LTI Impulse Response
The response of a discrete-timeLTI system to a unit impulse is used to find its filter, . Mathematically, the solution to any input of the discrete-time LTI can be expressed as the convolution
Similarly, the solution to any input of the continuous-time LTI can be expressed as the convolution of its
Proof
We proceed by a simple proof, taking and a filter :
y(n) &= (x \textasteriskcentered h)(n) \\ &= \sum^{\infty}_{k=-\infty}x(k)h(n-k) \\ &= \sum^{\infty}_{k=-\infty}\delta(k)h(n-k) \\ &= h(n), \end{align}$$ since $\delta(k) \neq 0 \iff k = 0$. # Examples >[!example] (Discrete-Time) Impulse Response of a Recursive Filter > >A causal, discrete-time LTI system $H$ is characterized by the recursive difference equation $y(n) = \alpha y(n-N)+x(n)$. We can determine its impulse response by observing that for $x(n)=\delta(n)$, >$$\begin{align} >h(n) &= \sum^{\infty}_{k=-\infty}\delta(n-k)[\alpha h(n-N)+\delta(n)] \\ \\ >&= \alpha h(n-N) + \delta(n). >\end{align}$$ >We quickly see that since $H$ is causal, then $h(n) = 0$ for all $n < 0$. Therefore, $h(0) = \delta(0)=1$. By observation, we find $h(N) = \alpha$, $h(2N) = \alpha^2$, and so on. Therefore, we can establish that >$$h(n) = \begin{cases}\alpha^{\frac{n}{N}} \text{ if } n \mod{N} \equiv 0 \\ 0 \text{ otherwise}\end{cases}.$$ >Using the Kronecker delta, we can get rid of this piecewise nature so that >$$h(n) = \sum^{\infty}_{k=-\infty}\alpha^n \delta(n - kN).$$ >This is because the delta is nonzero only for $n=kN$, (i.e. a multiple of $N$).