Phase and amplitude spectrum of signal

In summary, the conversation is about calculating the amplitude and phase spectrum of a signal. The speaker suggests using the complex Fourier series to calculate the coefficients, and notes that the amplitude spectrum can be written as |Fn| and the phase spectrum as Θn. The speaker also mentions the importance of including the value at n=0, and suggests evaluating the integral using l'Hôpital's rule. There is also discussion about how to calculate the amplitude and phase spectrum of a time-shifted signal, with the conclusion that the new spectrum will be the old one multiplied by a complex spiral.
  • #1
etf
179
2
Hi!

1. Homework Statement


My task is to calculate amplitude and phase spectrum of this signal:
postavka1.jpg


Homework 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.

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?

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  • #2
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 [itex] dt [/itex] when setting up your integral, and I'm also assuming that several times when you wrote [itex] Fn [/itex] you actually meant [itex] F_n [/itex]. Other than stuff like that, it looks good to me.
 
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  • #3
What about the value at n=0 ?
 
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  • #4
There must be a dc term too.
 
  • #5
lazyaditya said:
What about the value at n=0 ?

lazyaditya said:
There must be a dc term too.

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

(Hint: l'Hôpital to the rescue)

[Edit: btw, lazyaditya, recall that n goes from [itex] -\infty [/itex] to [itex] \infty [/itex]. So n = 0 is in the middle there somewhere. :)]
 
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  • #6
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
lazyaditya said:
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 .

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 [itex] F_0 [/itex] specifically. It turns out in this case the answer you get is the same either way (such as using the non-rigorous [itex] \frac{\sin x}{x} =1 [/itex], when x approaches 0) in this particular problem, but the rigorous approach is better in general.]
 
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  • #8
@collinsmark
Yes, I forgot dt and I didn't know how to write "n" in index :)
@https://www.physicsforums.com/members/lazyaditya.394087/
I didn't notice that for n=0, Fn is undefined. With substitution n=0 in Fn I got F0=E*tau/T.

 
  • #9
And that's general rule when Fn is undefined for some value n, I put n in original formula and calculate?
 
  • #10
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
Yes, you can use it.
 
  • #12
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
Yes it is correct and you are not bothering anyone .
 
  • #14
etf said:
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)?

Changing [itex] f(t) \rightarrow f(t + t_1)[/itex] 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 [itex] e^{j n \{ \mathrm{something} \} t_1} [/itex]. In other words, [itex] F_n \rightarrow F_n e^{j n \{ \mathrm{something} \} t_1} [/itex]. I'll let you work out what that something is.
 
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  • #15
I think I got it :) I created new thread with some problem which involve time shift...
 

Related to Phase and amplitude spectrum of signal

1. What is the difference between phase and amplitude in a signal?

The phase of a signal refers to the timing or position of the signal in relation to a reference point. It is measured in degrees or radians. The amplitude, on the other hand, represents the strength or intensity of the signal. It is measured in units such as volts or decibels.

2. How is the phase spectrum of a signal calculated?

The phase spectrum of a signal is calculated by taking the inverse tangent of the ratio of the imaginary part to the real part of the signal's Fourier transform. This results in a phase value for each frequency component of the signal.

3. What is the significance of the phase and amplitude spectrum in signal analysis?

The phase and amplitude spectrum provide important information about the characteristics of a signal. The amplitude spectrum can reveal the frequency components present in a signal, while the phase spectrum can indicate any time delays or phase shifts that may occur in the signal.

4. How does the phase and amplitude spectrum change with signal processing?

Signal processing techniques such as filtering or modulation can alter the phase and amplitude spectrum of a signal. Filtering can change the amplitude spectrum by attenuating certain frequencies, while modulation can affect the phase spectrum by shifting the phase of the signal.

5. How can the phase and amplitude spectrum be used in practical applications?

The phase and amplitude spectrum are used in various fields such as telecommunications, audio and image processing, and medical imaging. In these applications, they can provide insight into the quality and characteristics of a signal, as well as assist in identifying and removing unwanted noise or interference.

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