Solution:
See Fig. 7-8 and 7-9.

Solution:
\begin{eqnarray*} \frac{2\pi}{0.16\pi} &=& \frac{200}{16}=\frac{25}{2}. \\ x[n+25] &=& 3\cos(0.16\pi(n+25)-1)=3\cos(0.16\pi n+4\pi-1)=x[n]. \end{eqnarray*}

Solution:
$\{(3)(2),(3)(6)+(8)(2),(8)(6)+(4)(2),(4)(6)\} =\{\underline{6},34,56,24\}$.

$w[n]=4x[n]+x[n-1]$

$y[n]=2w[n]+3w[n-1]$

Solution:
Systems in cascade (series): Convolve impulse responses. $\{\underline{4},1\}*\{\underline{2},3\}=\{\underline{8},14,3\}$.

Solution:
$\{\underline{1},-5,6\}*(3)^n\;u[n] $
          $=3^n\;u[n]-5(3)^{n-1}\;u[n-1]+6(3)^{n-2}\;u[n-2] $
          $=\{\underline{1},-2\}$.

Solution:
All MA systems are stable; see Section 7-4.5. All systems with geometric impulse responses with poles inside the unit circle are stable; see Section 7-4.6.

Solution:
$\displaystyle {\cal Z}\{\{\underline{1},-2\}+(2)^n\;u[n]\} $
     $\displaystyle =(1-2{{\bf z}}^{-1})+\frac{{{\bf z}}}{{{\bf z}}-2} $
     $\displaystyle =\frac{{{\bf z}}-2}{{{\bf z}}} \frac{{{\bf z}}-2}{{{\bf z}}-2}+\frac{{{\bf z}}}{{{\bf z}}-2} $
     $\displaystyle =\frac{2{{\bf z}}^2-4{{\bf z}}+4}{{{\bf z}}^2-2{{\bf z}}}. $

$y[n]+2y[n-1]=4x[n]+5x[n-1]$

Solution:
Read off $\displaystyle {\bf H} ({\bf z})=\frac{4{\bf z}+5}{{\bf z}+2} $ and $\displaystyle {\bf X} ({\bf z}) = \frac{{\bf z}^2+3{\bf z}+2}{{\bf z}^2}. $ $\displaystyle {\bf Y} ({\bf z}) = {\bf H} ({\bf z}){\bf X} ({\bf z})=\frac{4{\bf z}+5}{{\bf z}+2}\frac{({\bf z}+1)({\bf z}+2)}{{\bf z}^2} $
          $\displaystyle =\frac{({\bf z}+1)(4{\bf z}+5)}{{\bf z}^2}=4+9{\bf z}^{-1}+5{\bf z}^{-2}, $
so $y[n]=\{\underline{4},9,5\}$.

Solution:
\begin{eqnarray*} j(3-j4)^n\;u[n]-j(3+j4)^n\;u[n] &=& 2(5)^n\cos\left(-0.927n+{{\pi}\over 2}\right)\;u[n] \\ &=& 2(5)^n\sin(0.927n)\;u[n]. \end{eqnarray*}

Solution:
The same answer can be derived in different ways and in different forms.
$\displaystyle\frac{{\bf z}-1}{{\bf z}-2}=\frac{{\bf z}}{{\bf z}-2}-\frac1{{\bf z}-2}$.
An inverse $\displaystyle{\bf z}$-transform gives $\displaystyle(2)^n\;u[n]-(2)^{n-1}\;u[n-1]$.
$\displaystyle\frac{{\bf z}-1}{{\bf z}-2}=1+\frac1{{\bf z}-2}$.
An inverse $\displaystyle{\bf z}$-transform gives $\delta[n]+(2)^{n-1}\;u[n-1]$. Same! (Try it.)

Solution:
$$ \frac{({\bf z}-3)({\bf z}-4)}{({\bf z}-1)({\bf z}-2)}=\frac{{\bf z}^2-7{\bf z}+12}{{\bf z}^2-3{\bf z}+2}= \frac{{\bf Y} ({\bf z})}{{\bf X} ({\bf z})}. $$ Cross-multiply: $({\bf z}^2-3{\bf z}+2)\;{\bf Y} ({\bf z})=({\bf z}^2-7{\bf z}+12)\;{\bf X} ({\bf z})$.
An inverse $\displaystyle{\bf z}$-transform gives $$ y[n]-3y[n-1]+2y[n-2]=x[n]-7x[n-1]+12x[n-2]. $$

Solution:
\begin{eqnarray*} {\bf X} ({\bf z}) &=& \frac{{\bf z}^2-3{\bf z}+2}{{\bf z}^2}. \\ {\bf Y} ({\bf z}) &=& {\bf H} ({\bf z}){\bf X} ({\bf z}) \\ &=& \frac{{\bf z}^2-7{\bf z}+12}{{\bf z}^2-3{\bf z}+2}\frac{{\bf z}^2-3{\bf z}+2}{{\bf z}^2} =1-7{\bf z}^{-1}+12{\bf z}^{-2}. \end{eqnarray*} An inverse $\displaystyle{\bf z}$-transform gives $y[n]=\{\underline{1},-7,12\}$.

Solution:
$$ \frac{{\bf H} ({\bf z})}{\bf z}=\frac{({\bf z}-3)({\bf z}-4)}{{\bf z}({\bf z}-1)({\bf z}-2)}= \frac{6}{\bf z}-\frac{6}{{\bf z}-1}+\frac1{{\bf z}-2}. $$ Then multiply by ${\bf z}$. Computation of partial fraction residues: $$ \frac{(0-3)(0-4)}{(0-1)(0-2)}=6; \qquad \frac{(1-3)(1-4)}{1(1-2)}=-6; \qquad\frac{(2-3)(2-4)}{2(2-1)}=1. $$ An inverse ${\bf z}$-transform gives $h[n]=6\delta[n]-6u[n]+(2)^n\;u[n]$.

Solution:
\begin{eqnarray*} {\bf X} ({\bf z}) &=& \frac{\bf z}{{\bf z}-4}\frac{{\bf z}-3}{{\bf z}-3}-\frac{\bf z}{{\bf z}-3}\frac{{\bf z}-4}{{\bf z}-4}= \frac{\bf z}{({\bf z}-3)({\bf z}-4)}. \\ {\bf Y} ({\bf z}) &=& {\bf H} ({\bf z})\;{\bf X} ({\bf z})= \frac{({\bf z}-3)({\bf z}-4)}{({\bf z}-1)({\bf z}-2)}\frac{\bf z}{({\bf z}-3)({\bf z}-4)} \\ &=&\frac{\bf z}{({\bf z}-1) ({\bf z}-2)}=\frac{\bf z}{{\bf z}-2}-\frac{\bf z}{{\bf z}-1}. \end{eqnarray*} An inverse $\displaystyle{\bf z}$-transform gives $y[n]=(2)^n\;u[n]-u[n]$.

{$h[n]=\{\underline{1},2\}$}

Solution:
\begin{eqnarray*} {\bf H} (e^{j\Omega}) &=& 1+2e^{-j\Omega}. \\ {\bf H}(e^{j2\pi/3}) &=& 1+2\left(-\frac1{2}-j\frac{\sqrt{3}}{2}\right) =-j\sqrt{3}=\sqrt{3}e^{-j\pi/2}. \end{eqnarray*} The output is $\sqrt{3}\cos(\frac{2\pi}{3}n-\frac{\pi}{2})=\sqrt{3}\sin(\frac{2\pi}{3}n)$.

$h[n]=(\frac1{3})^n\;u[n]$

Solution:
\begin{eqnarray*} {\bf H}(e^{j\Omega}) &=& \frac1{1-\frac1{3}e^{-j\Omega}}. \\ {\bf H}(e^{j\pi}) &=& \frac1{1-\frac1{3}e^{-j\pi}}=\frac1{1-\frac1{3}(-1)}=\frac3{4}\;. \end{eqnarray*} Output is $\frac3{4}\cos(\pi n)$.

$y[n]-2y[n-1]+y[n-2]=x[n]-3x[n-1]+x[n-2]$

Solution:
\begin{eqnarray*} {\bf H} (e^{j\Omega}) &=& \frac{1-3e^{-j\Omega}+e^{-j2\Omega}}{1-2e^{-j\Omega}+e^{-j2\Omega}} =\frac{e^{j\Omega}-3+e^{-j\Omega}}{e^{j\Omega}-2+e^{-j\Omega}} =\frac{2\cos(\Omega)-3}{2\cos(\Omega)-2}. \\ {\bf H} (e^{j\pi/3}) &=& \frac{2(\frac1{2})-3}{2(\frac1{2})-2}=2. \end{eqnarray*} The output is $2\cos(\frac{\pi}{3}n)$.

$y[n]-2y[n-1]+y[n-2]=x[n]-3x[n-1]+x[n-2]$

Solution:
\begin{eqnarray*} {\bf H} (e^{j\Omega}) &=& \frac{1-3e^{-j\Omega}+e^{-j2\Omega}}{1-2e^{-j\Omega}+e^{-j2\Omega}} =\frac{e^{j\Omega}-3+e^{-j\Omega}}{e^{j\Omega}-2+e^{-j\Omega}} =\frac{2\cos(\Omega)-3}{2\cos(\Omega)-2}\;. \\ {\bf H} (e^{j\pi/2}) &=& \frac{2(0)-3}{2(0)-2}=\frac3{2}\;. \end{eqnarray*} The output is $\frac3{2}\cos(\frac{\pi}{2}n)$.

Solution:
See Section 7-13.

Solution:
This is obvious by inspection, but it can be computed as follows: The form of the DTFS is \begin{eqnarray*} x[n] &=& {\bf x}_0+{\bf x}_1e^{j\pi/2n} +{\bf x}_2e^{j\pi n}+{\bf x}_3e^{j3\pi/2n}. \\ {\bf x}_0 &=& \frac1{4}(4+2+0+2)=2. \\ {\bf x}_1 &=& \frac1{4}(4-j2+0+j2)=1. \\ {\bf x}_2 &=& \frac1{4}(4-2+0-2)=0. \\ x[n] &=& 2+e^{j\pi/2n}+e^{j3\pi/2n}=2+2\cos\left(\frac{\pi}{2}n\right), \end{eqnarray*} since $e^{j3\pi/2n}=e^{-j\pi/2n}$ and ${\bf x}_3={\bf x}_1^*=1$.

$\displaystyle h[n]=\frac{\sin\frac{\pi}{3}n)}{\pi n}$

Solution:
The DTFS has the form ${\bf x}_0+{\bf x}_1e^{j\pi/2n}+{\bf x}_2e^{j\pi n}+{\bf x}_3e^{j3\pi/2n}$. ${\bf x}_0=2$. $h[n]$ is a brick-wall lowpass filter with cutoff frequency $\Omega_0=\frac{\pi}{3}<\frac{\pi}{2}$. Only ${\bf x}_0$ passes.

$\displaystyle h[n] =\frac{\sin(\frac{\pi}{10}n)}{\pi n}2\cos\left(\frac{3\pi}{10}n\right)$

Solution:
The DTFS has the form ${\bf x}_0+{\bf x}_1e^{j\pi/2n}+{\bf x}_2e^{j\pi n}+{\bf x}_3e^{j3\pi/2n}$. $h[n]$ is a brick-wall bandpass filter that passes $\frac{2\pi}{10}<|\Omega|<\frac{4\pi}{10}<\frac{\pi}{2}\;$, so nothing passes.

Solution:
Keep (a) in mind when dealing with discrete-time frequency $\Omega$.

Solution:
See Section 7-14 for all of these.

Solution:
Plug into \begin{eqnarray*} {\bf X}(e^{j\Omega}) &=&\sum_{n=-\infty}^{\infty}x[n]\;e^{-j\Omega n}: \\ {\bf X} (e^{j\Omega}) &=& 4e^{j2\Omega}+2e^{j\Omega}+1+2e^{-j\Omega}+4e^{-j2\Omega} =8\cos(2\Omega)+4\cos(\Omega)+1. \end{eqnarray*} Check: Setting $\Omega=0$ gives $4+2+1+2+4=8+4+1$.

Solution:
$$ {\bf X}({\bf z})=\frac{\bf z}{{\bf z}-\frac1{2}}. $$ Setting ${\bf z}=e^{j\Omega}$ gives $$ {\bf X}(e^{j\Omega})=\frac{e^{j\Omega}}{e^{j\Omega}-\frac1{2}}\;. $$

Solution:
Let $y[n]=\{1,2,\underline{7},2,1\}$. $y[n]$ is even.
Since $7>1+2+2+1$, $\angle{\bf Y} (e^{j\Omega})]=0$.
Since $x[n]=y[n+1]$, $\angle{\bf X} (e^{j\Omega})=\angle{\bf Y} (e^{j\Omega})+\Omega=\Omega$
using entry #2 of Table 7-7, which is

2. Time shift

$\displaystyle x[n-n_0]$

$\displaystyle {\bf X}(e^{j\Omega})\;e^{-jn_0\Omega}$

Solution:
$$ {\bf X}[4]=\sum_{n=0}^7x[n]\;(-1)^n=7-8+1-4+0-0+0-0=-4. $$

Solution:
\begin{eqnarray*} {\bf X} [0] &=& (16+8+12+4)=40. \\ {\bf X} [1] &=& (16-j8-12+j4)=4-4j. \\ {\bf X} [2] &=& (16-8+12-4)=16. \\ {\bf X} [3] &=& (16+j8-12-j4)=4+4j={\bf X}^*[1]. \end{eqnarray*}

Solution:
\begin{eqnarray*} {\bf X} [0] &=& (12+8+4+8)=32. \\ {\bf X} [1] &=& (12-j8-4+j8)=8. \\ {\bf X} [2] &=& (12-8+4-8)=0. \\ {\bf X} [3] &=& (12+j8-4-j8)=8. \end{eqnarray*} Why are none of these complex numbers? The periodic extension of $\{12,8,4,8\}$ is $\{\dots, 12,8,4,8,\underline{12},8,4,8,12,8,4,8,\dots\}$ is even. Any Fourier transform of any kind of a real and even function is itself real and even.