Zero State and Zero Input Responses from Steady State Response

Question

Let a linear, time-invariant, causal discrete linear system be described by the following difference equation:

$$y[n] = 0.9y[n-1] + 0.1x[n]$$

Assuming that $y[-1]=2$ and $x[n]=20\cos(\Omega n)u[n]$, $\Omega = \omega T_s = 0.2\pi$, find the total response, identifiying the terms related to the zero state and zero input responses.

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I've found the transfer function of the system to be: $$H(z) = 0.1\frac{z}{z-0.9}$$

and from it I've determined the magnitude and phase of the steady-state response: $$|H \left(e^{j \Omega}\right)| = \frac{0.1}{\sqrt{1.81-1.8\cos(\Omega)}}$$ $$\angle H \left(e^{j \Omega}\right) = -\arctan\left[\frac{-\sin(\Omega)}{1-0.9\cos(\Omega)} \right]$$

Therefore the steady-state response of the system is: $$y_{ss}[n] = 20 \frac{0.1}{\sqrt{1.81-1.8\cos(\Omega)}} \cos\left(\Omega n -\arctan\left[\frac{-\sin(\Omega)}{1-0.9\cos(\Omega)} \right] \right)$$

$$\therefore \boxed{y_{ss}[n] = 3.363\cos(0.2\pi n + 1.138)}$$

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But how do I find the total response (zero state and zero input) from the steady-state response?

Answer 1

Some books prefer different notations, but I will assume that zero-state response is the one due to pure input $x[n] = \cos(0.2 \pi n) u[n]$ under system zero-initial conditions, while the zero-input response is the one with $x[n] = 0$ and initial state being $y[-1] = 2$.

Now, first the zero-input ($x[n] = 0$ ) response, purely due to the nonzero initial condition ($y[-1] = 2$), of the system can be found by a simple recursion of the causal system as:

$$ \begin{align} y[0] &= 0.9 \cdot 2 \\ y[1] &= 0.9^2 \cdot 2 \\ y[2] &= 0.9^3 \cdot 2 \\ ... &= ... \\ y[n] &= 2 \cdot 0.9^{n+1} u[n] \\ \end{align} $$

which yields for the zero input response (for $n \geq 0$) as

$$\boxed{ y_{zi}[n] = 2 \cdot (0.9)^{n+1} u[n] }$$

Now we shall find the zero-state response, assuming zero initial conditions. This will be lengthy because of the input form $x[n] = 20 \cos(0.2 \pi n) u[n]$, we can follow the following approach.

Let $X(w)$ be the Fourier transform of the input, and $H(w)$ be the Freuqency response of the system described by the above differential equation, then the output $y[n]$ will have the Fourier transform given by $$Y(\omega) = H(\omega) X(\omega).$$

Now $$H(\omega) = \frac{0.1}{1- 0.9 e^{-j \omega} } $$ and you have to find out what $X(w)$ is, then multiply them to find out $Y(\omega)$ and perform an inverse Fourier transform, after a simple partial fraction expansion. Denoting this solution as $y_{zs}[n]$, then your total solution is

$$y[n] = y_{zi}[n] + y_{zs}[n]$$

Now, try this on your own and see if you can complete the solution... Then write a matlab program to check your result.