What Is the Amplitude of the Third Harmonic in a 200kHz Square Wave?

[bengaltiger14]

Harmonics of a Square Wave...

Homework Statement

A 200kHz square wave with a fundamental frequency amplitude of 15V. Find amplitude of the third harmonic.

Homework Equations

This is what I do not know. I know this is a Fourier transform problem but just don't know what to do with it.

The Attempt at a Solution

I just took 15V and substracted (1/3) from it to get 14.6667 Volts.

 

[denverdoc]

You need to do the transform, or at least look it up. The amplitude coefficients are expressed as a function of harmonic. Post the FT and we can help if still confused.

 

[m2003]

I would like to clarify a few things about harmonics and square waves in order to provide a more accurate response to the given content.

Firstly, harmonics refer to the frequencies that are multiples of the fundamental frequency. In the case of a square wave, the fundamental frequency is the frequency at which the square wave repeats itself. In this case, it is 200kHz.

Secondly, the amplitude of a wave is the maximum displacement of the wave from its equilibrium position. In the given problem, the fundamental frequency amplitude of the square wave is 15V, meaning that the wave oscillates between +15V and -15V.

To find the amplitude of the third harmonic, we need to understand that the third harmonic has a frequency that is three times the fundamental frequency. So, the third harmonic in this case would have a frequency of 3*200kHz = 600kHz.

Now, according to Fourier analysis, any periodic waveform can be represented as a sum of sinusoidal waves of different frequencies and amplitudes. So, to find the amplitude of the third harmonic, we need to find the amplitude of the sinusoidal wave with a frequency of 600kHz.

Using the formula for amplitude of a sinusoidal wave, A = (2/T)*∫f(t)sin(2πft)dt, where T is the period of the wave, we can calculate the amplitude of the third harmonic.

Since the square wave repeats itself every 1/200kHz = 5μs, the period of the wave is 5μs. Substituting the values in the formula, we get A = (2/5μs)*∫f(t)sin(2π*600kHz*t)dt.

Now, the integral of a square wave is a triangular wave, and the amplitude of a triangular wave is equal to the amplitude of the square wave divided by the number of harmonics. So, in this case, the amplitude of the triangular wave would be 15V/3 = 5V.

Substituting this value in the formula, we get A = (2/5μs)*∫5V*sin(2π*600kHz*t)dt = 2V.

Therefore, the amplitude of the third harmonic of the given square wave is 2V. This means that the third harmonic would have a peak amplitude of +2V and a minimum amplitude of -2V, with