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I have a system given by $$y[n] - \frac{1}{4} y[n-1] - \frac{1}{8} y[n-2] =3x[n] $$

I'm asked to determine the impulse response of both an anti-causal and a causal Linear Shift Invariant System characterized by this difference equation.

The impulse response to the causal system is $$2\left(\frac{1}{2}\right)^{n}u[n]+\left(-\frac{1}{4}\right)^{n}u[n]$$

I found this without using Z transform or Fourier Transform but by solving the difference equation using the initial condition $h[n]=0$ for $n<0$.

How can I find the anti-causal impulse response without using Z-transform or Fourier Transform. What boundary conditions do I use now?

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  • $\begingroup$ Hint: a causal system has an impulse response which is a right-handed signal of the form $h[n<0]=0$. Hence, an anti-causal system will have a left-handed impulse response of the form $h[n>0]=0$. $\endgroup$
    – havakok
    Jan 26 at 7:17
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You've already found the solutions of the characteristic equation as $\lambda_1=\frac12$ and $\lambda_2=-\frac14$, so you know that the solution must have the form

$$y[n]=c_1\left(\frac12\right)^n+c_2\left(-\frac14\right)^n\tag{1}$$

For the anti-causal solution we know that $y[n]=0$ for $n>0$. The constants $c_1$ and $c_2$ are simply determined by making sure that the difference equation is satisfied for $x[n]=\delta[n]$, given that we require $y[n]=0$ for $n>0$:

$$\begin{align}n=2:\quad&&&-\frac18y[0]&=0\\n=1:\quad&&-\frac14y[0]&-\frac18y[-1]&=0\\n=0:\quad&y[0]&-\frac14 y[-1]&-\frac18y[-2]&=3\end{align}\tag{2}$$

From the first equation in $(2)$ we see that the impulse response actually starts at $n=-1$, not at $n=0$. The constants $c_1$ and $c_2$ can be determined from the second and third equation, which simplify to

$$y[-1]=0\quad\textrm{and}\quad -\frac18 y[-2]=3\tag{3}$$

Plugging $(3)$ into $(1)$ finally gives the actual values of $c_1$ and $c_2$, which I'm sure you can figure out yourself. The final anti-causal solution has the form

$$y[n]=\left[c_1\left(\frac12\right)^n+c_2\left(-\frac14\right)^n\right]u[-n-1]\tag{4}$$

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  • $\begingroup$ I got it....thanks a lot Matt L....I confused anti causal signal with non causal signal! $\endgroup$
    – Orpheus
    Jan 26 at 12:49

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