DT Convolution

Let $h[n]$ be the impulse response of a linear and time-invariant(LTI) system. If the signal $x[n]$ is input to the system, the output signal from the system is given by:$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]$
This operation is called convolution and we say that the signal $y[n]$ is the convolution of the signal $x[n]$ and the signal $h[n]$ and we denote this by $y[n] = x[n]* h[n]$. Thus,

(1)
\begin{align} y[n]=x[n]*h[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k] \end{align}

To compute the signal $y[n]$, perform the following steps:

1. Think of $x[n]$ and $h[n]$ as signals $x[k]$ and $h[k]$ respectively, i.e., with the independent variable being $k$ instead of $n$.
2. Flip $h[k]$ about the Y -axis to obtain the signal $h[-k]$.
3. To compute the signal $y[n]$ for a ﬁxed value of $n$, shift the signal $h[-k]$ by n units to the right to obtain the signal $h[n-k]$. When n is negative, this amounts to shifting the signal $h[-k]$ to the left, but mathematically it is equivalent to shift right by a negative number.
4. Compute $w_n[k] = x[k]h[n-k]$, i.e., $w_n[k]$ is the product of the signals $x[k]$ and $h[n-k]$.
5. Compute $y[n] =\sum_{k}w_n[k] =\sum_{k}x[k]h[n-k]$ by summing the values of $w_n[k]$ for all values of $k$.

This will give you the value of the signal $y[n]$ for one value of $n$. Repeat this procedure for every integer value of $n$, i.e., $n \in$ […,-3,-2,-1,0,1,2,3,…] to obtain the full signal $y[n]$. In practice, it will be easy to start with large negative values of $n$ and increase $n$.

page revision: 7, last edited: 15 Aug 2016 02:12