Question 3.13: Zero-Input Response of a Second-Order System with Real Roots......
Zero-Input Response of a Second-Order System with Real Roots
The LTID system described by the difference equation
y[n+2]-0.6 y[n+1]-0.16 y[n]=5 x[n+2]
has input x[n]=4^{-n} u[n] and initial conditions y[-1]=0 \text { and } y[-2]=25 / 4. Determine the zero-input response y_0[n]. The zero-state response of this system is considered later, in Ex. 3.21.
The system equation in operator notation is
\left(E^2-0.6 E-0.16\right) y[n]=5 E^2 x[n]
The characteristic polynomial is
\gamma^2-0.6 \gamma-0.16=(\gamma+0.2)(\gamma-0.8)
The characteristic equation is
(\gamma+0.2)(\gamma-0.8)=0
The characteristic roots are \gamma_1=-0.2 \text { and } \gamma_2=0.8. The zero-input response is
y_0[n]=c_1(-0.2)^n+c_2(0.8)^n (3.24)
To determine arbitrary constants c_1 \text { and } c_2, we set n=−1 and −2 in Eq. (3.24), then substitute y_0[-1]=0 \text { and } y_0[-2]=25 / 4 to obtain†
\left.\begin{array}{c}0=-5 c_1+\frac{5}{4} c_2 \\\frac{25}{4}=25 c_1+\frac{25}{16} c_2\end{array}\right\} \quad \Longrightarrow \quad \begin{array}{l}c_1=\frac{1}{5} \\c_2=\frac{4}{5}\end{array}
Therefore,
y_0[n]=\frac{1}{5}(-0.2)^n+\frac{4}{5}(0.8)^n \quad n \geq 0
The reader can verify this solution by computing the first few terms using the iterative method (see Exs. 3.11 and 3.12).