# Fourier Analysis - Can someone check my answer!?

#### Oracle

Permabanned
If have a square periodic waveform where the period is 0.3 times the width of the pulse.

I understand this wavefore to be even therefore i'm going to derive an expression (An) in order to find the ampiltude of the fundamental and the harmonics

The voltage at the peak is 12v

So, let me see if I have this right sorry of for the long maths and or bad formatting

An = 1/Pi (integrale with pi/03 and -pi/0.3)12cos (ntheta)dtheta

An = 12/Pi [sin(ntheta)/n](integrale Pi/0.3 and -Pi/0.3

An = 12/nPi [ sin(nPi/0.3) - sin(-nPi/0.3)]

An = 12/nPi [ sin (nPi/0.3) + sin (nPi/0.3)] as sin(theta) = -sin(-theta)

An = 12/nPi [2sin (nPi/0.3)]

An = 24/nPi sin (nPi/0.3) volts

So taking the 2nd harmonic of this waveform would give

An = (24/2Pi)sin(2Pi/0.3) = 3.31 volts

God, I hope thats right, but please correct me if i'm wrong

Thanks

#### Mindriot

Associate
Yup that all seems right to me, the second harmonic has a value of 3.3079~=3.31

At least the integration is, can't remember any fourier series stuff (unfortunately will need it in a big way in about a month for my exam!)

#### Oracle

Permabanned
OP
So the period of the waveform is whats used as the pi/ -Pi/ in the integration

If the waveform was purely even i.e. had a period of 1ms, would you just use Pi or you would you alter it!?

Edit: - Its just that on one of my examples the period and pulse are both 0.5ms. Using these in the equation gives zero values for the fundamental and harmonics.

But in the example, the pulse marked by 2pi and -2pi and 2 has been used in the integration

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Suspended
Oracle said:
If have a square periodic waveform where the period is 0.3 times the width of the pulse.
I don't quite understand you here - are you saying that the value of the function is 12V between -0.3 and 0.3, and zero in the ranges [-0.6, -0.3] and [0.3, 0.6], and then periodic with period 1.2?

The Fourier coefficient A_n for an even function f(x) with period 2L is given by

A_n = 1/pi (Int between -L and L) cos(n pi x/L) f(x) dx.

So if your waveform is as I have described above, then the integral you should be doing is

A_n = 1/pi (Int between -0.3 and 0.3) cos(n pi x/0.6) dx.

Edit: Actually, if I remember correctly then a square wave is only made up of odd harmonics - so you must be doing something wrong to get a nonzero amplitude for the second harmonic.

Can you explain what you mean when you say that "the period is 0.3 times the width of the pulse"? What do you mean when you say "period"?

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#### Oracle

Permabanned
OP
Edit: Actually, if I remember correctly then a square wave is only made up of odd harmonics - so you must be doing something wrong to get a nonzero amplitude for the second harmonic.

Can you explain what you mean when you say that "the period is 0.3 times the width of the pulse"? What do you mean when you say "period"?

Right imagine a simple square wave - the peak lasts for 0.1ms (at 12v) and the trough lasts for 0.3ms (at 0v) - by saying the period is 0.3 times, i mean in length, not voltage, and yes you are right, only the odd harmonics should have voltage!

Dammit

Suspended
Oracle said:

Right imagine a simple square wave - the peak lasts for 0.1ms (at 12v) and the trough lasts for 0.3ms (at 0v) - by saying the period is 0.3 times, i mean in length, not voltage, and yes you are right, only the odd harmonics should have voltage!

Dammit
Ah, okay. So I think you have a waveform that is 12V in the range [-0.05, 0.05] for a total length of 0.1, and is 0V in the ranges [-0.2, -0.05] and [0.05, 0.2], for a total trough length of 0.3.

Admittedly the trough is then 3 times the length of the peak, rather than 0.3, but I can't make it work both ways!

So the integral that you need to be doing is

A_n = 1/pi (Int between -0.05 and 0.05) 12 cos(n pi x/0.2) dx

You're actually integrating between -0.2 and 0.2, but the waveform is zero for most of it so you can ignore it all except for the bit in the range [-0.05, 0.05]. In this case you'll also get even harmonics - this is because the "only odd harmonics" rule is only true for waveforms where the peak is the same length as the trough.

#### panthro

Soldato
what are you using fourier analysis for? In my last job I used fast fourier transform.......mind numbingly boring!

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Suspended
Among other things, you can use it to isolate spectral components from a waveform, or use it to construct complicated waveforms from simple harmonics. It has loads of applications in electronics, and apparently to optical imaging as well. One of my lecturers invented a new method for generalising Fourier transforms (which are like a continuous version of Fourier series) and is now minted because it turned out to have loads of applications to brain scanning.

#### panthro

Soldato
I used FFT for vibration and conditional mointoring. Reducing the vibration in very large electric motors. MathCAD actually turned out to have a useful application after uni!

Suspended
How did that work, then? Do you decompose the vibration to find the major harmonics so you know what frequencies you have to damp, or something like that?

#### Oracle

Permabanned
OP
Ok, can we take this one:-

A voltage squarewave signal alternates between 0 and 10v with a period of 1ms (freq = 1kHz).

'An' Derivation as follows:- (Simplified)

An = 1/pi (Int: Pi/2 and -Pi/2)10 cos(ntheta)dthets

An = 10/Pi [sin (nPi/2) + sin (nPi/2)]

An = 10/Pi [2sin (nPi/s)]

An = 20/nPi sin (nPi/2) volts

So F = 6.37v
2 = 0
3 = -2.12v
4 = 0
5 = 1.27v
6 = 0

Firstly are these correct!?

Secondly - assuming a 1Ohm resistance can you calculate the average power in the wave form:-

10v(sqrd.) = 100v (Av = 50v(sqrd.)) so, Vrms = 7.071

Pav = Vrms2 / R or 7.071(sqrd)/1 = 50watts

How do I work out Pdc and Pac??

Is Pdc 25 watts!? If so, why!?

whats Pac and then I guess you add them together to get Pav

Thanks so far

Permabanned
OP
anyone!?