Question

Reread(iii) in the remarks on page 293find the zero-input and the zero-slate response for the IVP in problem 40.

Short answer

the zero-input and the zero-slate response Is

$y\left(s\right)=\frac{6}{\left(s-3\right)\left(s^2-4s\right)}-\frac{6}{\left(s+1\right)\left(s^2-4s\right)}+\frac{s-5}{s^2-4s}$

The zero input response is ${y_0}_{}\left(t\right)=L^{-1}\left\{\frac{1}{s^2-4s}\right\}$

Thus, the zero-state response is$y_1\left(t\right)=\frac{5}{4}+\frac{27}{20}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t}$.

Step-by-step solution

Step1: definition of zero input.

The zero-input solution is the system's response to the initial conditions when the input is set to zero.

$y\left(s\right)=\frac{6}{\left(s-3\right)\left(s^2-4s\right)}-\frac{6}{\left(s+1\right)\left(s^2-4s\right)}+\frac{s-5}{s^2-4s}$

The transfer function is

$W\left(s\right)=\frac{1}{s^2-4s}$

The zero input response is

${y_0}_{}\left(t\right)=L^{-1}\left\{\frac{1}{s^2-4s}\right\}$

Decompose the fraction using the partial fraction technique.

$$\begin{gathered} \frac{s-5}{\left(s-4\right)}=\frac{A}{S}+\frac{B}{S-4} \\ \begin{array}{l}s-5=As-4A+Bs\\ s-5=\left(A+B\right)s-4A\end{array} \end{gathered}$$

Comparing coefficient on both sides, we get.

$\frac{\begin{array}{c}A+B=1\\ -4A=-5\end{array}}{\begin{array}{c}A+B=1\\ A=\frac{5}{4}\end{array}}$

Substitute the value of A in equation $A+B=1$, we get

$\begin{array}{c}\frac{5}{4}+B=1\\ B=-\frac{1}{4}\end{array}$

Substituting A and B in the equation(2) , we get.

$\frac{s-5}{s\left(s-4\right)}=\frac{5}{4S}+\frac{B}{(S-4)}$

The equation(1) becomes

$\begin{array}{c}{y_0}_{}\left(t\right)=L^{-1}\left\{\frac{5}{4s}\right\}-L^{-1}\left\{\frac{1}{(s-4)}\right\}\\ =\frac{5}{4}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{4}{L}^{-1}\left\{\frac{1}{s-4}\right\}\\ =\frac{5}{4}-\frac{1}{4}{e}^{4t}\\ {y_0}_{}\left(t\right)=\frac{5}{4}-\frac{1}{4}{e}^{4t}\end{array}$

Thus, the zero-input response is ${y_0}_{}\left(t\right)=\frac{5}{4}-\frac{1}{4}{e}^{4t}$.

$\begin{array}{c}{}_{\mathbf{1}}(t)=L^{-1}\left\{\frac{6}{(s-3)(s^2-4s)}\right\}-\left\{\frac{3}{(s+1)(s^2-4s)}\right\}\\ =L^{-1}\left\{\frac{6}{(s-3)(s^2-4s)}\right\}-L^{-1}\left\{\frac{3}{(s+1)(s^2-4s)}\right\}\end{array}$

${}^{-1}\left\{\frac{6}{(s-3)(s^2-4s)}\right\}$

Step2:  fraction using partial fraction.

In arithmetic, a fraction is a number expressed as a quotient in which a numerator is divided by a denominator. Both are integers in a simple fraction.

The fraction using partial fraction

$\begin{array}{rcl}& & \frac{6}{(s-3)(s^2-4s)}\\ \left\{\frac{6}{(s-3)(s^2-4s)}\right\}& =& \frac{A}{s}+\frac{B}{s-4}+\frac{C}{s-3}\\ & =& \frac{A(S-4)(S-3)B(S-3)+C(S-4)}{S(S-4)(S-3)}\end{array}$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

data-custom-editor="chemistry" $\begin{array}{l}3=A{s}^2-3As-4A+B{s}^2+Bs+C{s}^2-4Cs\\ 3=(A+B+C)s^2+(B-4C-3A)S-AA\end{array}$

Comparing coefficient on both sides, we get.

$\begin{array}{l}A+B+C=0\\ B-4C-3A=0\\ \frac{-4A=0}{\begin{array}{c}A+B+C=0\\ B-4C-3A=0\\ A=-\frac{3}{4}\end{array}}\end{array}$

Based on the above system, we get

$A=-\frac{3}{4},\frac{3}{20}\text{and}c=\frac{3}{5}$

Substitute A,B and c and in the equation(3) , we get.

$\frac{6}{(s+1)\left(s\right(s-4\left)\right)}=-\frac{3}{4}+\frac{3}{20\left(s-4\right)}+\frac{3}{5\left(s+1\right)}$

The equation

$$\begin{gathered} =L^{-1}\left\{\frac{1}{2s}\right\}+L^{-1}\left\{\frac{3}{2(s-4)}\right\}-L^{-1}\left\{\frac{2}{(s-3)}\right\} \\ =L^{-1}\left\{\frac{3}{4s}\right\}-L^{-1}\left\{\frac{3}{20(S-4)}\right\}-L^{-1}\left\{\frac{3}{5(s+1)}\right\} \\ =\frac{5}{4}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{27}{20}{L}^{-1}\left\{\frac{1}{s-4}\right\}-2L^{-1}\left\{\frac{1}{s-4}\right\}-\frac{3}{5}{L}^{-1}\left\{\frac{1}{s-(-1)}\right\} \\ =\frac{5}{4}+\frac{27}{20}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t} \end{gathered}$$

$y_1\left(t\right)=\frac{5}{4}+\frac{27}{20}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t}$

Thus, the zero-state response is$y_1\left(t\right)=\frac{5}{4}+\frac{27}{20}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t}$

Hence,

The zero-input response is${y_0}_{}\left(t\right)=L^{-1}\left\{\frac{1}{s^2-4s}\right\}$

Thus, the zero-state response is$y_1\left(t\right)=\frac{5}{4}+\frac{27}{20}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t}$