Switching builds the waveform. The switches can only connect the load to a few fixed voltages (+V_dc, 0, −V_dc for a full bridge). The output is a staircase or a train of pulses made from those values.

Fourier analysis splits the waveform into a fundamental and harmonics. The fundamental is the sine at the wanted frequency, and it does the useful work. The harmonics are the unwanted remainder. Total harmonic distortion (THD) measures how large they are compared with the fundamental.

The load filters the current. Each harmonic of the voltage drives a current I_n = V_n/|Z_n|. An inductive load has an impedance that grows with n, so the current is much closer to a sine than the voltage. Power adds up harmonic by harmonic: P = ΣI_n,rms² R.

Each section has a lab with sliders. Change one thing at a time and watch the waveforms, the spectrum, and the numbers. Below each lab are worked examples that open on the correct answer. Move your answer to see what fails, change the givens to see how the answer changes, and ask questions about the problem. Activities check your reading of the labs. The design challenges ask you to meet a full specification.

All labs use ideal switches and a series RL load, and they sum harmonics through n = 300. The textbook sums only a few terms by hand, so a THD may differ from the book by a few tenths of a percent. The Fourier values come from a numerical analysis of the actual waveform, not from a formula, so the labs also work for cases the book does not cover.

Tick switches to close them, choose the direction of the load current, and read the explanation. The load is an inductor in series with a resistor, so its current has to keep flowing when the switches change. Try every preset, then switch the current direction and try again.

Feedback diodes. A transistor conducts in one direction. An inductive load can force current the other way, for instance right after the voltage reverses. An antiparallel diode across each switch gives that current a path and returns the stored energy to the source. Many power modules include these diodes.

Shoot-through and blanking time. Real switches take time to turn off. If the upper switch of a leg is still conducting when the lower one turns on, the dc source is shorted. Controllers therefore add a short blanking time (dead time), during which both switches in a leg are off. During that interval the diodes carry the load current, and the output voltage is whatever the current direction dictates, not necessarily what you intended.

The half bridge. Two switches and a split dc source (two equal capacitors) give outputs of ±V_dc/2. It needs half the switches, but each open switch must block the full V_dc and the output amplitude is halved.

For a series RL load, the current in each half-cycle is the sum of a final value ±V_dc/R and an exponential with time constant τ = L/R. In steady state the current is symmetric: it swings between −I_max and +I_max. Matching the end of one half-cycle to the start of the next gives

If T/2τ is large the current has time to reach nearly ±V_dc/R, so it looks like the voltage. If T/2τ is small the current ramps and hardly curves: it is nearly triangular. The same current also shows you which devices work: transistors when the current and voltage have the same sign, feedback diodes when they differ.

A square wave with no dc component has only odd harmonics:

This method is more general: it works for any load whose impedance you can write at each harmonic, and for any waveform. Its price is a sum, but because |Z_n| grows with n, a handful of terms is usually enough.

Total harmonic distortion is THD = √(Σ_n≥2 V_n,rms²)/V_1,rms (or the same with currents). For the square-wave voltage it is always 48.3 %. The load inductance does not change that, but it does reduce the current THD.

Try these scenarios: (1) make the load almost purely resistive and then very inductive, and watch T/2τ, the shape of the current, and the THD. (2) Look at the colored bands: which devices carry current when the current is negative while the voltage is positive? (3) Raise the frequency and see how the current THD responds.

The switching sequence closes S1 and S2 for the middle of the positive half, closes S1 and S3 (or S2 and S4) during the zero intervals, and closes S3 and S4 for the middle of the negative half. The Fourier series of the notched wave has coefficients

Two consequences follow directly. The fundamental is V₁ = (4V_dc/π) cos α, so α adjusts the amplitude. And harmonic n is zero whenever nα = 90°: α = 30° removes the 3rd, α = 18° the 5th, α ≈ 12.9° the 7th.

One angle can do only one job well. Choosing α to remove the 3rd harmonic fixes the fundamental at 0.866 of its square-wave value. If you also need to control amplitude, V_dc must be adjustable, for example with a dc-dc converter ahead of the inverter.

Try: (1) α = 30° and read V₃. (2) Slide α slowly from 0 and watch the voltage spectrum: which bars shrink first, and which grow? (3) Find the minimum of the current THD curve. Is it at one of the harmonic-elimination angles?

A comparator compares a reference (the wanted sine, amplitude V_m,sine) with a carrier (a triangle at the switching frequency, amplitude V_m,tri). Two ratios define the scheme:

Bipolar switching turns S1 and S2 on when the reference exceeds the carrier (output +V_dc), and S3 and S4 on otherwise (−V_dc). Unipolar switching drives the two legs from the reference and from its inverse, so the output steps between +V_dc and 0 in the positive half and between −V_dc and 0 in the negative half.

For m_a ≤ 1 the fundamental is exactly V₁ = m_aV_dc, so m_a gives you linear amplitude control, even from an unregulated dc source. The harmonics cluster around m_f and its multiples for bipolar PWM, and around 2m_f for unipolar PWM. Make m_f an odd integer for bipolar, and an even integer for unipolar, so the waveform keeps half-wave symmetry.

Above m_a = 1 (overmodulation) the output is no longer linear and low-order harmonics reappear. The fundamental tops out at 4V_dc/π when the output has become a square wave.

Try: (1) vary m_a from 0.2 to 1 and watch the fundamental follow it. (2) Move m_f from 7 to 41 and watch the current THD fall and the carrier frequency rise. (3) Switch to unipolar and compare the spectrum cluster. (4) Push m_a to 1.4 and read the third and fifth harmonics. The two lower charts take a moment: they scan the whole range of m_f and m_a for you.

If the reference is an audio signal and the carrier is around 250 kHz, a PWM bridge becomes an audio amplifier, and a low-pass filter recovers the audio. Because the transistors are switches, a class D amplifier can approach 100 % efficiency in theory, and roughly 75 % or more in practice, compared with a theoretical limit of 78.5 % (and about 20 % in practice) for a conventional class AB design. This is why they appear where size and heat matter, from phones to concert stacks.

In the six-step scheme each switch is on for 180°, and a switching event occurs every 60°. The two switches in the same leg are never on together. The line-to-line voltages are 120°-wide pulses of ±V_dc with zero intervals between them. A wye-connected load sees line-to-neutral voltages with six steps per cycle, at ±V_dc/3 and ±2V_dc/3.

The Fourier series contains only harmonics n = 6k ± 1 (5, 7, 11, 13, …). Triplen harmonics cancel in a balanced three-phase system, and even harmonics are absent by symmetry. For the fundamental, the line-to-neutral peak is V_LN,1 = 2V_dc/π, and the THD of both line-to-neutral and line-to-line voltage is about 31 %. The amplitude cannot be adjusted except through V_dc.

Each leg has its own sine reference, 120° apart, compared with one shared carrier. The upper switch of a leg is on when its reference exceeds the carrier, and the lower is on otherwise. Choosing m_f as an odd multiple of 3 minimizes the harmonics in the line-to-line voltage. For m_a ≤ 1 the line-to-line fundamental is (√3/2)·m_aV_dc, which is where the 0.866 in the text's Table 8-8 comes from.

The synchronous speed of an induction motor with p poles is ω_s = 2ω/p, so changing the electrical frequency changes the speed. But flux in the air gap is proportional to V/f. If you lower f and keep V, the flux rises, the iron saturates, and the motor overheats. The remedy is constant volts per hertz: lower V in proportion to f. At low frequency the stator resistance drops a significant share of the voltage, so many drives add a small boost there.

A PWM inverter suits this job well, because m_a adjusts the voltage with a fixed dc link (an ac-ac converter with a dc link). A six-step inverter can do it only if the dc link voltage can be adjusted. The lab below uses a 230 V, 60 Hz, 4-pole motor with a standard equivalent circuit.

Try: (1) at 30 Hz, compare constant V/f with constant V. (2) Go down to 10 Hz with constant V/f, then switch on the boost. (3) Reduce the dc link voltage until the PWM m_a exceeds 1 at 60 Hz.

In the three-phase sine PWM of the previous section, each leg compares its own sine reference with a shared triangle. In the linear range (m_a ≤ 1) the output of each leg follows its reference, so the phase-voltage amplitude is m_aV_dc/2 and the line-to-line fundamental is

A six-step inverter reaches 1.103V_dc from the same link, so sine PWM uses only 78.5 % of what the bridge can give. Beyond m_a = 1 the signals exceed the carrier peak, pulses vanish, and low-order harmonics return (overmodulation).

Each leg has two states, so a three-phase bridge has 2³ = 8 switch states. Write a state as (a b c), where 1 means the upper switch of that leg is on. Six of the states put a voltage on the load, and two short it:

The three phase voltages can be written as a single rotating complex number, the space vector. A balanced three-phase sine set with peak V_m is a vector of length V_m rotating at the output frequency. The six active vectors are the corners of a hexagon. SVPWM makes any reference vector inside the hexagon by averaging neighboring vectors over each switching period T_s.

Here m = V_m/(V_dc/2) (the same m_a used in sine PWM), and α′ is the angle of the reference measured from the nearer active vector, V_n, in its sector. T₁ goes to V_n, T₂ to V_n+1, and T₀ is shared between the zero vectors 000 and 111. Because T₀ cannot be negative, the largest linear reference is the inscribed circle of the hexagon: V_m = V_dc/√3, which means m = 2/√3 = 1.155.

Drag the angle to move the reference vector around the hexagon, or press Play. Watch the sector, the dwell times, and the gate pattern change. Then raise m: at which angle does the vector first leave the green circle? Try m = 1.0 (the limit of sine PWM, the grey circle) and m = 1.155.

If the dwell times are placed symmetrically in each period (zero vectors at the ends and middle, as in the gate pattern above) the duty ratio of every leg equals 0.5 + 0.5·(v_ref,x/(V_dc/2)) with the common-mode signal

added to all three sine references. That is a triangular wave at three times the output frequency with peak m_a/4. The same signal in all three legs cancels in the line-to-line voltages, so the load sees the same fundamental, but the peaks of the modulating signals fall from m_a to (√3/2)·m_a. That is why the linear range extends to m_a = 1.155.

A simpler signal with the same purpose is third-harmonic injection: add (m_a/6)·sin 3θ to each reference. It has the same linear limit, 1.155, and slightly more ripple than SVPWM.

The injected signal does appear as a common-mode voltage at three times the output frequency. It does not reach the load current of a balanced wye load, but it does matter for bearing currents and EMI filter design.

Try: (1) at m_a = 0.9, switch among the three methods: what changes in the line-to-line voltage, and what in the current THD? (2) Raise m_a to 1.1 with sine PWM, then switch to SVPWM. (3) Read the modulating-signal chart: does any signal cross the carrier peak? (4) Look at the two sweep charts at the bottom. They take a moment, because they scan m_a for all three methods.

Put k H bridges in series, each with its own isolated dc source V_dc. Each bridge contributes +V_dc, 0, or −V_dc, so the total has 2k + 1 levels. Give bridge j a notch angle α_j and the output is a staircase with Fourier coefficients

The modulation index M_i compares the fundamental with that of a square wave of amplitude kV_dc. With k angles you can set M_i and eliminate k − 1 harmonics: the equations Σcos(mα_j) = 0 for each chosen m, together with the M_i equation. They are nonlinear, so they are solved numerically, usually by Newton-Raphson iteration. The lab does this for you, and it will tell you when no solution exists. There often are none: a particular set of harmonics can be eliminated only over certain ranges of M_i.

Pattern swapping. The bridge with the smallest angle conducts the longest and supplies the most energy. If the sources are batteries, that battery runs down first. Rotating the angle assignment from cycle to cycle makes every source deliver the same average power.

Try: (1) k = 2 with α = 20° and 40° (set by hand). (2) Switch to “Solve” with M_i = 0.8: which harmonics vanish? (3) k = 5, “skip triplens”, M_i = 0.8. (4) Slide M_i with k = 3 in the “lowest odd” mode and watch where solutions disappear. (5) Turn on pattern swapping.

The diode-clamped converter needs one dc source, divided by a stack of capacitors. Each leg has twice as many switches as it has levels minus one, and clamp diodes tie the junctions to the capacitor taps. Closing a window of adjacent switches connects the leg to one of the capacitor taps. Because the open switches share the dc voltage, each blocks only V_dc/(levels − 1), instead of the full V_dc.

The next lab rounds a sine wave to the nearest of N levels, a simple way to see how the THD falls as you add levels, and what the extra levels cost in switches, sources, and diodes. (This is an illustration of the idea; real converters use the selective-elimination angles above, or PWM on top of a few levels.)

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