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h(t) = 6 sinc? (2t) y(t) X2. Consider this system: «(t) = 10 sine(5t) Evaluate Y (S) at f = 1 Hz. (a) 1 (b) 3 (05 (d) 7 (e) y
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Answer #1

Answer:- The Fourier transform of 10*sinc(5t) is-

2 * rect(t/5)

The Fourier transform of 6*sinc2(2t) is-

3* trilt 2)

at f = 1 Hz-

2 * rectt/5) = 2
3* trilt 2) = 1.5

Convolution in time domain is equivalent to multiplication in frequency domain. Hence the result is-

Yit = 1) = 2 x 1.5 = 3

So the option b is the correct answer.

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