can you answer all the parts and do them using Matlab only the ones said to be done on Matlab and the rest need to be done on a piece of paper with all the steps done. Two functions are given as$,$ assume $x[n]$&$h[n]$ as input and impulse response respectively

$x[n]=-5[n]-2[n-1]-2[n-2]+7[n-3]+5[n-4]+2[n-5]$

$h[n]=-2[n]-2[n-2]+3[n-3]-2[n-4]$

a$)$ Plot the signals of $x[n]$&$h[n{]}_{using\text{M}\text{a}\text{t}\text{l}\text{a}\text{b}or\text{o}\text{t}\text{h}\text{e}\text{r}\text{t}\text{o}\text{o}\text{l}\text{s}}$

b$)$ Using graphical representation method $($flip one function$),$ just like we have done the

convolution in the continuous time domain

$y[n]={\displaystyle \underset{=-}{\overset{}{}}}x[]*h[n-]={\displaystyle \underset{=-}{\overset{}{}}}h[]*x[n-]$

c$)$ Table method

d$)$ Simple convolution of impulse vs$.$ impulse functions. For an example

$[n-3]*[n-2]=[n-5]$

e$)$ Using conversion of signals in time to DTFT

i$.$

$Y(e^j)={\displaystyle \underset{n=-}{\overset{}{}}}y[n]e^{-jn},[=\frac{2f}{fs}]$

For an example, $F\{[n-3]\}=e^{j3}$

f$)$ Using Z$-$transforms method.

i$.,Y(z)=x(z)*H(z),z=e^j$

g$)$ Use of "conv" in the Matlab to verify the output of your answers $($of course you need

to show the code of your program that you used in the Matlab or Octave$)$

h$)$ Verify that the length of the signal after the convolution is

length$($input$)+$ length$($impulse response