This is my typed up of solutions to first exam, EE 420 based on corrections after key solution is presented to class by the instructor.
The exam was not very hard, about average difficulty, but there were two questions (3 and 4) where one needs to be extra careful. I did make few silly mistakes here and there which I know I should not have done. I think class average was 70/100.
(a) Let operator \(T\left [ \cdot \right ] \equiv{\sum \limits _{n-5}^{n+N}}x\left ( k\right ) \), hence we need to show that, for any \(\alpha ,\beta \), the following\begin{equation} \alpha T\left [ x_{1}\left ( n\right ) \right ] +\beta T\left [ x_{2}\left ( n\right ) \right ] \overset{?}{=}T\left [ \alpha x_{1}\left ( n\right ) +\beta x_{2}\left ( n\right ) \right ] \tag{1} \end{equation} Start by finding the RHS of (1). Let \[ \hat{x}\left ( n\right ) =\alpha x_{1}\left ( n\right ) +\beta x_{2}\left ( n\right ) \] Hence\begin{align} T\left [ \hat{x}\left ( n\right ) \right ] & ={\sum \limits _{n-5}^{n+N}}\hat{x}\left ( k\right ) \nonumber \\ & ={\sum \limits _{n-5}^{n+N}}\left ( \alpha x_{1}\left ( k\right ) +\beta x_{2}\left ( k\right ) \right ) \nonumber \\ & ={\sum \limits _{n-5}^{n+N}}\alpha x_{1}\left ( k\right ) +{\sum \limits _{n-5}^{n+N}}\beta x_{2}\left ( k\right ) \ \ \ \ \ \text{By linearity of summation}\nonumber \\ & =\alpha{\sum \limits _{n-5}^{n+N}}x_{1}\left ( k\right ) +\beta{\sum \limits _{n-5}^{n+N}}x_{2}\left ( k\right ) \ \ \ \ \text{since }\alpha ,\beta \text{ do not depend on }k\tag{2} \end{align}
Now consider LHS of (1), which is\[ \alpha T\left [ x_{1}\left ( n\right ) \right ] +\beta T\left [ x_{2}\left ( n\right ) \right ] =\alpha{\sum \limits _{n-5}^{n+N}}x_{1}\left ( k\right ) +\beta{\sum \limits _{n-5}^{n+N}}x_{2}\left ( k\right ) \] We see that is the same as (2). Hence LHS is the same as RHS in (1). Hence \(T\left [ \cdot \right ] \) is linear
(2) Delay the input \(x\left ( n\right ) \) by an amount \(\beta \) and see if a delayed output by the same amount is the same or not.
Let delayed input be \(x\left ( n-\beta \right ) \), hence the output is\[ T\left [ x\left ( n-\beta \right ) \right ] ={\sum \limits _{k=n-5}^{n+N}}x\left ( k-\beta \right ) \] Let \(k^{\prime }=k-\beta \), so when \(k=n-5\), then \(k^{\prime }=\left ( n-5\right ) -\beta \), and when \(k=n+N\), then \(k^{\prime }=\left ( n+N\right ) -\beta \), hence the above becomes\[ T\left [ x\left ( n-\beta \right ) \right ] ={\sum \limits _{k^{\prime }=\left ( n-5\right ) -\beta }^{\left ( n+N\right ) -\beta }}x\left ( k^{\prime }\right ) \] Since \(k^{\prime }\) is dummy variable, we can rename it to be \(k\)\begin{equation} T\left [ x\left ( n-\beta \right ) \right ] ={\sum \limits _{k=\left ( n-5\right ) -\beta }^{\left ( n+N\right ) -\beta }}x\left ( k\right ) \tag{1} \end{equation} Now let us consider what a delayed output will be. The output due to \(x\left ( n\right ) \) is \(T\left [ x\left ( n\right ) \right ] ={\sum \limits _{k=n-5}^{n+N}}x\left ( k\right ) \), hence a delayed output is where we delay any occurrence of \(n\) in the RHS of the above to become \(n-\beta \), hence a delayed output is\begin{equation}{\sum \limits _{k=\left ( n-\beta \right ) -5}^{\left ( n-\beta \right ) +N}}x\left ( k\right ) \tag{2} \end{equation} We see that (1) and (2) are the same, hence shift invariant. (3)A system is causal if its output at time \(n\) does not depend on future values of the input. i.e on values larger than \(n\). From the definition\[ y\left ( n\right ) ={\sum \limits _{n-5}^{n+N}}x\left ( k\right ) \] We see that \(y\left ( n\right ) \) will depend on future values on the input \(x\left ( n\right ) \) only if \(N>0\). Therefore, the system is causal for \(N\leq 0\), and not causal otherwise.
(4)To show stability, use the BIBO approach. Let \(M\) the absolute value of the largest possible value of the input which will be finite. Hence the output magnitude\begin{align*} \left \vert y\left ( n\right ) \right \vert & =\left \vert{\sum \limits _{n-5}^{n+N}}x\left ( k\right ) \right \vert \\ & \leq{\sum \limits _{n-5}^{n+N}}\left \vert x\left ( k\right ) \right \vert \\ & \leq{\sum \limits _{n-5}^{n+N}}M\\ & =M{\sum \limits _{n-5}^{n+N}}1\\ & =M\left ( \left ( N+n\right ) -\left ( n-5\right ) +1\right ) \\ & =M\left ( N+6\right ) \end{align*}
Hence if \(N<\infty \), then \(\left \vert y\left ( n\right ) \right \vert <\infty \) and the system is BIBO stable.
Use graphical approach
First region: \(n<0\rightarrow y\left ( n\right ) =0\) since no overlapping
second region: \(0\leq n\leq 20\) i.e. partial overlap from the left end of \(x\left ( n\right ) \)\[ y\left ( n\right ) ={\sum \limits _{k=0}^{n}}a^{k}=\frac{1-a^{n+1}}{1-a}\] Third region: \(20\leq n\leq 40\)\begin{align*} y\left ( n\right ) & ={\sum \limits _{k=n-20}^{20}}a^{k}={\sum \limits _{k=0}^{20}}a^{k}-{\sum \limits _{k=0}^{n-19}}a^{k}\\ & =\frac{1-a^{21}}{1-a}-\frac{1-a^{n-20}}{1-a}\\ & =\frac{\left ( 1-a^{21}\right ) -\left ( 1-a^{n-20}\right ) }{1-a}\\ & =\frac{a^{n-20}-a^{21}}{1-a} \end{align*}
Fourth region: \(n>40\rightarrow y\left ( n\right ) =0\) since no overlapping.
\begin{align*} H\left ( e^{j\omega }\right ) & ={\sum \limits _{n=-\infty }^{\infty }}h\left ( n\right ) e^{-j\omega n}\\ & =h\left ( -1\right ) e^{-j\omega \left ( -1\right ) }+h\left ( 1\right ) e^{-j\omega \left ( 1\right ) }+h\left ( 3\right ) e^{-j\omega \left ( 3\right ) }\\ & =-e^{j\omega }+e^{-j\omega }-e^{-3j\omega } \end{align*}
Collect complex exponential with same coefficients\begin{align*} H\left ( e^{j\omega }\right ) & =\left ( -e^{j\omega }-e^{-3j\omega }\right ) +e^{-j\omega }\\ & =-e^{-j\omega }\left ( e^{2j\omega }+e^{-2j\omega }\right ) +e^{-j\omega }\\ & =-e^{-j\omega }\left ( 2\cos 2\omega \right ) +e^{-j\omega }\\ & =e^{-j\omega }\left ( 1-2\cos 2\omega \right ) \end{align*}
Hence \[ \left \vert H\left ( e^{j\omega }\right ) \right \vert =\left \vert 1-2\cos 2\omega \right \vert \] And\[ phase\left ( H\left ( e^{j\omega }\right ) \right ) =-\omega \] Now carefully plotting these we obtain
Maximum frequency in the signal can be seen to be \(20Hz\). Hence Nyquist frequency is twice this, which is \(40Hz\).
However we are told that sampling frequency \(f_{s}=\frac{1}{T}=20Hz\). Hence this system is undersampled.
Another observation to make: Since \(X\left ( f\right ) \) is not zero at the maximum frequency \(20Hz\). one should use a sampling frequency slightly over Nyquist.
Now, to plot the DTFT of the sampled signal. Make copies of \(X\left ( f\right ) \) centered at multiplies of the the sampling frequency. We obtain the following diagram
