Date due and handed in April 6,2010
Write state variable description of the following 2 systems. For what values of \(k\) will the system be stable?
This system has 2 integrators, hence it is of order 2. Hence we need 2 state variables. Assign a state variable as the output of each integrator
Hence\begin{align*} x_{1}^{\prime } & =-3x_{1}+u_{1}+x_{2}\\ x_{2}^{\prime } & =x_{2}+kx_{1}+u_{2} \end{align*}
and \(y=x_{1},\)Hence\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{pmatrix} & =\overset{A}{\overbrace{\begin{pmatrix} -3 & 1\\ k & 1 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\overset{B}{\overbrace{\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} }}\begin{pmatrix} u_{1}\\ u_{2}\end{pmatrix} \\ y & =\overset{C}{\overbrace{\begin{pmatrix} 1 & 0 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} \end{align*}
To find what values of \(k\) the system is stable, the eigenvalues of the \(A\) matrix are found and the \(K\) range which makes these values negative is the range of value needed.\[ \left \vert A-\lambda I\right \vert =\left \vert \begin{pmatrix} -3-\lambda & 1\\ k & 1-\lambda \end{pmatrix} \right \vert =\left ( 1-\lambda \right ) \left ( -3-\lambda \right ) -k \] Hence the characteristic equation is\[ \lambda ^{2}+2\lambda -k-3=0 \] and the roots are \begin{align*} \lambda _{1} & =-1+\sqrt{k+4}\\ \lambda _{2} & =-1-\sqrt{k+4} \end{align*}
consider \(\lambda _{1}\). For this root to be stable, then \(\sqrt{k+4}<1\) or \(k<-3\)
consider \(\lambda _{2}\). This root is stable for any value of \(k\) since when \(k+4<0\) then it is stable since real part is already negative, and when \(k+4>0\) then it is stable also.
Hence we conclude that the system is stable for \(k<-3\)
To find the ODE:
From \(x_{1}^{\prime }=-3x_{1}+u_{1}+x_{2}\) we obtain \(x_{1}^{\prime \prime }=-3x_{1}^{\prime }+u_{1}^{\prime }+x_{2}^{\prime }\). Substitute the value of \(x_{2}^{\prime }\) from above, we obtain \(x_{1}^{\prime \prime }=-3x_{1}^{\prime }+u_{1}^{\prime }+x_{2}+kx_{1}+u_{2}\), but \(x_{2}=\).\(x_{1}^{\prime }+3x_{1}-u_{1}\), hence\begin{align*} x_{1}^{\prime \prime } & =-3x_{1}^{\prime }+u_{1}^{\prime }+x_{1}^{\prime }+3x_{1}-u_{1}+kx_{1}+u_{2}\\ & =-2x_{1}^{\prime }+x_{1}\left ( 3+k\right ) -u_{1}+u_{1}^{\prime }+u_{2} \end{align*}
since \(x_{1}=y\) we obtain\[ y^{\prime \prime }=-2y^{\prime }+y\left ( 3+k\right ) -u_{1}+u_{1}^{\prime }+u_{2}\]
This system has 2 integrators, hence it is of order 2. Hence we need 2 state variables. Assign a state variable as the output of each integrator
Hence\begin{align*} x_{1}^{\prime } & =-\frac{3}{4}x_{1}+u_{1}+x_{2}\\ x_{2}^{\prime } & =-\frac{3}{4}x_{2}+kx_{1} \end{align*}
and \(y=x_{1},\)Hence\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{pmatrix} & =\overset{A}{\overbrace{\begin{pmatrix} -\frac{3}{4} & 1\\ k & -\frac{3}{4}\end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\overset{B}{\overbrace{\begin{pmatrix} 1\\ 0 \end{pmatrix} }}u_{1}\\ y & =\overset{C}{\overbrace{\begin{pmatrix} 1 & 0 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} \end{align*}
To find what values of \(k\) the system is stable, the eigenvalues of the \(A\) matrix are found and the \(K\) range which makes these values negative is the range of value needed.\[ \left \vert A-\lambda I\right \vert =\left \vert \begin{pmatrix} -\frac{3}{4}-\lambda & 1\\ k & -\frac{3}{4}-\lambda \end{pmatrix} \right \vert =\left ( -\frac{3}{4}-\lambda \right ) \left ( -\frac{3}{4}-\lambda \right ) -k \] Hence the characteristic equation is\[ \lambda ^{2}+\frac{3}{2}\lambda -k+\frac{9}{16}=0 \] and the roots are \begin{align*} \lambda _{1} & =-\frac{3}{4}-\sqrt{k}\\ \lambda _{2} & =-\frac{3}{4}+\sqrt{k} \end{align*}
For \(\lambda _{1}\), all values of \(k\) will result in stable root. For \(\lambda _{2}\), \(\sqrt{k}<\frac{3}{4}\) or \(k<\frac{9}{16}\) or \(k<0.562\,5\)
Hence \(k<\frac{9}{16}\) or \(k<0.562\,5\) is the range of \(k\) for stability.
To find the ODE: From \(x_{1}^{\prime }=-\frac{3}{4}x_{1}+u_{1}+x_{2}\), we obtain \(x_{1}^{\prime \prime }=-\frac{3}{4}x_{1}^{\prime }+u_{1}^{\prime }+x_{2}^{\prime }\) Substitute the value of \(x_{2}^{\prime }\) from above, we obtain \(x_{1}^{\prime \prime }=-\frac{3}{4}x_{1}^{\prime }+u_{1}^{\prime }-\frac{3}{4}x_{2}+kx_{1}\) but \(x_{2}=x_{1}^{\prime }+\frac{3}{4}x_{1}-u_{1}\), hence\begin{align*} x_{1}^{\prime \prime } & =-\frac{3}{4}x_{1}^{\prime }+u_{1}^{\prime }-\frac{3}{4}\left ( x_{1}^{\prime }+\frac{3}{4}x_{1}-u_{1}\right ) +kx_{1}\\ & =-\frac{3}{4}x_{1}^{\prime }+u_{1}^{\prime }-\frac{3}{4}x_{1}^{\prime }-\frac{9}{16}x_{1}+\frac{3}{4}u_{1}+kx_{1}\\ & =-\frac{3}{2}x_{1}^{\prime }+x_{1}\left ( k-\frac{9}{16}\right ) +u_{1}^{\prime }+\frac{3}{4}u_{1} \end{align*}
since \(x_{1}=y\) we obtain\[ y^{\prime \prime }+\frac{3}{2}y^{\prime }-y\left ( k-\frac{9}{16}\right ) =u_{1}^{\prime }+\frac{3}{4}u_{1}\]
\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{pmatrix} & =\overset{A}{\overbrace{\begin{pmatrix} 0 & 1\\ -2 & -3 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\overset{B}{\overbrace{\begin{pmatrix} 0\\ 1 \end{pmatrix} }}u_{1}\\ y & =\overset{C}{\overbrace{\begin{pmatrix} c_{1} & c_{2}\end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\left [ d\right ] \ u_{1} \end{align*}
\[ \left ( j\omega I-A\right ) =j\omega \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} -\begin{pmatrix} 0 & 1\\ -2 & -3 \end{pmatrix} =\begin{pmatrix} j\omega & -1\\ 2 & j\omega +3 \end{pmatrix} \] Hence\begin{align*} \left ( j\omega I-A\right ) ^{-1} & =\begin{pmatrix} j\omega & -1\\ 2 & j\omega +3 \end{pmatrix} ^{-1}=\frac{1}{\left ( j\omega \right ) \left ( j\omega +3\right ) +2}\begin{pmatrix} j\omega +3 & 1\\ -2 & j\omega \end{pmatrix} \\ & =\frac{1}{-\omega ^{2}+3j\omega +2}\begin{pmatrix} j\omega +3 & 1\\ -2 & j\omega \end{pmatrix} \end{align*}
To find \(e^{At}\) use the eigenvalue method.\[ \left \vert A-\lambda I\right \vert =\begin{vmatrix} -\lambda & 1\\ -2 & -3-\lambda \end{vmatrix} =\lambda ^{2}+3\lambda +2 \] Hence the roots of \(\lambda ^{2}+3\lambda +2=0\) are found to be \(\lambda _{1}=-1\) and \(\lambda _{2}=-2\). Hence the 2 equations to solve are\begin{align*} e^{\lambda _{1}t} & =\beta _{0}+\beta _{1}\lambda _{1}\\ e^{\lambda _{2}t} & =\beta _{0}+\beta _{1}\lambda _{2} \end{align*}
or\begin{align*} e^{-t} & =\beta _{0}-\beta _{1}\\ e^{-2t} & =\beta _{0}-2\beta _{1} \end{align*}
Solving we obtain\begin{align*} \beta _{0} & =2e^{-t}-e^{-2t}\\ \beta _{1} & =e^{-t}-e^{-2t} \end{align*}
Hence\begin{align*} e^{At} & =\beta _{0}I+\beta _{1}A\\ & =\left ( 2e^{-t}-e^{-2t}\right ) \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} +\left ( e^{-t}-e^{-2t}\right ) \begin{pmatrix} 0 & 1\\ -2 & -3 \end{pmatrix} \\ & = \end{align*}
Hence\[ e^{At}=\begin{pmatrix} \left ( 2e^{-t}-e^{-2t}\right ) & \left ( e^{-t}-e^{-2t}\right ) \\ -2\left ( e^{-t}-e^{-2t}\right ) & -e^{-t}+2e^{-2t}\end{pmatrix} \]
First need to find \(H\left ( j\omega \right ) \). We start from the system equations\begin{align} x^{\prime } & =Ax+Bu\tag{1}\\ y & =Cx+Du\tag{2} \end{align}
Let \(u=e^{j\omega t}\), hence the state particular solution is \begin{equation} x_{p}\left ( t\right ) =X\left ( j\omega \right ) e^{j\omega t}\tag{3} \end{equation} And \begin{equation} y_{p}\left ( t\right ) =H\left ( j\omega \right ) e^{j\omega t}\tag{4} \end{equation} From (1) and (3) , we obtain\begin{align} j\omega X\left ( j\omega \right ) e^{j\omega t} & =AX\left ( j\omega \right ) e^{j\omega t}+Be^{j\omega t}\nonumber \\ j\omega X\left ( j\omega \right ) & =AX\left ( j\omega \right ) +B\nonumber \\ \left ( j\omega I-A\right ) X\left ( j\omega \right ) & =B\nonumber \\ X\left ( j\omega \right ) & =\left ( j\omega I-A\right ) ^{-1}B\tag{5} \end{align}
and from (2) and (4) we obtain\begin{align*} H\left ( j\omega \right ) e^{j\omega t} & =CX\left ( j\omega \right ) e^{j\omega t}+De^{j\omega t}\\ H\left ( j\omega \right ) & =CX\left ( j\omega \right ) +D \end{align*}
Substitute (5) into the above\[ H\left ( j\omega \right ) =C\left ( j\omega I-A\right ) ^{-1}B+D \] From part(a) we found \(\left ( j\omega I-A\right ) ^{-1}\), hence the above becomes\begin{align*} H\left ( j\omega \right ) & =\begin{pmatrix} c_{1} & c_{2}\end{pmatrix} \frac{1}{-\omega ^{2}+3j\omega +2}\begin{pmatrix} j\omega +3 & 1\\ -2 & j\omega \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix} +d\\ & =\frac{1}{-\omega ^{2}+3j\omega +2}\begin{pmatrix} \left ( j\omega +3\right ) c_{1}-2c_{2} & c_{1}+c_{2}j\omega \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix} +d\\ & =\frac{\left ( c_{1}+c_{2}j\omega \right ) }{-\omega ^{2}+3j\omega +2}+d\\ & =\frac{\left ( c_{1}+c_{2}j\omega \right ) +d\left ( -\omega ^{2}+3j\omega +2\right ) }{-\omega ^{2}+3j\omega +2}\\ & =\frac{\left ( c_{1}+2d-d\omega ^{2}\right ) +j\left ( c_{2}\omega +3d\omega \right ) }{\left ( -\omega ^{2}+2\right ) +3j\omega } \end{align*}
Hence\begin{align*} \left \vert H\left ( j\omega \right ) \right \vert ^{2} & =\frac{\left ( c_{1}+2d-d\omega ^{2}\right ) ^{2}+\left ( c_{2}\omega +3d\omega \right ) ^{2}}{\left ( -\omega ^{2}+2\right ) ^{2}+9\omega ^{2}}\\ & =\frac{d^{2}\omega ^{4}+5d^{2}\omega ^{2}+4d^{2}-2d\omega ^{2}c_{1}+6d\omega ^{2}c_{2}+4dc_{1}+\omega ^{2}c_{2}^{2}+c_{1}^{2}}{\omega ^{4}+5\omega ^{2}+4} \end{align*}
Now, from diagram, at \(\omega =0\,\) we have \(\left \vert H\left ( j\omega \right ) \right \vert ^{2}=1\), hence\begin{equation} 1=d^{2}+dc_{1}+\frac{1}{4}c_{1}^{2}\tag{6} \end{equation} And at \(\omega =1\,\) we have \(\left \vert H\left ( j\omega \right ) \right \vert ^{2}=0\) hence\[ 0=\frac{10d^{2}+2dc_{1}+6dc_{2}+c_{2}^{2}+c_{1}^{2}}{10}\] Or\begin{equation} 0=10d^{2}+2dc_{1}+6dc_{2}+c_{2}^{2}+c_{1}^{2}\tag{7} \end{equation} And at \(\omega =-1\,\) we have \(\left \vert H\left ( j\omega \right ) \right \vert ^{2}=0\) but this will not add new equation. So need to look at the limit as \(\omega \rightarrow \infty \) \[ \left \vert H\left ( j\omega \right ) \right \vert ^{2}=\frac{d^{2}+\frac{5d^{2}}{\omega ^{2}}+\frac{4d^{2}}{\omega ^{4}}-\frac{2dc_{1}}{\omega ^{2}}+\frac{6dc_{2}}{\omega ^{2}}+\frac{4dc_{1}}{\omega ^{4}}+\frac{c_{2}^{2}}{\omega ^{2}}+\frac{c_{1}^{2}}{\omega ^{4}}}{1+\frac{5}{\omega ^{2}}+\frac{4}{\omega ^{4}}}\] Hence we see that as \(\omega \rightarrow \infty \), \(\left \vert H\left ( j\omega \right ) \right \vert ^{2}\rightarrow d^{2}\), hence \(d=0\) since \(\left \vert H\left ( j\omega \right ) \right \vert \rightarrow 0\) in the limit. So now we know \(d\), we have 2 equations and 2 unknowns to solve for from (6) and (7). Re write (6) and (7) again by setting \(d=0\) we obtain\begin{equation} 1=\frac{1}{4}c_{1}^{2}\tag{6} \end{equation} \begin{equation} 0=c_{2}^{2}+c_{1}^{2}\tag{7} \end{equation} Hence \(c_{1}=2\) and \(c_{2}=2j\) therefore, the system now looks like\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{pmatrix} & =\overset{A}{\overbrace{\begin{pmatrix} 0 & 1\\ -2 & -3 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\overset{B}{\overbrace{\begin{pmatrix} 0\\ 1 \end{pmatrix} }}u_{1}\\ y & =\overset{C}{\overbrace{\begin{pmatrix} 2 & 2j \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} \end{align*}
To find \(h\left ( t\right ) \), Let the input be \(\delta \left ( t\right ) \), and find \(y\left ( t\right ) \). From the system equation\[ y_{p}\left ( t\right ) ={\displaystyle \int \limits _{t_{0}}^{t}} Ce^{A\left ( t-\tau \right ) }Bu\left ( \tau \right ) d\tau \] Let \(u\left ( \tau \right ) =\delta \left ( t\right ) \), so the above becomes\begin{align*} h\left ( t\right ) & ={\displaystyle \int \limits _{t_{0}}^{t}} Ce^{A\left ( t-\tau \right ) }B\delta \left ( \tau \right ) d\tau \\ & =Ce^{A\left ( t\right ) }B\qquad t\geq 0 \end{align*}
But we found \(e^{A\left ( t\right ) }\) in part (b), hence\begin{align*} h\left ( t\right ) & =\begin{pmatrix} 2 & 2\ j \end{pmatrix}\begin{pmatrix} \left ( 2e^{-t}-e^{-2t}\right ) & \left ( e^{-t}-e^{-2t}\right ) \\ -2\left ( e^{-t}-e^{-2t}\right ) & -e^{-t}+2e^{-2t}\end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix} \\ & =2e^{-t}-2e^{-2t}-2j\left ( e^{-t}-2e^{-2t}\right ) \end{align*}
To check for stability\[ \left \vert A-\lambda I\right \vert =\left \vert \begin{pmatrix} -\lambda & 1\\ -2 & -3-\lambda \end{pmatrix} \right \vert =\left ( -\lambda \right ) \left ( -3-\lambda \right ) +2 \] Hence\[ \lambda ^{2}+3\lambda +2=0 \] The roots are \(-1,-2\) and since they are both negative, hence the system is stable.
