What is the Sampling Period Range for Recoverable Signals?

AI Thread Summary
The discussion focuses on determining the appropriate sampling period T for recovering the signal y(t) generated by the convolution of two band-limited signals x1(t) and x2(t). The frequency limits of x1(t) and x2(t) are given, with x1 having a maximum frequency of 1000Π and x2 of 2000Π. To ensure y(t) is recoverable from the sampled signal yp(t), the Nyquist rate must be applied, which indicates that the sampling frequency should be at least twice the maximum frequency present in y(t). Participants emphasize the importance of understanding the convolution theorem and how to analyze the frequency components of y(t) to find fmax. The conversation highlights the need for clarity on the implications of the sampling theorem in relation to the specific signal characteristics.
satchmo05
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Homework Statement



The signal y(t) is generated by convolving a band limited signal x1(t) with another band limited signal x2(t) that is y(t)=x1(t)*x2(t) where:

--> X1(jω)=0 for|ω| > 1000Π
--> X2(jω)=0 for|ω| >2000Π

Impulse train sampling is performed on y(t) to obtain:
--> yp(t)= [summation from n = (−∞,∞)] y(nT)δ(t− nT)

Specify the range of values for sampling period T which ensures that y(t) is recoverable from yp(t).

Homework Equations


All of the equations that I would are most likely showing.

The Attempt at a Solution


My thoughts were to plug in (nT) for every t in both x1(t) and x2(t) and then take the Fourier transform of that, cut of the edges where the transforms are equal to zero and then that is where I go blank...

I imagine that is the right implementation to start the problem with, but please correct me if I am wrong. Thank you in advance to all who may be able to help - it is much appreciated!
 
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Hint: convolution theorem.
 
I can see where convolution comes into play, but how can I implement the CTF transforms that are given?
 
What sets the lower limit on the sampling rate if you want to be able to recover the original signal?
 
The Nyquist rate, to sample at the perfect rate (without aliasing/oversampling to occur) - it would be = 2fmax
 
I am pretty confused at what you're trying to hint at here. I appreciate the help, but my mind is still blank.
 
It's kind of hard to say anything without giving away the answer. Think about Y(jω). Where is it zero? Can you deduce fmax from that information?
 
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