Question

24. A signal x[n] has a DTFT, X(F) = 5 drcl(F,5). What is its signal energy?

drcl(t,N)=sin(pi×N×t)÷(N×sin(pi×t))

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Answer #1

I have solved this question bye two methods:-

1.I have derived DTFT of a discrete pulse which is from -N to +N. Then i have compared this DTFT with our given X(f) and find x(n). When you find x(n) then calculation of energy of this discrete input signal x(n) is very easy process.

2. In second method, i iave used a property of Energy theom. Where ESDF is energy spectral density which is equal to square of modules of X(f). DTFT is continuous signal in case of frequency domain from -π to +π. Our X(f) is periodic from -2.5 to +2.5 then it is integrated for this period.

You can verify your answer from these two methods.

teyw - iw Sol? ↑ fin) N -Jwn DTFT of fro a Fews - fons en Elie n NEN N 2N put mantn -2 NWT F(w) = Seww(m-N) JUN 2N - jum 2 MU

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drcl(t,N)=sin(pi×N×t)÷(N×sin(pi×t)) 24. A signal x[n] has a DTFT, X(F) = 5 drcl(F,5). What is its signal...
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