Section E.3 Laplace Transforms
💡 Derivation of \(\lap{u_c(t)}\).
Start with the definition of the Laplace transform:
\begin{equation*}
\lap{u_c(t)} = \int_{0}^{\infty} u_c(t) \, e^{-st} \, dt.
\end{equation*}
Recall that:
\begin{equation*}
u_c(t) =
\begin{cases}
0, \amp t \lt c \\
1, \amp t \ge c
\end{cases}
\end{equation*}
which means the integrand is zero until \(t=c\) and equals \(e^{-st}\) afterward.
That trims our integral down since:
\begin{equation*}
\lap{u_c(t)}
= \os{\large t\ \lt\ c}{\overline{\int_0^{c} (0)\ e^{-st}\ dt}} + \os{\large t\ \ge\ c}{\overline{\int_c^{\infty} (1)\ e^{-st}\ dt}}
= \int_c^{\infty} e^{-st}\ dt
\end{equation*}
Evaluate the improper integral as a limit:
\begin{align*}
\lap{u_c(t)} \amp = \lim_{b \to \infty} \int_{c}^{b} e^{-st}\ dt\\
\amp = \lim_{b \to \infty} \left[ \frac{-1}{s} e^{-st} \right]_{c}^{b} = \lim_{b \to \infty} \left( \frac{-1}{s}\big(e^{-sb} - e^{-sc}\big) \right)
\end{align*}
As long as \(s \gt 0\text{,}\) the term \(e^{-sb}\) goes to zero as \(b \to \infty\text{,}\) leaving:
\begin{equation*}
\lap{u_c(t)} = \frac{e^{-cs}}{s}.
\end{equation*}
💡 Derivation of \(\lap{f(t)\ u_c(t)}\).
Start with the Laplace transform definition:
\begin{equation*}
\lap{f(t) u_c(t)} = \int_{0}^{\infty} f(t) u_c(t) e^{-st}\ dt.
\end{equation*}
Since \(u_c(t) = 0\) for \(t \lt c\text{,}\) the integral simplifies to:
\begin{equation*}
\lap{f(t)\cdot u_c(t)} = \int_c^\infty f(t)\ e^{-st}\ dt.
\end{equation*}
This new integral looks a lot like the Laplace transform of \(f(t)\text{,}\) but it starts at \(t = c\) instead of \(t = 0\text{.}\) Luckily, we can shift the graph of \(f(t)\ e^{-st}\) to the left \(c\) units to force the integral start back at \(t = 0\text{.}\)
Note: Shifting a function \(c\) units to the left means adding \(c\) to all of the \(t\) inputs:
\begin{equation*}
\text{before shift:}\quad f(t)\ e^{-st}\qquad\text{after shift:}\quad f(t + c)\ e^{-s(t + c)}
\end{equation*}
Using this idea leads to the desired transform:
\begin{align*}
\int_{c}^{\infty} f(t)\ e^{-st}\ dt
\amp \os{\large\DLBb \text{A}}{=} \int_{0}^{\infty} f(t + c)\ e^{-s(t + c)}\ dt\\
\amp \os{\large\DLBb \text{B}}{=} \int_0^\infty f(t + c) \left(e^{-st} \cdot e^{-sc}\right)\ dt\\
\amp \os{\large\DLBb \text{C}}{=} e^{-cs} \int_0^\infty f(t + c) e^{-st}\ dt\\
\amp = e^{-cs} \lap{f(t + c)}
\end{align*}
Therefore,
\begin{equation*}
\lap{f(t) \, u_c(t)} = e^{-cs} \lap{f(t + c)}.
\end{equation*}
\(\DLBb \text{A}, \text{B}, \text{C}\) Details.
\(\DLBb \text{A}\ \): Shifting \(f(t)e^{-st}\) to the left \(c\) units makes it
\begin{equation*}
f(t + c)\ e^{-s(t + c)}
\end{equation*}
and starts the integral at \(t=0\text{.}\)
\(\DLBb \text{B}\ \): Exponential Property \(e^{-s(t + c)} = e^{-st} e^{-sc}\)
\(\DLBb \text{C}\ \): Factor out the constant, \(e^{-sc}\text{,}\) as we are integrating in \(t\text{.}\)
Derivation of \(\lap{f(t-c) u_c(t)}\).
In terms of the definition of the Laplace transform, we want to show:
\begin{align*}
\os{\ds\lap{f(t-c) u_c(t)}}{\os{\ds\downarrow}{\overline{\int_{0}^{\infty} f(t-c) u_c(t) e^{-st}\ dt}}}
\amp = e^{-cs} \os{\ds \lap{f(t)}}{\os{\ds\downarrow}{\overline{\int_{0}^{\infty} f(t) e^{-st}\ dt}}}
\end{align*}
To show this, we will start with the integral on the left and show it leads to the integral on the right.
\begin{align*}
\int_{0}^{\infty} f(t-c) u_c(t) e^{-st}\ dt
\amp \os{\large\DLBb \text{A}}{=} \int_{c}^{\infty} f(t-c) e^{-st}\ dt\\
\amp \os{\large\DLBb \text{B}}{=} \int_{0}^{\infty} f(t)\ e^{-s(t+c)}\ dt \\
\amp \os{\large\DLBb \text{C}}{=} e^{-cs} \int_0^\infty f(t) e^{-st} \, du
\end{align*}
Thus, we have shown
\begin{equation*}
\lap{f(t-c) u_c(t)} = e^{-cs} F(s)
\end{equation*}
where \(F(s) = \lap{f(t)}\text{.}\)
\(\DLBb \text{A}, \text{B}, \text{C}\) Details.
\(\DLBb\text{A}\ \): \(u_c(t)=0\) on \([0,c)\) makes the lower limit of the integral \(c\text{.}\)
\(\DLBb\text{B}\ \): Shifting \(f(t-c)e^{-st}\) to the left \(c\) units makes it \(f(t)\ e^{-s(t + c)}\) and starts the integral at \(t=0\text{.}\)
\(\DLBb\text{C}\ \): Use the exponential property \(e^{-s(t + c)} = e^{-st} e^{-sc}\) and factor out the constant, \(e^{-sc}\text{,}\) as we are integrating in \(t\text{.}\)
🗺️ A Roadmap Overview.
\begin{equation*}
\small\ul{\qquad\ \textbf{Original Domain}\ \qquad}
\end{equation*}
\begin{equation*}
\small\DLBa\textbf{2️⃣ Laplace Domain}
\end{equation*}
\begin{gather*}
\small \os{\vphantom{m}}{ y'' + 4y = g(t) }\\
\small y(0) = 0,\ y'(0) = 0 \\
\small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\
\\
\small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\
\small y(t) = \left(\frac14 - \frac18\cos\big(2(t - 1)\big)\right)u_1(t)\\
\small \qquad - \left(\frac14 - \frac18\cos\big(2(t - 2)\big)\right)u_2(t)
\end{gather*}
\begin{gather*}
\small \underrightarrow{\text{1️⃣ Forward}}\\
\small \text{Apply}\ \laplacesym\\
\\
\\
\\
\\
\small \text{Apply}\ \laplacesym^{-1}\\
\small \overleftarrow{\text{3️⃣ Backward}}
\end{gather*}
\begin{align*}
\amp\small\DLBa s^2Y(s) + 4Y(s) = \frac{e^{-s}}{s} - \frac{e^{-2s}}{s}\\
\amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\
\amp\small\DLBa Y(s) = \frac{1}{s^2 + 4} \left( \frac{e^{-s}}{s} - \frac{e^{-2s}}{s} \right)\\
\amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\
\amp\small\DLBa Y(s) = F(s)e^{-s} - F(s)e^{-2s}\\
\amp\small\text{where}\ \DLBa F(s) = \frac14 \left( \frac{1}{s} - \frac{s}{s^2 + 4} \right)
\end{align*}
🔎 Here are the Details.
Decompose:
\begin{equation*}
F(s) = \frac{1}{s(s^2 + 4)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 4}\text{.}
\end{equation*}
Multiply by \(s(s^2 + 4)\) and collect the \(s^2\text{,}\) \(s\text{,}\) and free terms:
\begin{equation*}
1 = A(s^2 + 4) + (Bs + C)s = (A + B)s^2 + Cs + 4A\text{.}
\end{equation*}
Matching coefficients gives the system:
\begin{equation*}
A+B=0,\quad C=0,\quad 4A=1
\end{equation*}
which has solution \(A=\sfrac14, B=-\sfrac14, C=0\text{.}\) Thus:
\begin{equation*}
F(s) = \frac14\cdot\frac{1}{s} - \frac14\cdot\frac{s}{s^2 + 4}
\end{equation*}
and so the inverse is:
\begin{equation*}
f(t) = \ilap{F(s)} = \frac14 - \frac14\cos(2t) = \frac14\left(1 - \cos(2t)\right)\text{.}
\end{equation*}
🔎 Here are the details.
Decompose:
\begin{equation*}
F(s) = \frac{3}{s^2(s + 2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + 2}\text{.}
\end{equation*}
Multiply through and collect the \(s^2\text{,}\) \(s\text{,}\) and free terms:
\begin{equation*}
3 = As(s + 2) + B(s + 2) + Cs^2 = (A+C)s^2 + (2A+B)s + 2B\text{.}
\end{equation*}
Matching coefficients gives the system:
\begin{equation*}
A+C=0,\quad 2A+B=0,\quad 2B=3
\end{equation*}
which has solution \(A=-\sfrac34, B=\sfrac32, C=\sfrac34\text{.}\) Thus:
\begin{equation*}
F(s) = \frac{-\sfrac34}{s} + \frac{\sfrac32}{s^2} + \frac{\sfrac34}{s + 2}
\end{equation*}
and
\begin{equation*}
f(t) = \ilap{F(s)} = -\frac34 + \frac32t + \frac34e^{-2t}\text{.}
\end{equation*}
🔎 Partial Fraction Details.
\begin{equation*}
F_2(s):\qquad\frac{1}{s(2s + 1)} = \frac{A}{s} + \frac{B}{2s + 1}
\end{equation*}
Multiplying through and grouping \(s\) terms:
\begin{align*}
1 \amp = A(2s + 1) + Bs = 2As + A + Bs = (2A + B)s + A
\end{align*}
Leads to:
\begin{equation*}
2A + B = 0,\quad A = 1 \quad\Rightarrow\quad A = 1,\quad B = -2
\end{equation*}
and so
\begin{equation*}
F_2 = \frac{1}{s} - \frac{2}{2s + 1}
\end{equation*}
\begin{equation*}
F_3(s):\qquad F_3(s) = \frac{1}{s^2(2s + 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{2s + 1}
\end{equation*}
Multiplying through and grouping \(s\) terms:
\begin{align*}
1 = As(2s + 1) + B(2s + 1) + Cs^2
\amp = 2As^2 + As + 2Bs + B + Cs^2\\
\amp = (2A + C)s^2 + (A + 2B)s + B
\end{align*}
Leads to:
\begin{equation*}
2A + C = 0,\quad A + 2B = 0,\quad B = 1 \quad\Rightarrow\quad A = -2,\ B = 1,\ C = 4
\end{equation*}
Thus,
\begin{equation*}
F_3(s) = \frac{-2}{s} + \frac{1}{s^2} + \frac{4}{2s + 1}
\end{equation*}
🔎 Partial Fraction Details.
For \(F_1(s)\text{,}\) the general form is:
\begin{equation*}
\frac{1}{(s + 1)^2(s - 5)} = \frac{A}{s + 1} + \frac{B}{(s + 1)^2} + \frac{C}{s - 5}
\end{equation*}
Multiplying through:
\begin{align*}
1 \amp = A(s + 1)(s - 5) + B(s - 5) + C(s + 1)^2
\end{align*}
Selecting values of \(s\) finds \(A\text{,}\) \(B\text{,}\) and \(C\text{:}\)
|
\(s = -1:\)
|
\(1 = A(0) + B(-6) + C(0)\)
|
\(\implies\)
|
\(B = -\frac16\)
|
|
\(s = 5:\)
|
\(1 = A(0) + B(0) + C(36)\)
|
\(\implies\)
|
\(C = \frac{1}{36}\)
|
|
\(s = 0:\)
|
\(1 = A(-5) + B(-5) + C(1)\)
|
\(\implies\)
|
\(1 = -5A + \frac56 + \frac{1}{36}\)
|
|
\(A = -\frac{1}{36}\)
|
With these constants, we can write \(F_1\) as:
\begin{equation*}
F_1(s) = \frac{-\frac{1}{36}}{s + 1} - \frac{\frac{1}{6}}{(s + 1)^2} + \frac{\frac{1}{36}}{s - 5}
\end{equation*}
Now, for \(F_2(s)\text{,}\) we have:
\begin{equation*}
F_2(s) = \frac{1}{s(s + 1)^2(s - 5)} = \frac{A}{s} + \frac{B}{s + 1} + \frac{C}{(s + 1)^2} + \frac{D}{s - 5}
\end{equation*}
Multiplying through:
\begin{align*}
1 \amp = A(s + 1)^2(s - 5) + Bs(s + 1)(s - 5) + Cs(s - 5) + Ds(s + 1)^2
\end{align*}
Selecting values of \(s\) finds \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\text{:}\)
| \(s = 0:\) | \(1 = A(-5) + B(0) + C(0) + D(0)\) | \(\implies\) | \(A = -\frac{1}{5}\) |
| \(s = -1:\) | \(1 = A(0) + B(6) + C(0) + D(0)\) | \(\implies\) | \(B = \frac{1}{6}\) |
| \(s = 5:\) | \(1 = A(0) + B(0) + C(0) + D(180)\) | \(\implies\) | \(D = \frac{1}{180}\) |
| \(s = 1:\) | \(1 = A(-16) + B(-8) + C(-4) + D(4)\) |
To find \(C\text{,}\) we plug in the values of \(A\text{,}\) \(B\text{,}\) and \(D\text{:}\)
\begin{equation*}
1 = \frac{16}{5} - \frac{4}{3} + C(-4) + \frac{1}{45} \implies
C = \frac{2}{9}
\end{equation*}
With these constants, we can write \(F_2\) as:
\begin{equation*}
F_2(s) = \frac{-\frac{1}{5}}{s} + \frac{\frac{1}{6}}{s + 1} + \frac{\frac{2}{9}}{(s + 1)^2} + \frac{\frac{1}{180}}{s - 5}
\end{equation*}
You have attempted of activities on this page.
