Laplace Transform 1A8080

1. **Problem Statement:** Find the Laplace Transform $\mathcal{L}\{f(t)\}$ of the piecewise function: $$ f(t) = \begin{cases} t^3 & m+1 < t < m+2 \\ 4 - t^2 & m+2 < t < m+4 \\ 1 & \text{otherwise} \end{cases} $$ where $m$ is the last digit of your student ID. 2. **Recall the Laplace Transform definition:** $$ \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) dt $$ Since $f(t)$ is piecewise, split the integral accordingly: $$ \mathcal{L}\{f(t)\} = \int_0^{m+1} e^{-st} \cdot 1 \, dt + \int_{m+1}^{m+2} e^{-st} t^3 \, dt + \int_{m+2}^{m+4} e^{-st} (4 - t^2) \, dt + \int_{m+4}^{\infty} e^{-st} \cdot 1 \, dt $$ 3. **Calculate each integral separately:** **Integral 1:** $$ I_1 = \int_0^{m+1} e^{-st} dt = \left[-\frac{e^{-st}}{s}\right]_0^{m+1} = \frac{1 - e^{-s(m+1)}}{s} $$ **Integral 2:** $$ I_2 = \int_{m+1}^{m+2} e^{-st} t^3 dt $$ Use integration by parts repeatedly or known formulas. Let’s use the formula for Laplace transform of $t^n$ shifted by $a$: Rewrite $I_2$ as: $$ I_2 = \int_{m+1}^{m+2} t^3 e^{-st} dt $$ Integration by parts 4 times yields: General formula: $$ \int t^n e^{-st} dt = -\frac{t^n e^{-st}}{s} + \frac{n}{s} \int t^{n-1} e^{-st} dt $$ Apply stepwise: - For $n=3$: $$ I_2 = -\frac{t^3 e^{-st}}{s} + \frac{3}{s} \int t^2 e^{-st} dt $$ - For $\int t^2 e^{-st} dt$: $$ = -\frac{t^2 e^{-st}}{s} + \frac{2}{s} \int t e^{-st} dt $$ - For $\int t e^{-st} dt$: $$ = -\frac{t e^{-st}}{s} + \frac{1}{s} \int e^{-st} dt = -\frac{t e^{-st}}{s} - \frac{1}{s^2} e^{-st} + C $$ Putting all together: $$ I_2 = -\frac{t^3 e^{-st}}{s} + \frac{3}{s} \left(-\frac{t^2 e^{-st}}{s} + \frac{2}{s} \left(-\frac{t e^{-st}}{s} - \frac{1}{s^2} e^{-st} \right) \right) + C $$ Simplify: $$ I_2 = e^{-st} \left(-\frac{t^3}{s} - \frac{3 t^2}{s^2} - \frac{6 t}{s^3} - \frac{6}{s^4} \right) + C $$ Evaluate definite integral from $t=m+1$ to $t=m+2$: $$ I_2 = \left[ e^{-st} \left(-\frac{t^3}{s} - \frac{3 t^2}{s^2} - \frac{6 t}{s^3} - \frac{6}{s^4} \right) \right]_{m+1}^{m+2} $$ **Integral 3:** $$ I_3 = \int_{m+2}^{m+4} e^{-st} (4 - t^2) dt = 4 \int_{m+2}^{m+4} e^{-st} dt - \int_{m+2}^{m+4} t^2 e^{-st} dt $$ Calculate each part: - First part: $$ 4 \int_{m+2}^{m+4} e^{-st} dt = 4 \left[-\frac{e^{-st}}{s}\right]_{m+2}^{m+4} = \frac{4}{s} \left(e^{-s(m+2)} - e^{-s(m+4)}\right) $$ - Second part $\int t^2 e^{-st} dt$: Use integration by parts similarly: $$ \int t^2 e^{-st} dt = e^{-st} \left(-\frac{t^2}{s} - \frac{2 t}{s^2} - \frac{2}{s^3} \right) + C $$ Evaluate from $m+2$ to $m+4$: $$ \left[ e^{-st} \left(-\frac{t^2}{s} - \frac{2 t}{s^2} - \frac{2}{s^3} \right) \right]_{m+2}^{m+4} $$ So, $$ I_3 = \frac{4}{s} \left(e^{-s(m+2)} - e^{-s(m+4)}\right) - \left[ e^{-st} \left(-\frac{t^2}{s} - \frac{2 t}{s^2} - \frac{2}{s^3} \right) \right]_{m+2}^{m+4} $$ **Integral 4:** $$ I_4 = \int_{m+4}^{\infty} e^{-st} dt = \left[-\frac{e^{-st}}{s}\right]_{m+4}^{\infty} = \frac{e^{-s(m+4)}}{s} $$ 4. **Combine all parts:** $$ \mathcal{L}\{f(t)\} = I_1 + I_2 + I_3 + I_4 $$ Substitute all expressions: $$ = \frac{1 - e^{-s(m+1)}}{s} + \left[ e^{-st} \left(-\frac{t^3}{s} - \frac{3 t^2}{s^2} - \frac{6 t}{s^3} - \frac{6}{s^4} \right) \right]_{m+1}^{m+2} + \frac{4}{s} \left(e^{-s(m+2)} - e^{-s(m+4)}\right) - \left[ e^{-st} \left(-\frac{t^2}{s} - \frac{2 t}{s^2} - \frac{2}{s^3} \right) \right]_{m+2}^{m+4} + \frac{e^{-s(m+4)}}{s} $$ 5. **Final simplified expression:** $$ \boxed{ \begin{aligned} \mathcal{L}\{f(t)\} = & \frac{1 - e^{-s(m+1)}}{s} + e^{-s(m+2)} \left(-\frac{(m+2)^3}{s} - \frac{3 (m+2)^2}{s^2} - \frac{6 (m+2)}{s^3} - \frac{6}{s^4} \right) \\ & - e^{-s(m+1)} \left(-\frac{(m+1)^3}{s} - \frac{3 (m+1)^2}{s^2} - \frac{6 (m+1)}{s^3} - \frac{6}{s^4} \right) + \frac{4}{s} \left(e^{-s(m+2)} - e^{-s(m+4)}\right) \\ & - e^{-s(m+4)} \left(-\frac{(m+4)^2}{s} - \frac{2 (m+4)}{s^2} - \frac{2}{s^3} \right) + e^{-s(m+2)} \left(-\frac{(m+2)^2}{s} - \frac{2 (m+2)}{s^2} - \frac{2}{s^3} \right) + \frac{e^{-s(m+4)}}{s} \end{aligned} } $$ This expression fully characterizes the Laplace Transform of the given piecewise CPU load function.