Question

A system is described by the differential equation \( \frac{d^2y(t){dt^2} + 6\frac{dy(t)}{dt} + 5y(t) = x(t) \). Impulse response of the system is

• A. \(h(t) = \frac{1}{4}[e^{-2t} - e^{-3t}]u(t)\)
• B. \(h(t) = [e^{-t} - e^{-5t}]u(t)\)
• C. \(h(t) = \frac{1}{4}[e^{-t} - e^{-5t}]u(t)\)
• D. \(h(t) = \frac{1}{4}[e^{-3t} - e^{-2t}]u(t)\)

Hint:

You can instantly cross-verify the correctness of your partial fraction step by using the difference gap rule: when you have a fraction form of \(\frac{1}{(S+a)(S+b)}\), the constant multiplier out front is always exactly \(\frac{1}{b-a}\). Here, \(\frac{1}{5-1} = \frac{1}{4}\)!

Answer 1

The Correct Option is C

Understanding the Concept: The impulse response \(h(t)\) of a continuous-time Linear Time-Invariant (LTI) system is the system output when the input is a Dirac delta function, \(x(t) = \delta(t)\), assuming all initial conditions are zero. The most elegant way to solve this is by applying the Laplace Transform, which maps differential equations into algebraic equations: \[ \mathcal{L}\left\{\frac{d^n y(t)}{dt^n}\right\} = S^n Y(S) \quad \text{(for zero initial conditions)} \] The system transfer function \(H(S)\) is defined as the ratio of the output transform to the input transform: \[ H(S) = \frac{Y(S)}{X(S)} \] Taking the inverse Laplace transform of \(H(S)\) yields the time-domain impulse response \(h(t)\).

Step 1: Convert the differential equation to the S-domain via Laplace transform.
Given equation: \[ \frac{d^2y(t)}{dt^2} + 6\frac{dy(t)}{dt} + 5y(t) = x(t) \] Applying the unilateral Laplace Transform under zero initial state conditions: \[ S^2 Y(S) + 6S Y(S) + 5Y(S) = X(S) \] Factoring out \(Y(S)\) on the left-hand side: \[ (S^2 + 6S + 5)Y(S) = X(S) \]

Step 2: Determine the Transfer Function \(H(S)\).
\[ H(S) = \frac{Y(S)}{X(S)} = \frac{1}{S^2 + 6S + 5} \] Find the roots of the quadratic polynomial in the denominator to factor it: \[ S^2 + 6S + 5 = S^2 + 5S + S + 5 = S(S + 5) + 1(S + 5) = (S + 1)(S + 5) \] Substituting this back into \(H(S)\): \[ H(S) = \frac{1}{(S + 1)(S + 5)} \]

Step 3: Expand using Partial Fractions.
We express \(H(S)\) as a sum of simpler fractions with unknown coefficients \(A\) and \(B\): \[ \frac{1}{(S + 1)(S + 5)} = \frac{A}{S + 1} + \frac{B}{S + 5} \] Using the standard Heaviside cover-up method to calculate constants: For coefficient \(A\) (at pole \(S = -1\)): \[ A = \left. \frac{1}{S + 5} \right|_{S = -1} = \frac{1}{-1 + 5} = \frac{1}{4} \] For coefficient \(B\) (at pole \(S = -5\)): \[ B = \left. \frac{1}{S + 1} \right|_{S = -5} = \frac{1}{-5 + 1} = \frac{1}{-4} = -\frac{1}{4} \] Substituting these computed coefficients back gives: \[ H(S) = \frac{1}{4}\left( \frac{1}{S + 1} - \frac{1}{S + 5} \right) \]

Step 4: Take the Inverse Laplace Transform to find \(h(t)\).
Using the standard Laplace pair formula \(\mathcal{L}^{-1}\left\{\frac{1}{S + a}\right\} = e^{-at}u(t)\): \[ h(t) = \mathcal{L}^{-1}\{H(S)\} = \frac{1}{4}\left( e^{-t}u(t) - e^{-5t}u(t) \right) \] Factoring out the common unit step signal \(u(t)\): \[ h(t) = \frac{1}{4}[e^{-t} - e^{-5t}]u(t) \] This precisely corresponds to option (C).