Question 11.4: Find a closed-form expression for the output y(n) of the sys......

The initial state vector was determined, from the given initial output conditions, in Example 11.2 as

q_{1}(0)=\frac{17}{3},\quad q_{2}(0)=\frac{16}{3}

The characteristic values, as determined in Example 11.3, are

\lambda_{1}=1+j\sqrt{2}\quad\mathrm{and}\quad\lambda_{2}=1-j\sqrt{2}

The transition matrix is given by

A^{n}=c_{0}I+c_{1}A

=c_{0}\left[{1 \ \ 0\atop0 \ \ 1}\right]+c_{1}\left[{2 \ -3}\atop1 \ \ \ \ 0\right]=\left [ \begin{matrix} c_{0}&+2c_{1}&-3c_{1} \\ &c_1&c_0\end{matrix} \right ]

where

{\left[\begin{array}{l}{c_{0}}\\ {c_{1}}\end{array}\right]}={\left[\begin{array}{l}{1\lambda_{1}}\\ {1\lambda_{2}}\end{array}\right]}^{-1}{\left[\begin{array}{l}{\lambda_{1}^{n}}\\ {\lambda_{2}^{n}}\end{array}\right]}

=\frac{j}{2\sqrt{2}}\left [ \begin{matrix} 1-j\sqrt{2}-1-j\sqrt{2} \\ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \end{matrix} \right ] \left [ \begin{matrix} (1+j{\sqrt{2}})^{n} \\ (1-j{\sqrt{2}})^{n} \end{matrix} \right ]

=\frac{j}{2\sqrt{2} } \left [ \begin{matrix}(1-j{\sqrt{2}})(1+j{\sqrt{2}})^{n}+(-1-j{\sqrt{2}})(1-j{\sqrt{2}})^{n} \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad -(1+j{\sqrt{2}})^{n}+(1-j{\sqrt{2}})^{n} \end{matrix} \right ]

A^{n}={\frac{j}{2{\sqrt{2}}}} \left [ \begin{matrix} -(1+j{\sqrt{2}})^{(n+1)}+(1-j{\sqrt{2}})^{(n+1)}\qquad\qquad3(1+j{\sqrt{2}})^{n}-3(1-j{\sqrt{2}})^{n}\\ -(1+j{\sqrt{2}})^{n}+(1-j{\sqrt{2}})^{n}\ 3(1+j{\sqrt{2}})^{(n-1)}-3(1-j{\sqrt{2}})^{(n-1)} \end{matrix} \right ]

As a check on A^{n}, verify that A^{n} = I with n = 0 and A^{n} = A with n = 1.

The zero-input component of the state vector is

q_{z i}(n)=A^{n}q(0)={\frac{j}{6{\sqrt{2}}}}\left [ \begin{matrix} (31-j17\sqrt2)(1+j\sqrt2)^{n}-(31+j17\sqrt2)(1-j\sqrt2)^{n} \\ (-1-j16{\sqrt{2}})(1+j{\sqrt{2}})^{n}+(1-j16{\sqrt{2}})(1-j{\sqrt{2}})^{n} \end{matrix} \right ]

Using the fact that the sum a complex number and its conjugate is twice the real part of either of the numbers, we get

q_{z i}(n)=\left [ \begin{matrix} {\frac{17}{3}}({\sqrt{3}})^{n}\cos(\tan^{-1}({\sqrt{2}})n)-{\frac{31}{3{\sqrt{2}}}}({\sqrt{3}})^{n}\sin(\tan^{-1}({\sqrt{2}})n)) \\ {\textstyle\frac{16}{3}}({\sqrt3})^{n}\cos(\tan^{-1}({\sqrt2})n)+{\textstyle\frac{1}{3{\sqrt2}}}({\sqrt3})^{n}\sin(\tan^{-1}({\sqrt2})n)) \end{matrix} \right ]

The zero-input response y_{z i}\left(n\right) is given by

C A^{n}q(0)={\left[1-2\right]} \left [ \begin{matrix} {\textstyle\frac{17}{3}}(\sqrt{3})^{n}\cos(\tan^{-1}(\sqrt{2})n)-{\textstyle\frac{31}{3{\sqrt{2}}}}(\sqrt{3})^{n}\sin(\tan^{-1}(\sqrt{2})n)) \\ {\textstyle\frac{16}{3}}(\sqrt{3})^{n}\cos(\tan^{-1}(\sqrt{2})n)+{\textstyle\frac{1}{3\sqrt{2}}}(\sqrt{3})^{n}\sin(\tan^{-1}(\sqrt{2})n)) \end{matrix} \right ]

=(-5(\sqrt{3})^{n}\cos(\tan^{-1}(\sqrt{2})n)-\frac{11}{\sqrt{2}}(\sqrt{3})^{n}\sin(\tan^{-1}(\sqrt{2})n))u(n)

The first four values of the zero-input response y_{z i}(n) are

y_{z i}(0)=-5,\quad y_{z i}(1)=-16,\quad y_{z i}(2)=-17,\quad y_{z i}(3)=14

The zero-state component of the state vector is

q_{z s}(n)=\sum_{m=0}^{n-1}A^{n-1-m}B x(m)

The convolution-summation, A^{n-1}u(n-1)*B x(n), can be evaluated, using the shift theorem of convolution (Chap. 4), by evaluating A^{n}u(n)*B x(n) first and then replacing n by n – 1.

B(x)=\left [ \begin{matrix} 1 \\ 0 \end{matrix} \right ] u(n)=\left [ \begin{matrix} u(n) \\ 0 \end{matrix} \right ]

A^n ∗ Bx(n) =\frac{j}{2\sqrt{2} }\left [ \begin{matrix} (-(1+j{\sqrt{2}})^{(n+1)}+(1-j{\sqrt{2}})^{(n+1)})*u(n) \\ \ \ \ \ \ \ \ \ \ \ \ \ (-(1+j{\sqrt{2}})^{n}+(1-j{\sqrt{2}})^{n})*u(n) \end{matrix} \right ]

Since the first operand of the convolutions is the sum of two complex conjugate expressions and the convolution of p(n) and u(n) is equivalent to the sum of the first n + 1 values of p(n), we get

Replacing n = n – 1, we get

The zero-state response is given by multiplying the state vector with the C vector and adding the input signal as

The first four values of the zero-state response y_{zs}(n) are

y_{z s}(0)=2,\quad y_{z s}(1)=3,\quad y_{z s}(2)=3,\quad y_{z s}(3)=0

Adding the zero-input and the zero-state components, we get the total response of the system as

y(n)\,=\,1.5-4.5(\sqrt{3})^{n}\cos(\tan^{-1}(\sqrt{2})n)

-\,\frac{10}{\sqrt{2}}(\sqrt{3})^{n}\sin(\tan^{-1}(\sqrt{2})n),\quad n=0,1,2,…

The first four values of the total response y(n) are

y(0)=-3,\quad y(1)=-13,\quad y(2)=-14,\quad y(3)=14