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The auto-correlation function Rx(τ) of the signal x(t) = V sinωt is given by
- a)
- b)V2cosωτ
- c)V2cos2ωτ
- d)2V2cos2ωτ
Correct answer is option 'A'. Can you explain this answer?
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The auto-correlation function Rx(τ) of the signal x(t) = V sin&ome...
Rx(T) =E [x(t)x(t+T)]
=E[vsin(wt)vsin(wt+wT)]
=E[v^2sin(wt)sin(wt+wT)]
=v^2 E[sin(wt)sin(wt+wT)]
=v^2 E[(cos(wT)-cos(2wt+wT))/2]
now E[cos(2wt+wT)]=0 as average of cosine function over one cycle is zero
and E[(cos(wT)]=coswT. as it is a constant
so we get v^2( cos(wT))/2 ie option a
=E[vsin(wt)vsin(wt+wT)]
=E[v^2sin(wt)sin(wt+wT)]
=v^2 E[sin(wt)sin(wt+wT)]
=v^2 E[(cos(wT)-cos(2wt+wT))/2]
now E[cos(2wt+wT)]=0 as average of cosine function over one cycle is zero
and E[(cos(wT)]=coswT. as it is a constant
so we get v^2( cos(wT))/2 ie option a
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