# Convolution of Input signal and Impulse response

I can solve the output if the input is

$$x[n] = \delta[n+1] + \delta[n] + 2\delta[n-1] + \delta[n-2]$$

and the impulse response is

$$h[n]= 3\delta[n] − 2 \delta[n−1] − \delta[n−2] + \delta[n−3]$$

I calculate the convolution between the strings [1 1 2 1] and [3 -2 -1 1].

But I got a problem in computing the output when the input is $$x[n] = (\frac{1}{4})^{-n} \left( \delta[n +1] − \delta[n −4] \right)$$

and the impulse response is

$$h[n] = \delta[n] −\delta[n −5].$$

I took different strings but actually could not find what should be input and impulse response string and why?

• Please use Latex to format your equations. Your question is really unreadable and nobody will be able to help you. Especially the signal $s[n]$ that gives you trouble is totally unreadable. Sep 8, 2018 at 19:28

In the second case, which caused you the trouble, you should first use the sifting property of the impulse to find out the input signal $x[n]$ and then proceed with convolution.

sifting property is: $$f[n] \delta[n-d] = f[d]\delta[n-d]$$

Apply this to simplify your input signal $x[n]$ as follows:

\begin{align} x[n] &= (\frac{1}{4})^{-n} \left(\delta[n+1] - \delta[n-4] \right) \\ & = (\frac{1}{4})^{-n} \delta[n+1] - (\frac{1}{4})^{-n} \delta[n-4] \\ & = (\frac{1}{4})^{-(-1)} \delta[n+1] - (\frac{1}{4})^{-(4)} \delta[n-4] \\ x[n] & = \frac{1}{4} \delta[n+1] - 256 ~ \delta[n-4] \\ \end{align}

So, your input sequence is $x[n] = \frac{1}{4} \delta[n+1] - 256 ~\delta[n-4]$

Convolving this with the impulse response of $$h[n] = \delta[n] - \delta[n-5]$$ is easy, this time by utilizing the shifting property of the impulse which is:

$$f[n] \star \delta[n-d] = f[n-d]$$

Where $\star$ is th convolution. Then apply distribution of convolution over addition to get your result as:

\begin{align} y[n] &= x[n] \star h[n] \\ & = x[n] \star (\delta[n] - \delta[n-5] ) \\ &= x[n] - x[n-5] \\ &= (\frac{1}{4} \delta[n+1] - 256 ~\delta[n-4]) - (\frac{1}{4} \delta[n-4] - 256~ \delta[n-9]) \\ &= \frac{1}{4} \delta[n+1] - 256~ \delta[n-4] - \frac{1}{4} \delta[n-4] + 256~ \delta[n-9] \\ y[n] &= \frac{1}{4} \delta[n+1] - (256~ \delta[n-4] + \frac{1}{4}) \delta[n-4] + 256~ \delta[n-9] \\ \end{align}

I hope this helps.