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Question

# A continuous time signal x(t) is defined as x(t) = 2 + cos(50πt) is sampled with sampling interval ts=0.025 sec and passed through an ideal low pass filter whose frequency response is as shown in figure below The spectrum of output will be

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Solution

## The correct option is A Given, x(t)=2+cos(50πt) Frequency of signal ωm=50π rad/sec Ts=0.025sec ∴ sampling frequency ωs=2πTs=80π rad/sec then, X(jω)=4πδ(ω)+π[ω+50π+δ(ω−50π)] Let the sampled signal be represented as Xs(jω), when Xp(jω) is given as Xs(jω)=1Ts∑∞n=−∞X(j(ω−nωs)) Xs(jω)=40∑∞n=−∞[4πδ(ω−80π)]+πδ(ω−50π−80πn)−πδ(ω+50π−80πn) now, the sampled input X(jω) is passed through a low passed filter having cut-off frequency at ω=40π. Therefore the output Y(jω) will contain only the components which are less than ω=40π Now, by putting Ts=0.025 , we will get

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