Impulse Response of an LTI System
Consider the differential equation of a second-order system:
\[ \frac{d^2 y(t)}{dt^2} - \frac{dy(t)}{dt} - 6y(t) = x(t) \]
We want to find the impulse response \(h(t)\). For an LTI system, the impulse response is the output when the input is a unit impulse \( \delta(t) \).
So we set:
\[ x(t)=\delta(t) \]
Step 1: Take Laplace Transform
Applying Laplace transform with zero initial conditions:
\[ s^2Y(s) - sY(s) - 6Y(s) = X(s) \]
Factor \(Y(s)\):
\[ Y(s)(s^2 - s - 6) = X(s) \]
Step 2: Find the Transfer Function
\[ H(s)=\frac{Y(s)}{X(s)} \]
\[ H(s)=\frac{1}{s^2 - s - 6} \]
Factor the denominator:
\[ s^2 - s - 6 = (s-3)(s+2) \]
\[ H(s)=\frac{1}{(s-3)(s+2)} \]
Step 3: Partial Fraction Expansion
\[ \frac{1}{(s-3)(s+2)}=\frac{A}{s-3}+\frac{B}{s+2} \]
Multiply both sides:
\[ 1=A(s+2)+B(s-3) \]
Solving gives:
- \(A=\frac{1}{5}\)
- \(B=-\frac{1}{5}\)
Thus:
\[ H(s)=\frac{1}{5}\frac{1}{s-3}-\frac{1}{5}\frac{1}{s+2} \]
Step 4: Inverse Laplace Transform
Using the standard result:
\[ \mathcal{L}^{-1}\left(\frac{1}{s-a}\right)=e^{at}u(t) \]
We obtain:
\[ h(t)=\frac{1}{5}e^{3t}-\frac{1}{5}e^{-2t} \]
Step 5: Adjust for Non-Causal System
The system is given as neither causal nor stable.
- Causal systems use \(u(t)\)
- Anti-causal systems use \(u(-t)\)
The region of convergence lies between the poles:
\[ -2 < Re(s) < 3 \]
Thus the impulse response becomes:
\[ h(t)=\frac{1}{5}e^{3t}u(-t)-\frac{1}{5}e^{-2t}u(t) \]
Result
\[ h(t)=\frac{1}{5}e^{3t}u(-t)-\frac{1}{5}e^{-2t}u(t) \]
Explanation:
- \(e^{3t}u(-t)\) → grows for negative time (non-causal part)
- \(e^{-2t}u(t)\) → decays for positive time
- Region of convergence lies between poles, so the system is neither causal nor stable