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'''Solution''' | '''Solution''' | ||
| − | + | To begin, we can write express the ideally sampled signal, x_s(t), as | |
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | + | x_s(t)&=\text{comb}_T \left \{ x(t) \right \}\\ | |
| + | &= x(t)\sum_{n=-\infty}^{\infty} \delta(t-kT) | ||
\end{align} | \end{align} | ||
| − | </math> | + | </math> |
| + | |||
| + | We can then get the spectrum as | ||
| + | |||
| + | <math> \begin{align} | ||
| + | X_s(f) &= \text{CTFT} \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ | ||
| + | &= \frac{1}{T} \text{rep}_{\frac{1}{T}} \left \{ X(f) \right \}\\ | ||
| + | &=\frac{1}{T} \sum_{k=-\infty}^{\infty}X(f-k\frac{1}{T}) | ||
| + | \end{align}</math> | ||
| + | |||
| − | + | The interpolation filter should filter out the copies of the original spectrum while preserving the one at baseband. Conceptually, the simplest filter is | |
<math> | <math> | ||
| − | H_r(f)=\begin{cases} T \mbox{, where } |f| \leq \frac{1}{ | + | H_r(f)=\begin{cases} T \mbox{, where } |f| \leq \frac{1}{2T} \\ 0 \mbox{, else}\end{cases} |
</math> | </math> | ||
<math> | <math> | ||
| − | h_r(t) = sinc\left ( \frac{t}{T} \right ) | + | h_r(t) = \text{sinc}\left ( \frac{t}{T} \right ) |
</math> | </math> | ||
| Line 36: | Line 46: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | z(t) &= | + | z(t) &= \text{comb}_{T_s} \{ x(t) \} \ast \text{sinc} \left ( \frac{t}{T} \right ) \\ |
| − | &=\left ( \sum_{n=-\infty}^{\infty} x(nT) \delta (t-nT) \right )\ast sinc \left ( \frac{t}{T} \right ) \\ | + | &=\left ( \sum_{n=-\infty}^{\infty} x(nT) \delta (t-nT) \right )\ast \text{sinc} \left ( \frac{t}{T} \right ) \\ |
| − | &=\sum_{n=-\infty}^{\infty} x(nT) sinc \left (\frac{t-nT }{T}\right) \\ | + | &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (\frac{t-nT }{T}\right) \\ |
\end{align} | \end{align} | ||
</math> | </math> | ||
| Line 50: | Line 60: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | z(mT) &=\sum_{n=-\infty}^{\infty} x(nT) sinc \left (\frac{mT-nT }{T}\right) \\ | + | z(mT) &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (\frac{mT-nT }{T}\right) \\ |
| − | &=\sum_{n=-\infty}^{\infty} x(nT) sinc \left (m-n \right) \mbox{, where } sinc(m-n) = \begin{cases} 1 \mbox{, where } m=n \\ 0 \mbox{, else}\end{cases} \\ | + | &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (m-n \right) \mbox{, where } \text{sinc}(m-n) = \begin{cases} 1 \mbox{, where } m=n \\ 0 \mbox{, else}\end{cases} \\ |
&= x(mT) | &= x(mT) | ||
\end{align} | \end{align} | ||
| Line 60: | Line 70: | ||
'''Solution''' | '''Solution''' | ||
| − | The original signal, x(t), must be bandlimited, so that X(f) = 0 for all f>1/2T. Otherwise, there | + | The original signal, x(t), must be bandlimited, so that X(f) = 0 for all f>1/2T. Otherwise, there might be aliasing. |
---- | ---- | ||
| Line 67: | Line 77: | ||
a) Write a formula for a piece-wise constant interpolation of the samples. | a) Write a formula for a piece-wise constant interpolation of the samples. | ||
| + | |||
| + | '''Solution''' | ||
| + | |||
| + | The basic reconstruction segment we will use for the piece-wise constant interpolation is | ||
| + | |||
| + | <math>s_R(t) = \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right )</math> | ||
| + | |||
| + | This segment will be weighted and shifted for each sample point. For example, the nth sample point would produce | ||
| + | |||
| + | <math>\begin{align} | ||
| + | s_{R,n} (t) &= x(nT) \text{rect}\left( \frac{t-\frac{T}{2} - nT}{T} \right)\\ | ||
| + | &= x(nT) \left ( \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \delta(t-nT) \right ) \\ | ||
| + | &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \left (x(nT) \delta(t-nT)\right) | ||
| + | \end{align}</math> | ||
| + | |||
| + | Extending this for all n, we get | ||
| + | |||
| + | <math>\begin{align} | ||
| + | x_R(t) &= \sum_{n=-\infty}^{\infty} \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \left (x(nT) \delta(t-nT)\right) \\ | ||
| + | &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \sum_{n=-\infty}^{\infty} \left (x(nT) \delta(t-nT)\right) \\ | ||
| + | &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \text{comb}_T \left \{ x(t) \right \} | ||
| + | \end{align}</math> | ||
| + | |||
| + | The last line helps with part b. | ||
b) Derive the relationship between the Fourier transform of the interpolation you wrote in 2a) and the Fourier transform of x(t). (Do not simply write down the formula; show how to derive it.) | b) Derive the relationship between the Fourier transform of the interpolation you wrote in 2a) and the Fourier transform of x(t). (Do not simply write down the formula; show how to derive it.) | ||
| + | |||
| + | '''Solution''' | ||
| + | |||
| + | <math>\begin{align} | ||
| + | X_R(f) &= \text{CTFT} \left \{ \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \right \} CTFT \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ | ||
| + | &= e^{-j2\pi f \frac{T}{2}} T \text{sinc}(Tf) \frac{1}{T} rep_{\frac{1}{t}}\left \{ X(f) \right \} \\ | ||
| + | & = e^{-j \pi f T} \text{sinc}(Tf) rep_{\frac{1}{t}}\left \{ X(f) \right \} | ||
| + | \end{align}</math> | ||
c) Is the interpolation you wrote in 2a) band-limited? Explain. | c) Is the interpolation you wrote in 2a) band-limited? Explain. | ||
| + | |||
| + | '''Solution''' | ||
| + | |||
| + | No, it is not band-limited. For one, the rep function replicates the spectrum every 1/T, and the sinc is not bandlimited. | ||
---- | ---- | ||
| Line 79: | Line 125: | ||
a) Obtain the Fourier transform X(f) of the signal and sketch the graph of |X(f)|. | a) Obtain the Fourier transform X(f) of the signal and sketch the graph of |X(f)|. | ||
| + | |||
| + | '''Solution''' | ||
| + | |||
| + | To solve this, we will use the transform of a sinc, as well as the scaling and shifting properties: | ||
| + | |||
| + | <math>\begin{align} | ||
| + | \text{sinc}(t) &\leftrightarrow \text{rect}(f) \\ | ||
| + | x(at) &\leftrightarrow \frac{1}{|a|} X(af) \\ | ||
| + | x(t-t_0) &\leftrightarrow e^{-j2\pi f t_0}X(f) | ||
| + | \end{align}</math> | ||
| + | |||
| + | In our case, t_0=5 and a=2, so | ||
| + | |||
| + | <math>\begin{align} | ||
| + | X(f) &= CTFT \left \{7 \text{sinc}\left ( \frac{x-5}{2} \right ) \right \} \\ | ||
| + | &= 7\cdot 2 e^{-j2\pi f 5} \text{rect}(2f) \\ | ||
| + | &= 14 e^{-j10\pi f} \text{rect}(2f) \\ | ||
| + | \end{align}</math> | ||
b) What is the Nyquist rate <math>f_0</math> for this signal? | b) What is the Nyquist rate <math>f_0</math> for this signal? | ||
| + | |||
| + | '''Solution''' | ||
| + | |||
| + | The Nyquist rate is 1/2 Hz (twice the maximum non-zero frequency). | ||
c) Let | c) Let | ||
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Sketch the graph of <math>|X_s(f)| </math>. | Sketch the graph of <math>|X_s(f)| </math>. | ||
| + | |||
| + | |||
| + | '''Solution''' | ||
| + | |||
| + | <math>\begin{align} | ||
| + | X_s(f) &= \text{CTFT} \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ | ||
| + | &= \frac{1}{T} \text{rep}_{\frac{1}{T}} \left \{ X(f) \right \} \\ | ||
| + | &= 3f_0 \text{rep}_{3f_0} \left \{ X(f) \right \} \\ | ||
| + | &= \frac{3}{2} \text{rep}_{\frac{3}{2}} \left \{ 14 \text{rect}( 2f )\right \} \\ | ||
| + | &= 21\sum_{k=-\infty}^{\infty} \text{rect}\left ( 2f - \frac{3}{2}k\right ) | ||
| + | \end{align}</math> | ||
d) Let | d) Let | ||
| Line 98: | Line 177: | ||
Write a mathematical expression for the Fourier transform <math>{\mathcal X}_d(\omega) </math> of <math>x_d[n]= x(nT)</math> and sketch the graph of <math>|{\mathcal X}_d(\omega)| </math>. | Write a mathematical expression for the Fourier transform <math>{\mathcal X}_d(\omega) </math> of <math>x_d[n]= x(nT)</math> and sketch the graph of <math>|{\mathcal X}_d(\omega)| </math>. | ||
| + | '''Solution''' | ||
| + | |||
| + | The relationship between the DTFT of <math>x_d[n] </math> and the CTFT of <math>x_s(t)</math> is | ||
| + | |||
| + | <math>\mathcal{X}_d(\omega) = X_s\left (\frac{\omega F_s}{2\pi} \right )</math> | ||
| + | |||
| + | Therefore | ||
| + | |||
| + | <math>\begin{align} | ||
| + | \mathcal{X}_d(\omega) &= 35\sum_{k=-\infty}^{\infty} \text{rect}\left ( \frac{5\omega}{2\pi} - \frac{5}{2}k\right ) \\ | ||
| + | &= 35 \sum_{k=-\infty}^{\infty} \text{rect}\left ( \frac{5}{2} \frac{1}{2\pi} \left ( 2\omega- 2 \pi k \right )\right ) \\ | ||
| + | &= 35 \text{rep}_{2\pi} \left \{ \text{rect} \left (\frac{5}{2\pi} \omega \right ) \right \} | ||
| + | \end{align}</math> | ||
---- | ---- | ||
== Discussion == | == Discussion == | ||
You may discuss the homework below. | You may discuss the homework below. | ||
*write comment/question here | *write comment/question here | ||
| + | |||
| + | - I believe there is an error in solution three with the restated time scaling property | ||
---- | ---- | ||
[[2014_Fall_ECE_438_Boutin|Back to ECE438, Fall 2014, Prof. Boutin]] | [[2014_Fall_ECE_438_Boutin|Back to ECE438, Fall 2014, Prof. Boutin]] | ||
Latest revision as of 19:16, 18 December 2014
Contents
Homework 3 Solution, ECE438, Fall 2014
Question 1
Let x(t) be a continuous-time signal and let y[n]=x(nT) be a sampling of that signal with period T>0. We would like to interpolate the samples (i.e., "connect the dots") in order to try to recover x(t).
a) Derive a formula for a band-limited interpolation of the samples (i.e., an expression for a continuous signal z(t) in terms of the samples y[n]). (Do not simply write down the formula; show how to derive it.)
Solution
To begin, we can write express the ideally sampled signal, x_s(t), as
$ \begin{align} x_s(t)&=\text{comb}_T \left \{ x(t) \right \}\\ &= x(t)\sum_{n=-\infty}^{\infty} \delta(t-kT) \end{align} $
We can then get the spectrum as
$ \begin{align} X_s(f) &= \text{CTFT} \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ &= \frac{1}{T} \text{rep}_{\frac{1}{T}} \left \{ X(f) \right \}\\ &=\frac{1}{T} \sum_{k=-\infty}^{\infty}X(f-k\frac{1}{T}) \end{align} $
The interpolation filter should filter out the copies of the original spectrum while preserving the one at baseband. Conceptually, the simplest filter is
$ H_r(f)=\begin{cases} T \mbox{, where } |f| \leq \frac{1}{2T} \\ 0 \mbox{, else}\end{cases} $
$ h_r(t) = \text{sinc}\left ( \frac{t}{T} \right ) $
Performing the convolution in the time domain
$ \begin{align} z(t) &= \text{comb}_{T_s} \{ x(t) \} \ast \text{sinc} \left ( \frac{t}{T} \right ) \\ &=\left ( \sum_{n=-\infty}^{\infty} x(nT) \delta (t-nT) \right )\ast \text{sinc} \left ( \frac{t}{T} \right ) \\ &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (\frac{t-nT }{T}\right) \\ \end{align} $
b) Show that your interpolation is equal to the original signal at all sample points.
Solution
Looking at the mth sample point
$ \begin{align} z(mT) &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (\frac{mT-nT }{T}\right) \\ &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (m-n \right) \mbox{, where } \text{sinc}(m-n) = \begin{cases} 1 \mbox{, where } m=n \\ 0 \mbox{, else}\end{cases} \\ &= x(mT) \end{align} $
c) Under what circumstances is your interpolation equal to the original signal x(t)? Explain.
Solution
The original signal, x(t), must be bandlimited, so that X(f) = 0 for all f>1/2T. Otherwise, there might be aliasing.
Question 2
Again, we consider a continuous-time signal x(t) and a sampling y[n]=x(nT) of that signal.
a) Write a formula for a piece-wise constant interpolation of the samples.
Solution
The basic reconstruction segment we will use for the piece-wise constant interpolation is
$ s_R(t) = \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) $
This segment will be weighted and shifted for each sample point. For example, the nth sample point would produce
$ \begin{align} s_{R,n} (t) &= x(nT) \text{rect}\left( \frac{t-\frac{T}{2} - nT}{T} \right)\\ &= x(nT) \left ( \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \delta(t-nT) \right ) \\ &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \left (x(nT) \delta(t-nT)\right) \end{align} $
Extending this for all n, we get
$ \begin{align} x_R(t) &= \sum_{n=-\infty}^{\infty} \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \left (x(nT) \delta(t-nT)\right) \\ &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \sum_{n=-\infty}^{\infty} \left (x(nT) \delta(t-nT)\right) \\ &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \text{comb}_T \left \{ x(t) \right \} \end{align} $
The last line helps with part b.
b) Derive the relationship between the Fourier transform of the interpolation you wrote in 2a) and the Fourier transform of x(t). (Do not simply write down the formula; show how to derive it.)
Solution
$ \begin{align} X_R(f) &= \text{CTFT} \left \{ \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \right \} CTFT \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ &= e^{-j2\pi f \frac{T}{2}} T \text{sinc}(Tf) \frac{1}{T} rep_{\frac{1}{t}}\left \{ X(f) \right \} \\ & = e^{-j \pi f T} \text{sinc}(Tf) rep_{\frac{1}{t}}\left \{ X(f) \right \} \end{align} $
c) Is the interpolation you wrote in 2a) band-limited? Explain.
Solution
No, it is not band-limited. For one, the rep function replicates the spectrum every 1/T, and the sinc is not bandlimited.
Question 3
Let
$ x(t)=7 \text{sinc } ( \frac{t-5}{2} ). $
a) Obtain the Fourier transform X(f) of the signal and sketch the graph of |X(f)|.
Solution
To solve this, we will use the transform of a sinc, as well as the scaling and shifting properties:
$ \begin{align} \text{sinc}(t) &\leftrightarrow \text{rect}(f) \\ x(at) &\leftrightarrow \frac{1}{|a|} X(af) \\ x(t-t_0) &\leftrightarrow e^{-j2\pi f t_0}X(f) \end{align} $
In our case, t_0=5 and a=2, so
$ \begin{align} X(f) &= CTFT \left \{7 \text{sinc}\left ( \frac{x-5}{2} \right ) \right \} \\ &= 7\cdot 2 e^{-j2\pi f 5} \text{rect}(2f) \\ &= 14 e^{-j10\pi f} \text{rect}(2f) \\ \end{align} $
b) What is the Nyquist rate $ f_0 $ for this signal?
Solution
The Nyquist rate is 1/2 Hz (twice the maximum non-zero frequency).
c) Let
$ T = \frac{1}{3 f_0}. $
Write a mathematical expression for the Fourier transform $ X_s(f) $ of
$ x_s(t)= \text{ comb}_T \left( x(t) \right). $
Sketch the graph of $ |X_s(f)| $.
Solution
$ \begin{align} X_s(f) &= \text{CTFT} \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ &= \frac{1}{T} \text{rep}_{\frac{1}{T}} \left \{ X(f) \right \} \\ &= 3f_0 \text{rep}_{3f_0} \left \{ X(f) \right \} \\ &= \frac{3}{2} \text{rep}_{\frac{3}{2}} \left \{ 14 \text{rect}( 2f )\right \} \\ &= 21\sum_{k=-\infty}^{\infty} \text{rect}\left ( 2f - \frac{3}{2}k\right ) \end{align} $
d) Let
$ T = \frac{1}{5 f_0}. $
Write a mathematical expression for the Fourier transform $ {\mathcal X}_d(\omega) $ of $ x_d[n]= x(nT) $ and sketch the graph of $ |{\mathcal X}_d(\omega)| $.
Solution
The relationship between the DTFT of $ x_d[n] $ and the CTFT of $ x_s(t) $ is
$ \mathcal{X}_d(\omega) = X_s\left (\frac{\omega F_s}{2\pi} \right ) $
Therefore
$ \begin{align} \mathcal{X}_d(\omega) &= 35\sum_{k=-\infty}^{\infty} \text{rect}\left ( \frac{5\omega}{2\pi} - \frac{5}{2}k\right ) \\ &= 35 \sum_{k=-\infty}^{\infty} \text{rect}\left ( \frac{5}{2} \frac{1}{2\pi} \left ( 2\omega- 2 \pi k \right )\right ) \\ &= 35 \text{rep}_{2\pi} \left \{ \text{rect} \left (\frac{5}{2\pi} \omega \right ) \right \} \end{align} $
Discussion
You may discuss the homework below.
- write comment/question here
- I believe there is an error in solution three with the restated time scaling property
