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QUESTION
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\item[(a)]Sketch the locus in the $\omega$-plane of the point
$$\omega=\frac{1-z}{z(1+z)(2+z)}$$ as $z$ moves down the imaginary
axis with an indentation to the right to avoid 0.

\item[(b)]
Hence by using Nyquist theory find the values of $k>0$ for which
the closed loop system with $$A(s)=\frac{1-s}{(1+s)(1+2s)},\ \
B(s)=\frac{k}{s}$$ is stable.

\item[(c)]
Check your answer by looking for the poles of the transfer
function

$f=\frac{A}{(1+AB)}$ in the right half plane directly.

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ANSWER
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\item[(a)]
$$ \ $$ $\begin{array}{l} \ds w=\frac{1-z}{z(1+z)(2+z)}
\end{array}
\ \ \
\begin{array}{c}
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\end{array}$

Parameterise $z=iy,\ \infty>y\geq \varepsilon$ and $-\varepsilon
\geq y>-\infty$
\begin{eqnarray*}
w&=&\frac{(1-iy)(1-iy)(2-iy)}{iy(1+y^2)(4+y^2)}\\
&=&-i\frac{(1-y^2-2iy)(2-iy)}{y(1+y^2)(4+y^2)}\\
&=&\frac{-i(2-2y^2-2y^2-4iy-iy+iy^3)}{y(1+y^2)(4+y^2)}\\
&=&i\frac{4y^2-2}{y(1+y^2)(4+y^2)}+\frac{y^2-5}{(1+y^2)(4+y^2)}
\end{eqnarray*}
As $|y|\to \infty, w \approx 4y^{-3}i+y^{-2}$

At $y=\pm \sqrt{5}$, Re$w=0$, Im$w=\pm \frac{1}{3\sqrt{5}}$

At $y=\pm \frac{1}{\sqrt{2}}$, Im$w=0$, Re$w=-\frac{2}{3}$ As
$|y|\to 0,\ w\approx -\frac{i}{2}y^{-1}-\frac{5}{4}$

 Parameterise
$z=\varepsilon e^{i \theta},\ \frac{\pi}{2}\geq \theta \geq
-\frac{\pi}{2}\\ w\approx \frac{1}{2z}=\frac{1}{2\varepsilon}e^{-i
\theta}$

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\item[(b)]
$A(s)=\frac{1-s}{(1+s)(1+2s)}\ B(s)=\frac{k}{s}\
h(S)=A(s)B(s)=k(s)$ The loop system is stable $\Leftrightarrow
h(s)$ does not wind around $-1\Leftrightarrow w(s)$ does not wind
around $-\frac{1}{k} \Leftrightarrow -\frac{1}{k}<-\frac{2}{3}
\Leftrightarrow \frac{1}{k}>\frac{2}{3} \Leftrightarrow
k<\frac{3}{2}$(assuming $k>0$)

\item[(c)]
The system is stable if the transfer function
$$f(s)=\frac{A(s)}{1+A(s)B(s)}=\frac{s(1-s)}{s(1+s)(2+s)+(1-s)k}$$
has two poles in the right half plane.

Poles of $f(s)$ are zeros of $s(1+s)(2+s)+(1-s)k$. Let us find the
marginal value of $k$ that gives $f(s)$ a pole just on the
imaginary axis, i.e. $s=iy$.

We solve  $iy(1+iy)(2+iy)+(1-iy)k=0$ for $y$ and $k.$


i.e., $iy(1-y^2-k)+(k-3y^2)=0$

The solution $y=0,\ k=0$ is not the one we are after. The other
solution is $y^2=\frac{1}{2},\ k=\frac{3}{2}$ as in 1(b). One
would still have to show that Re$(s)>0$ for $k>\frac{3}{2}$.

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