Homework Seven

Figure out what happens if your sampled signal, x(t), has frequency components only for $\frac{1}{2}{f}_s<f<f_s$. 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:

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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:

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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 $\frac{1}{2}{f}_s<f<f_s$, we get the following signal, $X(f)$:

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Figure 3: Preparing signal to be sent and sampled.

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

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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:

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

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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 $f_s=\frac{1}{T}$ (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

$H(f)=\{\begin{array}{ll}T,& |f|<\frac{1}{2T}\\ 0,& \text{ else }\end{array}$

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

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

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

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

We learned in class that $x^{\prime }(t)={\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(nT)\delta (t-nT)$ and, thus, we must convolve $x^{\prime }(t)\text{ with }h(t)$ to find the function of $x(t)$ desired:

$x(t)=x^{\prime }(t)*h(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(nT)\delta (t-nT)*\left[sinc\left(\frac{t}{T}\right)\right]=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(nT)sinc\left(\frac{t-nT}{T}\right)$

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

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