Question

A linear system at rest is subject to an input signal \(r(t) = 1 - e^{-t}\). The response of the system for t>0 is given by \(c(t) = 1 - e^{-2t}\). The transfer function of the system is:

• A. \( (s+2)/(s+1) \)
• B. \( 2(s+1)/(s+2) \)
• C. \( (s+1)/(s+2) \)
• D. \( (s+1)/2(s+2) \)

Hint:

Transfer Function. \(H(s) = C(s)/R(s)\), where C(s) and R(s) are the Laplace transforms of the output and input signals, respectively. Use standard Laplace transform pairs and properties. \(\mathcal{L\{1\ = 1/s\), \(\mathcal{L\{e^{-at\ = 1/(s+a)\).

Answer 1

The Correct Option is B

The transfer function \(H(s)\) of a linear time-invariant (LTI) system is the ratio of the Laplace transform of the output (\(C(s)\)) to the Laplace transform of the input (\(R(s)\)), assuming zero initial conditions
$$ H(s) = \frac{C(s)}{R(s)} $$ First, find the Laplace transform of the input signal \(r(t) = 1 - e^{-t}\), for \(t>0\)
Using standard Laplace transforms: \( \mathcal{L}\{1\} = \frac{1}{s} \) and \( \mathcal{L}\{e^{-at}\} = \frac{1}{s+a} \)
$$ R(s) = \mathcal{L}\{1 - e^{-t}\} = \mathcal{L}\{1\} - \mathcal{L}\{e^{-t}\} = \frac{1}{s} - \frac{1}{s+1} = \frac{(s+1) - s}{s(s+1)} = \frac{1}{s(s+1)} $$ Next, find the Laplace transform of the output signal \(c(t) = 1 - e^{-2t}\), for \(t>0\)
$$ C(s) = \mathcal{L}\{1 - e^{-2t}\} = \mathcal{L}\{1\} - \mathcal{L}\{e^{-2t}\} = \frac{1}{s} - \frac{1}{s+2} = \frac{(s+2) - s}{s(s+2)} = \frac{2}{s(s+2)} $$ Now, calculate the transfer function: $$ H(s) = \frac{C(s)}{R(s)} = \frac{2 / (s(s+2))}{1 / (s(s+1))} $$ $$ H(s) = \frac{2}{s(s+2)} \times \frac{s(s+1)}{1} $$ $$ H(s) = \frac{2(s+1)}{s+2} $$ This matches option (2)