Solution:
The ${\bf s}$-domain circuit equivalent of the capacitor is impedance $\frac1{{\bf s} C}$ in series with voltage source $\displaystyle \frac{v(0^-)}{\bf s}=\frac1{\bf s}. $
The step input becomes voltage source
${\cal L}\{u(t)\}=\frac1{\bf s}$.

Let ${\cal L}\{i(t)\}={\bf I} ({\bf s})$. Applying KVL gives $\displaystyle \frac1{\bf s}-{\bf I} ({\bf s})\;R-{\bf I} ({\bf s})\;\frac1{{\bf s} C}-\frac1{\bf s}=0, $
which leads to ${\bf I} ({\bf s})=0$.
The capacitor voltage is then $\displaystyle {\bf I} ({\bf s})\;\frac1{{\bf s} C}+\frac1{\bf s}=\frac1{\bf s}, $ whose inverse Laplace transform is $u(t)$.

Solution:
By voltage division,
$\displaystyle {\bf Y}({\bf s})=\frac{1/({\bf s}/3)}{1+1/({\bf s}/3)}\; {\bf X}({\bf s})=\frac3{{\bf s}+3}\; {\bf X}({\bf s}) $
          $\displaystyle =\frac3{{\bf s}+3}\frac{{\bf s}+3}{({\bf s}+3)^2+3^2} =\frac3{({\bf s}+3)^2+3^2}, $
using entry #14 of Table 3-2.
Then $y(t)={\cal L}^{-1}\{\bf Y ({\bf s})\}=e^{-3t}\sin(3t)\;u(t)$ using entry #13 of Table 3-2, which are:

13	$\displaystyle e^{-at}\sin(\omega_0 t)\;u(t)$		$\displaystyle \frac{\omega_0}{({\bf s}+a)^2+\omega_0^2}$
14	$\displaystyle e^{-at}\cos(\omega_0 t)\;u(t)$		$\displaystyle \frac{{\bf s}+a}{({\bf s}+a)^2+\omega_0^2}$

Solution:
Electromechanical analogues:
  spring$\to$inductor with inductance $\frac1{k}$;
  mass$\to$capacitor with capacitance $m$.
${\bf s}$-domain circuits:
  inductor$\to$impedance $\frac{\bf s}{k}$;
  capacitor$\to$impedance $\frac1{{\bf s} m}$; $\to{\cal L}\{\delta(t)\}=1$.
Capacitor voltage $\displaystyle {\cal L}\{v_{\rm c} (t)\}={\bf V}({\bf s})=1\frac{1/({\bf s} m)}{1/({\bf s} m)+{\bf s}/k}=\frac{k/m}{{\bf s}^2+k/m}. $
Then
$\displaystyle v_{\rm c}(t)=\sqrt{k/m}\sin(\sqrt{k/m}\,\,t)\;u(t) $
is the mass velocity using entry #9 in Table 3-2, which is

9

$\displaystyle \sin(\omega_0 t)\;u(t)$

$\displaystyle\frac{\omega_0}{{\bf s}^2+\omega_0^2} $

Solution:
See Table 4-3, which shows
 

Solution:
Noting units immediately eliminates (a) and (b). See Table 4-3, which shows
 

Solution:
Compare Fig. 4-20 with 4-21(b) (both shown below):
 

Solution:
This is the only true advantage of integrators.

Solution:
All of these are presented in Sections 4-8 to 4-10.

Solution:
${\bf H} ({\bf s})=\frac{b}{{\bf s}-a}$. Hence Eq. (4.101) is $\displaystyle {\bf Q}({\bf s})=\frac{{\bf H} ({\bf s})}{1+K{\bf H} ({\bf s})}=\frac{b/({\bf s}-a)}{1+Kb/({\bf s}-a)}=\frac{b}{({\bf s}-a)+Kb}, $
whose pole is $a-Kb$, which is in the left half-plane if $a-Kb<0$. See Section 4-8.3.

Solution:
$\displaystyle {\bf Q}({\bf s}) = \frac{{\bf H} ({\bf s})}{1+K\bf H({\bf s})} $
          $\displaystyle = \frac{({\bf s}+3)/({\bf s}^2+4)}{1+K({\bf s}+3)/({\bf s}^2+4)} $
          $\displaystyle =\frac{{\bf s}+3}{{\bf s}^2+4+K{\bf s}+3K} $
          $\displaystyle =\frac{{\bf s}+3}{{\bf s}^2+K{\bf s}+(3K+4)}. $
The roots of the denominator ${\bf s}^2+K{\bf s}+(3K+4)$ lie in the left half-plane if and only if $K>0$ and $(3K+4)>0$, which combine to just $K>0$. See Section 4-8.3.

Solution:
Taking the Laplace transform of the LCCDE gives ${\bf Y} ({\bf s})\;({\bf s}+a)=b\;{\bf X} ({\bf s})$. The transfer function is $\displaystyle {\bf H} ({\bf s})=\frac{{\bf Y} ({\bf s})}{{\bf X} ({\bf s})}=\frac{b}{{\bf s}+a} $
and the impulse response is $h(t)=be^{-at}\;u(t)$. $\displaystyle {\bf Q}({\bf s})=\frac{{\bf H} ({\bf s})}{1+K{\bf H} ({\bf s})} =\frac{b/({\bf s}+a)}{1+Kb/({\bf s}+a)}=\frac{b}{({\bf s}+a)+Kb}. $
$Kb=a$ speeds up $h(t)$ by a factor of 2.

Solution:
Using the final-value theorem Eq. (3.32), $\displaystyle \lim_{t\to\infty}y_{\rm step}(t) = \lim_{{\bf s}\to 0}{\bf s}\;{\bf Y} _{\rm step}({\bf s}), $
$\displaystyle {\bf Y} _{\rm step}({\bf s}) = {\bf H} ({\bf s})\;{\bf U}({\bf s}) =\frac{{\bf s}^2+a{\bf s}+b}{{\bf s}^2+c{\bf s}+d} \frac1{\bf s}. $
Cancelling $\frac{\bf s}{\bf s}$ leads to $\displaystyle \lim_{t\to\infty}y_{\rm step}(t) ={\bf H} (0)= \left.\frac{{\bf s}^2+a{\bf s}+b}{{\bf s}^2+c{\bf s}+d}\right|_{{\bf s}=0}=\frac{b}{d}. $
See Example 3-4. This is a very useful result in control theory.

Solution:
$\displaystyle {\bf H} ({\bf s})={\cal L}\{\cos(6t)\;u(t)\}=\frac{\bf s}{{\bf s}^2+36}. $
Eq. (4.101) is
$\displaystyle {\bf Q}({\bf s}) = \frac{{\bf H} ({\bf s})}{1+K{\bf H}({\bf s})} =\frac{{\bf s}/({\bf s}^2+36)}{1+K{\bf s}/({\bf s}^2+36)} $
          $\displaystyle = \frac{\bf s}{({\bf s}^2+36)+K{\bf s}}=\frac{\bf s}{{\bf s}^2+K{\bf s}+36}=\frac{\bf s}{({\bf s}+4)({\bf s}+9)} $
if $K=13$.

Solution:
$\displaystyle {\bf H} ({\bf s})={\cal L}\{\cos(6t)\;u(t)\}=\frac{\bf s}{{\bf s}^2+36}. $
Eq. (4.101) is
$\displaystyle {\bf Q}({\bf s})=\frac{{\bf H} ({\bf s})}{1+K{\bf H} ({\bf s})}=\frac{{\bf s}/({\bf s}^2+36)}{1+K{\bf s}/({\bf s}^2+36)} $
          $\displaystyle =\frac{\bf s}{({\bf s}^2+36)+K{\bf s}}=\frac{\bf s}{{\bf s}^2+K{\bf s}+36} $
is critically damped if $K=2\sqrt{36}=12$. Then ${\bf s}^2+K{\bf s}+36={\bf s}^2+12{\bf s}+36=({\bf s}+6)^2$.

Solution:
$\displaystyle {\bf H} ({\bf s})={\cal L}\{\cos(6t)\;u(t)\}=\frac{\bf s}{{\bf s}^2+36}. $
Eq. (4.101) is
$\displaystyle {\bf Q}({\bf s}) = \frac{{\bf H}({\bf s})}{1+{\bf G}({\bf s}){\bf H}({\bf s})} $
         $\displaystyle =\frac{{\bf s}/({\bf s}^2+36)} {1+(K_1+K_2/{\bf s}){\bf s}/({\bf s}^2+36)} $
         $\displaystyle = \frac{\bf s}{({\bf s}^2+36)+(K_1+K_2/{\bf s}){\bf s}} $
         $\displaystyle =\frac{\bf s}{{\bf s}^2+K_1{\bf s}+(36+K_2)}. $
For ${\bf Q}({\bf s})$ with poles at ${\bf s}=8$ and ${\bf s}=9$, $\displaystyle \frac{\bf s}{({\bf s}+8)({\bf s}+9)}=\frac{\bf s}{{\bf s}^2+17{\bf s}+72}, $
which leads to $K_1=17$ and $K_2=36$.

Solution:
Eq. (4.101) is
$\displaystyle {\bf Q}({\bf s}) = \frac{{\bf H}({\bf s})}{1+K{\bf H} ({\bf s})} =\frac{1/{\bf s}}{1+K/{\bf s}}=\frac1{{\bf s}+K}, $
   $\displaystyle q(t) = {\cal L}^{-1} \left\{ \frac1{{\bf s}+K} \right\}=e^{-Kt}\;u(t). $
So $K=2$.

Solution:
$\displaystyle {\bf H} ({\bf s})={\cal L}\{\sin(6t)\;u(t)\}=\frac{6}{{\bf s}^2+36}. $
Eq. (4.101) is $\displaystyle {\bf Q}({\bf s})=\frac{{\bf H} ({\bf s})}{1+K{\bf H} ({\bf s})}= \frac{6/({\bf s}^2+36)}{1+K6/({\bf s}^2+36)}= \frac{6}{({\bf s}^2+36)+6K} $

has poles $\pm j\sqrt{36+6K}$ if $36+6K>0$ and $\pm\sqrt{-36-6K}$ if $36+6K<0$. Either way, the closed-loop system is unstable.

Solution:
$\displaystyle {\bf H} ({\bf s})={\cal L}\{\sin(6t)\;u(t)\}=\frac{6}{{\bf s}^2+36}. $
Eq. (4.101) is
$\displaystyle {\bf Q}({\bf s}) = \frac{{\bf H} ({\bf s})}{1+{\bf G} ({\bf s})\;{\bf H} ({\bf s})} $
         $\displaystyle =\frac{6/({\bf s}^2+36)}{1+(K_2{\bf s})6)/({\bf s}^2+36)} $
         $\displaystyle = \frac{6}{({\bf s}^2+36)+6K_2{\bf s}}= \frac{6}{{\bf s}^2+6K_2{\bf s}+36}=\frac{6}{({\bf s}+6)^2} $
if $K_2=2$.

Solution:
$\displaystyle {\bf H} ({\bf s})={\cal L}\{\sin(6t)\;u(t)\}=\frac{6}{{\bf s}^2+36}. $
Eq. (4.101) is
$\displaystyle {\bf Q}({\bf s}) = \frac{{\bf H} ({\bf s})}{1+{\bf G} ({\bf s})\;{\bf H} ({\bf s})} $
         $\displaystyle = \frac{6/({\bf s}^2+36)}{1+(K_2{\bf s})6/({\bf s}^2+36)} $
         $\displaystyle = \frac{6}{({\bf s}^2+36)+6K_2{\bf s}}= \frac{6}{{\bf s}^2+6K_2{\bf s}+36}=\frac{6}{({\bf s}+4)({\bf s}+9)} $
if $K_2=\frac{13}{6}$.

Solution:
Eq.(4.101) is $\displaystyle {\bf Q}({\bf s})=\frac{{\bf H}({\bf s})}{1+K{\bf H} ({\bf s})} = \frac{1/{\bf s}}{1+K/{\bf s}}=\frac1{{\bf s}+K}. $
The pole is in the left half-plane for $K>0$.