Let's follow the instructions **step-by-step** (using the selected values from the crossed boxes in the table): - **4th digit of student ID:** 3 → \( K_D = 2.5 \) - **5th digit of student ID:** 2 → \( T_1 = .3 \) s We are asked to **draw the ramp response of a DT1 system** with these values. --- ## 1. **DT1 System Transfer Function** A DT1 system has the transfer function: \[ G(s) = \frac{K_D}{T_1 s + 1} \] For a **ramp input** (\( r(t) = t \)), the Laplace transform is \( R(s) = \frac{1}{s^2} \). The output in Laplace domain: \[ Y(s) = G(s) \cdot R(s) = \frac{K_D}{T_1 s + 1} \cdot \frac{1}{s^2} \] --- ## 2. **Find the Time Response** Let's inverse Laplace: \[ Y(s) = \frac{K_D}{T_1} \cdot \frac{1}{s^2 (s + \frac{1}{T_1})} \] Using partial fractions: \[ \frac{1}{s^2 (s + a)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + a} \] where \( a = \frac{1}{T_1} \). Solving for \( A, B, C \): \[ \frac{1}{s^2 (s + a)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + a} \] \[ 1 = A s (s + a) + B (s + a) + C s^2 \] Let \( s = \): \( 1 = B \cdot a \implies B = \frac{1}{a} = T_1 \) Let \( s = -a \): \( 1 = C a^2 \implies C = \frac{1}{a^2} = T_1^2 \) For \( A \): Expand and collect coefficients: \[ 1 = A s (s + a) + B (s + a) + C s^2 \] Plug in \( B = T_1, C = T_1^2 \): \[ 1 = A s (s + a) + T_1 (s + a) + T_1^2 s^2 \] Expand: \[ 1 = A (s^2 + a s) + T_1 s + T_1 a + T_1^2 s^2 \] \[ 1 = (A + T_1^2) s^2 + (A a + T_1) s + T_1 a \] Match constant term: \( T_1 a = 1 \Rightarrow a = \frac{1}{T_1} \) (as before) Match \( s \) coefficient: \( A a + T_1 = \implies A = -\frac{T_1}{a} = -T_1^2 \) Match \( s^2 \) coefficient: \( A + T_1^2 = \implies A = -T_1^2 \) (consistent!) So, the partial fractions: \[ \frac{1}{s^2 (s + a)} = -T_1^2 \frac{1}{s} + T_1 \frac{1}{s^2} + T_1^2 \frac{1}{s + a} \] Multiply by \( \frac{K_D}{T_1} \): \[ Y(s) = K_D \left[ -T_1 \frac{1}{s} + \frac{1}{s^2} + T_1 \frac{1}{s + a} \right] \] Inverse Laplace: - \( \mathcal{L}^{-1}\left[ \frac{1}{s^2} \right] = t \) - \( \mathcal{L}^{-1}\left[ \frac{1}{s} \right] = 1 \) - \( \mathcal{L}^{-1}\left[ \frac{1}{s + a} \right] = e^{-a t} \) So: \[ y(t) = K_D \left[ -T_1 \cdot 1 + t + T_1 (1 - e^{-t/T_1}) \right] \] \[ y(t) = K_D \left[ t - T_1 e^{-t/T_1} \right] \] --- ## 3. **Plug in the Values** Given: - \( K_D = 2.5 \) - \( T_1 = .3 \) s \[ y(t) = 2.5 \left[ t - .3 \cdot e^{-t/.3} \right] \] --- ## 4. **Graphing Instructions** - **Time axis:** \( 5\,\text{cm} \equiv 1\,\text{sec} \) - **Output axis:** \( 2\,\text{cm} \equiv 1 \) - **Label \( K_D \) (slope for large \( t \)) and \( T_1 \) (time constant) on the graph.** --- ## 5. **Key Points for Plot** - At \( t = \): \( y() = 2.5[ - .3 \cdot 1] = -.75 \) - As \( t \to \infty \): \( e^{-t/.3} \to \), \( y(t) \to 2.5 t \) - The curve starts at \( -.75 \), then rises and approaches a straight line with slope \( 2.5 \). --- ### Example Table of Values | \( t \) (s) | \( e^{-t/.3} \) | \( y(t) \) | |-------------|------------------|--------------------| | | 1 | -.75 | | .2 | .5134 | 2.5[.2 - .3*.5134] ≈ 2.5[.2 - .154] ≈ 2.5*.046 ≈ .115 | | .5 | .1889 | 2.5[.5 - .3*.1889] ≈ 2.5[.5 - .0567] ≈ 2.5*.443 ≈ 1.108 | | 1. | .0357 | 2.5[1. - .3*.0357] ≈ 2.5[1. - .0107] ≈ 2.5*.989 ≈ 2.472 | | 2. | .00127 | 2.5[2. - .3*.00127] ≈ 2.5[2. - .00038] ≈ 2.5*1.99962 ≈ 4.999 | --- ## 6. **How to Draw** 1. Draw time axis, label every 5 cm as 1 sec. 2. Draw output axis, label every 2 cm as 1. 3. Plot the starting point at \((, -.75)\). 4. Plot points for \( t = .2, .5, 1., 2. \) as above. 5. Draw a tangent for large \( t \), which is a straight line with slope \( K_D = 2.5 \). 6. Mark the time constant \( T_1 = .3 \) s (the time it takes for the exponential to decay to \( 1/e \)). --- ## 7. **Final Answer (Summary and Formula)** **Ramp response of the DT1 system:** \[ \boxed{ y(t) = 2.5 \left[ t - .3 \, e^{-t/.3} \right] } \] - Initial value: \( y() = -.75 \) - Slope for large \( t \): \( 2.5 \) - Time constant: \( T_1 = .3 \) s --- ### **Graph (Description)** - The curve starts at \( y() = -.75 \), rises quickly, and for large \( t \), becomes a straight line with slope \( 2.5 \). - Mark \( K_D = 2.5 \) as the tangent slope. - Mark \( T_1 = .3 \) s on the time axis to show the time constant. --- If you want a plotted graph, let me know!