Example 1
Consider the function
\[f(t) = e^{3t}t + e^{\pi t} u_\pi.\]We can look at the table and see that
\[\mathcal{L}(e^{\alpha t}g(t)) = G(s-\alpha)\qquad \text{where}\qquad \mathcal{L}(g) = G(s).\]Then
\[\mathcal{L}(t) = \frac{1}{s^2},\qquad \text{so }\qquad \mathcal{L}(e^{3t}t) = \frac{1}{(s-3)^2}.\]Similarly, we need to write
\[e^{\pi t}u_\pi = g(t-\pi) u_\pi,\]which means
\[g(t-\pi) = e^{\pi t}\qquad\text{and}\qquad g\left((t+\pi) -\pi \right) = e^{\pi(t+\pi)} = e^{\pi^2+\pi t}.\]So
\[g(t) = g(t+\pi-\pi) = e^{\pi^2} e^{\pi t},\]and \(\mathcal{L}(g) = e^{\pi^2} \frac{1}{s-\pi}\)
So \(\mathcal{L}(e^{\pi t} u_\pi) = \mathcal{L}(g(t-\pi) u_\pi)\\ =e^{-\pi s}\mathcal{L}(g)\\ = e^{-\pi s} e^{\pi^2} \frac{1}{s-\pi}\\ = e^{\pi^2 -\pi s} \frac{1}{s-\pi}.\)
Combining these two facts gives
\[\mathcal{L}\left(e^{3t}t + e^{\pi t}u_\pi\right) = \frac{1}{(s-3)^2} + \frac{e^{\pi^2-\pi s}}{s-\pi}.\]Example 2
Consider the function
\[f(t) = 7e^{3t} + 2t u_{9} + 2t^2 u_{5}.\]We wish to use the table but in order to do that we need to include terms like $g(t-c)u_c$ instead of just $g(t)u_c$.
That means, we want
\[2tu_9 = g(t-9)u_9\\ t^2 u_5 = h(t-5)u_5.\]This means that
\[g(t-9) = 2t\\ h(t-5) = 2t^2.\]If we use a dummy variable of $x = t-9$ and $t = x+9$ for $g(t-9)$, we get
\[g(x) = g(t-9) = 2t= 2(x+9) = 2x + 18,\]and since $x$ was a dummy variable we can replace $x$ by $t$ in the extreme ends
\[g(x) = 2x+18 \rightsquigarrow g(t) = 2t+18.\]Then
\[G(s) = \mathcal{L}(g) = \mathcal{L}\left(2t+18 \right) = \frac{2}{s^2}+ \frac{18}{s}.\]Similarly, if we set $x = t-5$ and $t = x+5$ for the $h$ function gives
\[h(x) = h(t-5) =2t^2 = 2(x+5)^2 = 2x^2 +20x + 50.\]And so
\[h(t) = 2t^2+20t + 50.\]Then
\[H(s) = \mathcal{L}(h) = \mathcal{L}(2t^2+20t+5) = \frac{4}{s^3} + \frac{20}{s^2}+ \frac{5}{s}\]Hence
\[\mathcal{L}(f) = \mathcal{L}\left(7e^{3t} + 2t u_9 +t^2 u_5 \right)\\ = 7\mathcal{L}\left(e^{3t} \right) + \mathcal{L}(g(t-9)u_9 ) + \mathcal{L}(h(t-5)u_5)\\ = \frac{7}{s-3} + e^{-9s} G(s) + e^{-5s}H(s)\]where $G$ and $H$ are defined above.
Example 3
Taking the Laplace transform of
\[f(t) = \left\{ \begin{array}{l} 3t^2 - 9t &: 0\le t< 3\\ 0 &: 3\le t<2\pi\\ \sin(t) &: \pi \le t \end{array} \right..\]Below is a graph of the function $f$.
The first thing we do is to rewrite the function $f$ in terms of Heaviside functions $u_c$. We also make no distinction between the function $u_0(t)$ and the function which is identically 1.
We will handle this by treating each different part of the function $f$ one at a time (which is why the image above has red, blue and green sections).
The red curve, which is $f(t)$ for $0\le t< 2$, is the function $3t^2-9t$. The Heaviside functions can be viewed as turning on and off a function at specified times. That is $g(t)(u_a - u_b)$ is the function which is $g(t)$ for $a\le t< b$ and $0$ for $0\le t< a$ and 0 for $t\ge b$. Hence the red curve is represented by
\[\left(3t^2-9t\right)\left(u_0 - u_3\right) = (3t^2-9t)(1-u_3).\]The blue curve is just zero so we can represent that by $0(u_3-u_{2\pi}) = 0$.
The green curve is a sine curve, which is turned on after time $t = 2\pi$ and is never turned off. Therefore, the green part of the curve is represented by
\[\sin(t) u_{2\pi}.\]Hence
\[f(t) = (3t^2 - 9t)(1-u_{3}) + \sin(t) u_{2\pi}.\]In order to easily use the table of Laplace transforms, we must have terms of the form \(g(t-c)u_c.\)
We now write
\[f(t) = \left(3t^2-9t\right) + (3t^2-9t)u_3 + \sin(t) u_{2\pi}\\ = f_1(t) - f_2(t)u_3 + f_3(t)u_{2\pi}\]The Laplace transform of $f_1$ can be read off as
\[\mathcal{L}(f_1) = \mathcal{L}(3t^2-9t) = \frac{6}{s^{3}}- \frac{9}{s^2}.\]We must now write $f_2(t) = g(t-3)$ in order to easily use the table. If we set $x = t-3$ as a dummy variable, then $t = x+3$ and so we can write
\[g(x) = g(t-3)\\ =f_2(t)\\ = (3t^2 - 9t)\\ = 3(x+3)^2 - 9(x+3)\\ = 3(x^2+9x+6) - 9(x+3)\\ = 3x^2 + 27 x +18 - 9x - 27\\ = 3x^2+18x - 9.\]Again, since $x$ is a dummy variable we can replace $x$ by $t$ at the extreme ends of the above equality
\[g(t) = 3t^2+18t - 9.\]The Laplace transform of this function $g$ can be found in the table and is simply
\[G(s) = \mathcal{L}(g) = \mathcal{L}(3t^2+18t-9) = \frac{6}{s^3}+ \frac{18}{s^2}- \frac{9}{s}\]Hence
\[\mathcal{L}(f_2(t)u_3) = \mathcal{L}(g(t-3)u_3) = e^{-3s} G(s) = e^{-3s} \left(\frac{6}{s^3}+ \frac{18}{s^2}- \frac{9}{s} \right).\]Lastly, to handle the sine term we write
\[f_3(t) = h(t-2\pi)\]and we do a similar change of variable as above. We set $x = t-2\pi$ and then
\[h(x) = h(t-2\pi)\\ = f_3(t)\\ =\sin(t)\\ = \sin(x+2\pi )\\ =\sin(x),\]and so $h(t) = \sin(t)$ and
\[H(s) = \mathcal{L}(h(t)) = \mathcal{L}(\sin(t)) = \frac{1}{s^2+1}.\]That means
\[\mathcal{L}(f_3(t)u_{2\pi}) = \mathcal{L}(h(t-2\pi) u_{2\pi}) = e^{-2\pi s}H(s) = e^{-2\pi s}\frac{1}{s^2+1}.\]Combining all the work that we’ve done yields:
\[\mathcal{L}({f}) = \frac{6}{s^{3}}- \frac{9}{s^2} - e^{-3s} \left(\frac{6}{s^3}+ \frac{18}{s^2}- \frac{9}{s} \right) + \frac{e^{-2\pi s}}{s^2+1}\]