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The lecture (page 8) shows output voltage spectrum with sinusoidal modulation of duty cycle of a dc-dc converter.

I am wondering where the hamonics and sidebands come from.

For hamonics, I read somewhere that because converter power stage is non-linear so it generates hamonics. I can see it from the nonlinear static control-to-output characteristic of buck-boost converter below.

However, I don't see where switching hamonics and sidebands come from.

Could anyone explain why these hamonics and sidebands exist here.

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  • \$\begingroup\$ It comes from the ripple i.e. The triangular waveform around the sine wave. \$\endgroup\$ – user110971 Dec 6 '16 at 18:08
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First of all, an unmodulated switching waveform is rectangular in nature and, as per any basic analysis of a square wave, it contains harmonics of the fundamental switching frequency: -

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So, a square wave contains a series of sinewave harmonics starting at the fundamental switching frequency and extending, theoretically to infinity. Next, look at the top picture in the image below. This is the general case for a rectangular wave with the duty cycle as a variable: -

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It tells us that harmonics can arise at all integer multiples of the fundamental switching frequency. As a side note, the special symmetrical case of a square wave happens to only contain odd numbered harmonics.

The question shows a sinusoidal base frequency altering (modulating) the PWM duty cycle in order to generate a PWM sinewave. This will produce sidebands either side of every generated harmonic. The distance in Hz from the centre of the harmonic to the centre of either sideband is twice the modulating frequency.

Analysing the sidebands is the real tricky bit to understand; you have to start by considering the formula for the n\$_{th}\$ harmonic: -

a\$_n\$ is proportional to \$\dfrac{2}{n} sin(n\pi d)\$

And, if d (the duty cycle) approaches zero (or unity) the "sin" term becomes zero i.e. the harmonics greatly reduces in amplitude. This happens at twice the modulating frequency i.e. all the harmonics are amplitude modulated at twice the frequency of the modulating waveform.

Because there are DC terms involved, this boils down to exactly the same analysis as a regular AM broadcast: -

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The wiki link above shows the full math behind AM and sidebands but it basically boils down to any one of the following trig identities: -

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So, sum and difference frequencies are produced and it is these sum and difference frequencies that create the sidebands.


New section that hopefully demonstrates that a simple PWM circuit (based around an AD8605 op-amp, a 100 kHz triangle wave and a 1 kHz sine wave) produces sidebands at +/- 2 kHz from the harmonics of the basic PWM switching waveform: -

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There are wider sidebands too and these are also at intervals of +/- 2 kHz. These are most likely due to my circuit being an imperfect modulator. So, using a much faster op-amp and, band pass filtering the resultant PWM at 100 kHz, it can be clearly seen that it is a classic case of amplitude modulation at twice the modulation frequency: -

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    \$\begingroup\$ Now that's a little tricky to get to the bottom of mathematically. I'll start by reminding you that the sidebands are due to the PWM modulating waveform. If there were no modulating waveform i.e. the duty cycle was fixed, then there would be no sidebands. What you then have to consider is that the extremes of duty cycle are caused by the peaks of the modulating waveform. At these extremes the formula for the first harmonic contains the value sin(pi.d). So, if d is nearly zero or nearly 1, then sin(pi.d) is a value close to zero. If d is 0.5 then sin (pi/2) is 1........ \$\endgroup\$ – Andy aka Dec 6 '16 at 19:08
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    \$\begingroup\$ ..... thus, the 1st harmonic is amplitude modulated by the modulating frequency. This is AM in exactly the same way as in radio circuits and, I've just realized I've made a small error in my answer regarding the position of the sideband relative to the harmonic - I am correcting it now - I meant to say the side band is shifted from the harmonic by twice the modulating frequency. Hold on..... \$\endgroup\$ – Andy aka Dec 6 '16 at 19:11
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    \$\begingroup\$ So, if you can follow the maths behind multiplication of sin(x) by sin(y) you will find that the answer involves cos(x+y) and cos(x-y). Where x is the harmonic and y is twice the modulating frequency. This leads to two frequency components at x+y and x-y. But, because it isn't true 4 quadrant multiplication i.e. there is a dc term involved there is also a term at just x. So, three frequency components; one at x, one at x+y and one at x-y. You might need to think about this for a while if you haven't done amplitude modulation theory before. \$\endgroup\$ – Andy aka Dec 6 '16 at 19:18
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    \$\begingroup\$ Can you see that the PWM waveform is being duty cycle modulated (in your question)? The "d" part is the duty cycle and when the modulation amplitude is high, d = (nearly) 1. When the modulation amplitude is low, d = (nearly) 0. Sine of pi *1 = 0 because pi represents 180 degrees. Sin of pi*0 is clearly zero. Sidebands are the result of any two waveforms multiplied together providing they are different frequencies. \$\endgroup\$ – Andy aka Dec 6 '16 at 20:53
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    \$\begingroup\$ Because harmonics reach minima and maxima twice every cycle of the modulating signal, the effective frequency of that modulation is twice that of the basic modulation signal. For links see the AM wiki article I linked in my answer. \$\endgroup\$ – Andy aka Dec 6 '16 at 21:06
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If you have untuned circuits, you'll simply see the fast-switching signals.

If you have sharply tuned circuits, with resonant frequencies right on a "harmonic", the circuit will respond to the energy in that "harmonic".

If you have sharply tuned circuits, with resonant frequencies halfway between "harmonics", you'll get a messy response, with the ringing perhaps starting and then stopping and perhaps reversing in phase.

If you have low-Q broadly tuned circuits, you'll get a messy response, looking like a bandpass filter being stimulated by a square wave.

Summary: if you go looking for "harmonics" with narrow-band filters, you will think you have harmonics. Radio frontends could be upset with harmonics, but what really is the problem is the strong broadband transients, because that energy causes momentary overload of the first amplifier, and no radio signal gets thru until the amplifier recovers from the overload.

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