Given x(n)={2,3,4,5}and h(n)={5,2,3,4}

i) By time domain method

y(n)=

$\begin{matrix} [2&5&4&3] [5]\\ |3&2&5&4||2|\\ |4&3&2&5||3|\\ [5&4&3&2][4] \end{matrix}$

$\begin{matrix} [10+&10+&12+&12]\\ |15+&4+&15+&16|\\ |20+&6+&6+&20|\\ [25+&8+&9+&8] \end{matrix}$

$=\begin{matrix} [44]\\ |50|\\ |52|\\ [50] \end{matrix}$

y(n)={44,50,52,50}

ii) By frequency domain method

Calculate DFT of x(n)

X(k) =

$\begin{matrix} [1&1&1&1] [2]\\ |1&-j&-1&+j||3|\\ |1&-1&1&-1||4|\\ [1&+j&-1&-j][5] \end{matrix}$

=

$\begin{matrix} 14\\ -2+2j\\ -2\\ -2-2j \end{matrix}$

Calculate DFT of h(n)

H(k) =

$\begin{matrix} [1&1&1&1] [5]\\ |1&-j&-1&+j||2|\\ |1&-1&1&-1||3|\\ [1&+j&-1&-j][4] \end{matrix}$

=

$\begin{matrix} 14\\ 2+2j\\ 2\\ 2-2j \end{matrix}$

We know that convolution in time domain is equivalent to multiplication in frequency domain

Y(k)=X(k).H(k)

={196,-8,-4,-8}

Now calculate IDFT , to get y(n)

y(n) =

$\frac{1}{N}$* $\begin{matrix} [1&1&1&1] [196]\\ |1&+j&-1&-j||-8|\\ |1&-1&1&-1||-4|\\ [1&-j&-1&+j][-2] \end{matrix}$

=$\frac{1}{4}$*

$\begin{matrix} 176\\ 200\\ 208\\ 200 \end{matrix}$

y(n)={44,50,52,50}

iii) Linear Convolution

y(n)={10,19,32,50,34,31,20}