7 - Sampling a Signal

Figure out what happens if your sampled signal, x(t), has frequency components only for $\textstyle {\frac {1}{2}}f_{s}<f<f_{s}$. Can you recover the original signal from it? If so, find the expression for x(t) in terms of x(nt)

In this problem we are going to sample a signal as shown below where $\textstyle f_{s}=1/T$

When we sample the signal at $\textstyle f_{s}$ The signal repeats every $\textstyle f_{s}$ yielding

Now to get the original signal back all we need to do is us a filter to eliminate everything we don't want. I this case depending on what we want we can either take the centered frequency which is usually done in a receiver since a broad casted signal is usually broken up into a positive and negative component as seen in the original figure or we can just take the broken apart signal that we started with as seen below

Notice if $\textstyle f>{\frac {1}{2}}f_{s}$ we couldn't get the original signal after sampling because the signals would overlap as seen below

Now that we can see that the original signal is recoverable lets look at x(t) in terms of x(nt). To do this all we need to do is take the Fourier Transform of H(f) for my example I will use the first case H(f) = ${\begin{cases}1,&{\mbox{ }}|f|<{\frac {1}{2T}}\\0,&{\mbox{ else }}\end{cases}}$

$h(t)={\mathcal {F}}^{-1}[H(f)]=\int _{-{\frac {1}{2T}}}^{\frac {1}{2T}}e^{j2\pi ft}\,df={\frac {e^{j2\pi ft}}{j2\pi t}}{\Bigg |}_{-{\frac {1}{2T}}}^{\frac {1}{2T}}={\frac {1}{\pi t}}\left[{\frac {e^{j\pi t/T}-e^{-j\pi t/T}}{j2}}\right]$

Now using Euler's Identity $sin(\theta )={\frac {e^{j\theta }-e^{-j\theta }}{2j}}$ yields

${\frac {1}{\pi t}}\left[{\frac {e^{j\pi t/T}-e^{-j\pi t/T}}{j2}}\right]={\frac {1}{\pi t}}sin({\frac {\pi t}{T}})$

from class we know x'(t) = $\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)$ So to get x(t) in terms of x(nt) all we have to do is convolve x'(t) and h(t).

x(t) = x'(t)*h(t) = $\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)*{\frac {1}{\pi t}}sin\left({\frac {\pi t}{T}}\right)=\sum _{n=-\infty }^{\infty }x(nT){\frac {1}{\pi t}}sin\left({\frac {\pi (t-nT)}{T}}\right)$