The Laplace Transform

Brown University Applied Mathematics

Notatation and Basic Properties

The Laplace Transform applied to a function $f(t)$ will look like the following:
$\mathcal{L}[f(t)]=\int_0^{\infty } e^{-\lambda t} f(t) \,dt = F(\lambda)$
Shorthand notation is usually denoted in 3 common ways:
$\mathcal{L}[f(t)] = F^{L} = Laplace(f(t))$
In general, your function of time $f(t)$ will be transformed into a function of lambda $F(\lambda)$:
$f(t)$ =>$F^{L}(\lambda)$

Please note the following properties of the Laplace Transform:
Always remember that the Laplace Transform is only valid for t>0.
Constants can be pulled out of the Laplace Transform:
$\mathcal{L}[af(t)] = a\mathcal{L}[f(t)]$ where a is a constant
Also, the Laplace of a sum of multiple functions can be split up into the sum of multiple Laplace Transforms:
$\mathcal{L}[g(t)+f(t)] = \mathcal{L}[g(t)]+\mathcal{L}[f(t)]$

There are 5 rules that you should memorize about the Laplace Transform:

1. Convolution Rule
We will denote the convolution of 2 functions f and g as the following:
$(f * g) = (g * f) = \int_{0}^{t} f(\tau) g(t-\tau)\mathrm{d} \tau$
When we apply the Laplace Transform to the convolution of 2 functions we obtain the following result:
$\mathcal{L}[f * g] = F^{L}(\lambda)G^{L}(\lambda)$

2. Derivative Rule
Given a derivative (n) of a function f, denoted by $f^{n}$, the Laplace Transform will be the following:
$\mathcal{L}[f^{n}] = \lambda^{n}F^{L}(\lambda)-\sum_{k=1}^{n}\lambda^{n-k}f^{k-1}(+0)$

3. Similarity Rule
Given a function that has a constant $a$ multiplied by t in a function:
$\mathcal{L}[f(at)] = \frac{1}{a}F^{L}(\frac{\lambda}{a}),$ such that $a>0$

4. Shift Rule
Given a function shifted by a certain amount multiplied by a shifted Heaviside Function:
$\mathcal{L}[H(t-a)g(t-a))] = e^{-a\lambda}G^{L}(\lambda)$

5. Attenuation rule
Given an exponential function multiplied by an exponential function, where a is a constant:
$\mathcal{L}[e^{-at}f(t)] = F^{L}(\lambda+a)$

Note that the most important rules that we will use are #1, #2, and #4, however it is a good idea to learn all of the rules.

Common Laplace Transforms to Memorize

Please note that for all examples, we will assume t >0

1. $f(t)=1$
$\mathcal{L}[f(t)]=\mathcal{L}[1]=\int_{0}^{\infty} 1e^{-t\lambda}\mathrm{d}t$
Recall that the Laplace Transform is actually an infinte integral. If we evaluate this improper integral, we obtain:
$\frac{-1}{\lambda}e^{-\lambda t}|_{t=0}^{t=\infty}$
As $t -> \infty$ we notice that the term $e^{-\lambda t} -> 0$
Thus when we double substitute in our values, we obtain:
$\frac{-1}{\lambda}e^{-\lambda t}|_{t=0}^{t=\infty} = \frac{1}{\lambda}$
Here is what you should memorize:
$\mathcal{L}[1] = \frac{1}{\lambda}$

2. $f(t)=H(t)$
Recall the that Heaviside function is actually just the value 1 for t>0, and that the Laplace Transform is only valied for t>0
Referring to example 1, we obtain the following result:
$\mathcal{L}[H(t)] = \frac{1}{\lambda}$

3. $f(t)=t$
$\mathcal{L}[f(t)]=\mathcal{L}[t]=\int_{0}^{\infty} te^{-t\lambda}\mathrm{d}t$
Using Integration by Parts:
$\int_{0}^{\infty} te^{-t\lambda}\mathrm{d}t = \int_{0}^{\infty} \frac{-\mathrm{d}}{\mathrm{d}\lambda}e^{-t\lambda}\mathrm{d}t = \frac{-\mathrm{d}}{\mathrm{d}\lambda} \int_{0}^{\infty} e^{-t\lambda}\mathrm{d}t = \frac{-\mathrm{d}}{\mathrm{d}\lambda}(\frac{1}{\lambda}) = \frac{1}{\lambda^2}$
$\mathcal{L}[t] = \frac{1}{\lambda^2}$

4. $f(t)=t^n$
We can generalize example 3 to obtain the following result:
$\mathcal{L}[t^n] = \frac{n!}{\lambda^n+1}$

5. $f(t)=e^at$
$\mathcal{L}[e^at] = \int_{0}^{\infty} e^{at}e^{-t\lambda}\mathrm{d}t = \int_{0}^{\infty} e^{-(\lambda - a)t}\mathrm{d}t = \frac{1}{\lambda - a}$
By the similarity rule. So:
$\mathcal{L}[e^at] = \frac{1}{\lambda - a}$

6. $f(t)=\sin(kt)$
It is necesarry to Recall Euler's Formula for this example:
$e^{jt} = \cos(t)+j\sin(t)$ where j is a variable such that $j^2 = -1$
$\sin(t) = Im (e^{jt}$)
$\cos(t) = Re (e^{jt}$)
Where Im and Re are tools to extract Real parts of the answer (without any terms multiplied by j), and Imaginary Parts of the answer (multiplied by j)
With our equation, we obtain
$\mathcal{L}[f(t)]=\mathcal{L}[\sin(kt)] = \int_{0}^{\infty} \sin(kt)e^{-t\lambda}\mathrm{d}t = Im\int_{0}^{\infty} e^{jkt}e^{-t\lambda}\mathrm{d}t$
Notice that we can extract just the Imaginary part of $e^{jkt}$, which will just give us $\sin(kt)$
$ = Im \frac{1}{\lambda - jk}$
We can now multiply by the complex conjugate of the denominator to bring the imaginary term to the numerator:
$ = Im \frac{1}{\lambda - jk} ( \frac{\lambda + jk}{\lambda + jk}) = Im \frac{\lambda + jk}{\lambda^2 + k^2} = \frac{k}{\lambda^2 + k^2}$
$\mathcal{L}[\sin(t)] = \frac{k}{\lambda^2 + k^2}$

7. $f(t)=\cos(kt)$
Similarly, Cosine will yield the real part of the same fraction:
$\mathcal{L}[\cos(kt)] = Re \frac{\lambda + jk}{\lambda^2 + k^2} = \frac{\lambda}{\lambda^2 + k^2}$

Rule Example Problems

Example 1:
Find the Laplace Transform of the Convolution of $f(t) = t$ and $g(t) = H(t)$
$\mathcal{L}[t] = \frac{1}{\lambda^2}$ and $\mathcal{L}[H(t)] = \frac{1}{\lambda}$
So by Rule #1 Convolution, we obtain:
$\mathcal{L}[t * H(t)] = \frac{1}{\lambda^2}(\frac{1}{\lambda}) = \frac{1}{\lambda^3}$

Example 2:
Proove the Derivative Rule (Hint, think about the Fundamental Theorem of Calculus)
From the Fundamental Theorem of Calculus, we know that any function can be represented as the derivative of another function.
We already prooved that
(1) $\mathcal{L}[t] = \frac{1}{\lambda^2}$
and we know that
(2) $ (\frac{t^2}{2})' = t$
By using the derivative rule on equation (2) we obtain:
$\mathcal{L}[(t^2/2)'] = \mathcal{L}[\frac{(t^2)'}{2}] = \frac{1}{2}\mathcal{L}[(t^2)'] = \frac{1}{2}(\lambda 2!/\lambda^3 - f(0))$
and we know that f(0) = 0, so canceling 2 with 2! and one Lambda term,
$\mathcal{L}[(t^2/2)'] = \frac{1}{\lambda^2}$
Thus
$\mathcal{L}[t] = \frac{1}{\lambda^2} = \mathcal{L}[(t^2/2)']$
and the derivative rule holds true. QED.

Example 3:
$\mathcal{L}[cosh(2t)]$ where cosh is the cosine hyperbolic function, $\cosh(at) = \frac{e^{at} + e^{-at}}{2}$
$\mathcal{L}[cosh(2t)] = \mathcal{L}[\frac{e^{at} + e^{-at}}{2}] = \frac{1}{2}(\frac{1}{\lambda - 2} + \frac{1}{\lambda+2}) = \frac{\lambda}{\lambda^2+4}$
This is an interesting feature - recall that
$\mathcal{L}[\cos(kt)] = \frac{\lambda}{\lambda^2 + k^2} = \mathcal{L}[cosh(2t)]$

Application to Piecewise Continuous Functions

Consider the following piecewise continuous function:
$ f(t) = \begin{cases} t &\mbox{if } 0 \leq t \leq 2 \\1 &\mbox{if } 2 \leq t \leq 3 \\4-t &\mbox{if } 3 \leq t \leq 4 \\0 &\mbox{if } 4 \leq t \\ \end{cases}$
From the last section, we know how s to create piecewise continuous functions using the Heaviside function in specific intervals. Doing this yields the following result:
$f(t) = t[H(t)-H(t-1)]+1[H(t-1)-H(t-3)]+(4-t)[H(t-3)-H(t-4)]$
$= tH(t) - tH(t-1)+H(t-1) - H(t-3)+ (4-t)H(t-3)-(4-t)H(t-4)$
$=tH(t) - (t-1)H(t-1) + (3-t)H(t-3) + (t-4)H(t-4)$
Now we can transform all the parts seperately and add them to reveal the answer:
$\mathcal{L}[tH(t)] = \frac{1}{\lambda^2}$ and $\mathcal{L}[(t-1)H(t-1)] = \frac{1}{\lambda^2}e^{-\lambda}$
Notice that when we have a function shifted by the same amount as a Heaviside function multiplied together, we get the Laplace Transform of the function multiplied by an exponential shifting function. Or in general:
$\mathcal{L}[f(t-a)H(t-a)] = F^{L}(\lambda)e^{-a\lambda}$
By using the above rule, we obtain the final answer:
$\mathcal{L}[f(t)] = \frac{1}{\lambda^2} - \frac{1}{\lambda^2}e^{-\lambda} - \frac{1}{\lambda^2}e^{-3\lambda} + \frac{1}{\lambda^2}e^{-4\lambda}$

Delta Dirac Function

A notable function discovered by Mathematician Paul Dirac is the Delta Dirac Function. Dirac solved the following function in 1920, and it was not proven rigorously until 1938 by Sobelev.
$H'(t) = \delta(t)$
Why isn't the derivative of the Heaviside function just 0? Recall the definition of a derviative and apply it here:
$\frac{\mathrm{d}}{\mathrm{d}t}H(t)$ at point zero blows up because $\lim_{x->0} \frac{H(h)-H(0)}{h} = \frac{0-0.5}{0} => \infty$
From Paul Dirac, we know that:
$\mathcal{L}[\delta(t)] = 1$ because $\mathcal{L}[\delta(t)] = \int_{0}^{\infty} \delta(t)e^{-\lambda t}\mathrm{d}t = e^{-\lambda 0} = 1$
Antoher interesting feature of the Delta Dirac Function is the following:
$\int \delta(t)f(t)\mathrm{d}t = f(0)$

Using the MuPAD Laplace Solver

To call the Laplace solver, we must first set up how we want the input function to look and which order the variables should come in. We do this by calling laplace::addpattern and setting up our input function. Then we can follow the outline we set up and simply type "laplace".

The general form for this will be:
[ laplace::addpatern(pat, t, s, res)
"::addpatern" tells MuPAD that we want to set up our input pattern, "pat" represents the function, "t" is the independent variable, "s" is our Laplace independent variable, and "res" is an internal MuPAD command to let it know which solver to use.
The general form for solving the laplace will then be:
[ laplace(function, t, `& lambda;`)
This follows the same outline, just with the solver now implicitly stated.

Continuous Functions

laplace::addpattern(pat, t, s, res)

laplace(sin(t),t,`λ`)

laplace(exp(t),t,`λ`)

laplace(heaviside(t-3),t,`λ`)

Piecewise Functions

f(t):=piecewise([0<=t<=2,0],[2<=t<=3,t-3],[3<=t<=4,5-t],[4<=t<=5,5])

piecewise([t in Dom::Interval([0], [2]), 0], [t in Dom::Interval([2], [3]), t - 3], [t in Dom::Interval([3], [4]), 5 - t], [t in Dom::Interval([4], [5]), 5])

assume(t>0)

f2(t):=(t-3)*(heaviside(t-2)-heaviside(t-3))+(5-t)*(heaviside(t-3)-heaviside(t-4))+5*(heaviside(t-4)-heaviside(t-5))

f3(t):=simplify(f2(t))

lapf:=laplace(f3(t),t,`λ`)

Convolution of 2 Functions

$f(t)=\int_{0}^{t} (t-\tau)^2 e^{\tau} \mathrm{d}\tau$
First we notice that we are taking the convolution of 2 functions:
$t^2$ and $e^t$
We can take the Laplace transform of the given function => equation of variable lambda, then use partial fractions to set it up for the inverse Laplace transform.

laplace::addpattern(pat, t, s, res)

a:=laplace(t^2,t,`λ`)

b:=laplace(exp(t),t,`λ`)

c:=a*b

partfrac(c,`λ`)