# Phase and amplitude spectrum of signal

1. Oct 2, 2014

### etf

Hi!

1. The problem statement, all variables and given/known data

My task is to calculate amplitude and phase spectrum of this signal:

2. Relevant equations

My idea is to calculate complex Fourier series of this signal, $$f(t)=\sum_{n=-\infty}^{n=+\infty}Fne^{j\frac{2n\pi t}{T}},$$ where $$Fn=\frac{1}{T}\int_{0}^{T}f(t)e^{-j\frac{2n\pi t}{T}}.$$Fn will be some complex number, which can be written as $$|Fn|e^{j\Theta n},$$ where $$|Fn|$$ is amplitude spectrum and $$\Theta n$$ is phase spectrum.

3. The attempt at a solution

I got $$Fn=\frac{E\tau }{T}\frac{\sin{(nw0\tau /2)}}{nw0\tau /2}e^{-jnw0(t1+\tau /2)},$$ where w0=2*pi/T. We see that phase spectrum is $$\Theta n=-nw0(t1+\tau /2)$$ and amplitude spectrum is $$|Fn|=\frac{E\tau }{T}\frac{\sin{(nw0\tau /2)}}{nw0\tau /2}.$$ Now for some values of $$\tau,$$ $$E$$ and $$T$$ I can plot amplitude and phase spectrum as function of n?

Last edited: Oct 2, 2014
2. Oct 2, 2014

### collinsmark

Yeah, that looks right to me. :)

The assumption of course is that f(t) is periodic, and that same period repeats itself from -infinity to infinity. (Otherwise you need to use the Fourier transform instead.)

By the way, notation wise, you forgot your $dt$ when setting up your integral, and I'm also assuming that several times when you wrote $Fn$ you actually meant $F_n$. Other than stuff like that, it looks good to me.

3. Oct 3, 2014

What about the value at n=0 ?

4. Oct 3, 2014

There must be a dc term too.

5. Oct 3, 2014

### collinsmark

Yes, that's right. Perhaps one could try to evaluate the limit of $\frac{\sin x}{x}$ as $x \rightarrow 0$.

(Hint: l'Hôpital to the rescue)

[Edit: btw, lazyaditya, recall that n goes from $-\infty$ to $\infty$. So n = 0 is in the middle there somewhere. :)]

Last edited: Oct 3, 2014
6. Oct 3, 2014

L'hospital rule can't be applied to discrete sequence and since Fourier series of a periodic signal is discrete in nature thus dc term need to be calculated by keeping n=0 in the equation used for calculating Fourier series coefficient .

7. Oct 3, 2014

### collinsmark

Correct, if you want to be mathematically rigorous about it, one cannot technically use l'Hopital's rule for a discrete value, as you say. This thread was posted in the engineering section though. We're not the most rigorous bunch.

[Edit, But yes, lazyaditya's method is preferred; it is mathematically better and less likely to get you into trouble in future problems. Treat n = 0 differently than all the other ns, if you would otherwise find yourself in a divide by 0 situation. Evaluate the integral substituting 0 for n in the original integral to produce $F_0$ specifically. It turns out in this case the answer you get is the same either way (such as using the non-rigorous $\frac{\sin x}{x} =1$, when x approaches 0) in this particular problem, but the rigorous approach is better in general.]

Last edited: Oct 3, 2014
8. Oct 3, 2014

### etf

@collinsmark
Yes, I forgot dt and I didn't know how to write "n" in index :)
I didn't notice that for n=0, Fn is undefined. With substitution n=0 in Fn I got F0=E*tau/T.

9. Oct 3, 2014

### etf

And that's general rule when Fn is undefined for some value n, I put n in original formula and calculate?

10. Oct 3, 2014

### etf

One more question: If I have $$|Fn|$$ and $$\Theta n$$ of some signal $$f(t),$$ and I want to calculate phase and amplitude spectrum of $$f(t+t1),$$ where t1 is some time shift, can I use phase and amplitude spectrum which I have already found for $$f(t)$$ to calculate phase and amplitude shift for $$f(t+t1)$$, or I have to start with calculation from beginning (Represent f(t+t1) in terms of complex Fourier series, calculate |Fn| etc)?

11. Oct 3, 2014

Yes, you can use it.

12. Oct 3, 2014

### etf

So if $$f(t)=\sum_{n=-\infty }^{n=\infty }Fne^{jnw0t},$$ $$f(t+t1)$$ will be $$\sum_{n=-\infty }^{n=\infty }Fne^{jnw0(t+t1)}$$ and it will have same amplitude and phase spectrum as $$f(t)$$? I'm sorry to bother you but I have exam very soon and I'm trying to learn it...

13. Oct 3, 2014

Yes it is correct and you are not bothering anyone .

14. Oct 3, 2014

### collinsmark

Changing $f(t) \rightarrow f(t + t_1)$ will cause a difference in the phase response. Specifically, the new result will be the old result multiplied by a complex spiral, sometimes called a corkscrew function, of the form $e^{j n \{ \mathrm{something} \} t_1}$. In other words, $F_n \rightarrow F_n e^{j n \{ \mathrm{something} \} t_1}$. I'll let you work out what that something is.

Last edited: Oct 3, 2014
15. Oct 3, 2014

### etf

I think I got it :) I created new thread with some problem which involve time shift...