System Step Response Basics

A stable first-order linear time-invariant (LTI) system's response to a step input can be modeled using a standard formula. The output signal, \( y(t) \), over time \( t \) is given by:

$ y(t) = K \left( 1 - e^{-t/\tau} \right) $

In this equation:

• \( y(t) \) represents the system's output value at a specific time \( t \).
• \( K \) signifies the final, steady-state value that the output approaches.
• \( \tau \) is the system's time-constant, a crucial parameter indicating how quickly the system responds to input changes. Specifically, it's the time required to reach about 63.2% of the total change from the initial value to the final value.

Determining Final Value (K)

The provided data includes measurements of the system's output at different times following a step input. We need to identify the steady-state output, which is the value the output settles to as time becomes very large.

Given Step Response Data:

Observing the data, we see that the output value reaches 3 and stays there as time approaches infinity (\( t \to \infty \)). Therefore, the final value gain of the system is \( K = 3 \).

Calculating Time-Constant (τ)

With the final value \( K \) known, we can use the step response equation \( y(t) = 3 \left( 1 - e^{-t/\tau} \right) \) and the other data points to calculate the time-constant \( \tau \). Let's rearrange the formula to solve for \( \tau \):

First, divide both sides by \( K \):

$ \frac{y(t)}{K} = 1 - e^{-t/\tau} $

Rearrange to isolate the exponential term:

$ e^{-t/\tau} = 1 - \frac{y(t)}{K} $

Now, apply the natural logarithm (\( \ln \)) to both sides:

$ \ln \left( e^{-t/\tau} \right) = \ln \left( 1 - \frac{y(t)}{K} \right) $

$ -\frac{t}{\tau} = \ln \left( 1 - \frac{y(t)}{K} \right) $

Finally, solve for \( \tau \):

$ \tau = -\frac{t}{\ln \left( 1 - \frac{y(t)}{K} \right)} $

Substitute \( K = 3 \) and use one of the data points (e.g., \( t = 1.6 \) sec, \( y(t) = 1.65 \)) to find \( \tau \):

$ \tau = -\frac{1.6}{\ln \left( 1 - \frac{1.65}{3} \right)} $

$ \tau = -\frac{1.6}{\ln(1 - 0.55)} $

$ \tau = -\frac{1.6}{\ln(0.45)} $

Using a calculator, \( \ln(0.45) \approx -0.798 \).

$ \tau \approx -\frac{1.6}{-0.798} \approx 2.005 \text{ sec} $

Let's verify with another point (e.g., \( t = 0.6 \) sec, \( y(t) = 0.78 \)):

$ \tau = -\frac{0.6}{\ln \left( 1 - \frac{0.78}{3} \right)} = -\frac{0.6}{\ln(1 - 0.26)} = -\frac{0.6}{\ln(0.74)} $

Using a calculator, \( \ln(0.74) \approx -0.301 \).

$ \tau \approx -\frac{0.6}{-0.301} \approx 1.99 \text{ sec} $

The results from different data points are very close to 2 seconds.

Conclusion on Time-Constant

Based on the analysis of the provided step response data and the standard first-order system model, the calculated time-constant \( \tau \) is approximately 2 seconds. This value represents the characteristic time for the system's response.