Calculating Energy in Square Wave Harmonics | Fourier Series Homework

[jk2007]

Homework Statement

What fraction of the energy of a square wave is in its fundamental? In its first five harmonics? First seven? Nine?

Homework Equations

the Fourier series of a square wave...

v(t) = ( 4V / pi ) ( sinw0t + 1/3sin3w0t + 1/5sin5w0t + ... )

also given a graph showing the frequency spectrum of pulses

at w0 4V/pi
at 3w0 1/3(4V/pi)
at 5w0 1/5(4V/pi)
at 7w0 1/7(4v/pi)
...

The Attempt at a Solution

Been working on this and wasn't able to find a solution... I'm not sure how I'm supposed to derive an equation for energy that I can use to solve the problem... I know that E is the infinitely bounded integral of (v^2)/R dt, but not sure how to apply this here with a Fourier transform made up of discrete pulses.

 

[The Electrician]

First off, note that the ratio of the energy in the fundamental to the energy in the square wave is the same as the ratios of the power into a given load.

Normalize your square wave--that is, give it an amplitude of 1 so that the power (into a 1 ohm load) is also 1. Now calculate the amplitude of the fundamental. Then the power (into a 1 ohm load) is the square of the RMS value of the fundamental. The power of several harmonics is the sum of the squares of the RMS values of those harmonics; the sum of the powers in other words.

Knowing all this you should be able to solve your problem.

 

[jk2007]

ok, i was more or less able to solve it, but I guess there is no way to calculate/sum the total energy without adding up each harmonic one at a time? I found that its impossible to match the answers in the key (close enough so I know they're right) without adding many harmonics to find the total energy... more than I'm willing to punch in the calculator.

 

[The Electrician]

It is possible. The total energy is the energy of the square wave, and the sum of the energies in the harmonics must equal that value. In the example I gave you, I selected the amplitude of the square wave to be 1 so that the total energy would be 1. If you want a mathematical value for the sum of the infinite number of harmonics, this formula, which you can find in various handbooks, should help: {1/3 + 1/5 +1/7 +...} = Pi^2/8.

By the way, I'm assuming the square is an ideal 50% duty cycle square wave without any dead time. You didn't say otherwise, so I assume that's the case.

Imagine you put the square wave through a perfect rectifier, without any diode voltage drops. Then you would get a DC voltage just equal to the peak (positive) voltage of the square wave. The negative half cycle would be changed to positive and would exactly fill in the space between the positive half cycles. A square wave (50% duty cycle) has an RMS value equal to its peak voltage, which is the same as the DC you would get if you rectified it (without any diode drops).