HW7

The plant is

$P\left ( s,\rho \right ) =\frac{\left ( 3+\rho \left [ 0,1\right ] \right ) s+\left ( 1+\rho \left [ -0.5,0.25\right ] \right ) }{s^{4}+\left ( 3+\rho \left [ -1,1\right ] \right ) s^{3}+\left ( 6+\rho \left [ -1,1\right ] \right ) s^{2}+\left ( 3+\rho \left [ -1,0\right ] \right ) s+\left ( 4+\rho \left [ -0.5,0.75\right ] \right ) }$

Part(a)

When

$\rho =0$

$\boxed{ P\left ( s\right ) =\frac{3s+1}{s^{4}+3s^{3}+6s^{2}+3s+4}}$ First we ﬁnd the closed loop transfer function. Using the following diagram

Replacing

$E$ in second equation with

$E$ from the ﬁrst equation

$\begin{align*} Y & =\left ( U-Y\right ) HP\\ & =UHP-YHP \end{align*}$

Hence

$Y\left ( 1+HP\right ) =UHP$ The closed loop transfer function is

$\frac{Y}{U}$ . From the above we obtain

$\frac{Y}{U}=G_{cl}\left ( s\right ) =\frac{HP}{1+HP}$ For unity feedback,

$H=1$ , the above reduces to

$\boxed{ G_{cl}\left ( s\right ) =\frac{P\left ( s\right ) }{1+P\left ( s\right ) } }$ This is stable if poles of

$G_{cl}\left ( s\right )$ are stable. This is the same as saying the zeros of denominator of

$G_{cl}\left ( s\right )$ all have negative real parts. Writing

$P\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) }$ then

$G_{cl}\left ( s\right ) =\frac{\frac{N\left ( s\right ) }{D\left ( s\right ) }}{1+\frac{N\left ( s\right ) }{D\left ( s\right ) }}=\frac{N\left ( s\right ) }{D\left ( s\right ) +N\left ( s\right ) }=\frac{3s+1}{s^{4}+3s^{3}+6s^{2}+3s+4+3s+1}=\frac{3s+1}{s^{4}+3s^{3}+6s^{2}+6s+5}$ We now need to check stability of the denominator of

$G_{cl}\left ( s\right )$ given by

$s^{4}+3s^{3}+6s^{2}+6s+5.$ Using Hurwitz matrix where

$s^{4}+3s^{3}+6s^{2}+6s+5\equiv a_{4}s^{4}+a_{3}s^{3}+a_{2}s^{2}+a_{1}s+a_{0}$ gives

$H=\begin{bmatrix} a_{1} & a_{3} & 0 & 0\\ a_{0} & a_{2} & a_{4} & 0\\ 0 & a_{1} & a_{3} & 0\\ 0 & a_{0} & a_{2} & a_{4}\end{bmatrix} =\begin{bmatrix} 6 & 3 & 0 & 0\\ 5 & 6 & 1 & 0\\ 0 & 6 & 3 & 0\\ 0 & 5 & 6 & 1 \end{bmatrix}$ Hence

$\Delta _{1}=6,\Delta _{2}=21,\Delta _{3}=27,\Delta _{4}=27$ . Since all

$\Delta _{i}>0$ then the

Hence closed loop system is stable. To verify, using the computer, the roots of

$s^{4}+3s^{3}+6s^{2}+6s+5$ are

$\{-0.296974,-0.296974,-1.20303,-1.20303\}$ . Since they are all negative, this veriﬁes system is stable as well.

Part(b)

Still using the unity compensator but now using the interval plant gives

$\begin{align*} G_{cl}\left ( s\right ) & =\frac{P\left ( s\right ) }{1+P\left ( s\right ) }=\frac{\frac{N\left ( s\right ) }{D\left ( s\right ) }}{1+\frac{N\left ( s\right ) }{D\left ( s\right ) }}=\frac{N\left ( s\right ) }{D\left ( s\right ) +N\left ( s\right ) }\\ & =\frac{\left ( 3+\rho \left [ 0,1\right ] \right ) s+\left ( 1+\rho \left [ -0.5,0.25\right ] \right ) }{s^{4}+\left ( 3+\rho \left [ -1,1\right ] \right ) s^{3}+\left ( 6+\rho \left [ -1,1\right ] \right ) s^{2}+\left ( 3+\rho \left [ -1,0\right ] \right ) s+\left ( 4+\rho \left [ -0.5,0.75\right ] \right ) +\left ( 3+\rho \left [ 0,1\right ] \right ) s+\left ( 1+\rho \left [ -0.5,0.25\right ] \right ) } \end{align*}$

The denominator polynomial from above is

$\begin{equation} \Delta \left ( s\right ) =s^{4}+\left ( 3+\rho \left [ -1,1\right ] \right ) s^{3}+\left ( 6+\rho \left [ -1,1\right ] \right ) s^{2}+\left ( 6+\rho \left [ -1,0\right ] +\rho \left [ 0,1\right ] \right ) s+\left ( 5+\rho \left [ -0.5,0.75\right ] +\rho \left [ -0.5,0.25\right ] \right ) \tag{1} \end{equation}$ But³

$\left ( 6+\rho \left [ -1,0\right ] +\rho \left [ 0,1\right ] \right ) s=\left ( 6+\rho \left [ -1,1\right ] \right ) s$ And

$\left ( 5+\rho \left [ -0.5,0.75\right ] +\rho \left [ -0.5,0.25\right ] \right ) =\left ( 5+\rho \left [ -1,1\right ] \right )$ Therefore (1) becomes

$\boxed{ \Delta \left ( s\right ) =s^{4}+\left ( 3+\rho \left [ -1,1\right ] \right ) s^{3}+\left ( 6+\rho \left [ -1,1\right ] \right ) s^{2}+\left ( 6+\rho \left [ -1,1\right ] \right ) s+\left ( 5+\rho \left [ -1,1\right ] \right ) }$ The above is the polynomial to examine for ﬁnding the maximum

$\rho$ . Notice when

$\rho =0$ we obtain

$s^{4}+3s^{3}+6s^{2}+6s+5$ as in part(a) which is stable. Note that if the nominal polynomial is not stable, then there will be no point in checking for robust stability. The four Kharitonov polynomials are from the above are

$\begin{align*} K_{1} & =\left ( 5-\rho \right ) +\left ( 6-\rho \right ) s+\left ( 6+\rho \right ) s^{2}+\left ( 3+\rho \right ) s^{3}+s^{4}\\ K_{2} & =\left ( 5+\rho \right ) +\left ( 6+\rho \right ) s+\left ( 6-\rho \right ) s^{2}+\left ( 3-\rho \right ) s^{3}+s^{4}\\ K_{3} & =\left ( 5+\rho \right ) +\left ( 6-\rho \right ) s+\left ( 6-\rho \right ) s^{2}+\left ( 3+\rho \right ) s^{3}+s^{4}\\ K_{4} & =\left ( 5-\rho \right ) +\left ( 6+\rho \right ) s+\left ( 6+\rho \right ) s^{2}+\left ( 3-\rho \right ) s^{3}+s^{4} \end{align*}$

We want to ﬁnd the maximum

$\rho$ such that the four polynomials above are still stable. We setup the Hurwitz matrix for each and determine the condition on

$\rho$ needed. For

$K_{1}$

$\begin{bmatrix} a_{1} & a_{3} & 0 & 0\\ a_{0} & a_{2} & a_{4} & 0\\ 0 & a_{1} & a_{3} & 0\\ 0 & a_{0} & a_{2} & a_{4}\end{bmatrix} =\begin{bmatrix} \left ( 6-\rho \right ) & \left ( 3+\rho \right ) & 0 & 0\\ \left ( 5-\rho \right ) & \left ( 6+\rho \right ) & 1 & 0\\ 0 & \left ( 6-\rho \right ) & \left ( 3+\rho \right ) & 0\\ 0 & \left ( 5-\rho \right ) & \left ( 6+\rho \right ) & 1 \end{bmatrix}$ Hence

$\Delta _{1}=\left ( 6-\rho \right ) >0$ which means

$\rho <6$ . And

$\Delta _{2}=21-2\rho >0$ hence

$\rho <10.5$ and

$\Delta _{3}=27+27\rho -3\rho ^{2}>0$ . Hence

$-0.9083<\rho <9.908$ .

$\Delta _{4}$ is the same as

$\Delta _{3}$ hence no new information is obtained from it. Therefore, from

$K_{1}$ we ﬁnd the following

$\left \{ \rho <6,\rho <10.5,-0.9083<\rho <9.908\right \}$ For

$K_{2}$

$\begin{bmatrix} \left ( 6+\rho \right ) & \left ( 3-\rho \right ) & 0 & 0\\ \left ( 5+\rho \right ) & \left ( 6-\rho \right ) & 1 & 0\\ 0 & \left ( 6+\rho \right ) & \left ( 3-\rho \right ) & 0\\ 0 & \left ( 5+\rho \right ) & \left ( 6-\rho \right ) & 1 \end{bmatrix}$ Hence

$\Delta _{1}=\left ( 6+\rho \right ) >0$ which means

$\rho >-6$ . And

$\Delta _{2}=21+2\rho >0$ hence

$\rho >-10.5$ and

$\Delta _{3}=27-27\rho -3\rho ^{2}>0$ . Hence

$-9.908<\rho <0.908$ .

$\Delta _{4}$ is the same as

$\Delta _{3}$ hence no new information is obtained from it. Therefore, from

$K_{2}$ we ﬁnd the following

$\left \{ \rho >-6,\rho >-10.5,-9.908<\rho <0.908\right \}$ For

$K_{3}$

$\begin{bmatrix} \left ( 6-\rho \right ) & \left ( 3+\rho \right ) & 0 & 0\\ \left ( 5+\rho \right ) & \left ( 6-\rho \right ) & 1 & 0\\ 0 & \left ( 6-\rho \right ) & \left ( 3+\rho \right ) & 0\\ 0 & \left ( 5+\rho \right ) & \left ( 6-\rho \right ) & 1 \end{bmatrix}$ Hence

$\Delta _{1}=\left ( 6-\rho \right ) >0$ which means

$\rho <6$ . And

$\Delta _{2}=21-20\rho >0$ hence

$\rho <1.05$ and

$\Delta _{3}=27-27\rho -21\rho ^{2}>0$ . Hence

$-1.936<\rho <0.66059$ .

$\Delta _{4}$ is the same as

$\Delta _{3}$ hence no new information is obtained from it. Therefore, from

$K_{3}$ we ﬁnd the following

$\left \{ \rho <6,\rho <1.05,-1.936<\rho <0.66059\right \}$ For

$K_{4}$

$\begin{bmatrix} \left ( 6+\rho \right ) & \left ( 3-\rho \right ) & 0 & 0\\ \left ( 5-\rho \right ) & \left ( 6+\rho \right ) & 1 & 0\\ 0 & \left ( 6+\rho \right ) & \left ( 3-\rho \right ) & 0\\ 0 & \left ( 5-\rho \right ) & \left ( 6+\rho \right ) & 1 \end{bmatrix}$ Hence

$\Delta _{1}=\left ( 6+\rho \right ) >0$ which means

$\rho >-6$ . And

$\Delta _{2}=21+20\rho >0$ hence

$\rho >-1.05$ and

$\Delta _{3}=27+27\rho -21\rho ^{2}>0$ . Hence

$-0.66059<\rho <1.956$ .

$\Delta _{4}$ is the same as

$\Delta _{3}$ hence no new information is obtained from it. Therefore, from

$K_{4}$ we ﬁnd the following

$\left \{ \rho >-6.,\rho >-1.05,-0.66059<\rho <1.956\right \}$ We now have found all the range for

$\rho$ from each polynomial. We put them together in order to determine the largest

$\rho$ allowed

$\begin{align*} K_{1} & \Rightarrow \left \{ \rho <6,\rho <10.5,-0.9083<\rho <9.908\right \} \\ K_{2} & \Rightarrow \left \{ \rho >-6,\rho >-10.5,-9.908<\rho <0.908\right \} \\ K_{3} & \Rightarrow \left \{ \rho <6,\rho <1.05,-1.936<\rho <0.66059\right \} \\ K_{4} & \Rightarrow \left \{ \rho >-6.,\rho >-1.05,-0.66059<\rho <1.956\right \} \end{align*}$

We see that the largest allowed positive

$\rho$ is

$\boxed{ \rho _{\max }=0.66 }$

Part(c)

Using

$\rho =\frac{1}{2}\rho _{\max }=\frac{1}{2}\left ( 0.908\right ) =0.454$ The plant becomes

$\begin{align*} p\left ( s\right ) & =\frac{\left ( 3+0.454\left [ 0,1\right ] \right ) s+\left ( 1+0.454\left [ -0.5,0.25\right ] \right ) }{s^{4}+\left ( 3+0.454\left [ -1,1\right ] \right ) s^{3}+\left ( 6+0.454\left [ -1,1\right ] \right ) s^{2}+\left ( 3+0.454\left [ -1,0\right ] \right ) s+\left ( 4+0.454\left [ -0.5,0.75\right ] \right ) }\\ & =\frac{\left ( 3+\left [ 0,0.454\right ] \right ) s+\left ( 1+\left [ -0.5\left ( 0.454\right ) ,0.25\left ( 0.454\right ) \right ] \right ) }{s^{4}+\left ( 3+\left [ -0.454,0.454\right ] \right ) s^{3}+\left ( 6+\left [ -0.454,0.454\right ] \right ) s^{2}+\left ( 3+\left [ -0.454,0\right ] \right ) s+\left ( 4+\left [ -0.5\left ( 0.454\right ) ,0.75\left ( 0.454\right ) \right ] \right ) }\\ & =\frac{\left ( 3+\left [ 0,0.454\right ] \right ) s+\left ( 1+\left [ -0.227,0.1135\right ] \right ) }{s^{4}+\left ( 3+\left [ -0.454,0.454\right ] \right ) s^{3}+\left ( 6+\left [ -0.454,0.454\right ] \right ) s^{2}+\left ( 3+\left [ -0.454,0\right ] \right ) s+\left ( 4+\left [ -0.227,0.3405\right ] \right ) } \end{align*}$

But

$a+\left [ b,c\right ] =\left [ a+b,a+c\right ]$ , hence we can simplify the above to

$p\left ( s\right ) =\frac{\left [ 3,3.454\right ] s+\left [ 0.773,1.1135\right ] }{s^{4}+\left [ 2.546,3.454\right ] s^{3}+\left [ 5.546,6.454\right ] s^{2}+\left [ 2.546,3\right ] s+\left [ 3.773,4.3405\right ] }$ Now the compensator is no longer unity but

$H\left ( s\right ) =\frac{1}{s}$ . Hence the closed loop transfer function is

$\begin{align*} G_{cl}\left ( s\right ) & =\frac{H\left ( s\right ) P\left ( s\right ) }{1+H\left ( s\right ) P\left ( s\right ) }\\ & =\frac{p\left ( s\right ) H\left ( s\right ) }{1+\left ( \frac{1}{s}\right ) \left ( \frac{\left [ 3,3.454\right ] s+\left [ 0.773,1.1135\right ] }{s^{4}+\left [ 2.546,3.454\right ] s^{3}+\left [ 5.546,6.454\right ] s^{2}+\left [ 2.546,3\right ] s+\left [ 3.773,4.3405\right ] }\right ) }\\ & =\frac{p\left ( s\right ) H\left ( s\right ) }{1+\left ( \frac{\left [ 3,3.454\right ] s+\left [ 0.773,1.1135\right ] }{s^{5}+\left [ 2.546,3.454\right ] s^{4}+\left [ 5.546,6.454\right ] s^{3}+\left [ 2.546,3\right ] s^{2}+\left [ 3.773,4.3405\right ] s}\right ) }\\ & =\frac{\left [ 3,3.454\right ] s+\left [ 0.773,1.1135\right ] }{s^{5}+\left [ 2.546,3.454\right ] s^{4}+\left [ 5.546,6.454\right ] s^{3}+\left [ 2.546,3\right ] s^{2}+\left [ 3.773,4.3405\right ] s+\left [ 3,3.454\right ] s+\left [ 0.773,1.1135\right ] } \end{align*}$

But

$\left [ 3.773,4.3405\right ] s+\left [ 3,3.454\right ] s=\left [ 6.773,7.7945\right ] s$ , and the above becomes

$G_{cl}\left ( s\right ) =\frac{\left [ 3,3.454\right ] s+\left [ 0.773,1.1135\right ] }{s^{5}+\left [ 2.546,3.454\right ] s^{4}+\left [ 5.546,6.454\right ] s^{3}+\left [ 2.546,3\right ] s^{2}+\left [ 6.773,7.7945\right ] s+\left [ 0.773,1.1135\right ] }$ The system is stable if the zeros of the denominator of

$G_{cl}\left ( s\right )$ are stable. The interval polynomial to check for robust stability is

$\boxed{ s^{5}+\left [ 2.546,3.454\right ] s^{4}+\left [ 5.546,6.454\right ] s^{3}+\left [ 2.546,3\right ] s^{2}+\left [ 6.773,7.7945\right ] s+\left [ 0.773,1.1135\right ] }$ The four Kharitonov polynomials from the above is

$\begin{align*} K_{1} & =0.773+6.773s+3s^{2}+6.454s^{3}+2.546s^{4}+s^{5}\\ K_{2} & =1.1135+7.7945s+2.546s^{2}+5.546s^{3}+3.454s^{4}+s^{5}\\ K_{3} & =1.1135+6.773s+2.546s^{2}+6.454s^{3}+3.454s^{4}+s^{5}\\ K_{4} & =0.773+7.7945s+3s^{2}+5.546s^{3}+2.546s^{4}+s^{5} \end{align*}$

Finding the real part of the roots of each polynomial gives

$\begin{align*} K_{1} & =\{-1.29,-1.29,-0.119,0.0806,0.0806\}\\ K_{2} & =\{-1.92,-1.92,-0.148,0.270,0.270\}\\ K_{3} & =\{-1.83,-1.83,-0.171,0.186,0.186\}\\ K_{4} & =\{-1.42,-1.42,-0.102,0.199,0.199\} \end{align*}$

2.7.3 problem 2

$G\left ( s\right ) =k\frac{3s+1}{s^{4}+s^{3}+as^{2}+s+b}$ Where

$12\leq a\leq 36,1\leq b\leq 2$ . The closed loop transfer function for unity feedback is

$\begin{align*} G_{cl}\left ( s\right ) & =\frac{G\left ( s\right ) }{1+G\left ( s\right ) }\\ & =\frac{k\frac{3s+1}{s^{4}+s^{3}+as^{2}+s+b}}{1+k\frac{3s+1}{s^{4}+s^{3}+as^{2}+s+b}}\\ & =\frac{k\left ( 3s+1\right ) }{\left ( s^{4}+s^{3}+as^{2}+s+b\right ) +k\left ( 3s+1\right ) }\\ & =\frac{k\left ( 3s+1\right ) }{s^{4}+s^{3}+as^{2}+s\left ( 1+3k\right ) +\left ( b+k\right ) } \end{align*}$

We need to ﬁnd the largest

$k$ such that the zeros of

$s^{4}+s^{3}+as^{2}+s\left ( 1+3k\right ) +\left ( b+k\right )$ remain stable. Writing this using uncertainties

$\Delta =s^{4}+s^{3}+\left [ 12,36\right ] s^{2}+s\left ( 1+3k\right ) +\left ( \left [ 1,2\right ] +k\right )$ The four Kharitonov polynomials are

$\begin{align*} K_{1} & =\left ( k+1\right ) +\left ( 1+3k\right ) s+36s^{2}+s^{3}+s^{4}\\ K_{2} & =\left ( k+2\right ) +\left ( 1+3k\right ) s+12s^{2}+s^{3}+s^{4}\\ K_{3} & =\left ( k+2\right ) +\left ( 1+3k\right ) s+12s^{2}+s^{3}+s^{4}\\ K_{4} & =\left ( k+1\right ) +\left ( 1+3k\right ) s+36s^{2}+s^{3}+s^{4} \end{align*}$

We want to ﬁnd the maximum

$k$ such that the four polynomials above are stable. We setup the Hurwitz matrix for each and determine the condition on

$k$ needed. For

$K_{1}$

$\begin{bmatrix} a_{1} & a_{3} & 0 & 0\\ a_{0} & a_{2} & a_{4} & 0\\ 0 & a_{1} & a_{3} & 0\\ 0 & a_{0} & a_{2} & a_{4}\end{bmatrix} =\begin{bmatrix} \left ( 1+3k\right ) & 1 & 0 & 0\\ 1+k & 36 & 1 & 0\\ 0 & \left ( 1+3k\right ) & 1 & 0\\ 0 & 1+k & 36 & 1 \end{bmatrix}$ Hence

$\Delta _{1}=1+3k>0$ which means

$k>\frac{-1}{3}$ . And

$\Delta _{2}=107k+35>0$ hence

$k>\frac{-35}{107}=-0.3271$ and

$\Delta _{3}=-9k^{2}+101k+34>0$ . Hence

$-0.3271<k<11.5493$ .

$\Delta _{4}$ is the same as

$\Delta _{3}$ hence no new information is obtained from it. Therefore, from

$K_{1}$ we ﬁnd the following

$\left \{ k>-0.333,k>-0.3271,-0.3271<k<11.5493\right \}$ Looking at

$K_{2}$

$\begin{bmatrix} \left ( 1+3k\right ) & 1 & 0 & 0\\ 2+k & 12 & 1 & 0\\ 0 & \left ( 1+3k\right ) & 1 & 0\\ 0 & 2+k & 12 & 1 \end{bmatrix}$ Hence

$\Delta _{1}=1+3k>0$ which means

$k>\frac{-1}{3}$ . And

$\Delta _{2}=35k+10>0$ hence

$k>-0.285714$ and

$\Delta _{3}=-9k^{2}+29k+9>0$ . Hence

$-0.285<k<3.50734$ .

$\Delta _{4}$ is the same as

$\Delta _{3}$ hence no new information is obtained from it. Therefore, from

$K_{2}$ we ﬁnd the following

$\left \{ k>-0.333,k>-0.285714,-0.285<k<3.50734\right \}$ Looking at

$K_{3}$

$\begin{bmatrix} \left ( 1+3k\right ) & 1 & 0 & 0\\ 2+k & 12 & 1 & 0\\ 0 & \left ( 1+3k\right ) & 1 & 0\\ 0 & 2+k & 12 & 1 \end{bmatrix}$ This is the same as

$K_{2}.$ Finally, looking at

$K_{4}$

$\begin{bmatrix} \left ( 1+3k\right ) & 1 & 0 & 0\\ 2+k & 12 & 1 & 0\\ 0 & \left ( 1+3k\right ) & 1 & 0\\ 0 & 2+k & 12 & 1 \end{bmatrix}$ This is the same as

$K_{1}.$ We now have found all the range for

$k$ from each polynomial. We put them together in order to determine the largest

$k$ allowed

$\begin{align*} K_{1} & \Rightarrow \left \{ k>-0.333,k>-0.3271,-0.3271<k<11.5493\right \} \\ K_{2} & \Rightarrow \left \{ k>-0.333,k>-0.285714,-0.285<k<3.50734\right \} \end{align*}$

We see the range of positive

$k$ values for robust stability is

$0<k<3.50734$ . Therefore

$k_{\max }=3.50734$

2.7.4 problem 3

$\begin{align*} G\left ( s\right ) & =\frac{s+1+\rho \left [ -1,1\right ] }{s^{2}+\left ( 2+\rho \left [ -1,1\right ] \right ) s+3}\\ H\left ( s\right ) & =\frac{s+1}{s-1} \end{align*}$

The closed loop transfer function is

$\begin{align*} G_{cl}\left ( s\right ) & =\frac{H\left ( s\right ) G\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) }\\ & =\frac{H\left ( s\right ) G\left ( s\right ) }{1+\left ( \frac{s+1}{s-1}\right ) \left ( \frac{s+1+\rho \left [ -1,1\right ] }{s^{2}+\left ( 2+\rho \left [ -1,1\right ] \right ) s+3}\right ) } \end{align*}$

The denominator of the

$G_{cl}\left ( s\right )$ becomes

$\begin{align*} \Delta & =\left ( s-1\right ) \left ( s^{2}+\left ( 2+\rho \left [ -1,1\right ] \right ) s+3\right ) +\left ( s+1\right ) \left ( s+1+\rho \left [ -1,1\right ] \right ) \\ & =\left ( s^{3}+\left ( 2+\rho \left [ -1,1\right ] \right ) s^{2}+3s\right ) -\left ( s^{2}+\left ( 2+\rho \left [ -1,1\right ] \right ) s+3\right ) +\left ( s^{2}+s+\rho \left [ -1,1\right ] s\right ) +\left ( s+1+\rho \left [ -1,1\right ] \right ) \\ & =s^{3}+\left ( 2+\rho \left [ -1,1\right ] \right ) s^{2}+3s-s^{2}-\left ( 2+\rho \left [ -1,1\right ] \right ) s-3+s^{2}+s+\rho \left [ -1,1\right ] s+s+1+\rho \left [ -1,1\right ] \\ & =s^{3}+\left ( 2+\rho \left [ -1,1\right ] \right ) s^{2}+3s-\rho \left [ -1,1\right ] s+\rho \left [ -1,1\right ] s-2+\rho \left [ -1,1\right ] \end{align*}$

But

$\rho \left [ -1,1\right ] s-\rho \left [ -1,1\right ] s=\rho \left [ -2,2\right ] s$ , hence

$\Delta =s^{3}+s^{2}\left ( 2+\rho \left [ -1,1\right ] \right ) +s\left ( 3+\rho \left [ -2,2\right ] s\right ) -2+\rho \left [ -1,1\right ]$ We need to ﬁrst check that the nominal polynomial is stable before checking for robust stability. When

$\rho =0$ the denominator becomes

$\Delta =s^{3}+2s^{2}+3s-2$ Since there is a sign change, then the nominal polynomial is not stable. This means the closed loop is not stable. Therefore no need to do robust stability. There is no

$\rho _{\max }$ .

2.7.5 problem 4

$p_{\lambda }=\left ( 1-\lambda \right ) p_{0}\left ( s\right ) +\lambda p_{1}\left ( s\right )$ For

$0\leq \lambda \leq 1$ . We are given that

$p_{\lambda =0}=p_{0}\left ( s\right )$ is stable and

$p_{\lambda =1}=p_{1}\left ( s\right )$ is stable. In other-words,

$p_{0}\left ( s\right ) ,p_{1}\left ( s\right )$ are both stable polynomials. For any value of

$0<\lambda <1$ we then have a sum of two stable polynomials, each being multiplied by a constant.

$\begin{equation} p_{\lambda }\left ( s\right ) =\left ( 1-\lambda \right ) p_{0}\left ( s\right ) +\lambda p_{1}\left ( s\right ) \tag{1} \end{equation}$ Let the zeros of

$p_{0}$ be

$r_{i},i=1\cdots n$ where

$n$ is the order of

$p_{0}\left ( s\right )$ and let the zeros of

$p_{1}\left ( s\right )$ be

$z_{j},j=1\cdots m$ where

$m~$ is the order of

$p_{1}\left ( s\right )$ . Using the fundamental theorem of algebra, we can write

$\begin{align*} p_{0}\left ( s\right ) & =\left ( s-r_{1}\right ) \left ( s-r_{2}\right ) \cdots \left ( s-r_{n}\right ) \\ p_{1}\left ( s\right ) & =\left ( s-z_{1}\right ) \left ( s-z_{2}\right ) \cdots \left ( s-z_{m}\right ) \end{align*}$

Equation (1) becomes

$\begin{equation} p_{\lambda }\left ( s\right ) =\left ( 1-\lambda \right ) \left ( s-r_{1}\right ) \left ( s-r_{2}\right ) \cdots \left ( s-r_{n}\right ) +\lambda \left ( s-z_{1}\right ) \left ( s-z_{2}\right ) \cdots \left ( s-z_{m}\right ) \tag{2} \end{equation}$ Now we will proof that

$p_{\lambda }\left ( s\right )$ can only have negative zeros. Proof is by contradiction. Assume that

$p_{\lambda }\left ( s\right )$ have a positive root, say

$\xi >0$ , then this root when substituted in (2) will result in zero by deﬁnition

$\begin{equation} p_{\lambda }\left ( \xi \right ) =\overset{\Delta _{1}}{\overbrace{\left ( 1-\lambda \right ) \left ( \xi -r_{1}\right ) \left ( \xi -r_{2}\right ) \cdots \left ( \xi -r_{n}\right ) }}+\overset{\Delta _{2}}{\overbrace{\lambda \left ( \xi -z_{1}\right ) \left ( \xi -z_{2}\right ) \cdots \left ( \xi -z_{m}\right ) }}\tag{3} \end{equation}$ Each term

$\left ( \xi -r_{i}\right )$ is therefore positive quantity, since each

$r_{i}$ is negative since

$p_{0}\left ( s\right )$ is stable. We also have

$\left ( 1-\lambda \right ) >0$ . Therefore the product shown as

$\Delta _{1}$ in (3) it a positive quantity.

Similarly, each term

$\left ( \xi -z_{j}\right )$ is positive quantity, since each

$z_{j}$ is negative since

$p_{1}\left ( s\right )$ is stable. We also have

$\lambda >0$ . Therefore the product shown as

$\Delta _{2}$ in (3) it a positive quantity.

This shows that (3) is the sum of two positive quantities. Hence the sum can not be zero. This contradicts our assumptions

$p_{\lambda }\left ( \xi \right ) =0$ due to assuming

$\xi >0$ . Therefore zeros of

$p_{\lambda }\left ( s\right )$ can not be positive.

Similarly, we show that

$\xi =0$ is also not possible root of

$p_{\lambda }(s)$ . Assume

$\xi =0$ is a root, then this leads to contradiction as above, since we will have positive quantities added to zero, which is not possible.

Therefore, the only possible choice left is for all zeros of

$p_{\lambda }(s)$ to be negative.

Hence

$p_{\lambda }\left ( s\right )$ is be stable for any

$0<\lambda <1$ . Therefore the engineer was correct. QED.

2.7.6 problem 5

The Lyapunov equation is

$\begin{align*} A^{T}P+PA & =-Q\\\begin{bmatrix} -2 & 0 & 0 & 0\\ 4 & -5 & 0 & 0\\ -3 & 2 & -1 & 0\\ 1 & -1 & 2 & -4 \end{bmatrix}\begin{bmatrix} p_{11} & p_{12} & p_{13} & p_{14}\\ p_{12} & p_{22} & p_{23} & p_{24}\\ p_{13} & p_{23} & p_{33} & p_{34}\\ p_{14} & p_{24} & p_{34} & p_{44}\end{bmatrix} +\begin{bmatrix} p_{11} & p_{12} & p_{13} & p_{14}\\ p_{12} & p_{22} & p_{23} & p_{24}\\ p_{13} & p_{23} & p_{33} & p_{34}\\ p_{14} & p_{24} & p_{34} & p_{44}\end{bmatrix}\begin{bmatrix} -2 & 4 & -3 & 1\\ 0 & -5 & 2 & -1\\ 0 & 0 & -1 & 2\\ 0 & 0 & 0 & -4 \end{bmatrix} & =-\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \end{align*}$

Hence

$\begin{multline} \begin{bmatrix} -4p_{11} & 4p_{11}-7p_{12} & 2p_{12}-3p_{11}-3p_{13} & p_{11}-p_{12}+2p_{13}-6p_{14}\\ 4p_{11}-7p_{12} & 8p_{12}-10p_{22} & 4p_{13}-3p_{12}+2p_{22}-6p_{23} & p_{12}-p_{22}+4p_{14}+2p_{23}-9p_{24}\\ 2p_{12}-3p_{11}-3p_{13} & 4p_{13}-3p_{12}+2p_{22}-6p_{23} & 4p_{23}-6p_{13}-2p_{33} & p_{13}-3p_{14}-p_{23}+2p_{24}+2p_{33}-5p_{34}\\ p_{11}-p_{12}+2p_{13}-6p_{14} & p_{12}-p_{22}+4p_{14}+2p_{23}-9p_{24} & p_{13}-3p_{14}-p_{23}+2p_{24}+2p_{33}-5p_{34} & 2p_{14}-2p_{24}+4p_{34}-8p_{44}\end{bmatrix} = \\ \begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{bmatrix} \end{multline}$ There are

$10$ unknowns. They are

$\allowbreak p_{11},p_{12},p_{13},p_{14},p_{22},p_{23},p_{24},p_{33},p_{34}$ . The 10 equations to solve are from the upper triangle above

$\begin{align} -4p_{11} & =-1\tag{1}\\ 4p_{11}-7p_{12} & =0\tag{2}\\ 2p_{12}-3p_{11}-3p_{13} & =0\tag{3}\\ p_{11}-p_{12}+2p_{13}-6p_{14} & =0\tag{4}\\ 8p_{12}-10p_{22} & =-1\tag{5}\\ 4p_{13}-3p_{12}+2p_{22}-6p_{23} & =0\tag{6}\\ p_{12}-p_{22}+4p_{14}+2p_{23}-9p_{24} & =0\tag{7}\\ 4p_{23}-6p_{13}-2p_{33} & =-1\tag{8}\\ p_{13}-3p_{14}-p_{23}+2p_{24}+2p_{33}-5p_{34} & =0\tag{9}\\ 2p_{14}-2p_{24}+4p_{34}-8p_{44} & =-1\tag{10} \end{align}$

From (1)

$p_{11}=\frac{1}{4}$ , substituting in (2) gives

$p_{12}=\frac{4}{7}\left ( \frac{1}{4}\right ) =$

$\frac{1}{7}$ , substituting these in (3) gives

$3p_{13}=2p_{12}-3p_{11}$ or

$p_{13}=\frac{2p_{12}-3p_{11}}{3}=\frac{2\left ( \frac{1}{7}\right ) -3\left ( \frac{1}{4}\right ) }{3}=-\frac{13}{84}$ . Substituting these in (4) gives

$6p_{14}=p_{11}-p_{12}+2p_{13}$ or

$p_{14}=\frac{p_{11}-p_{12}+2p_{13}}{6}=\frac{\frac{1}{4}-\frac{1}{7}-2\left ( \frac{13}{84}\right ) }{6}=-\frac{17}{504}$ .

From (5) we ﬁnd

$8p_{12}=10p_{22}-1$ hence

$p_{22}=\frac{8p_{12}+1}{10}=\frac{8\left ( \frac{1}{7}\right ) +1}{10}=\frac{3}{14}$ . From (6)

$4p_{13}-3p_{12}+2p_{22}=6p_{23}$ . hence

$p_{23}=\frac{4p_{13}-3p_{12}+2p_{22}}{6}=\frac{4\left ( -\frac{13}{84}\right ) -3\left ( \frac{1}{7}\right ) +2\left ( \allowbreak \frac{3}{14}\right ) }{6}=-\frac{13}{126}$ . From (7)

$p_{12}-p_{22}+4p_{14}+2p_{23}=9p_{24}$ , hence

$p_{24}=\frac{p_{12}-p_{22}+4p_{14}+2p_{23}}{9}=\frac{\frac{1}{7}-\allowbreak \frac{3}{14}+4\left ( -\frac{17}{504}\right ) +2\left ( -\frac{13}{126}\right ) }{9}=-\frac{26}{567}$ . And from (8),

$4p_{23}-6p_{13}-2p_{33}=-1$ , hence

$p_{33}=\frac{4p_{23}-6p_{13}+1}{2}=\frac{4\left ( -\frac{13}{126}\right ) -6\left ( -\frac{13}{84}\right ) +1}{2}=\frac{191}{252}$ . From (9),

$p_{13}-3p_{14}-p_{23}+2p_{24}+2p_{33}=5p_{34}$ , hence

$p_{34}=\frac{p_{13}-3p_{14}-p_{23}+2p_{24}+2p_{33}}{5}=\frac{-\frac{13}{84}-3\left ( -\frac{17}{504}\right ) -\left ( -\frac{13}{126}\right ) +2\left ( -\frac{26}{567}\right ) +2\left ( \allowbreak \frac{191}{252}\right ) }{5}=\frac{191}{648}$ , and ﬁnally from (10)

$2p_{14}-2p_{24}+4p_{34}=8p_{44}-1$ , hence

$p_{44}=\frac{2p_{14}-2p_{24}+4p_{34}+1}{8}=\frac{2\left ( -\frac{17}{504}\right ) -2\left ( -\frac{26}{567}\right ) +4\left ( \frac{191}{648}\right ) +1}{8}=\frac{4997}{18144}$ . Therefore the solution is

$\begin{align*} P & =\begin{bmatrix} p_{11} & p_{12} & p_{13} & p_{14}\\ p_{12} & p_{22} & p_{23} & p_{24}\\ p_{13} & p_{23} & p_{33} & p_{34}\\ p_{14} & p_{24} & p_{34} & p_{44}\end{bmatrix} =\begin{bmatrix} \frac{1}{4} & \frac{1}{7} & -\frac{13}{84} & -\frac{17}{504}\\ \frac{1}{7} & \frac{3}{14} & -\frac{13}{126} & -\frac{26}{567}\\ -\frac{13}{84} & -\frac{13}{126} & \frac{191}{252} & \frac{191}{648}\\ -\frac{17}{504} & -\frac{26}{567} & \frac{191}{648} & \frac{4997}{18144}\end{bmatrix} \\ & =\begin{bmatrix} 0.25 & 0.14286 & -0.15476 & -0.03373\\ 0.14286 & 0.21429 & -0.10317 & -4.5855\times 10^{-2}\\ -0.15476 & -0.10317 & 0.75794 & 0.29475\\ -0.03373 & -4.5855\times 10^{-2} & 0.29475 & 0.27541 \end{bmatrix} \end{align*}$

The associated

$V\left ( x\left ( t\right ) \right )$ function is

$\begin{align*} V\left ( x\left ( t\right ) \right ) & =X^{T}PX\\ & =\begin{bmatrix} x_{1} & x_{2} & x_{3} & x_{4}\end{bmatrix}\begin{bmatrix} \frac{1}{4} & \frac{1}{7} & -\frac{13}{84} & -\frac{17}{504}\\ \frac{1}{7} & \frac{3}{14} & -\frac{13}{126} & -\frac{26}{567}\\ -\frac{13}{84} & -\frac{13}{126} & \frac{191}{252} & \frac{191}{648}\\ -\frac{17}{504} & -\frac{26}{567} & \frac{191}{648} & \frac{4997}{18144}\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{bmatrix} \\ & =\frac{1}{4}x_{1}^{2}+\frac{2}{7}x_{1}x_{2}-\frac{13}{42}x_{1}x_{3}-\frac{17}{252}x_{1}x_{4}+\frac{3}{14}x_{2}^{2}-\frac{13}{63}x_{2}x_{3}-\frac{52}{567}x_{2}x_{4}+\frac{191}{252}x_{3}^{2}+\frac{191}{324}x_{3}x_{4}+\frac{4997}{18144}x_{4}^{2} \end{align*}$

$V\left ( x\left ( t\right ) \right ) =0.25x_{1}^{2}+0.286x_{1}x_{2}-0.309x_{1}x_{3}-0.067x_{1}x_{4}+0.214x_{2}^{2}-0.206x_{2}x_{3}-0.091x_{2}x_{4}+0.758x_{3}^{2}+0.589x_{3}x_{4}+0.275x_{4}^{2}$

Part(b)

EDU>> A=[-2 4 -3 1;0 -5 2 -1;0 0 -1 2;0 0 0 -4]

    -2     4    -3     1
     0    -5     2    -1
     0     0    -1     2
     0     0     0    -4

EDU>> syms x1 x2 x3 x4;
EDU>> P=lyap(A',eye(4))

         0.25      0.14286     -0.15476     -0.03373
      0.14286      0.21429     -0.10317    -0.045855
     -0.15476     -0.10317      0.75794      0.29475
     -0.03373    -0.045855      0.29475      0.27541

EDU>> x=[x1;x2;x3;x4];
EDU>> V=x.'*P*x;
EDU>> vpa(expand(V),3)

0.25*x1^2 + 0.286*x1*x2 - 0.31*x1*x3 - 0.0675*x1*x4 + 0.214*x2^2
    - 0.206*x2*x3 - 0.0917*x2*x4 + 0.758*x3^2 + 0.59*x3*x4 + 0.275*x4^2

2.7.7 problem 6

Part (a)

Assuming the problem is asking for the 2-norm. This is deﬁned as positive square root of the largest eigenvalue of

$AA^{T}$ . Therefore, we ﬁrst ﬁnd

$AA^{T}$ , then ﬁnd the eigenvalues, then pick the largest one in absolute terms, then take the square root.

$\begin{equation*} AA^{T} = \begin{bmatrix} 0.95 & 0.48 & 0.45\\ 0.23 & 0.89 & 0.01\\ 0.6 & 0.76 & 0.82 \end{bmatrix}\begin{bmatrix} 0.95 & 0.23 & 0.6\\ 0.48 & 0.89 & 0.76\\ 0.45 & 0.01 & 0.82 \end{bmatrix} = \begin{bmatrix} 1.3354 & 0.6502 & 1.3038\\ 0.6502 & 0.8451 & 0.8226\\ 1.3038 & 0.8226 & 1.61 \end{bmatrix} \end{equation*}$ Now we ﬁnd the eigenvalues.

$\begin{align*} p\left ( \lambda \right ) & =\left \vert \lambda I-AA^{T}\right \vert =\begin{vmatrix} \lambda -1.3354 & -0.6502 & -1.3038\\ -0.6502 & \lambda -0.8451 & -0.8226\\ -1.3038 & -0.8226 & \lambda -1.61 \end{vmatrix} \\ & =\lambda ^{3}-3.7905\lambda ^{2}+1.8398\lambda -0.1908 \end{align*}$

The roots of this polynomials are

$\lambda =0.14584,\lambda =0.40366,\lambda =3.241.$ Hence the largest eigenvalue is

$\lambda =3.241$ . Therefore the 2-norm is

$\boxed{\sqrt{3.241}=1.8003}$

Part(b)

EDU>> A=[0.95 0.48 0.45;0.23 0.89 0.01;0.6 0.76 0.82];
EDU>> norm(A,2)
ans =
1.8003

2.7.8 problem 7

$A=\begin{bmatrix} -1 & 2 & 1 & 3 & 7\\ 0 & -2 & -1 & 5 & -3\\ 0 & 0 & -3 & 4 & 0\\ 0 & 0 & 0 & -4 & 1\\ 0 & 0 & 0 & 0 & -5 \end{bmatrix}$ The robustness bound is given by

$\left \Vert \Delta A\right \Vert _{2}$ deﬁned as

$\boxed{ \left \Vert \Delta A\right \Vert _{2}=\frac{\lambda _{\min }\left [ Q\right ] }{2\lambda _{\max }\left [ P\right ] } }$ We ﬁrst need to solve the Lyapunov equation to ﬁnd

$P$ .

$A^{T}P+PA=-Q$ Using Matlab, and use

$Q=I_{5}$ gives the solution

$P$

EDU>> A=[-1 2 1 3 7;0 -2 -1 5 -3;0 0 -3 4 0;0 0 0 -4 1;0 0 0 0 -5]

A =

    -1     2     1     3     7
     0    -2    -1     5    -3
     0     0    -3     4     0
     0     0     0    -4     1
     0     0     0     0    -5

EDU>> P=lyap(A',eye(5))

P =

          0.5      0.33333     0.041667      0.66667      0.52778
      0.33333      0.58333    -0.033333      0.85278      0.35595
     0.041667    -0.033333      0.19167     0.076984      0.08006
      0.66667      0.85278     0.076984        1.768      0.83996
      0.52778      0.35595      0.08006      0.83996      0.79331