In this topic, you study the theory, derivation & solved examples for the impulse response of the Linear Time-Invariant (LTI) System.
When the system is linear as well as time-invariant, then it is called a linear time-invariant (LTI) system. When the input to LTI system is unit impulse $\delta (t)$ then the output of LTI system is known as impulse response $h(t)$. Consider an LTI system with impulse response $h(t)$ whose input and output are $x(t)$ and $y(t)$ respectively, shown in Figure 1.
Figure 1: LTI system
The ratio of the Laplace transform of output $y(t)$ to the Laplace transform of input $x(t)$ with zero initial conditions, is known as transfer function $H(s)$. So we write
\[H(s) = \frac{{Y(s)}}{{X(s)}}\]
Also
\[Y(s) = X(s)H(s)….(1)\]
Applying inverse laplace transform on Equation 1,
\[y(t) = x(t) * h(t)\]
Note:- * denotes convolution and LTI system is defined by impulse response in time domain and transfer function in frequency domain.
Example : The impulse response of a system is $h(t) = u(t)$ . What will be the output $y(t)$ for an input $\delta (t – 2)$, .
Solution: As we already discussed, the LTI system is defined by impulse response in the time domain and transfer function in the frequency domain.
\[Y(s) = X(s)H(s)\]
For the input $x(t) = \delta (t – 2)$ so applying laplace transfom gives
\[X(s) = {e^{ – 2s}}\]
Also the impulse response of a system is $h(t) = u(t)$ so applying laplace transfom gives
\[H(s) = \frac{1}{s}\]
Thus
\[Y(s) = {e^{ – 2s}} \cdot \frac{1}{s}\]
Applying Inverse laplace transfom gives
\[h(t) = u(t – 2)\]
