To find the time response expression for a standard first order system when subjected to a unit step function as the input, we need to follow several steps involving Laplace transforms. This process helps us understand how the system's output behaves over time after a sudden change in input.

First Order System Transfer Function

A standard first order system is characterized by its transfer function, which describes the relationship between the output and input in the Laplace domain. For a standard first-order system, the transfer function \(G(s)\) is given by:

\[G(s) = \frac{C(s)}{R(s)} = \frac{K}{Ts+1}\]

Where:

• \(C(s)\) is the Laplace transform of the output.
• \(R(s)\) is the Laplace transform of the input.
• \(K\) is the DC gain (for unity gain, \(K=1\)).
• \(T\) is the time constant of the system.

For a simplified standard first order system with unity gain, the transfer function is:

\[G(s) = \frac{1}{Ts+1}\]

Input Signal Definition

The question specifies a unit step function as the input. In the time domain, a unit step function, denoted as \(u(t)\), is defined as:

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

The Laplace transform of a unit step function, \(R(s)\), is:

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

Calculating Time Response

To find the time response expression, \(c(t)\), we first need to determine the output in the Laplace domain, \(C(s)\), by multiplying the system's transfer function \(G(s)\) by the input's Laplace transform \(R(s)\):

\[C(s) = G(s) \cdot R(s)\] \[C(s) = \left(\frac{1}{Ts+1}\right) \cdot \left(\frac{1}{s}\right)\] \[C(s) = \frac{1}{s(Ts+1)}\]

Next, we use partial fraction expansion to break down \(C(s)\) into simpler terms, which can then be easily transformed back into the time domain using inverse Laplace transforms.

Let: \[\frac{1}{s(Ts+1)} = \frac{A}{s} + \frac{B}{Ts+1}\]

To find the constants \(A\) and \(B\), we can multiply both sides by \(s(Ts+1)\): \[1 = A(Ts+1) + Bs\]

Step 1: Find A by setting \(s = 0\) \[1 = A(T \cdot 0 + 1) + B \cdot 0\] \[1 = A(1)\] \[A = 1\]

Step 2: Find B by setting \(Ts+1 = 0 \implies s = -\frac{1}{T}\) \[1 = A\left(T \left(-\frac{1}{T}\right) + 1\right) + B \left(-\frac{1}{T}\right)\] \[1 = A(-1 + 1) - \frac{B}{T}\] \[1 = A(0) - \frac{B}{T}\] \[1 = -\frac{B}{T}\] \[B = -T\]

Substitute the values of \(A\) and \(B\) back into the partial fraction expression for \(C(s)\): \[C(s) = \frac{1}{s} + \frac{-T}{Ts+1}\] \[C(s) = \frac{1}{s} - \frac{T}{T(s+\frac{1}{T})}\] \[C(s) = \frac{1}{s} - \frac{1}{s+\frac{1}{T}}\]

Finally, we apply the inverse Laplace transform to \(C(s)\) to obtain the time response expression \(c(t)\):

Using standard Laplace transform pairs:

Applying these rules to our expression for \(C(s)\): \[c(t) = \mathcal{L}^{-1}\left\{ \frac{1}{s} - \frac{1}{s+\frac{1}{T}} \right\}\] \[c(t) = \mathcal{L}^{-1}\left\{ \frac{1}{s} \right\} - \mathcal{L}^{-1}\left\{ \frac{1}{s+\frac{1}{T}} \right\}\] \[c(t) = 1 - e^{-\frac{t}{T}}\]

This expression represents the output of a standard first order system over time when a unit step function is applied as the input. This is a common and fundamental result in control systems engineering, describing the exponential rise of the output towards a steady-state value of 1.