Let's break down your problem and solve it step by step. --- ### **Given:** 1. **Input \( r(t) \):** \[ r(t) = \begin{cases} 3, & t \geq 0 \\ 0, & t < 0 \end{cases} \] (i.e., a step input of magnitude 3 at \( t = 0 \)) 2. **Transfer function \( T(p) \) (assuming \( p = s \) in Laplace):** \[ T(p) = \frac{24p + 208}{p^2 + 12p + 52} \] --- ## **Step 1: Find Laplace Transform of Input** For a step input of magnitude 3: \[ r(t) = 3u(t) \] \[ R(s) = 3 \cdot \frac{1}{s} = \frac{3}{s} \] --- ## **Step 2: Output in Laplace Domain** Output is: \[ Y(s) = T(s) \cdot R(s) = \frac{24s + 208}{s^2 + 12s + 52} \cdot \frac{3}{s} \] \[ Y(s) = \frac{3(24s + 208)}{s(s^2 + 12s + 52)} \] --- ## **Step 3: Partial Fraction Decomposition** Let’s write: \[ Y(s) = \frac{3(24s + 208)}{s(s^2 + 12s + 52)} \] Let’s expand numerator: \[ 24s + 208 = 24(s + 8.67) \text{ (but let's keep as is for now)} \] Let’s write: \[ \frac{3(24s + 208)}{s(s^2 + 12s + 52)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 12s + 52} \] Multiply both sides by denominator: \[ 3(24s + 208) = A(s^2 + 12s + 52) + (Bs + C)s \] \[ 3(24s + 208) = A(s^2 + 12s + 52) + Bs^2 + Cs \] Expand: \[ 3(24s + 208) = A s^2 + 12A s + 52A + B s^2 + C s \] Group terms by powers of \( s \): - \( s^2 \) terms: \( (A + B) s^2 \) - \( s \) terms: \( (12A + C) s \) - constants: \( 52A \) Set equal to left side, which is: - \( s \) term: \( 72s \) - constant: \( 624 \) - (no \( s^2 \) term) So: - \( s^2 \): \( (A + B) \) must be zero - \( s \): \( (12A + C) = 72 \) - const: \( 52A = 624 \) Solve for \( A \): \[ 52A = 624 \implies A = \frac{624}{52} = 12 \] Now, \( A + B = 0 \implies B = -A = -12 \) Now, \( 12A + C = 72 \implies 12 \times 12 + C = 72 \implies 144 + C = 72 \implies C = 72 - 144 = -72 \) So: - \( A = 12 \) - \( B = -12 \) - \( C = -72 \) So: \[ Y(s) = \frac{12}{s} + \frac{(-12)s - 72}{s^2 + 12s + 52} \] \[ Y(s) = \frac{12}{s} - \frac{12s + 72}{s^2 + 12s + 52} \] --- ## **Step 4: Inverse Laplace Transform** Let's write the denominator in completed square: \[ s^2 + 12s + 52 = (s + 6)^2 + 16 \] So: \[ Y(s) = \frac{12}{s} - \frac{12s + 72}{(s + 6)^2 + 16} \] Rewrite numerator: \( 12s + 72 = 12(s + 6) \) So: \[ Y(s) = \frac{12}{s} - \frac{12(s + 6)}{(s + 6)^2 + 16} \] Now, let's write: \[ Y(s) = \frac{12}{s} - \frac{12(s + 6)}{(s + 6)^2 + 16} \] We can now write the time-domain inverse Laplace: - \( \mathcal{L}^{-1} \left\{\frac{1}{s}\right\} = 1 \) - \( \mathcal{L}^{-1} \left\{\frac{s + a}{(s + a)^2 + \omega^2}\right\} = e^{-a t} \cos \omega t \) So, in our case: - \( a = 6 \) - \( \omega = 4 \) So, \[ y(t) = 12 \cdot 1 - 12 e^{-6t} \cos(4t) \] \[ y(t) = 12 \left[1 - e^{-6t} \cos(4t)\right] \quad \boxed{} \] --- ## **Step 5: Sketch the Output Waveform** - At \( t = 0 \): \( y(0) = 12[1 - 1] = 0 \) - As \( t \to \infty \): \( e^{-6t} \to 0 \implies y(\infty) = 12 \) - The output rises from 0, with a decaying oscillation (since cosine term), to 12. ### **Sketch:** - x-axis: \( t \) - y-axis: \( y(t) \) - Starts at 0, oscillates with decreasing amplitude, approaches 12 as \( t \to \infty \). --- ## **Step 6: Response to a Unit Step Input** **For unit step (\( r(t) = u(t) \)):** \[ R(s) = \frac{1}{s} \] \[ Y_{unit}(s) = T(s) \times R(s) = \frac{24s + 208}{s(s^2 + 12s + 52)} \] We can do partial fraction as before, but with \( 3 \to 1 \): Let’s just divide by 3: From earlier, for 3, numerator was \( 3(24s + 208) \), and partial fractions gave \( 12/s \), etc. So for 1, \[ Y_{unit}(s) = \frac{24s + 208}{s(s^2 + 12s + 52)} \] By analogy, the coefficients are divided by 3: - \( \frac{12}{s} \to \frac{4}{s} \) - \( -\frac{12(s+6)}{(s+6)^2 + 16} \to -\frac{4(s+6)}{(s+6)^2 + 16} \) So, \[ Y_{unit}(s) = \frac{4}{s} - \frac{4(s+6)}{(s+6)^2 + 16} \] So, \[ y_{unit}(t) = 4[1 - e^{-6 t} \cos 4 t] \] --- ### **Steady State Value:** \[ \lim_{t \to \infty} y_{unit}(t) = 4[1 - 0] = 4 \] --- ## **Step 7: Conclusion** - **For a step input of 3:** The system responds as: \[ y(t) = 12[1 - e^{-6 t} \cos 4 t] \] - Starts from 0, rises with a decaying oscillatory pattern to a steady state value of 12. - **For a unit step input:** The response is: \[ y_{unit}(t) = 4[1 - e^{-6 t} \cos 4 t] \] - Starts from 0, rises with a decaying oscillatory pattern to a steady state value of 4. **The system is underdamped (oscillatory) but stable (since all exponents are negative), with a steady-state gain of 4 for a unit step.** --- ### **Sketch (for both cases):** The waveform is an exponentially decaying cosine, starting at 0 at \( t=0 \), oscill