Sampling - HW7

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Homework #7 - Sampling


Problem Statement
Figure out what happens if your sampled signal, x(t), has frequency components only for fs2<f<fs. Can you recover the original signal from it? If so, find the expression for x(t) in terms of x(nT).

Solution
For frequency components fs2<f<fs, our sampled signal x(t) in the frequency domain, or X(f), is going to look like this.


After sampling with frequency fs, the signal is going to be shifted over by 1T, since fs=1T. It will look like this.


To recover the original signal x(t), we need to a use bandpass filter to filter out the parts we don't want, as indicated by the red line in the figure below. (Note: not to scale.)


This leaves us with the original signal, as shown below.


The transfer function of the bandpass filter that will accomplish this for us is H(f)={T, 12T<|f|<1T0, else 

To find the expression for h(t), or the expression for the bandpass filter in the time domain, we can take the inverse Fourier transform of H(f).

h(t)=1[H(f)]=1T12TTej2πftdf+12T1TTej2πftdf=Tej2πftj2πt|f=1T12T+Tej2πftj2πt|f=12T1T

h(t)=Tejπt/Tej2πt/Tj2(πt)+Tej2πt/Tejπt/Tj2(πt)=Tj2[ejπt/Tej2πt/T+ej2πt/Tejπt/Tπt]

h(t)=Tj2[ejπt/T+ejπt/T+ej2πt/Tej2πt/Tπt]=Tj2[ej2πt/Tej2πt/Tπt]Tj2[ejπt/Tejπt/Tπt]

Recall sin(θ)=1j2(ejθejθ), so

h(t)=Tj2[ej2πt/Tej2πt/Tπt]Tj2[ejπt/Tejπt/Tπt]=Tπtsin(2πtT)Tπtsin(πtT)=2sin(2πtT)2πtTsin(πtT)πtT

Also recall sinc(θ)=sin(πθ)πθ, so

h(t)=2sin(2πtT)2πtTsin(πtT)πtT=2sinc(2tT)sinc(tT)

Now that we know h(t), we can find x(t) by convolving the function for x(t) after sampling with h(t).

x(t)=n=x(nT)δ(tnT)*[2sinc(2tT)sinc(tT)]=n=x(nT)[2sinc(2(tnT)T)sinc(tnTT)]

Or, if you prefer, <math> x(t) = \sum_{n=-\infty}^\infty x(nT)\Bigg[\frac{2sin\Big(\frac{2\pi (t-nT)}{T}\Big)}{\frac{2\pi (t-nT)}{T}} \ - \ \frac{sin\Big(\frac{\pi (t-nT)}{T}\Big)}{\frac{\pi (t-nT)}{T}}\Bigg]