written 6.0 years ago by | • modified 6.0 years ago |
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}