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I have to find the Zero input response of the following system if $y[-2]= 0.5$

$$x[n] - 3y[n-1] - 4y[n-2] = 0$$ I am not sure if this is the correct way since every other example I have done had two initial conditions given.

My solution:

$-3y[n-1] - 4[n-2] = 0$ ( Homogeneous response equal to zero)

Let's set $y(n) = Ar^n$. We have:

$-r^{n-2}(3r+4)=0 \leftrightarrow -r^{n-2} = 0$ gives $r_1 =0$

$3r+4=0$ gives $r_2= -4/3$

Finding coefficient $A$

$y[-2] = A(-4/3)^{-2} =0.5 \leftrightarrow A(9/16) = 0.5 \leftrightarrow A= 8/9$

Zero input response $y[n] = \frac{8}{9}\left(-\frac{4}{3}\right)^n$

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  • $\begingroup$ Yes, your solution looks correct to me. Next time, please make an effort to correctly format your question! $\endgroup$
    – Jdip
    Nov 23, 2023 at 0:19

1 Answer 1

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Your solution is correct, but it's good to understand the reason why you only need one initial condition in this case: there are only two output values in the difference equation, and they are only one sample instant apart. So for $x[n]=0$ (zero input), you simply have

$$y[n-1]=-\frac43 y[n-2]\tag{1}$$

and you can write down the solution without doing much math:

\begin{align*} y[-1] &= \frac12\cdot\left(-\frac43\right) \\ y[0] &= \frac12\cdot\left(-\frac43\right)^2 \\ y[1] &= \frac12\cdot\left(-\frac43\right)^3 \\\vdots \end{align*}

I.e., we have

$$y[n]=\frac12\cdot\left(-\frac43\right)^{n+2}=\frac89\cdot\left(-\frac43\right)^{n}$$

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