz for dead time   b SVM to \(m = 0.6\), then DPWM   c overmodulate above base speed   d synchronise when \(m_f \le 21\)
Formulas

Key Formulas

QuantityRelationNotes
Amplitude modulation index\(m_a = \hat V_{ref}/\hat V_{tri}\)Linear to 1
Frequency modulation index\(m_f = f_{tri}/f_{out}\)Odd, \(\times3\)
Linear fundamental\(\hat V_1 = m_aV_{dc}\)Full bridge
Square-wave limit\(4V_{dc}/\pi\)27.3% above \(m_a = 1\)
SHE coefficients\(\hat V_n = \dfrac{4V_{dc}}{n\pi}\left[1-2\cos n\alpha_1+2\cos n\alpha_2\right]\)Solve numerically
SHE 5th/7th angles\(16.25^\circ,\ 22.07^\circ\)\(\hat V_1 = 1.188V_{dc}\)
Harmonic current\(I_n = V_n/(n\omega L)\)Why moving harmonics up helps
Utilisation ladder0.612 / 0.707 / 0.780SPWM / THI-SVM / six-step
SVM modulation index\(m = 1\) linear, \(1.103\) six-step10.3% overmodulation range
DPWM saving\(P_{sw}\times\tfrac23\)120° clamped per leg
DPWM optimumclamp at the current peakUp to 50% saving
Continuous/discontinuous crossover\(m \approx 0.6\)On ripple
Synchronisation rule\(m_f \le 21\)Else subharmonics
Dead-time error\(V_{dc}t_df_s\)Lower \(f_s\) at low speed
Pitfalls

Common Mistakes

  1. Assuming \(\hat V_1 = m_aV_{dc}\) holds above \(m_a = 1\). The relationship saturates towards \(4V_{dc}/\pi\) — Problem 1.

  2. Choosing an even carrier ratio. It admits even harmonics and a DC offset — Problem 2.

  3. Running asynchronously at low carrier ratio. Subharmonics appear below the fundamental and produce mechanical oscillation — Problem 2.

  4. Expecting SHE to remove harmonic energy. The 11th and 13th grew larger than the 5th and 7th were — Problem 3.

  5. Using SHE angles at the wrong modulation index. They are valid at one operating point only and need a table — Problem 3.

  6. Treating overmodulation as a free voltage gain. The last 10.3% costs 31% THD — Problem 4.

  7. Keeping a linear current regulator through overmodulation. The gain from \(m\) to \(V_1\) falls towards zero — Problem 4.

  8. Assuming DPWM always saves exactly a third. Placement relative to the current peak changes it from 15% to 50% — Problem 5.

  9. Using discontinuous modulation at low modulation index. Below \(m \approx 0.6\) the ripple is worse than continuous SVM — Problem 5.

  10. Fixing one carrier frequency for the whole speed range. Dead-time distortion at creep argues for lowering it, and subharmonics at high speed for synchronising it — Problem 6.

Looking Ahead

Six problems and the modulation scheme has stopped being a detail. The linear range ended at \(m_a = 1\) with 27.3% still on the table; selective elimination nulled the fifth and seventh exactly and made the eleventh and thirteenth larger, which turned out to be a good trade; overmodulation bought the last 10.3% at the price of the whole spectrum; and discontinuous modulation cut switching loss by a third using the same zero-sequence freedom that third-harmonic injection had already used for voltage.

That freedom has now been used twice for two different purposes, which suggests the carrier-based view is not the natural one. Treating the three references as one rotating vector, and the inverter's eight states as six vectors and two zeros, turns the whole problem into geometry: the reference is a point, the states are vertices, and modulation is the question of which vertices to dwell on and for how long. The \(0.707\) utilisation that injection produced becomes, in that view, simply the radius of the largest circle that fits inside a hexagon.

Next: Set 28 — Space Vector Modulation, where the eight states are placed on a hexagon, dwell times are computed from a reference vector, the seven-segment switching sequence is constructed, and the 15.5% advantage over sinusoidal PWM is rederived as a purely geometric fact.