Homework Seven

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).

Nick Christman

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Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:

Figure 1: Original signal for which we want to analyze.

In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:

Figure 2: Signal separated into positive and negative aspects.

To "send the signal" we must essentially move it to a higher frequency and, remembering that x(t) has frequency components for $\textstyle {\frac {1}{2}}f_{s}<f<f_{s}$, we get the following signal, $\textstyle X(f)$:

Figure 3: Preparing signal to be sent and sampled.

Finally, we need to sample the signal at a rate of $\textstyle f_{s}={\frac {1}{T}}$. Theoretically, this leads to the following frequency plot:

Figure 4: Signal being sampled at a rate of 1/T.

In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal and filter out any high frequency components:

Figure 5: The use of a lowpass filter to obtain the original signal as shown in Figure 1.

Figure 6: Original signal obtained using a lowpass filter.

So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of $\textstyle f_{s}={\frac {1}{T}}$ (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5 & 6).

Our next task is to find an expression of x(t) in terms of x(nT). To accomplish this, recall from class that the transfer function of the lowpass filter is

$H(f)={\begin{cases}T,&{\mbox{ }}|f|<{\frac {1}{2T}}\\0,&{\mbox{ else }}\end{cases}}\!$

So to find the transfer function in the time-domain we simply need to find the inverse Fourier Transform of H(f):

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

By Euler's Identiy, however, we can see that

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

Recall that $\textstyle sinc(\theta )={\frac {sin(\pi \theta )}{\pi \theta }}$. Thus,

${\frac {sin\left({\frac {\pi t}{T}}\right)}{\frac {\pi t}{T}}}=sinc\left({\frac {t}{T}}\right)\Longrightarrow h(t)=sinc\left({\frac {t}{T}}\right)$

We learned in class that $\textstyle x'(t)=\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)$ and, thus, we must convolve $\textstyle x'(t){\mbox{ with }}h(t)$ to find the function of $\textstyle x(t)$ desired:

$x(t)=x'(t)*h(t)=\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)*\left[sinc\left({\frac {t}{T}}\right)\right]=\sum _{n=-\infty }^{\infty }x(nT){\mbox{ }}sinc\left({\frac {t-nT}{T}}\right)$

Therefore, the original signal (in terms of $\scriptstyle x(nT)$) is

$x(t)=\sum _{n=-\infty }^{\infty }x(nT){\mbox{ }}sinc\left({\frac {t-nT}{T}}\right)=\sum _{n=-\infty }^{\infty }{\frac {x(nT){\mbox{ }}sin\left({\frac {\pi (t-nT)}{T}}\right)}{\frac {\pi (t-nT)}{T}}}$