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
2 e-3t u(t)
Understanding the System and Impulse Response
The problem asks for the impulse response of a continuous-time system described by a differential equation. The system is linear, time-invariant (LTI), and initially at rest. The governing differential equation is given as:
$$ \rm \frac{dy(t)}{dt}+3y(t)=2x(t) $$
Here, x(t) represents the input voltage and y(t) represents the output voltage. The impulse response, denoted by h(t), is the output of the system when the input x(t) is the Dirac delta function, \(\delta(t)\). Since the system is initially at rest, all initial conditions (like initial voltage or derivatives) are zero.
Finding the Impulse Response using Laplace Transform
To find the impulse response h(t), we replace x(t) with \(\delta(t)\) in the system's differential equation and solve for y(t) (which will be h(t) in this case).
The equation becomes:
$$ \rm \frac{dh(t)}{dt}+3h(t)=2\delta(t) $$
We will use the Laplace transform method to solve this differential equation. The Laplace transform converts the differential equation in the time domain to an algebraic equation in the frequency domain (s-domain).
Step-by-Step Calculation
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Laplace Transform of the Equation:
Take the Laplace transform of both sides of the equation. We use the properties:
- \(\mathcal{L}\left\{\frac{dh(t)}{dt}\right\} = sH(s) - h(0^-)\)
- \(\mathcal{L}\{h(t)\} = H(s)\)
- \(\mathcal{L}\{\delta(t)\} = 1\)
$$ sH(s) - 0 + 3H(s) = 2(1) $$
$$ sH(s) + 3H(s) = 2 $$
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Solve for H(s):
Factor out \(H(s)\):
$$ H(s)(s + 3) = 2 $$
Isolate \(H(s)\):$$ H(s) = \frac{2}{s+3} $$
This \(H(s)\) is the transfer function of the system. -
Inverse Laplace Transform:
Now, we find the inverse Laplace transform of \(H(s)\) to get the impulse response \(h(t)\). Recall the standard Laplace transform pair:
$$ \mathcal{L}^{-1}\left\{\frac{a}{s+b}\right\} = ae^{-bt}u(t) $$
Comparing our \(H(s) = \frac{2}{s+3}\) with the standard form, we have \(a=2\) and \(b=3\). Therefore, the inverse Laplace transform is:$$ h(t) = \mathcal{L}^{-1}\left\{\frac{2}{s+3}\right\} = 2e^{-3t}u(t) $$
The term \(u(t)\) represents the unit step function, indicating that the impulse response is causal (starts at t=0).
Comparing with Options
The calculated impulse response is \(2e^{-3t}u(t)\). Let's compare this with the given options:
- Option 1: \(3e^{-2t}\) - Incorrect.
- Option 2: \(\rm \frac{1}{3}e^{-2t}u(t)\) - Incorrect.
- Option 3: \(2 e^{-3t} u(t)\) - Matches the calculated result.
- Option 4: \(2e^{-3t}\) - Incorrect (missing the unit step function \(u(t)\)).
- Option 5: Not specified.
Conclusion
The impulse response of the given continuous-time system is \(2e^{-3t}u(t)\), which corresponds to Option 3.
Important Questions from Continuous Time LTI Systems
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The output of the system is
