Solution:
The gain function for $\omega\gg\omega_\rm c $ is $$ M(\omega)=|\bf H (\omega)|\approx\frac{\omega_\rm c ^n}{\omega^n} $$ (see Eq. (6.42) and Eq. (6.44)). Hence, $$ M(10^6)\approx\frac{(10^3)^3}{(10^6)^3}=10^{-9}. $$

Solution:
The gain function for $\omega\gg\omega_\rm c $ is $$ M(\omega)=|\bf H (\omega)|\approx\frac{\omega_\rm c ^n}{\omega^n} $$ (see Eq. (6.42) and Eq. (6.44)). Here $$ 10^{-6}\approx\frac{100^n}{10,000^n}=10^{-2n}, $$ which requires $n=3$.

Solution:
$\log_{10}(2)\approx 0.3$, so $10^{0.3}\approx 2$.
So an octave is 0.3 decades,
and (20 dB/decade)(0.3 decade/octave) $=6$ dB/octave.
Both measure gain rolloff.

Solution:
See Eq. (6.60).

Solution:
See Section 6-5.3.

Solution:
See Eq. (6.63).

Solution:
See Section 6-5.3.

Solution:
$\omega_\rm c =\frac1{2}(18+12)=15$ and $\omega_d=\frac1{2}(18-12)=3$. $\displaystyle h(t)=\frac{\sin(\omega_dt)}{\pi t}\;2\cos(\omega_\rm c t)=\frac{\sin(3t)}{\pi t} \;2\cos(15t). $

Solution:
The filter has poles at $\{e^{\pm j135^\circ }\}$ (see Section 6-9.3).
The transfer function is $\displaystyle {\bf H} ({\bf s})=\frac1{({\bf s}-e^{j135^\circ })({\bf s}-e^{-j135^\circ })}= \frac1{{\bf s}^2-{\bf s}2\cos(135^\circ )+1} $
          $\displaystyle =\frac1{{\bf s}^2+\sqrt{2}{\bf s}+1}. $
Equate ${\bf H} (\bf s)=\frac{{\bf Y} (\bf s)}{{\bf X} (\bf s)}$, cross-multiply, and take ${\cal L}^{-1}$ to get the differential equation $\displaystyle {\bf Y} ({\bf s})\;({\bf }s^2+\sqrt{2}\;{\bf s}+1) ={\bf X} ({\bf s})\to\frac{d^2y}{dt^2}+\sqrt{2}\;\frac{dy}{dt}+y=x. $

Solution:
The filter has zeros $\{\pm j100\}$ and poles $\{-0.01\pm j100\}$ (see Section 6-7.1).
Transfer function
$\displaystyle {\bf H} ({\bf s})= \frac{({\bf s}-j100)({\bf s}+j100)}{({\bf s}+0.01-j100)({\bf s}+0.01+j100)} $
          $\displaystyle = \frac{{\bf s}^2+100^2}{{\bf s}^2+0.02{\bf s}+100^2+0.01^2} $
          $\displaystyle \approx\frac{{\bf s}^2+100^2}{{\bf s}^2+0.02{\bf s}+100^2}\;. $
Equate ${\bf H} ({\bf s}) =\frac{{\bf Y} ({\bf s})}{{\bf X} ({\bf s})}$, cross-multiply, and take ${\cal L}^{-1}$ to get the differential equation \begin{eqnarray*} &&{\bf Y} ({\bf s})\;({\bf s}^2+0.02{\bf s}+100^2+0.01^2)={\bf X} ({\bf s})\;({\bf s}^2+100^2) \\ &&\quad\to \frac{d^2y}{dt^2}+0.02\frac{dy}{dt}+100^2y=\frac{d^2x}{dt^2}+100^2x. \end{eqnarray*} ($-0.01$ could be replaced with any negative number close to zero.)

Solution:
A Butterworth lowpass filter removes high-frequency noise.

Solution:
A resonator filter enhances the harmonics of a periodic signal.

Solution:
A comb filter removes the harmonics of an interfering periodic signal.

Solution:
A notch filter removes an interfering sinusoidal signal.

Solution:
A comb filter is a cascade connection of notch filters, each of which removes a harmonic of an interfering periodic signal. See Section 6-7.2.

Solution:
The local oscillator frequency is 670-455=215 kHz, which frequency converts 670 kHz down to 455 kHz, in the passband of the IF filter.

Solution:
The spectrum of each of the four signals is 20 kHz wide, but one of them can be at baseband:

Signal #1	Signal #2	Signal #3	Signal #4
0<|f|<10	10<|f|<30	30<|f|<50	50<|f|<70

Solution:
The $90^\circ $ phase difference between modulation and demodulation creates fading. See Example 6-16.

Solution:
A is a reason, since a crystal radio, consisting of just an envelope detector and a tuner (coil) can be used to receive an AM signal. B is a reason, since fading does not occur for AM (unless the transmitter and receiver are exactly 90 degrees out of phase). C is not a reason, since both AM and DSB are vulnerable to additive noise, e.g., thunderstorms. So the answer is D (A and B).

Solution:
See Section 6-12.9.

Solution:
$f_\rm s=2B=$ Nyquist rate does not suffice; see Fig. 6-71.

Solution:
The sampled signal has components at $\{\pm 300\pm 500k\}=\pm\{200,800,1200\dots\}$ Hz, for integers $k$. The lowpass filter has a cutoff frequency at $\frac{500}{2}=250$ Hz, leaving only 200 Hz.

100 samples/s

$\bf H (\omega)=\cases{.01, & $|\omega|<2\pi 50$\cr 0, & $|\omega|>2\pi 50$}$

Solution:
$4\cos(2\pi 60t)$ aliases to $4\cos(2\pi 40t)$ and adds to $3\cos(2\pi 40t)$, giving $7\cos(2\pi 40t)$.

Solution:
Nyquist frequency $=2({\rm maximum\ frequency})=2(10)(440{\rm Hz})=8800{\rm Hz}=8.8$ kHz.

Solution:
The sampled signal has components at $$ \{\pm 7\pm 20k\}=\pm\{3,7,10,13,\dots\} \rm Hz, $$ for integers $k$. The lowpass filter has cutoff frequency $\frac{10}{2}=5$ Hz, leaving $\pm 3$ Hz.

Solution:
Baseband: -12<f<12 Hz. First image: 8<f<32 Hz, where $20-12=8$ and $20+12=32$. So -8<f<8 kHz of baseband is unaffected by the first image. Recall that ''images'' are copies of the original (baseband) spectrum induced by sampling.

1 sample/s

$\bf H (\omega)=\cases{1/2, & $0<|\omega|<7\pi$\cr 0, & $|\omega|>7\pi$}$

Solution:
Sampling a 4-Hz signal at 1 sample/s creates doubled copies at $\pm\{0,1,2,3,\dots\}$ Hz. The filter eliminates 0 Hz (did you notice that?) and copies above 3.5 Hz.

Solution:
The spectrum of the sampled signal when sampled at 2000 samples/s:

A bandpass filter with passband 2<|f|<3 kHz will recover $x(t)$ from its samples. This is an example of bandpass sampling. The spectrum is really only 1 kHz wide.