0
$\begingroup$

I'm trying to compute the impulse response for the following system:

$$ y(n) = \frac{1}{2} \left( x(n) + x(n-1) \right) + \frac{1}{2} y(n-1) $$

and am told to assume that $y(-1)=0$.

My solution: Letting $x(n) = \delta(n)$, we have

$$ h(n) = \frac{1}{2}\left(\delta(n) + \delta(n-1)\right) + \frac{1}{2}h(n-1). $$

Plugging in a few values for $n$ (and assuming that $h(-1)=0$), I've obtained:

$$ h(0) = \frac{1}{2}, $$

$$ h(1) = \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} $$

and

$$ h(2) = \frac{1}{2} + \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}. $$

From this, I've observed that

$$ h(n) = \begin{cases} 0, & n<0 \\ \frac{1}{2}, & n=0 \\ \frac{1}{2} + \left(\frac{1}{2}\right)^n, & n > 0 \end{cases} $$

Does this seem correct? I feel like I made a mistake somewhere and that the expression for $h(n)$ should be in a simpler form.

Note: I computed the impulse response by taking the Fourier Transform and then plugging in values for $n$ and got something different than this. However, I believe that this discrepancy is due to the initial condition $y(-1)=0$.

$\endgroup$
1
$\begingroup$

Your values for $h(0)$, $h(1)$ and $h(2)$ are correct. However, the general formula isn't. It should be

$$h(n) = \begin{cases} 0, & n<0 \\ \frac{1}{2}, & n=0 \\ \frac{3}{2}\left(\frac{1}{2}\right)^n, & n > 0 \end{cases}$$

A simple way to solve such problems is to use the $\mathcal{Z}$-transform:

$$Y(z)=\frac{1}{2}X(z)(1+z^{-1})+\frac{1}{2}Y(z)z^{-1}$$

which gives

$$H(z)=\frac{Y(z)}{X(z)}=\frac{1}{2}\frac{1+z^{-1}}{1-\frac{1}{2}z^{-1}}= \frac{1}{2}\frac{1}{1-\frac{1}{2}z^{-1}}+\frac{1}{2}\frac{z^{-1}}{1-\frac{1}{2}z^{-1}}$$

With the transform pair

$$\frac{1}{1-az^{-1}}\Longleftrightarrow a^nu(n),\quad |a|<1$$

and with the property

$$x(n-1)\Longleftrightarrow z^{-1}X(z)$$

you arrive at

$$h(n)=\frac{1}{2}\left(\frac{1}{2}\right)^nu(n)+\frac{1}{2}\left(\frac{1}{2}\right)^{n-1}u(n-1)$$

which corresponds to above result.

$\endgroup$
  • $\begingroup$ Thanks for the clarification. I mistyped when writing my question. $\endgroup$ – Nathan Green Apr 25 '14 at 7:21
-1
$\begingroup$

The line $h[2] = \dots$ should be $h[2] = \frac{1}{2}h[1]$

You can use the z-transform to derive the inpulse response: $x[n] = \delta[n] \Rightarrow X(z) = 1$ $Y(z) = .5 (1+z^{-1}) + .5 Y(z) z^{-1} \Rightarrow Y(z) = .5 \frac{1+z^{-1}}{1-.5z^{-1}} = .5 \left( \frac{1}{1-.5z^{-1}} + z^{-1}\frac{1}{1-.5z^{-1}} \right)$ $\Rightarrow y[n] = .5 \left( .5^{n} u[n] + .5^{n-1} u[n-1] \right)$, Where $u[n]$ is the heaviside step function.

See wikipedia for transform pairs.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.