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Question

Let a continuous time signal be defined as g(t)=rect(4t)4δ(2t) then the Fourier transform of the signal g(f) is given as

(where sinc(f)=sinπfπf)

A
2sinc(4f)
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B
12sinc(f4)
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C
4sinc(f4)
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D
4sinc(4f)
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Solution

The correct option is B 12sinc(f4)
g(t)=rect(4t)4δ(2t)

=4rect(4t)δ(2t) (δ(t)=δ(t))

=2rect(4t) (δ(at)=1|a|δ(t))

thus g(t) is given as

now, rect(t)F.Tsinc(f)

2rect(t)F.T2sinc(f)

2rect(4t)F.T2.14sinc(f4) (scaling property)

2rect(4t)F.T12sinc(f4)

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