Question 11.8: Find the zero-input, zero-state, transient, steady-state, an......
Find the zero-input, zero-state, transient, steady-state, and complete responses of the system governed by the difference equation
y(n)=2x(n)-x(n-1)+3x(n-2)+{\frac{9}{20}}y(n-1)-{\frac{1}{20}}y(n-2)
with the initial conditions y(–1) = 3 and y(–2) = 2 and the input x(n) = u(n), the unit-step function.
The corresponding state-space model is
A=\left [ \begin{matrix} 0.4500-0.0500 \\ 1.0000 \ \ \ \ \ \ \ \ \ \ \ 0 \end{matrix} \right ], \ \ \ B=\left [ \begin{matrix} 1 \\ 0 \end{matrix} \right ] ,
C=\left[-0.1000\ 2.9000\right],\quad D=2
We need the initial state vector values q_{1}(0){\mathrm{~and~}}q_{2}(0) to start the iteration.
Therefore, we have to derive these values from the initial output conditions y(–1) = 3 and y(–2) = 2. Using the state equations, we get
q_{1}(-1)=0.45q_{1}(-2)-0.05q_{2}(-2)
q_{2}(-1)=q_{1}(-2)
Using the output equations, we get
y(-2)=-0.1q_1(-2)+2.9q_2(-2)=2
y(-1)=-0.1q_{1}(-1)+2.9q_{2}(-1)=3
Solving these equations, we find q_1(–1) = 0.4360 and q_2(–1) = 1.0495. Now,
q_{1}(0)=0.45(0.4360)-1.0495(0.05)=0.1437
q_{2}(0)=q_{1}(-1)=0.4360
The initial state vector is
q(0)=\left[\begin{array}{l}{{0.1437}}\\ {{0.4360}}\end{array}\right]
(z I-A)=\left [ \begin{matrix} z-0.45 \ \ 0.05 \\ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ z \end{matrix} \right ] and (z I-A)^{-1}=\left [ \begin{matrix} {\frac{z}{z^{2}-0.45z+0.05}}\,-{\frac{0.05}{z^{2}-0.45z+0.05}} \\ \begin{array}{r l}{{\frac{1}{z^{2}-0.45{z}+0.05}}}&{{}{\frac{z-0.45}{z^{2}-0.45{z}+0.05}}\,}\end{array} \end{matrix} \right ]
As a check on (z I-A)^{-1} , we use the initial value theorem of the z-transform to verify that
\operatorname*{lim}_{z\rightarrow\infty}z(z I-A)^{-1}=I=A^{0}
The transform of the zero-input component of the state vector is
Q_{z i}(z)=z(z I-A)^{-1}q(0)
=z\left [ \begin{matrix} {\frac{z}{z^{2}-0.45z+0.05}}\,-{\frac{0.05}{z^{2}-0.45z+0.05}} \\ \begin{array}{r l}{{\frac{1}{z^{2}-0.45{z}+0.05}}}&{{}{\frac{z-0.45}{z^{2}-0.45{z}+0.05}}\,}\end{array} \end{matrix} \right ]\left [ \begin{matrix} 0.1437 \\ 0.4360 \end{matrix} \right ]
=z\left [ \begin{matrix} {\frac{0.1437z-0.0218}{z^{2}-0.45+0.05}} \\ {\frac{0.4360z -0.0525}{z^{2}-0.45z+0.05}} \end{matrix} \right ]
Y_{zi}(z) = \left [ \begin{matrix} -0.1 & 2.9 \end{matrix} \right ] z \ \left [ \begin{matrix} {\frac{0.1437z-0.0218}{z^{2}-0.45+0.05}} \\ {\frac{0.4360z -0.0525}{z^{2}-0.45z+0.05}} \end{matrix} \right ] =z\frac{1.25z-0.1500}{z^2-0.45z+0.05},
which is the same as that obtained in Chap. 9 Example 9.11.
The transform of the zero-state component of the state vector is
Q_{z s}(z)=(z I-A)^{-1}B X(z)=z\left [ \begin{matrix} {\frac{z}{z^{2}-0.45z+0.05}}\,-{\frac{0.05}{z^{2}-0.45z+0.05}} \\ \begin{array}{r l}{{\frac{1}{z^{2}-0.45{z}+0.05}}}&{{}{\frac{z-0.45}{z^{2}-0.45{z}+0.05}}\,}\end{array} \end{matrix} \right ]\left [ \begin{matrix} \frac{z}{z-1} \\ 0 \end{matrix} \right ] =z\left [ \begin{matrix} \frac{z}{(z-1)(z^{2}-0.45z+0.05)} \\ \frac{1}{(z-1)(z^{2}-0.45z+0.05)} \end{matrix} \right ]
Y_{z s}(z)=C(z I-A)^{-1}B X(z)=z\left[-0.1~2.9\right]\left [ \begin{matrix} \frac{z}{(z-1)(z^{2}-0.45z+0.05)} \\ \frac{1}{(z-1)(z^{2}-0.45z+0.05)} \end{matrix} \right ]
=z{\frac{-0.1z+2.9}{(z-1)(z^{2}-0.45z+0.05)}}
Adding the direct input component, we get
Y_{z s}(z)=z\left(\frac{-0.1z+2.9}{z^{3}-1.45z^{2}+0.5z-0.05}+\frac{2}{z-1}\right)=z\frac{2z^{2}-z+3}{z^{3}-1.45z^{2}+0.5z-0.05},
which is the same as that obtained in Chap. 9 Example 9.11.