Let a causal LTI system be governed by the following differential equation \(\rm y(t) + \frac{1}{4} \frac{dy}{dt} = 2x(t)\) where x(𝑡) and y(𝑡) are the input and output respectively. Its impulse response is
To find the impulse response, denoted by \(h(t)\), of a causal Linear Time-Invariant (LTI) system described by a differential equation, we set the input \(x(t)\) to the Dirac delta function, \(\delta(t)\), and the output \(y(t)\) becomes the impulse response \(h(t)\). We also assume zero initial conditions.
The given differential equation is:
\(\rm y(t) + \frac{1}{4} \frac{dy}{dt} = 2x(t)\)
Substitute \(x(t) = \delta(t)\) and \(y(t) = h(t)\):
\(\rm h(t) + \frac{1}{4} \frac{dh}{dt} = 2\delta(t)\)
Solving using Laplace Transform
We use the Laplace transform to solve this differential equation. Let \(H(s)\) be the Laplace transform of \(h(t)\). The Laplace transform properties we need are:
- \(\mathcal{L}\{h(t)\} = H(s)\)
- \(\mathcal{L}\{\frac{dh}{dt}\} = sH(s) - h(0^-)\)
- \(\mathcal{L}\{\delta(t)\} = 1\)
Since the system is causal and we assume zero initial conditions for the impulse response calculation, \(h(0^-) = 0\).
Applying the Laplace transform to the differential equation:
\(\rm H(s) + \frac{1}{4} [sH(s) - h(0^-)] = 2(1)\)
\(\rm H(s) + \frac{1}{4} sH(s) = 2\)
Calculating the Transfer Function H(s)
Factor out \(H(s)\) from the equation:
\(\rm H(s) \left( 1 + \frac{s}{4} \right) = 2\)
Combine the terms inside the parenthesis:
\(\rm H(s) \left( \frac{4+s}{4} \right) = 2\)
Solve for \(H(s)\):
\(\rm H(s) = \frac{2 \times 4}{s+4}\)
\(\rm H(s) = \frac{8}{s+4}\)
This \(H(s)\) is the transfer function of the system.
Finding the Impulse Response h(t)
To find the impulse response \(h(t)\), we take the inverse Laplace transform of \(H(s)\):
\(\rm h(t) = \mathcal{L}^{-1}\{H(s)\} = \mathcal{L}^{-1}\left\{ \frac{8}{s+4} \right\}\)
We know that the inverse Laplace transform of \(\frac{1}{s+a}\) is \(e^{-at}u(t)\), where \(u(t)\) is the unit step function.
Therefore:
\(\rm h(t) = 8e^{-4t}u(t)\)
This represents the impulse response of the given causal LTI system.
Important Questions from Continuous Time LTI Systems
The continuous time system described by the equation y(t) = x(t2) comes under the category of -
A continuous time LTI system is described by
\(\dfrac{d^2y(t)}{dt^2} + 4 \dfrac{dy(t)}{dt} + 3y(t) = 2 \dfrac{dx(t)}{dt} + 4x(t)\)
Assuming zero initial conditions, the response y(t) of the above system for the input x(t) = e-2t u(t) is given by
Consider a continuous-time system with input x(t) and output y(t) given by
y(t) = x(t)cos(t)
This system is
Let an input x(t) = 2 sin(10πt) + 5 cos(15πt) + 7 sin(42πt) + 4 cos(45πt) is passed through an LTI system having an impulse response
\(\rm h(t) = 2 \left( \frac{\sin (10 \pi t)}{\pi t} \right) \cos (40 \pi t)\)
The output of the system is
A continuous-time system that is initially at rest is described by
\(\rm \frac{dy(t)}{dt}+3y(t)=2x(t)\),
where 𝑥(𝑡) is the input voltage and 𝑦(𝑡) is the output voltage. The impulse response of the system is
