# Step-by-Step Solution Let's break down the problem: ## **Step 1: Understanding the First Order System Response** A **first order system** to a unit step input has the following time response: \[ y(t) = K \left(1 - e^{-t/\tau}\right) \] Where: - \( K \) = steady-state gain (final value) - \( \tau \) = time constant --- ## **Step 2: Identifying the Steady-State Value (K)** Look at the graph: - As \( t \to \infty \), \( y(t) \to 1 \) - So, **\( K = 1 \)** --- ## **Step 3: Finding the Time Constant (\( \tau \))** The **time constant (\( \tau \))** is the time at which the response reaches **63.2%** of its final value. \[ y(\tau) = K \left(1 - e^{-1}\right) = K(1 - .368) = .632 K \] From the graph: - Final value ≈ 1 - 63.2% of 1 ≈ .632 Find where \( y(t) = .632 \) on the graph: - This occurs at **\( t \approx 2 \) seconds** So, **\( \tau = 2 \) seconds**. --- ## **Step 4: Writing the System Equation** Substitute \( K = 1 \) and \( \tau = 2 \) into the general equation: \[ y(t) = 1 - e^{-t/2} \] --- ## **Step 5: Final Answers** ### **Time Constant** \[ \boxed{2 \ \text{seconds}} \] ### **System Equation** \[ \boxed{y(t) = 1 - e^{-t/2}} \] --- ## **Summary Table** | Parameter | Value | |-----------------|---------------------| | Time constant (\(\tau\)) | 2 seconds | | System Equation | \(y(t) = 1 - e^{-t/2}\) | --- Let me know if you need more explanation on any step!