Fourier Coefficients

Thread Starter

RdAdr

Joined May 19, 2013
214
I can't seem to find how the Fourier coefficients change with frequency.

For given signals, I can see how they change once I calculate the Fourier series.

But I want something more in a general sense. Can someone give me directions? Name of a book maybe.
 

WBahn

Joined Mar 31, 2012
24,684
Having a hard time figuring out what you are getting at.

Could you walk through a simple example?
 

RBR1317

Joined Nov 13, 2010
489
I can see how they change once I calculate the Fourier series.
Fourier coefficients are determined by the shape of the waveform over one complete cycle. Does the waveform shape over one complete cycle change with frequency? Once you calculate the Fourier series for a given signal, how would the coefficients change?
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Having a hard time figuring out what you are getting at.

Could you walk through a simple example?
For the Fourier series of a signal that has value E for T/2 and then value -E for the other half of the period, the Fourier coefficients are of the form:

constant/k

So, the amplitudes fall with k. But they do not depend on the frequency. As the frequency of the periodic signal changes, the Fourier coefficients stay the same.

For the Fourier series of an half-wave sinusoidal, the Fourier coefficients fall with 1/k^2, but again frequency does not play a role.

The Fourier coefficients of a triangular wave, however, do depend on frequency.

So, I am wondering if there is a theorem or something that explain how the amplitudes of the Fourier coefficients change with frequency.
How they change with k, I know. There is a theorem that based on the bounded variation of the derivatives of the signal, it tells you that they fall with 1/k or 1/k^2 or 1/k^3 or ...


I am talking about the simple an,bn coefficients. Not the complex ones obtained from mixing these simple up.
 
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Thread Starter

RdAdr

Joined May 19, 2013
214
Fourier coefficients are determined by the shape of the waveform over one complete cycle. Does the waveform shape over one complete cycle change with frequency? Once you calculate the Fourier series for a given signal, how would the coefficients change?
I'm not sure I am following you.

Once I calculate, the Fourier coefficients either change with frequency or not.


I observe though that the amplitudes of the complex Fourier coefficients do not depend on the frequency for all the waveforms I saw. The Fourier coefficients an, bn related to cos(2pi*n*f*t) and sin(2pi*n*f*t) do depend sometimes and when they mix together to form the complex coefficient, maybe they cancel one another.

But I would like to see a theorem or something like this.


For example,for a triangular wave, this is the Fourier coefficient corresponding to the cosine, for 50 Hz and 500 Hz:

http://www.wolframalpha.com/input/?i=1/(2*pi-50)*(cos(x*50)-1)/(x^2*50)

http://www.wolframalpha.com/input/?i=1/(2*pi-500)*(cos(x*500)-1)/(x^2*500)

We see that as x increases, they fall. But as the frequency increases, they change too.
 
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WBahn

Joined Mar 31, 2012
24,684
I think you need to step back and re-evaluate where you are getting your coefficients for the triangle waveform from.

If nothing else, the formulas you give are not dimensionally sound -- you a frequency in Hz subtracted from 2pi, which is in radians.
 

JSCV

Joined Oct 3, 2015
22
My favorite class at school back in the day's... (not)
Half year ago i wished i paid more attention to it.....
 

WBahn

Joined Mar 31, 2012
24,684
I'm not sure I am following you.

Once I calculate, the Fourier coefficients either change with frequency or not.
Then you are doing something wrong.

The coefficients come about by integrating over a specific time interval. That means that the time dependence is integrated away.

Show your calculation (not just what you think are the results) of a set of coefficients that change with frequency.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Then you are doing something wrong.

The coefficients come about by integrating over a specific time interval. That means that the time dependence is integrated away.

Show your calculation (not just what you think are the results) of a set of coefficients that change with frequency.
I've attached a picture. We did it in class, so I am pretty sure is correct.

We see that both Fourier coefficients depend on the fundamental frequency. Only when they mix up to form the magnitude of the complex Fourier coefficient, the fundamental frequency vanishes. And I see that this magnitude falls with 1/k^2, where k can only be odd.

PS: I think I might wrote wrong in the wolfram alpha. Not 50 Hz. That was actually the angular frequency which is measured in radians/sec, not Hz. My mistake. The picture is correct. In wolfram, I took a=1, E=1/2.
 

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Thread Starter

RdAdr

Joined May 19, 2013
214
My favorite class at school back in the day's... (not)
Half year ago i wished i paid more attention to it.....
When I took the class, I was also confused.
But now, after two years, when I came back to it, it appeared much easier.
 

RBR1317

Joined Nov 13, 2010
489
Only when they mix up to form the magnitude of the complex Fourier coefficient, the fundamental frequency vanishes.
Not sure that is a correct statement for the picture shown.
Look at just the denominator of the Fourier coefficient: a(T-a)(k²ω²)

Let a=cT and ω=1/T where c is a constant (and just ignore the radian scaling factor.)

Then, a(T-a)(k²ω²) = cT(T-cT)(k²)(1/T²) = cT²(1-c)(k²)/T² = c(1-c)k²

For the sin(kωa) & cos(kωa) terms, kωa=k(1/T)(cT)=kc

So the Fourier coefficient, sin() or cos(), has no frequency dependence.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Yes. You are correct.

Is there a formal proof that shows that the Fourier coefficients don't depend on the frequency?
 

WBahn

Joined Mar 31, 2012
24,684
I've attached a picture. We did it in class, so I am pretty sure is correct.

We see that both Fourier coefficients depend on the fundamental frequency. Only when they mix up to form the magnitude of the complex Fourier coefficient, the fundamental frequency vanishes. And I see that this magnitude falls with 1/k^2, where k can only be odd.

PS: I think I might wrote wrong in the wolfram alpha. Not 50 Hz. That was actually the angular frequency which is measured in radians/sec, not Hz. My mistake. The picture is correct. In wolfram, I took a=1, E=1/2.
Let's look at this a bit more closely.

The Ω in their equations is the fundamental frequency (what we usually use ω for). Thus

\[
\Omega \: = \: \frac{2\pi}{T}
\]

We can write 'a' as a fraction of T.

\[
a \: = \: \eta T
\]

With these, we can write Ck as

\[
C_k \: = \: \frac{2E}{a(T-a)} \cdot \frac{\cos(k \Omega a) - 1}{k^2 \Omega^2}
\]

\[
C_k \: = \: \frac{2E}{\eta T (T - \eta T)} \cdot \frac{\cos(k \frac{2\pi}{T} \eta T) - 1}{k^2 \( \frac{2\pi}{T}\)^2}
\]

\[
C_k \: = \: \frac{2E}{\eta (1 - \eta) T^2} \cdot \frac{\cos(2\pi k \eta) - 1}{\( \frac{4 \pi^2 k^2}{T^2}\)}
\]

\[
C_k \: = \: \frac{E}{2 \eta (1 - \eta) \pi^2 k^2} \cdot \[ \cos \(2\pi k \eta\) - 1 \]
\]

As you can see, no trace of the frequency remains, only parameters that define the shape of the waveform, not it's temporal extent. This is as expected, since any temporal information is integrated away.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Yes. I understood.
But is there a formal proof somewhere that they don't depend on frequency, no matter what the waveform of x(t) is?


PS: we used big omega for Fourier series, and then little omega for Fourier transform. When we got from the series to the transform, we noted big omega = big delta small omega. And when big omega was going to 0, big delta small omega was going to small delta small omega.
 

WBahn

Joined Mar 31, 2012
24,684
Yes. I understood.
But is there a formal proof somewhere that they don't depend on frequency, no matter what the waveform of x(t) is?
Frequency is nothing more than a scaling on the time axis. When you integrate over time, scaling on the time axis is integrated away.

The only way the frequency could enter into the coefficients is if you have a waveform defined such that the shape of the waveform itself is frequency dependent.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Ok. By integrating over time, the time dependance goes away.

But we still divide the result of this integration by T. If we were divided by T^2, then we would get something dependent.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
If I was evaluating a constant from 0 to T, then I would get T, and by dividing it by T, I get 1. So no T.

If I was evaluating t from 0 to T, then I would get T^2/2, and dividing it by T, I get T/2. So I have frequency dependance. I have this t at that triangular waveform. But somehow the cosine and sine intervene so that when I divide by T, the frequency dependance goes away.
 

WBahn

Joined Mar 31, 2012
24,684
If I was evaluating a constant from 0 to T, then I would get T, and by dividing it by T, I get 1. So no T.

If I was evaluating t from 0 to T, then I would get T^2/2, and dividing it by T, I get T/2. So I have frequency dependance. I have this t at that triangular waveform. But somehow the cosine and sine intervene so that when I divide by T, the frequency dependance goes away.
If you have a function

x(t) = t

then the units of x(t) itself are time and so your waveform amplitude is in units of time.

Think of it this way -- the waveform shape is frequency dependent because the lower the frequency the bigger the amplitude.

But a function of that form is almost always due to sloppiness. For instance, if this were a voltage waveform, then you want voltage as a function to time and this x(t) would not do it. So instead you need a function of the form

v(t) = A·t

where A is a scaling coefficient that has units of volt/second.

Now when you do the integration your time units will go away as expected.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Aa, I got it.

The two axis when I integrate are Volts and seconds. When I integrate, I actually calculate the area. So the area has units of V*s.
Then I divide by T to get only V. So no time dependance.

Thanks.
 
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