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Given a sequence x[n], to generate the sequence y[n] = x[3 – 4n].Which one of the following procedures would be correct?
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- First delay x[n] by 3 sample to generate z1[n], then pick every 4th smmple of z1[n] to generate z2[n], and then finally time reverse z2[n] to obtain y[n]
- First advancex[n] by 3 samples to generate z1[n], then pick every 4th sample of z1[n] to generate z2[n], and then finally time reverse z2[n] to obtain y[n]
- First pick every fourth sample of x[n] to generate v1[n], time-reverse v1[n] to obtain v2[n], and finally advance v2[n] by 3 sample to obtain y[n]
- First pick every fourth sample of x[n] to generate v1[n], time-reverse v1[n] to obtain v2[n], and finally delay v2[n] by 3 samples to obtain y[n]
Correct Option: D
On picking up fourth sample of x [n], we get
V1 [n] = x [4n]
On time reversing V1 [n], we get
V2 [n] = x [– 4n]
Now on delaying V2 [n] by 3 samples, we get
y[n] = x [3 – 4n]
