Power percentage, square wave, Fourier series

[Jd303]

Homework Statement

What is the percentage of power (out of the total power) contained up to the third harmonic (power in DC component, a1 , a-1 , a2 , a-2 , a3 , a-3 ) of the square waveform shown above? (the duty cycle = D = τ/T0= 0.5)

Homework Equations

The Attempt at a Solution

Hey all,
The following question refers to the attached diagram. I thought the question was simple enough due to the amplitude being 1, and the value of D being 0.5.

The simplified formula I calculated for the answer is 2/(k*pi) for odd harmonics (as even values of k result in a value of 0)

-However when adding 2/pi and 2/(3*pi) i get 0.8488 and hence 84.9%

However the multiple choice answers are 96.7, 95, 72.5 and 73.3%

I am not sure whether this is a simple mistake or a large misunderstanding of the theory.

Any help would be much appreciated!

 

[berkeman]

Homework Statement

What is the percentage of power (out of the total power) contained up to the third harmonic (power in DC component, a1 , a-1 , a2 , a-2 , a3 , a-3 ) of the square waveform shown above? (the duty cycle = D = τ/T0= 0.5)

Homework Equations

The Attempt at a Solution

Hey all,
The following question refers to the attached diagram. I thought the question was simple enough due to the amplitude being 1, and the value of D being 0.5.

The simplified formula I calculated for the answer is 2/(k*pi) for odd harmonics (as even values of k result in a value of 0)

-However when adding 2/pi and 2/(3*pi) i get 0.8488 and hence 84.9%

However the multiple choice answers are 96.7, 95, 72.5 and 73.3%

I am not sure whether this is a simple mistake or a large misunderstanding of the theory.

Any help would be much appreciated!

The waveform looks like it also has a DC component...?

 

[Jd303]

Sorry only just starting out this topic, would the DC component be 0.5? How does this component chang emy calculations?

 

[berkeman]

Sorry only just starting out this topic, would the DC component be 0.5? How does this component chang emy calculations?

This article should be of help: http://www.informit.com/articles/article.aspx?p=1374896&seqNum=7

 

[berkeman]

BTW, that article was the first hit on the list of my Google search for square wave harmonics amplitude.

 

[Jd303]

Yes I have actually looked at that page, but it just isn't clicking, here is my understanding.
-The zero harmonic is the DC component and hence 0.5
-Even harmonics have a value of 0
-Odd harmonics have a value of 2/(pi*n)
-Total amplitude is 1
-So percentage power should be ((2/pi) + 2/(3*pi))/1 = 0.8488

 

[berkeman]

Yes I have actually looked at that page, but it just isn't clicking, here is my understanding.
-The zero harmonic is the DC component and hence 0.5
-Even harmonics have a value of 0
-Odd harmonics have a value of 2/(pi*n)
-Total amplitude is 1
-So percentage power should be ((2/pi) + 2/(3*pi))/1 = 0.8488

That page is listing voltage component values. How is the power related to the voltage?

 

[Jd303]

P = (V^2)/R
Hence power percentage would be 0.8488^2.!
Hopefully I have finally gotten that one right! Thanks for your persistence with me

 

[berkeman]

 

[fbash]

i think its
(2/pi)^2 + (2/3pi)^2
divided by
(2/pi)^2 + (2/3pi)^2 +(2/5pi)^2 + (2/7pi)^2

which gives you 95%