Here D is the duty ratio, and Q1. G2. 03. and 04 represent the four switching tr

Question

Here D is the duty ratio, and Q1. G2. 03. and 04 represent the four switching transistors. Switches Q1 and G2 open and close alternately, ss do 03 and 04. From part one of the term paper: EA = DEH and Es = (1 -D)EH. Hence Eu - EH(2D-1) . If D=0.5 and the switching frequency is 103 kHz. the output waveform will be: Note that this voltage contains a zero DC output plus harmonics. Why is the DC output zero? If the DC supply voltage is 10D V. the output waveform contains a fundamental sinusoidal component (first harmonic) which peak amplitude is 127 V . see below: Why is the peak amplitude 127 V? Theoretically, a low pass filter can be designed to keep the fundamental waveform and block the higher harmonics. Practically, however, it does not work well because the third harmonic is not that far apart from the first.

Explanation / Answer

The answar to the first part: If any waveform have the property of "half wave symmetry" then it does not contains any DC part. This is simply because the average value comes out to zero.

Answar to the second part:

If you compute the fourier transform of a square wave signal like you provided here with Vp as the peak dc voltage. the its components are:

(4/pi)* (VpSin(wt))

+ (4/pi)* (1/3)*VpSin(3wt)

+ (4/pi)* (1/5)*VpSin(5wt)

+ (4/pi)* (1/7)*VpSin(7wt)

+ ...... to infinity

4/pi = 4/3.1416 = 1.2732

This is the reason you are seeing a peak voltage of 127 Volt.

There may be a confusion that you are providing only 100 V DC so the maximum voltage should be 100 V. But if you sum up all the frequency component at any point of time you will get the exact wave form i.e the square wave only. Because when the fundamental frequency component is at its peak some other frequency component will be having negative value.

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