Processing math: 100%

3.7 HW 7

This HW in one PDF

3.7.1 Problem 1

SOLUTION:

3.7.1.1 Part(a)

$$\begin{equation} GH=\frac{K}{s\left ( s^{2}+s+4\right ) } \tag{1} \end{equation}$$ The poles of the open loop are at $s=0$ and $s=-0.5\pm j1.94$. So we draw $\Gamma$ which encloses all the poles in the RHS making sure we avoid the pole at $s=0$ by making small circle. Here is the result. (we do not care about open loop poles in the LHS).

We now start by mapping each segment from $\Gamma$ to $\Gamma _{GH}$. Starting with segment 2. We see that

$$\lim _{s\rightarrow 0}\frac{K}{s\left ( s^{2}+s+4\right ) }=\lim _{s\rightarrow 0}\frac{K}{s}=\infty e^{-j\theta }$$ Where $\theta$ goes from $+90^{0}$ to $-90^{0}$ in $\Gamma$. This means on $\Gamma _{GH}$ segment (2) will map to a very large circle which goes from $-90^{0}$ to $+90^{0}$ (anti-clockwise). We update the plot after making each segment so we see the progress.

We now go to segment 4 in $\Gamma$.

$$\lim _{s\rightarrow \infty }\frac{K}{s\left ( s^{2}+s+4\right ) }=\lim _{s\rightarrow \infty }\frac{K}{s^{3}}=0e^{-j3\theta }$$ Where $\theta$ goes from $-90^{0}$ to $+90^{0}$ on $\Gamma$. This means on $\Gamma _{GH}$ segment (4) will map to a very small circle which goes from $+270^{0}$ to $-270^{0}$ on $\Gamma _{GH}$.  This is 1.5 circle rotation in clockwise that goes from around $540^{0}$. updating the plot gives

We now do segment 1. For this we need to find the real and imaginary part of $GH$ since we need the axis crossings. From (1)

$$GH=\frac{K}{s^{3}+s^{2}+4s}=\frac{K}{\left ( j\omega \right ) ^{3}+\left ( j\omega \right ) ^{2}+4\left ( j\omega \right ) }=\frac{K}{-j\omega ^{3}-\omega ^{2}+4j\omega }=\frac{K}{j\left ( 4\omega -\omega ^{3}\right ) -\omega ^{2}}$$ Multiply numerator and denominator by complex conjugate denominator gives $$\begin{align*} GH & =\frac{K}{j\left ( 4\omega -\omega ^{3}\right ) -\omega ^{2}}\frac{\left ( -j\left ( 4\omega -\omega ^{3}\right ) -\omega ^{2}\right ) }{\left ( -j\left ( 4\omega -\omega ^{3}\right ) -\omega ^{2}\right ) }\\ & =\frac{K\left ( -j\left ( 4\omega -\omega ^{3}\right ) -\omega ^{2}\right ) }{\left ( j\left ( 4\omega -\omega ^{3}\right ) -\omega ^{2}\right ) \left ( -j\left ( 4\omega -\omega ^{3}\right ) -\omega ^{2}\right ) }\\ & =\frac{-jK\left ( 4\omega -\omega ^{3}\right ) -K\omega ^{2}}{\omega ^{6}-7\omega ^{4}+16\omega ^{2}} \end{align*}$$

Hence

$$\begin{align*} \operatorname{Re}\left ( GH\right ) & =\frac{-K}{\omega ^{4}-7\omega ^{2}+16}\\ \operatorname{Im}\left ( GH\right ) & =\frac{-K\left ( 4\omega -\omega ^{3}\right ) }{\omega ^{6}-7\omega ^{4}+16\omega ^{2}}=\frac{K\left ( \omega ^{2}-4\right ) }{\omega \left ( \omega ^{4}-7\omega ^{2}+16\right ) } \end{align*}$$

To find the crossing. $\Gamma _{GH}$ will cross the real axis when $\operatorname{Im}\left ( GH\right ) =0$, hence

$$\begin{align*} \frac{K\left ( \omega ^{2}-4\right ) }{\omega \left ( \omega ^{4}-7\omega ^{2}+16\right ) } & =0\\ K\left ( \omega ^{2}-4\right ) & =0\\ \omega ^{2}-4 & =0\\ \omega ^{2} & =4\\ \omega & =\pm 2\text{ rad/sec} \end{align*}$$

Then

$$\begin{align*} GH\left ( j2\right ) & =\frac{K}{\left ( j2\right ) ^{3}+\left ( j2\right ) ^{2}+4\left ( j2\right ) }\\ & =\frac{K}{\left ( -j8\right ) +\left ( -4\right ) +\left ( j8\right ) }\\ & =\frac{-1}{4}K \end{align*}$$

So $\Gamma _{GH}$ will cross the real axis at $-0.25K$.

Since $K>0$, this will be somewhere on the negative real axis. To find where $\Gamma _{GH}$ crosses the imaginary axis we set $\operatorname{Re}\left ( GH\right ) =0$, hence

$$\frac{-K}{\omega ^{4}-7\omega ^{2}+16}=0$$ Which gives $\omega =\pm \infty$. When $\omega =\pm \infty$ then $GH\left ( \pm \infty \right ) =0^{\pm }$. This means the crossing at origin. This makes sense, as the small circle around the origin shrinks to zero size.

To find where segment (1) maps to, we see that for $\omega =+\infty$, $\operatorname{Re}\left ( GH\right ) <0$ and $\operatorname{Im}\left ( GH\right ) >0$, so it starts in first quadrant. $\omega =0^{+},\operatorname{Re}\left ( GH\right ) <0$ and $\operatorname{Im}\left ( GH\right ) <0$, which means segment (1) starts in the first quadrant but ends in the third quadrant.

Now for segment (3). we see that $\omega =0^{-},\operatorname{Re}\left ( GH\right ) <0$ and $\operatorname{Im}\left ( GH\right ) >0$, which means segment (3) starts in the first quadrant. $\omega =-\infty$ then $\operatorname{Re}\left ( GH\right ) <0$ and $\operatorname{Im}\left ( GH\right ) <0$ which means segment (3) in $\Gamma _{GH}$ starts in quadrant 1 and ends up in quadrant 3. There will be crossing at the real axis at $-0.25K$. Therefore the plot now looks like this

As the circle around $s=0$ shrinks to zero, $\Gamma _{GH}$ becomes as follows

We are now ready to answer the final question about stability and $K$. Since open loop has zero poles in RHS, then we need to have zero net clock wise encirclements around $-1$ for the closed loop to be stable. Only condition that will meet that, is to keep the crossing point $-0.25K$ to the right of $-1$. This means we need $0.25K<1$. This insures zero clockwise encirclements. This means

$$K<4$$ To verify, Routh table is used to determined $K$ using the closed loop transfer function. The closed loop is given by $$\begin{align*} T & =\frac{GH}{1+GH}\\ & =\frac{K}{s\left ( s^{2}+s+4\right ) +K}\\ & =\frac{K}{s^{3}+s^{2}+4s+K} \end{align*}$$

Hence the Routh table is

$s^{3}$	$1$	$4$
$s^{2}$	$1$	$K$
$s^{1}$	$4-K$
$s^{0}$	$K$

For no sign change in first column, we need $K>0$ and $4-K>0$. Which means $K<4$ as was found above. Verified OK.

3.7.1.2 Part(b)

$$\begin{equation} GH=\frac{K\left ( s+1\right ) }{s^{2}\left ( s+2\right ) }=\frac{K\left ( s+1\right ) }{s^{3}+2s^{2}} \tag{1} \end{equation}$$

The poles of the open loop are at $s=0$ (two poles) and $s=-2$. So we draw $\Gamma$ which encloses all the poles in the RHS making sure we avoid the poles at $s=0$ by making small circle. Here is the result. (we do not care about open loop pole in the LHS and about the zeros of the open loop).

We now start by mapping each segment from $\Gamma$ to $\Gamma _{GH}$. Starting with segment 2. We see that

$$\lim _{s\rightarrow 0}\frac{K\left ( s+1\right ) }{s^{3}+2s^{2}}=\lim _{s\rightarrow 0}\frac{K}{s^{2}}=\infty e^{-2j\theta }$$ Where $\theta$ goes from $+90^{0}$ to $-90^{0}$ in $\Gamma$. This means on $\Gamma _{GH}$ segment (2) will map to a very large circle which goes from $-180^{0}$ to $+180^{0}$. This is a full circle in the anti-clockwise. We update the plot after making each segment so we see the progress.

We now go to segment 4 in $\Gamma$.

$$\lim _{s\rightarrow \infty }\frac{K\left ( s+1\right ) }{s^{3}+2s^{2}}=\lim _{s\rightarrow \infty }\frac{Ks}{s^{3}}=\lim _{s\rightarrow \infty }\frac{K}{s^{2}}=0e^{-j2\theta }$$ Where $\theta$ goes from $-90^{0}$ to $+90^{0}$ on $\Gamma$. This means on $\Gamma _{GH}$ segment (4) will map to a very small circle which goes from $+180^{0}$ to $-180^{0}$ on $\Gamma _{GH}$.  This is basically a full circle in clockwise around zero. updating the plot gives

We now do segment 1. For this we need to find the real and imaginary part of $GH$ since we need the axis crossings. From (1)

$$GH=\frac{K\left ( s+1\right ) }{s^{3}+2s^{2}}=\frac{K\left ( j\omega +1\right ) }{\left ( j\omega \right ) ^{3}+2\left ( j\omega \right ) ^{2}}=\frac{K\left ( j\omega +1\right ) }{-j\omega ^{3}-2\omega ^{2}}$$ Multiply numerator and denominator by complex conjugate denominator gives $$\begin{align*} GH & =\frac{K\left ( j\omega +1\right ) }{-j\omega ^{3}-2\omega ^{2}}\frac{\left ( j\omega ^{3}-2\omega ^{2}\right ) }{\left ( j\omega ^{3}-2\omega ^{2}\right ) }\\ & =\frac{j\left ( -K\omega ^{3}\right ) -\left ( K\omega ^{4}+2K\omega ^{2}\right ) }{\omega ^{4}\left ( \omega ^{2}+4\right ) } \end{align*}$$

Hence

$$\begin{align*} \operatorname{Re}\left ( GH\right ) & =\frac{-\left ( K\omega ^{4}+2K\omega ^{2}\right ) }{\omega ^{4}\left ( \omega ^{2}+4\right ) }\\ \operatorname{Im}\left ( GH\right ) & =\frac{-K\omega ^{3}}{\omega ^{4}\left ( \omega ^{2}+4\right ) }=\frac{-K}{\omega \left ( \omega ^{2}+4\right ) } \end{align*}$$

To find the crossing. $\Gamma _{GH}$ will cross the real axis when $\operatorname{Im}\left ( GH\right ) =0$, hence

$$\begin{align*} \frac{-K}{\omega \left ( \omega ^{2}+4\right ) } & =0\\ \omega & =\pm \infty \text{ rad/sec} \end{align*}$$

Therefore

$$\begin{align*} GH\left ( j\infty \right ) & =\lim _{\omega \rightarrow \infty }\frac{K\left ( j\omega +1\right ) }{-j\omega ^{3}-2\omega ^{2}}\\ & =\lim _{\omega \rightarrow \infty }\frac{j\omega }{-j\omega ^{3}}\\ & =\lim _{\omega \rightarrow \infty }\frac{j}{-j\omega ^{2}}\\ & =0^{-} \end{align*}$$

So $\Gamma _{GH}$ will cross the real axis at $0$. Which is where the small circle shrinks to zero. To find where $\Gamma _{GH}$ crosses the imaginary axis we set $\operatorname{Re}\left ( GH\right ) =0$, hence

$$\begin{align*} \frac{-\left ( K\omega ^{4}+2K\omega ^{2}\right ) }{\omega ^{4}\left ( \omega ^{2}+4\right ) } & =0\\ K\omega ^{4}+2K\omega ^{2} & =0\\ \omega ^{2}\left ( \omega ^{2}+2\right ) & =0 \end{align*}$$

Hence

$$\omega =0$$ And $$\omega =\pm i\sqrt{2}$$ Since $\omega$ has to be real. Then $\omega =0$ is only used. When $\omega =0$ then $GH\left ( j0\right ) =\lim _{\omega \rightarrow 0}\frac{K\left ( j\omega +1\right ) }{-j\omega ^{3}-2\omega ^{2}}=\lim _{\omega \rightarrow 0}\frac{K}{-2\omega ^{2}}=\infty$ . Hence $\Gamma _{GH}$ will cross the imaginary axis at $\infty .$

To find which quadrants segment (1) maps to, we see that for positive $\omega$ then $\operatorname{Re}\left ( GH\right )$ is negative and $\operatorname{Im}\left ( GH\right )$ is negative (since $K>0$). Therefore segment (1) maps to third quadrant. And for segment (3), we see that negative $\omega$ then $\operatorname{Re}\left ( GH\right )$ is negative and $\operatorname{Im}\left ( GH\right )$ is positive (since $K>0$). Therefore segment (3) maps to first quadrant

Therefore the plot now looks like this

After the small circle shrinks to zero, and since the crossing on the real axis was found at $0$ then the final plot looks like

We are now ready to answer the final question about stability and $K$. Since open loop has zero poles in RHS, then we need to have zero net clockwise encirclements around $-1$ for the closed loop to be stable. We see that there is no encirclements around $-1$. No matter what $K>0$ value is. So the closed loop is stable for all positive $K$. To verify, Routh table is used to determined $K$ using the closed loop transfer function. The closed loop is given by

$$\begin{align*} T & =\frac{GH}{1+GH}\\ & =\frac{K\left ( s+1\right ) }{s^{3}+2s^{2}+K\left ( s+1\right ) } \end{align*}$$

Hence the Routh table is

$s^{3}$	$1$	$K$
$s^{2}$	$2$	$K$
$s^{1}$	$\frac{2K-K}{2}=\frac{K}{2}$
$s^{0}$	$K$

For no sign change in first column, we need $K>0$ and $\frac{K}{2}$ $>0$. Which is always true since $K$ is positive. Verified OK.

3.7.2 Problem 2

SOLUTION:

Since there is zero open loop poles in RHP, we need zero net clockwise encirclements around $-1$. This means we need to keep point $-0.5$ to the right of $-1$ and keep the point $-2$ to the left of $-1$. In other words, we need to satisfy

$$\begin{align*} 0.5K & <1\\ 2K & >1 \end{align*}$$

Or

$$K<2$$ And $$K>\frac{1}{2}$$ Hence the range of stable closed loop is for $K$ to meet the following requirement $$0.5<K<2$$

3.7.3 Problem 3

SOLUTION:

3.7.3.1 Part(a)

When $K_{v}=0$ the open loop is $GH=K\frac{1}{s^{2}}$. This has no poles in RHP and two poles at zero. Hence $\Gamma$ is

We start at segment $2.$

$$\lim _{s\rightarrow 0}\frac{K}{s^{2}}=\lim _{\varepsilon \rightarrow 0}\frac{K}{\left ( \varepsilon e^{j\theta }\right ) ^{2}}=\infty e^{-2j\theta }$$ As $\theta$ goes from $+90^{0}$ to $-90^{0}$ on $\Gamma$ segment $2$ goes from $-180^{0}$ to $+180$ on $\Gamma _{GH}$ with $\infty$ radius in anti-clockwise. Hence the plot now looks as follows

For segment $4$ in $\Gamma$. We have

$$\lim _{s\rightarrow \infty }\frac{K}{s^{2}}=\lim _{R\rightarrow \infty }\frac{K}{\left ( Re^{j\theta }\right ) ^{2}}=0e^{-2j\theta }$$ $\theta$ goes from $-90^{0}$ to $+90^{0}$ on $\Gamma$, segment $4$ goes from $+180^{0}$ to $-180$ on $\Gamma _{GH}$ with $0$ radius in clockwise. Hence the plot now looks as follows

Now

$$\frac{K}{s^{2}}=\frac{K}{\left ( j\omega \right ) ^{2}}=\frac{K}{-\omega ^{2}}$$ Hence $\operatorname{Re}\left ( GH\right ) =\frac{K}{-\omega ^{2}}$ and there is no imaginary part. Therefore, there is no crossing on the real axis. There is crossing on the imaginary axis at $\omega =\pm \infty$ which occurs at $GH=0$. This is when the small circle shrinks to zero size. It is clear that segment 1 will map to third quadrant and segment 3 will map to second quadrant since there is no crossing. The final plot is

After the small circle shrinks to zero, and since the crossing on the real axis was found at $0$ then the final plot looks like

3.7.3.2 Part(b)

Since open loop has zero poles in RHP, we need $\Gamma _{GH}$ to have zero net clockwise encirclements. Since $\Gamma _{GH}$ has no crossing on the real axis that depends on $K$ then $\Gamma _{GH}$ will remain as shown for any $K$. So closed loop is stable for all $K>0.$ To verify, we set the Routh table for the closed loop polynomial

The closed loop is given by

$$\begin{align*} T & =\frac{GH}{1+GH}\\ & =\frac{K}{s^{2}+K} \end{align*}$$

Hence the Routh table is

$s^{2}$	$1$	$K$
$s^{1}$	$0$	$0$
$s^{0}$	$0$

We see that there is no sign change in first column, no matter what $K$ is.

3.7.3.3 Part (c)

We first find the closed loop transfer function. Let $E\left ( s\right )$ be the error (just after the summing junction) then

$$\begin{align*} E & =\theta _{r}-\theta -K_{v}s\theta \\ \theta & =EK\frac{1}{s^{2}} \end{align*}$$

From the second equation $E=\frac{\theta s^{2}}{K}$, hence the first equation becomes

$$\begin{align*} \frac{\theta s^{2}}{K} & =\theta _{r}-\theta -K_{v}s\theta \\ \frac{\theta s^{2}}{K}+\theta +K_{v}s\theta & =\theta _{r}\\ \theta \left ( \frac{s^{2}}{K}+1+K_{v}s\right ) & =\theta _{r} \end{align*}$$

Therefore the closed loop transfer function $T\left ( s\right ) =\frac{\theta }{\theta _{r}}$ is

$$\begin{align*} \frac{\theta }{\theta _{r}} & =\frac{1}{\frac{s^{2}}{K}+1+K_{v}s}\\ & =\frac{K}{s^{2}+KK_{v}s+K} \end{align*}$$

We now find the open loop transfer function with unity feedback using the closed loop transfer function. Since $T\left ( s\right ) =\frac{G}{1+G}$ where $G\left ( s\right )$ is the closed loop transfer function, then letting $G\left ( s\right ) =\frac{N}{D}$ we have

$$\begin{align*} \frac{K}{s^{2}+KK_{v}s+K} & =\frac{G}{1+G}\\ & =\frac{\frac{N}{D}}{1+\frac{N}{D}}\\ & =\frac{N}{N+D} \end{align*}$$

Therefore $N=K$ and $N+D=s^{2}+KK_{v}s+K$ which means $D\left ( s\right ) =s^{2}+KK_{v}s$. Therefore the open loop transfer function is

$$G\left ( s\right ) =\frac{N}{D}=\frac{K}{s^{2}+KK_{v}s}$$ Hence $$\fbox{$GH=\frac{K}{s\left ( s+KK_v\right ) }$}$$ Therefore, the open loop has a pole at $s=0$ and at $s=-KK_{v}$.

Assuming positive gains, $-KK_{v}$ is in the LHP. Hence the open loop is stable. Which means the Nyquist plot should have zero net clockwise encirclement around $-1$ for the closed loop to be stable.

We now start by mapping each segment from $\Gamma$ to $\Gamma _{GH}$. Starting with segment 2. We see that

$$\lim _{s\rightarrow 0}\frac{K}{s^{2}+sKK_{v}}=\lim _{s\rightarrow 0}\frac{1}{sK_{v}}=\lim _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon e^{j\theta }K_{v}}=\infty e^{-j\theta }$$ Where $\theta$ goes from $+90^{0}$ to $-90^{0}$ in $\Gamma$. This means on $\Gamma _{GH}$ segment (2) will map to a half circle which goes from $-90^{0}$ to $+90^{0}$ in anti-clockwise.

We now go to segment 4 in $\Gamma$.

$$\lim _{s\rightarrow \infty }\frac{K}{s^{2}+sKK_{v}}=\lim _{s\rightarrow \infty }\frac{K}{s^{2}}=\lim _{R\rightarrow \infty }\frac{1}{Re^{2j\theta }}=0e^{-j2\theta }$$ Where $\theta$ goes from $-90^{0}$ to $+90^{0}$ on $\Gamma$. This means on $\Gamma _{GH}$ segment (4) will map to a very small circle which goes from $+180^{0}$ to $-180^{0}$ on $\Gamma _{GH}$.  This is basically a full circle in clockwise around zero. updating the plot gives

To find the intersections,

$$\begin{align*} GH & =\frac{K}{s^{2}+sKK_{v}}=\frac{K}{-\omega ^{2}+j\omega KK_{v}}=\frac{K}{\left ( -\omega ^{2}+j\omega KK_{v}\right ) }\frac{\left ( -\omega ^{2}-j\omega KK_{v}\right ) }{\left ( -\omega ^{2}-j\omega KK_{v}\right ) }\\ & =\frac{-K\omega ^{2}-j\omega K^{2}K_{v}}{K^{2}\omega ^{2}K_{v}^{2}+\omega ^{4}}=\frac{-K}{K^{2}K_{v}^{2}+\omega ^{2}}-j\frac{K^{2}K_{v}}{K^{2}\omega K_{v}^{2}+\omega ^{3}} \end{align*}$$

Hence $\operatorname{Re}\left ( GH\right ) =$ $\frac{-K}{K^{2}K_{v}^{2}+\omega ^{2}}$ and $\operatorname{Im}\left ( GH\right ) =-\frac{K^{2}K_{v}}{K^{2}\omega K_{v}^{2}+\omega ^{3}}$. When $\operatorname{Re}\left ( GH\right ) =0$, we get $\omega =\pm \infty$. When means $GH=0$ at this frequency. So $\Gamma _{GH}$ crosses the imaginary axis at the origin. When $\operatorname{Im}\left ( GH\right ) =0$, we also get $\omega =\pm \infty$ which also means $\Gamma _{GH}$ crosses the real axis at zero. By continuation, and since segment 1 must follow segment 4, then segment 1 maps to third quadrant in $\Gamma _{GH}$ and segment 3 must map to quadrant 2 in $\Gamma _{GH}$. As $\varepsilon \rightarrow 0$ the small circle become a point at origin and we get the final plot

Since the intersection is always at the origin, $\Gamma _{GH}$ will never move to the left passed $-1$ to make any encirclement around $-1$. We need at least one net encirclement for the closed loop to be unstable. Hence the closed loop is stable for all positive $K,K_{v}$.

To verify, we show Routh table for the closed loop found above, which is $\frac{K}{s^{2}+KK_{v}s+K}$. The Routh table is

$s^{2}$	$1$	$K$
$s^{1}$	$KK_{v}$	$0$
$s^{0}$	$K$

For positive $K,K_{v}$, we see that there can not be a sign change. Hence closed loop is stable for all $K,K_{v}$. (Note, I assume $K,K_{v}>0$. Verified this with instructor via email).

3.7.4 Problem 4

SOLUTION:

3.7.4.1 Part(a)

The open loop transfer function $T(s)$ is  $G_{A}\left ( s\right ) e^{-sT}G\left ( s\right ) G_{f}\left ( s\right )$. For $T=0$, we have

$$\begin{align*} T\left ( s\right ) & =G_{A}\left ( s\right ) G\left ( s\right ) G_{f}\left ( s\right ) \\ & =\frac{10}{s+1}\frac{3.15}{30s+1}\frac{1}{\frac{s^{2}}{9}+\frac{s}{3}+1} \end{align*}$$

The Nyquist plot is (using the program I wrote which shows $\Gamma$ and $\Gamma _{GH}$ side by side)

In the limit, as $R$ becomes very large we obtain

Here is also Matlab nyquist output (zoomed in version)

s=tf('s');
sys=10/(s+1)*(3.15)/(30*s+1)*1/(s^2/9+s/3+1);
sys =
850.5
----------------------------------------
90 s^4 + 363 s^3 + 1092 s^2 + 846 s + 27
nyquist([850.5],[90 363 1092 846 27])

3.7.4.2 Part(b)

For the gain margin, the $\Gamma _{GH}$ curve crosses the real axis at about $-0.41$. Therefore we need

$$0.41K_{\max }<1$$ For stability. Hence $$K_{\max }=\frac{1}{0.41}=2.439$$ In dB, the above becomes $$\begin{align*} gm & =20\log _{10}2.439\\ & =7.74\text{ db} \end{align*}$$

For the phase margin, we draw a unit circle and find the intersection with $\Gamma _{GH}$ and estimate the angle between the line from origin to the intersection and the $-180^{0}$ line. As follows

The angle seems to be approximately between $30^{0}$ and $35^{0}$. This is the phase margin. To get exact values, Matlab margin command can be used as follows

s=tf('s');
sys=10/(s+1)*(3.15)/(30*s+1)*1/(s^2/9+s/3+1);
sys =
850.5
----------------------------------------
90 s^4 + 363 s^3 + 1092 s^2 + 846 s + 27
>> [Gm,Pm,~,~] = margin(sys)
Gm =
2.3854
Pm =
35.6025

Converting the Gm value given in Matlab to dB, gives the result shown above. Matlab gives $35.6^{0}$ as the exact phase margin$.$

3.7.4.3 Part(c)

Let the open loop $GH$ when $T=0$ (which is what we analyzed in part (b)) be called $GH\left ( s\right )$ which can be written, in frequency domain as

$$\left . GH\left ( s\right ) \right \vert _{s=j\omega }=\left \vert GH\right \vert e^{j\theta }$$

Where both the magnitude $\left \vert GH\right \vert$ and the phase $\theta$ in the above, are functions of the frequency $\omega$. The above is polar representation of the complex quantity $GH\left ( j\omega \right )$. When $T>0$, then the open loop is now $e^{-sT}GH\left ( s\right )$, which can be written in frequency domain as

$$e^{-j\omega T}GH\left ( j\omega \right ) =\left \vert e^{-j\omega T}\right \vert \left \vert GH\right \vert e^{j\left ( \theta -\omega T\right ) }$$ In other words, magnitudes multiply and phases are added. But $\left \vert e^{-j\omega T}\right \vert =1$, so the above is $$e^{-j\omega T}GH\left ( j\omega \right ) =\left \vert GH\right \vert e^{j\left ( \theta -\omega T\right ) }$$ We see that the resulting open loop has the same magnitude as before, but its phase has change. We subtract angle $\omega T$ from the original phase $\theta$. subtract angle $\omega T$ is the same as rotating the complex vector representation clockwise $\omega T$. So this causes the whole Nyquist plot, which is a frequency plot $GH\left ( j\omega \right ) ,$ to just rotate by $\omega T$ clockwise (since negative angle) to what it was before. This makes $\Gamma _{GH}$ become closer to $-1$. This is illustrated in the following diagram

When $T=1$, the angle is $\omega$ radians. Since we found the phase margin to be $35^{0}$ or about $0.61$ radians, then the closed loop, which corresponds to the open $e^{-j\omega T}GH\left ( j\omega \right ) ,$ will have new phase reduced by $\omega$ radians. Since phase margin is measured at $0dB$ angle (or $\omega =1$ radian, or $57.3^{0}$). This is larger than the phase margin $35^{0}$. Therefore the new system is unstable. $\Gamma _{GH}$ will rotate and will cross over $-1$.

What the above shows, is that adding delays $e^{-sT}$ makes the system less stable (closer to becoming unstable). Delays causes the phase margin to reduce. We can find the amount of delay $T$ before the system becomes unstable.  We need $\omega T<35^{0}$ or $T<\frac{35^{0}}{57.3^{0}}<\allowbreak 0.611$ seconds. This is the maximum delay $T$ we can have before the closed loop phase margin is all used up and the system become unstable.

3.7.5 HW 7 key solution

;

;

;

;

;

;

;

;

;

;

;

;