Question

If the unit step response of a network is $(1-e^{-at})$, the unit impulse response is

• A. $ae^{-at}$
• B. $a^{-1}e^{-at}$
• C. $(1-a^{-1})e^{-at}$
• D. $(1-a)e^{-at}$

Hint:

Impulse response is always the time derivative of the step response.

Answer 1

The Correct Option is A

Step 1: The given step response is \( s(t) = 1 - e^{-at} \).
Step 2: Since the impulse response of an LTI system is the derivative of its step response, compute \( h(t) = \dfrac{d}{dt}\left(1 - e^{-at}\right) \).
Step 3: Differentiating term by term, the derivative of the constant \( 1 \) is \( 0 \), and the derivative of \( -e^{-at} \) is \( a e^{-at} \) (bringing down the exponent's coefficient \( -a \) with a sign flip). \[ \boxed{h(t) = ae^{-at}} \]

Answer 2

As an alternative to differentiating directly, use the Laplace-domain relationship between step and impulse response and check it against each option.

1. Option \( ae^{-at} \): The Laplace transform of the given step response is \( S(s) = \dfrac{1}{s} - \dfrac{1}{s+a} = \dfrac{a}{s(s+a)} \). Since \( S(s) = H(s)/s \) for a step input, the impulse response's transform is \( H(s) = sS(s) = \dfrac{a}{s+a} \), whose inverse Laplace transform is exactly \( ae^{-at} \), matching this option perfectly.
2. Option \( a^{-1}e^{-at} \): Its Laplace transform is \( \dfrac{1/a}{s+a} \), which does not equal \( \dfrac{a}{s+a} \) unless \( a^2 = 1 \); it fails to match \( H(s) \) in general.
3. Option \( (1-a^{-1})e^{-at} \): Its transform is \( \dfrac{1 - 1/a}{s+a} \), which again does not reduce to \( \dfrac{a}{s+a} \) for a general value of \( a \), so it is inconsistent with \( H(s) \).
4. Option \( (1-a)e^{-at} \): Its transform is \( \dfrac{1-a}{s+a} \), which likewise does not equal \( \dfrac{a}{s+a} \) except for a special coincidental value of \( a \), so it does not hold as a general identity.

Matching each candidate's Laplace transform against the one actually implied by \( sS(s) \) singles out only one consistent option.

So the correct answer is \( ae^{-at} \).