This lecture was all about sampling theory and Nyquist. Starting with an a periodic continuous time signal \(x\left ( t\right ) \) and its Fourier transform, we sample \(x\left ( t\right ) \) obtaining \(x\left [ n\right ] \), a sequence of numbers, then find the DTFT of \(x\left [ n\right ] .\) This summarizes the relation between them.
| time domain |
frequency domain |
| \(x\left ( t\right ) \) where \(t\) is time and in seconds \(x\left ( t\right ) \) is aperiodic in \(t\), and continuous \(-\infty <t<\infty \) |
\(X\left ( \Omega \right ) \) where \(\Omega \) is radian frequency (rad/sec) \(X\left ( \Omega \right ) \) is aperiodic in \(\Omega \), and continuous \(-\infty <\Omega <\infty \) |
| \(x\left ( t\right ) \) | \(X_{a}\left ( \Omega \right ) ={\int \limits _{-\infty }^{\infty }}x\left ( t\right ) e^{-j\Omega t}d\Omega \) |
| \(x\left ( t\right ) =\frac{1}{2\pi }{\int \limits _{-\infty }^{\infty }}X\left ( \Omega \right ) e^{j\Omega t}d\Omega \) |
\(X_{a}\left ( \Omega \right ) \) |
After sampling
| sequence domain |
frequency domain |
| \(x\left [ n\right ] \) where \(n\) are integers \(x\left [ n\right ] \) is aperiodic in \(n\), and discrete \(-\infty <n\in Integers<\infty \) |
\(X\left ( \omega \right ) \) where \(\omega \) is frequency (radians) \(X\left ( \omega \right ) \) is periodic in \(\omega \), and continuous, period = \(2\pi \) and \(-\infty <\omega <\infty \) |
| \(x\left [ n\right ] \) | \(X\left ( \omega \right ) ={\sum \limits _{n=-\infty }^{\infty }}x\left [ n\right ] e^{-j\omega n}\) |
| \(x\left [ n\right ] =\frac{1}{2\pi }{\int \limits _{-\pi }^{\pi }}X\left ( \omega \right ) e^{j\omega n}d\omega \) |
\(X\left ( \omega \right ) \) |
Main things to notice: In time domain \(t\rightarrow \Omega \), everything is continuous, and aperiodic. But once we sample, then the frequency domain becomes periodic. Notice that the unit of \(\Omega \) is rad/sec and units of \(\omega \) is radians only.
Next we derived the relation between \(X\left ( \Omega \right ) \) and \(X\left ( \omega \right ) ,\) the idea is this:
start with \(x\left ( t\right ) =\frac{1}{2\pi }{\int \limits _{-\infty }^{\infty }}X_{a}\left ( \Omega \right ) e^{j\Omega t}d\Omega \) and using the fact that at \(t=nT\), where \(T\) is the sampling period, we replace \(t\) by \(nT\) in the above and write
\[ x\left ( nT\right ) =x\left [ n\right ] =\frac{1}{2\pi }{\int \limits _{-\infty }^{\infty }}X_{a}\left ( \Omega \right ) e^{j\Omega nT}d\Omega \]
But \[ x\left [ n\right ] =\frac{1}{2\pi }{\int \limits _{-\pi }^{\pi }}X\left ( \omega \right ) e^{j\omega n}d\omega \]
So, now try to make the first expression above which involves \(\int \limits _{-\infty }^{\infty }\) looks like the second integral \(\int \limits _{-\pi }^{\pi }\). This is done using 2 tricks. First, break the integral \(\int \limits _{-\infty }^{\infty }\) into sums of integrals \(\cdots +{\int \limits _{-\frac{3\pi }{T}}^{-\frac{\pi }{T}}}\) +\(\int \limits _{-\frac{\pi }{T}}^{\frac{\pi }{T}}\) + \(\int \limits _{\frac{\pi }{T}}^{\frac{3\pi }{T}}\) +\(\cdots \) This results in
\[ x\left ( nT\right ) =x\left [ n\right ] =\frac{1}{2\pi }{\sum \limits _{r=-\infty }^{r=\infty }\int \limits _{\frac{\left ( 2r-1\right ) }{T}\pi }^{\frac{\left ( 2r+1\right ) }{T}\pi }}X_{a}\left ( \Omega \right ) e^{j\Omega nT}d\Omega \]
Ok, this sounds cool, but we have not done anything yet.
Next, is the second more important trick, let \(\Omega _{1}=\Omega -\frac{2\pi }{T}r\) (notice a minus sign here, in the lecture notes it was given as plus sign, but that would not work out as the final result was shown)
So now \(d\Omega =d\Omega _{1}\), and when \(\Omega =\frac{\left ( 2r-1\right ) }{T}\pi \), then \(\Omega _{1}=\frac{\left ( 2r-1\right ) }{T}\pi -\frac{2\pi }{T}r=\frac{\left ( 2r-1\right ) \pi -2\pi r}{T}=\frac{2r\pi -\pi -2\pi r}{T}=\frac{-\pi }{T}\) and when \(\Omega =\frac{\left ( 2r+1\right ) }{T}\pi \), then \(\Omega _{1}=\frac{\left ( 2r+1\right ) }{T}\pi -\frac{2\pi }{T}r=\frac{\left ( 2r+1\right ) \pi -2\pi r}{T}=\frac{2r\pi +\pi -2\pi r}{T}=\frac{+\pi }{T}\), hence the above integral becomes
\[ x\left ( nT\right ) =x\left [ n\right ] =\frac{1}{2\pi }{\sum \limits _{r=-\infty }^{r=\infty }\int \limits _{\frac{-\pi }{T}}^{\frac{+\pi }{T}}}X_{a}\left ( \Omega _{1}+\frac{2\pi }{T}r\right ) e^{j\left ( \Omega _{1}+\frac{2\pi }{T}r\right ) nT}d\Omega _{1}\]
But \(\Omega _{1}\) is a dummy variable, so rename it. We might as well rename it back to \(\Omega \), so the above becomes
\[ x\left ( nT\right ) =x\left [ n\right ] =\frac{1}{2\pi }{\sum \limits _{r=-\infty }^{r=\infty }\int \limits _{\frac{-\pi }{T}}^{\frac{+\pi }{T}}}X_{a}\left ( \Omega +\frac{2\pi }{T}r\right ) e^{j\Omega nT}e^{j2\pi rn}d\Omega \]
But in \(e^{j2\pi rn}\), we notice the exponent is always an integer (\(r\) is an integer, and so is \(n\)). Hence this is just \(1\), so the above becomes
\[ x\left ( nT\right ) =x\left [ n\right ] =\frac{1}{2\pi }{\sum \limits _{r=-\infty }^{r=\infty }\int \limits _{\frac{-\pi }{T}}^{\frac{+\pi }{T}}}X_{a}\left ( \Omega +\frac{2\pi }{T}r\right ) e^{j\Omega nT}d\Omega \]
But \(\omega =\Omega T\) (the frequency axis scaling for the discrete case is a collapsed version of the frequency axis scaling of the continuous case. The scaling is determined by \(T\)). Hence, using the above, then the integral becomes
\[ x\left ( nT\right ) =x\left [ n\right ] =\frac{1}{2\pi }{\sum \limits _{r=-\infty }^{r=\infty }}\frac{1}{T}{\int \limits _{-\pi }^{+\pi }}X_{a}\left ( \frac{\omega }{T}+\frac{2\pi }{T}r\right ) e^{j\omega n}d\omega \]
Notice the \(\frac{1}{T}\) coming out due to scaling effect. Now interchange the summation with the integration, we obtain
\[ x\left ( nT\right ) =x\left [ n\right ] =\frac{1}{2\pi }{\int \limits _{-\pi }^{+\pi }}\left \{ \frac{1}{T}{\sum \limits _{r=-\infty }^{r=\infty }}X_{a}\left ( \frac{\omega }{T}+\frac{2\pi }{T}r\right ) \right \} e^{j\omega n}d\omega \]
Compare the above to the original expression for the inverse DTFT of \(x\left [ n\right ] \) which is
\[ x\left [ n\right ] =\frac{1}{2\pi }{\int \limits _{-\pi }^{\pi }}X\left ( \omega \right ) e^{j\omega n}d\omega \]
We see immediately that
\[ X\left ( \omega \right ) =\frac{1}{T}{\sum \limits _{r=-\infty }^{r=\infty }}X_{a}\left ( \frac{\omega }{T}+\frac{2\pi }{T}r\right ) \]
But \(\frac{\omega }{T}=\Omega \,\), so the above becomes
\[ X\left ( \omega \right ) =\frac{1}{T}{\sum \limits _{r=-\infty }^{r=\infty }}X_{a}\left ( \Omega +\frac{2\pi }{T}r\right ) \]
Hence we see the final result, an important result, which is that the DTFT \(X\left ( \omega \right ) \) of the samples can be obtained from the Fourier transform \(X_{a}\) of the signal itself from which the samples are taken using sampling rate \(T\).
We just need to scale \(X_{a}\) and pick only \(X_{a}\) over the frequency \(\Omega \) range of \(-\pi \cdots \pi \) rad/sec, then divide the result by \(T\) to obtain \(-\pi \cdots \pi \) radians. Then make copies of these by shifting them left and right by \(2\pi \) at a time.
Hence, given \(X_{a}\) and \(T\) one can always generate \(X\left ( \omega \right ) \) (need to write small program to show this).