Sampling - HW7

Max Woesner

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Homework #7 - Sampling

Problem Statement
Figure out what happens if your sampled signal, $x(t)\!$, has frequency components only for ${\frac {f_{s}}{2}}<f<f_{s}\!$. 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 ${\frac {f_{s}}{2}}<f<f_{s}\!$, our sampled signal $x(t)\!$ in the frequency domain, or $X(f)\!$, is going to look like this.



After sampling with frequency $f_{s}\!$, the signal is going to be shifted over by ${\frac {1}{T}}\!$, since $f_{s}={\frac {1}{T}}\!$. 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 $H(f)={\begin{cases}T,&{\mbox{ }}{\frac {1}{2T}}<|f|<{\frac {1}{T}}\\0,&{\mbox{ else }}\end{cases}}\!$

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 $H(f)\!$.

$h(t)={\mathcal {F}}^{-1}[H(f)]=\int _{-{\frac {1}{T}}}^{-{\frac {1}{2T}}}Te^{j2\pi ft}df\ +\int _{\frac {1}{2T}}^{\frac {1}{T}}Te^{j2\pi ft}df={\frac {Te^{j2\pi ft}}{j2\pi t}}{\Bigg |}_{f=-{\frac {1}{T}}}^{-{\frac {1}{2T}}}+\ {\frac {Te^{j2\pi ft}}{j2\pi t}}{\Bigg |}_{f={\frac {1}{2T}}}^{\frac {1}{T}}\!$

$h(t)=T{\frac {e^{-j\pi t/T}-e^{-j2\pi t/T}}{j2(\pi t)}}\ +\ T{\frac {e^{j2\pi t/T}-e^{j\pi t/T}}{j2(\pi t)}}={\frac {T}{j2}}{\Bigg [}{\frac {e^{-j\pi t/T}-e^{-j2\pi t/T}+e^{j2\pi t/T}-e^{j\pi t/T}}{\pi t}}{\Bigg ]}\!$

$h(t)={\frac {T}{j2}}{\Bigg [}{\frac {-e^{j\pi t/T}+e^{-j\pi t/T}+e^{j2\pi t/T}-e^{-j2\pi t/T}}{\pi t}}{\Bigg ]}={\frac {T}{j2}}{\Bigg [}{\frac {e^{j2\pi t/T}-e^{-j2\pi t/T}}{\pi t}}{\Bigg ]}-\ {\frac {T}{j2}}{\Bigg [}{\frac {e^{j\pi t/T}-e^{-j\pi t/T}}{\pi t}}{\Bigg ]}\!$

Recall $sin(\theta )={\frac {1}{j2}}(e^{j\theta }-e^{-j\theta })\!$, so

$h(t)={\frac {T}{j2}}{\Bigg [}{\frac {e^{j2\pi t/T}-e^{-j2\pi t/T}}{\pi t}}{\Bigg ]}-\ {\frac {T}{j2}}{\Bigg [}{\frac {e^{j\pi t/T}-e^{-j\pi t/T}}{\pi t}}{\Bigg ]}={\frac {T}{\pi t}}sin{\Bigg (}{\frac {2\pi t}{T}}{\Bigg )}\ -\ {\frac {T}{\pi t}}sin{\Bigg (}{\frac {\pi t}{T}}{\Bigg )}={\frac {2sin{\Big (}{\frac {2\pi t}{T}}{\Big )}}{\frac {2\pi t}{T}}}\ -\ {\frac {sin{\Big (}{\frac {\pi t}{T}}{\Big )}}{\frac {\pi t}{T}}}\!$

Also recall $sinc(\theta )={\frac {sin(\pi \theta )}{\pi \theta }}\!$, so

$h(t)={\frac {2sin{\Big (}{\frac {2\pi t}{T}}{\Big )}}{\frac {2\pi t}{T}}}\ -\ {\frac {sin{\Big (}{\frac {\pi t}{T}}{\Big )}}{\frac {\pi t}{T}}}=2sinc{\Bigg (}{\frac {2t}{T}}{\Bigg )}-sinc{\Bigg (}{\frac {t}{T}}{\Bigg )}\!$

Now that we know $h(t)\!$, we can find $x(t)\!$ by convolving the function for $x(t)\!$ after sampling, or $\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)\!$, with $h(t)\!$.

$x(t)=\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)*{\Bigg [}2sinc{\Bigg (}{\frac {2t}{T}}{\Bigg )}-sinc{\Bigg (}{\frac {t}{T}}{\Bigg )}{\Bigg ]}=\sum _{n=-\infty }^{\infty }x(nT){\Bigg [}2sinc{\Bigg (}{\frac {2(t-nT)}{T}}{\Bigg )}\ -\ sinc{\Bigg (}{\frac {t-nT}{T}}{\Bigg )}{\Bigg ]}\!$

Or, if you prefer, $x(t)=\sum _{n=-\infty }^{\infty }x(nT){\Bigg [}{\frac {2sin{\Big (}{\frac {2\pi (t-nT)}{T}}{\Big )}}{\frac {2\pi (t-nT)}{T}}}\ -\ {\frac {sin{\Big (}{\frac {\pi (t-nT)}{T}}{\Big )}}{\frac {\pi (t-nT)}{T}}}{\Bigg ]}\!$

Alternatively, a different bandpass filter can be used that will simplify the math we have to do, such as the one indicated by the red line in the figure below. (Note: not to scale.)



Leaving us with the signal shown below.



The transfer function of this bandpass filter is $H(f)={\begin{cases}T,&{\mbox{ }}|f|<{\frac {1}{2T}}\\0,&{\mbox{ else }}\end{cases}}\!$

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 $H(f)\!$.

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

Recall again that $sin(\theta )={\frac {1}{j2}}(e^{j\theta }-e^{-j\theta })\!$, so

$h(t)={\frac {T}{j2}}{\Bigg [}{\frac {e^{j\pi t/T}-e^{-j\pi t/T}}{\pi t}}{\Bigg ]}={\frac {T}{\pi t}}sin{\Bigg (}{\frac {\pi t}{T}}{\Bigg )}\ ={\frac {sin{\Big (}{\frac {\pi t}{T}}{\Big )}}{\frac {\pi t}{T}}}\!$

Also recall again that $sinc(\theta )={\frac {sin(\pi \theta )}{\pi \theta }}\!$, so

$h(t)={\frac {sin{\Big (}{\frac {\pi t}{T}}{\bigg )}}{\frac {\pi t}{T}}}=sinc{\bigg (}{\frac {t}{T}}{\bigg )}\!$

Now that we know $h(t)\!$, we can find $x(t)\!$ by convolving the function for $x(t)\!$ after sampling, or $\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)\!$, with $h(t)\!$.

$x(t)=\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT)*{\Bigg [}sinc{\bigg (}{\frac {t}{T}}{\bigg )}{\Bigg ]}=\sum _{n=-\infty }^{\infty }x(nT)(sinc{\Bigg (}{\frac {t-nT}{T}}{\Bigg )}\!$

Or, if you prefer, $x(t)=\sum _{n=-\infty }^{\infty }x(nT){\Bigg [}{\frac {sin{\Big (}{\frac {\pi (t-nT)}{T}}{\Big )}}{\frac {\pi (t-nT)}{T}}}{\Bigg ]}\!$