Solution:
Multiply $\{7,2,6,4,5\}$ point-by-point by $\{0,\frac1{2},1,\frac1{2},0\}$.

Solution:
Increasing $N$ or $L$ does not reduce sidelobes.

Solution:
Increasing $N$ does not help resolve peaks. A window makes it harder.

Solution:
Increasing $N$ or using a window smoothes the spectrum.

Solution:
${\bf H} (e^{j0})=a+b+0-b-a=0$. ${\bf H} (e^{j\pi})=a-b+0-(-b)+(-a)=0$.

Solution:
${\bf H} (e^{j0})=a+b$+0+$b+a\ne 0$.
${\bf H} (e^{j\pi})=a-b-b+a\ne 0$.

Solution:
${\bf H}(e^{j0})=a+b+(-b)+(-a)=0$.
${\bf H} (e^{j\pi})=a-b+(-b)-(-a)\ne 0$.

Solution:
${\bf H} (e^{j0})=a+b+b+a\ne 0$.
${\bf H} (e^{j\pi})=a-b+b-a=0$.

Solution:
Using Eq. (9.74), $h[n]=Th(nT)=2e^{-2n}\;u[n]$.

Solution:
Using Eq. (9.84), $$ {\bf H} ({\bf z})={\bf H} \left({\bf s}=\frac{2}{T}\frac{{\bf z}-1}{{\bf z}+1}\right) =\frac1{({\bf z}-1)/({\bf z}+1)+1}= \frac{{\bf z}+1}{({\bf z}-1)+({\bf z}+1)}. $$ An inverse $\displaystyle{\bf z}$-transform gives $$ h[n]={\cal Z}^{-1}\left\{\frac{\bf z}{2{\bf z}}+\frac1{2{\bf z}}\right\}= \left\{\underline{\frac1{2}},\frac1{2}\right\}. $$

Solution:
Using Eq. (9.84), $$ {\bf H} ({\bf z})={\bf H}\left({\bf s}=\frac{2}{T}\frac{{\bf z}-1}{{\bf z}+1}\right)= \frac{{\bf z}-1}{{\bf z}+1}=\frac{{\bf Y} ({\bf z})}{{\bf X} ({\bf z})}. $$ Cross-multiply: $({\bf z}+1)\;{\bf Y} ({\bf z})=({\bf z}-1)\;{\bf X} ({\bf z})$.
An inverse ${\bf z}$-transform gives $y[n]+y[n-1]=x[n]-x[n-1]$.

Solution:
The prewarping formula Eq. (9.98) is $$ \omega_0=\frac{2}{T}\tan\frac{\Omega}{2} =\frac{2}{0.001}\tan\frac{\pi/2}{2}=2000. $$

Solution:
By Parseval's theorem, this is $$ 2\pi\sum_{n=-\infty}^{\infty}\left|h[n]-h_{\rm ideal}[n]\right|^2. $$ So set $h[n]=h_{\rm ideal}[n]$ where possible, which amounts to using a rectangular window.

Solution:
$$ h_{\rm IDEAL}[n]={{\sin({{\pi}\over 4}n)}\over{\pi n}}2\cos\left({{\pi}\over 2}n\right) =\{\cdots,-{1\over{\pi}},0,{1\over 2},0,-{1\over{\pi}},\cdots\}. $$ Truncate this.

Solution:
Use $h[n]=\{a,\underline{0},-a\}$ for some constant $a$.
Then ${\bf H} (e^{j0})={\bf H} (e^{j\pi})=0$ and ${\bf H} (e^{j\Omega})=ae^{j\Omega}-ae^{-j\Omega}=j2a\sin(\Omega)$.
Then $|{\bf H} (e^{j\pi/2})|=2a\sin(\pi/2)=1$ makes $a=\frac1{2}$.

Solution:
Rejects $0<\Omega<0.2\pi$ and $0.8\pi<\Omega<\pi$; passes $0.3\pi<\Omega<0.7\pi$.

Solution:
See Sections 9-10 and 9-11.

Solution:
Deleting every other sample of a sampled sinusoid doubles its frequency.

Solution:
Interpolation by a factor of 1000 removes all copies of spectrum of the sampled signal between 1 kHz and 1 MHz. This suffices: bandwidth 1 kHz $\ll1$ MHz.

Solution:
We need to separate the centers of the copies of the spectrum induced by sampling by 100 kHz, since baseband occupies $-10<f<10$ kHz and the first image occupies $S-10<f<S+10$ kHz, where $S$ is the sampling rate after upsampling. To reject all of the spectrum of the first image, we need $S-10=90\to S=100$ kHz, by upsampling by a factor of $\frac{100}{20}=5$.

$\bf\uparrow 3$

$\bf\downarrow 3$

Solution:
Upsampling inserts zeros, downsampling removes them. So output $=$ input!

$\bf\downarrow 3$

Solution:
300 Hz becomes $3(300)=900$Hz, which aliases to $(1000-900)=100$Hz.

$\bf\uparrow 4$

Solution:
200 Hz becomes $\frac1{4}(200)=50$ Hz, with copies at $\frac1{4}(1000-200)=200$ Hz, $\frac1{4}(1000+200)=300$ Hz, $\frac1{4}(2000-200)=450$ Hz, and $\frac1{4}(2000-200)=550$ Hz, which along with higher frequencies is filtered out by the reconstruction lowpass filter.

$\bf\downarrow 2$

$\bf\uparrow 2$

Solution:
$$ x[n]=\cos\left(2\right)\pi\frac{500}{1000}n=\cos(\pi n)=(-1)^n. $$ Downsampling by 2 removes all of the $-1$'s; upsampling by 2 replaces them with 0's.

Solution:
See Section 9-12.2.

Solution:
See Section 9-12.4.

Solution:
See Section 9-12.6.

Solution:
From Eq. (9.169) and Example 9-20, $$ \{\underline{1},2,3\}*\{3,2,\underline{1}\} =\{3,8,\underline{14},8,3\}. $$

Solution:
From Eq. (9.177) and Example 9-21, $$ \{\underline{1},2,3\}*\{6,5,\underline{4}\}=\{6,17,\underline{32},23,12\}. $$