System Transfer Function Derivation

To determine the transfer function of a linear time-invariant (LTI) system, we use the fundamental concept that for a system initially at rest, the transfer function \(H(s)\) is the ratio of the Laplace transform of its output \(Y(s)\) to the Laplace transform of its input \(X(s)\).

The formula for the transfer function is given by:

\[H(s) = \frac{Y(s)}{X(s)}\]

• \(Y(s)\) represents the Laplace transform of the system's output signal, \(y(t)\).
• \(X(s)\) represents the Laplace transform of the system's input signal, \(x(t)\).

Input Signal Properties

The problem states that the LTI system is subjected to a unit step input. A unit step function, commonly denoted as \(u(t)\) or \(x(t)\), is defined as:

\[u(t) = \begin{cases} 1 & \text{for } t \ge 0 \\ 0 & \text{for } t < 0 \end{cases}\]

The Laplace transform of the unit step input \(u(t)\) is a standard result:

\[X(s) = L\{u(t)\} = \frac{1}{s}\]

Output Response Transformation

The given output response of the system for \(t > 0\) is \(y(t) = te^{-t}\).

To find its Laplace transform \(Y(s)\), we can use known Laplace transform properties:

First, recall the Laplace transform of \(t\):

\[L\{t\} = \frac{1}{s^2}\]

Next, we apply the frequency shifting property (also known as the complex frequency shift theorem). This property states that if \(L\{f(t)\} = F(s)\), then \(L\{e^{at}f(t)\} = F(s-a)\).

In our case, \(f(t) = t\) and the exponential term is \(e^{-t}\), which means \(a = -1\).

Applying the frequency shifting property to \(y(t) = te^{-t}\):

\[Y(s) = L\{te^{-t}\} = \left. \frac{1}{s^2} \right|_{s \to s - (-1)} = \left. \frac{1}{s^2} \right|_{s \to s + 1}\]

Therefore, the Laplace transform of the output response is:

\[Y(s) = \frac{1}{{(s+1)^2}}\]

Transfer Function Derivation

Now, we can calculate the transfer function \(H(s)\) using the derived \(Y(s)\) and \(X(s)\):

\[H(s) = \frac{Y(s)}{X(s)}\]

Substitute the expressions for \(Y(s)\) and \(X(s)\):

\[H(s) = \frac{\frac{1}{{(s+1)^2}}}{\frac{1}{s}}\]

To simplify this complex fraction, we invert the denominator and multiply:

\[H(s) = \frac{1}{{(s+1)^2}} \times s\]

\[H(s) = \frac{s}{{(s+1)^2}}\]

Solution Options Comparison

Let's compare our derived transfer function with the given options to find the correct one:

• Option 1: \(\frac{1}{{{{\left( {s + 1} \right)}^2}}}\)
• Option 2: \(\frac{1}{{s{{\left( {s + 1} \right)}^2}}}\)
• Option 3: \(\frac{s}{{{{\left( {s + 1} \right)}^2}}}\)
• Option 4: \(\frac{1}{{s + 1}}\)

The calculated transfer function, \(H(s) = \frac{s}{{(s+1)^2}}\), perfectly matches Option 3.