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time and frequency domain relation

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eecs4ever

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Can some one prove that a signal can not be both limited in time, and be
bandlimited in frequency?

show that the following is impossible:

x(t) = 0 for all |t| > M for some finite M
and X(f) = 0 for all |f| > G for some finite G

where X(f) is the fourier transform of x(t).
 

"Proof: Assume that a signal which has finite support in both domains exists, and sample it faster than the Nyquist frequency. This finite number of time-domain coefficients should define the entire signal. Equivalently, the entire spectrum of the bandlimited signal should be expressible in terms of the finite number of time-domain coefficients obtained from sampling the signal. Mathematically this is equivalent to requiring that a (trigonometric) polynomial can have infinitely many zeros in bounded intervals since the bandlimited signal must be zero on an interval beyond a critical frequency which has infinitely many points. However, it is well-known that polynomials do not have more zeros than their orders due to the fundamental theorem of algebra. This contradiction shows that our original assumption that a time-limited and bandlimited signal exists is incorrect.

"

About this proof:
I followed until this sentence.

"Mathematically this is equivalent to requiring that a (trigonometric) polynomial can have infinitely many zeros in bounded intervals since the bandlimited signal must be zero on an interval beyond a critical frequency which has infinitely many points. "

Why is that?

Thanks for the help.
 

I agree, that is a bit difficult to understand; however, I don't think you need this to prove the theorem.

Physically (in the extreme cases), this theorem is saying that time domain dc signals look like impulses in the frequency domain and time domain impulses look like constants in the frequency domain.

Page 25 of this document has another way of proving this argument **broken link removed**

Check it out.

Best regards,
v_c
 

    eecs4ever

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