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Beware that, when fighting monsters, you yourself do not become a monster… for when you gaze long into the abyss. The abyss gazes also into you, Friedrich W. Nietzsche
The Laplace Transform of a function f(t), where t ≥ 0, is defined as $\mathcal{L}(f(t)) = \int_{0}^{∞} f(t)e^{-st}dt = F(s)$.
One of the most important properties of the Laplace Transform is linearity, which states: $\mathcal{L}(af(t)+bg(t)) = a\mathcal{L}(f(t))+b\mathcal{L}(g(t))$
We can extend the exponential shift formula to complex exponents. Let $a, b \in \mathbb{R}$, and consider the function $e^{(a+bi)t}$.
$$\mathcal{L}\{e^{(a+bi)t}\} = \int_{0}^{\infty} e^{(a+bi)t}e^{-st} dt$$Combine the exponents into the form $e^{-ct}$, where $c = s - a - ib$: $= \int_{0}^{\infty} e^{-(s-a-ib)t} dt$
Evaluate the integral: $\int_{0}^{∞} e^{-ct}dt = -\frac{1}{c}e^{-ct}\bigg|_{0}^{∞} =-\frac{1}{s -a -ib}e^{-(s -a -ib)t}\bigg|_{0}^{∞}$
Apply the limits: As $t \to \infty, e^{-(s -a -ib)t} = e^{-(s-a)t} \cdot e^{ibt} → 0$ 💡 assuming s > a. t = 0, $e^{-(s -a -ib)·0} = 1.$
💡 The $e^{ibt}$ part is just oscillating between -1 and 1 (it’s cos(bt) + i·sin(bt)), $|e^{ibt}|=1$, so it doesn’t affect whether the overall expression blows up or shrinks to zero. The key part is $e^{-(s-a)t}$. For this to approach zero as t approaches infinity, we need the exponent to be negative, i.e., -(s - a) < 0 ↭ s -a > 0 ↭ s > a
With s > a, the upper bound goes to 0, and the lower bound (t = 0) evaluates to 1: $\mathcal{L}(e^{(a+bi)t}) = -\frac{1}{s -a -ib}·0 +\frac{1}{s -a -ib}·1 = \frac{1}{s -a -ib}$
$\mathcal{L}(e^{(a+bi)t}) = \frac{1}{s-(a+ib)}$ valid for s > a, or alternatively, $e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}$.
Thus, the complex exponential function e(a + bi)t transforms into $\frac{1}{s-(a+bi)}$, demonstrating a direct application of the exponential shift formula with a complex shift.
Instead of using tedious integration by parts, we can use our new complex exponential transform alongside Euler’s Formula: $e^{iat} = cos(at) + isin(at), e^{-iat} = cos(at) - isin(at)$
From these, we can express the trigonometric functions in terms of exponentials: $cos(at) = \frac{e^{iat}+e^{-iat}}{2}, sin(at) = \frac{e^{iat}-e^{-iat}}{2i}$
$cos(at) =[\text{Euler's formula}] \frac{e^{iat}+e^{-iat}}{2}, \mathcal{L}(cos(at)) =[\text{By linearity of the Laplace Transform}, e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}] \frac{1}{2}(\frac{1}{s-ia}+\frac{1}{s+ia}) =$
Find a common denominator: $= \frac{1}{2}·\frac{(s+ia)+(s-ia)}{(s-ia)(s+ia)} = \frac{1}{2}\frac{s+ia+s-ia}{s^2+a^2} = \frac{1}{2}\frac{2s}{s^2+a^2} = \frac{s}{s^2+a^2}, cos(at) \leadsto \frac{s}{s^2+a^2}$ where s > 0.
$sin(at) =[\text{Euler's formula}] \frac{e^{iat}-e^{-iat}}{2i}, \mathcal{L}(sin(at)) =[\text{By linearity of the Laplace Transform}, e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}] \frac{1}{2i}(\frac{1}{s-ia}-\frac{1}{s+ia})$
Find a common denominator: $= \frac{1}{2i}\frac{(s+ia)-(s-ia)}{s^2+a^2} = \frac{1}{2i}\frac{2ia}{s^2+a^2} =[\text{The 2i terms cancel out gracefully:}] \frac{a}{s^2+a^2}$.
Thus, the Laplace Transform of sin(at) is: $sin(at) \leadsto \frac{a}{s^2+a^2}$. This result is also valid for s > 0.
IMPORTANT: When we used Euler’s formula to break down $\cos(at)$ and $\sin(at)$, we transformed them into a combination of $\mathcal{L}\{e^{iat}\}$ and $\mathcal{L}\{e^{-iat}\}$. If we look at the integral definition for just one of those complex components: $\mathcal{L}\{e^{iat}\} = \int_{0}^{\infty} e^{iat} e^{-st} dt = \int_{0}^{\infty} e^{-(s-ia)t} dt$
To evaluate this improper integral, we look at the limit as the upper bound approaches infinity: $\lim_{R \to \infty} \left[ \frac{e^{-(s-ia)t}}{-(s-ia)} \right]_{0}^{R} = \lim_{R \to \infty} \left( \frac{e^{-(s-ia)R}}{-(s-ia)} \right) - \left( \frac{1}{-(s-ia)} \right)$. For the integral to converge to a finite number, that first term must approach zero as $R \to \infty$, and this requires s > 0.
The following table summarizes the most frequent mappings used in system analysis.
| Function | Laplace Transform |
|---|---|
| Unit Step, u(t) | $\mathcal{L}(u(t)) = \frac{1}{s}, s > 0$ |
| Exponential, $e^{at}$ | $\frac{1}{s - a}, s > a$ |
| Complex Exp $e^{(a + bi)t}$ | $\mathcal{L}(e^{(a + bi)t}) = \frac{1}{s - (a + bi)}, s > a$ |
| Cosine $\cos(\omega t)$ | $\mathcal{L}(\cos(\omega t)) = \frac{s}{s^2 + \omega^2}, s > 0$ |
| Sine $\sin(\omega t)$ | $\mathcal{L}(\sin(\omega t)) = \frac{\omega}{s^2 + \omega^2}, s > 0$ |
| Power, $t^n$ | $\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}, s > 0$ |
| Delayed Step, $u(t-a)$ | $\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}, s > 0$ |
| Dirac Delta $\delta(t-a)$ | $\mathcal{L}(\delta(t-a)) = e^{-as}, a \geq 0$ |
| $\frac{1}{t}$ | $\mathcal{L}\left(\frac{1}{t}\right) = \text{not defined}$ |
| $e^{-bt} \cos(\omega t)$ | $\mathcal{L}(e^{-bt} \cos(\omega t)) = \frac{s + b}{(s + b)^2 + \omega^2}, s > -b$ |
| $e^{-bt} \sin(\omega t)$ | $\mathcal{L}(e^{-bt} \sin(\omega t)) = \frac{\omega}{(s + b)^2 + \omega^2}, s > -b$ |
| $e^{at}f(t)$ | $\mathcal{L}(e^{at}f(t)) = F(s-a)$ |
f you compress or expand a function in the time domain by a factor of $a$ (where $a > 0$), how does it affect the frequency domain ($s$-domain)?
Theorem: $\mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right)$
By definition, the Laplace transform of $f(at)$ is: $\mathcal{L}\{f(at)\} = \int_{0}^{\infty} e^{-st}f(at)dt$
We perform a change of variables to simplify the argument of the function. Let $u = at$. Then, $du = a \, dt \implies dt = \frac{du}{a}$.
The bounds remain the same: When $t = 0, u = 0$. When $t \to \infty, u \to \infty$.
Substituting these into our integral, we get: $\int_{0}^{\infty} e^{-s\left(\frac{u}{a}\right)} f(u) \frac{du}{a} = \frac{1}{a} \int_{0}^{\infty} e^{-\left(\frac{s}{a}\right)u} f(u) du$
Recognizing that the integral on the right is simply the definition of the Laplace Transform evaluated at $\frac{s}{a}$ instead of $s$, we get: $\mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right)$
Example Application: Find $\mathcal{L}\{\sin(bt)\}$
We know the base transform: $\mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1} = F(s)$.
Using the scaling property with $a = b$: $\mathcal{L}\{\sin(bt)\} = \frac{1}{b}F\left(\frac{s}{b}\right) = \frac{1}{b} \cdot \frac{1}{\left(\frac{s}{b}\right)^2+1}$
Multiply the numerator and denominator by $b^2$ to simplify: $= \frac{1}{b} \cdot \frac{b^2}{s^2+b^2} = \frac{b}{s^2+b^2}$
Just as multiplying by $t$ in the time domain leads to differentiation in the $s$-domain, dividing by $t$ leads to integration in the $s$-domain.
Theorem: $\mathcal{L}\left\{\frac{f(t)}{t}\right\} = \int_{s}^{\infty} F(u)du$
Derivation:
We aim to calculate $\mathcal{L}\left\{\frac{f(t)}{t}\right\} = \int_{0}^{\infty} e^{-st}\frac{f(t)}{t}dt$.
First, we use a clever calculus trick to express the term $\frac{e^{-st}}{t}$ as an integral over a dummy variable $u$: $\int_{s}^{\infty} e^{-ut}du = \left[ \frac{e^{-ut}}{-t} \right]_{s}^{\infty} = 0 - \left( \frac{e^{-st}}{-t} \right) = \frac{e^{-st}}{t}$
Now, substitute this integral representation back into the original Laplace definition: $\mathcal{L}\left\{\frac{f(t)}{t}\right\} = \int_{0}^{\infty} \left( \int_{s}^{\infty} e^{-ut}du \right) f(t)dt$
Next, interchange the order of integration (integrating with respect to $t$ first, then $u$): $= \int_{s}^{\infty} \left( \int_{0}^{\infty} e^{-ut}f(t)dt \right) du$
The inner integral is exactly the definition of the Laplace Transform $F(u)$. $= \int_{s}^{\infty} F(u)du$
Example Application: Find $\mathcal{L}\left\{\frac{\sin(t)}{t}\right\}$
We know $f(t) = \sin(t) \implies F(s) = \frac{1}{s^2+1}$. Replacing $s$ with our dummy variable $u$, we integrate:
$$\mathcal{L}\left\{\frac{\sin(t)}{t}\right\} = \int_{s}^{\infty} \frac{1}{u^2+1}du = \Big[ \tan^{-1}(u) \Big]_{s}^{\infty}$$Evaluate the limits (noting that as $u \to \infty, \tan^{-1}(u) \to \frac{\pi}{2}$): $\mathcal{L}\left\{\frac{\sin(t)}{t}\right\} =[\cdots] = \tan^{-1}(\infty) - \tan^{-1}(s) = \frac{\pi}{2} - \tan^{-1}(s)$
The goal of the Inverse Laplace Transform is to map a function from the $s$-domain back to the time domain ($t$). This usually involves algebraically manipulating $F(s)$ into recognizable forms from the Laplace table.
Problem: Find $\mathcal{L}^{-1} \left\{ \frac{1}{s(s+3)} \right\}$
Step 1: Decompose the fraction. $\frac{1}{s(s+3)} = \frac{A}{s} + \frac{B}{s+3}$
Combine into a single fraction to solve for the numerators: $\frac{A(s+3) + Bs}{s(s+3)} = \frac{(A + B)s + 3A}{s(s+3)}$
Step 2: Solve for constants.
Match the numerator coefficients to the original fraction ($1 = 0s + 1$):
Step 3: Apply the inverse transform.
$$\mathcal{L}^{-1} \left\{ \frac{1/3}{s} - \frac{1/3}{s+3} \right\} = \frac{1}{3}\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - \frac{1}{3}\mathcal{L}^{-1}\left\{\frac{1}{s+3}\right\}$$Using $1 \leadsto \frac{1}{s}$ and $e^{at} \leadsto \frac{1}{s-a}$: $= \frac{1}{3}(1) - \frac{1}{3}e^{-3t} = \frac{1}{3}(1 - e^{-3t})$
Problem: Find $\mathcal{L}^{-1} \left\{ \frac{2}{(s+1)^3} \right\}$
Recall two key properties: Polynomial rule: $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$ and Exponential shift: $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$
Looking at our target, it matches a polynomial where $n=2$ (since the denominator is $s^{2+1} = s^3$ and the numerator is $2! = 2, f(t) = \frac{2}{s^3}$). It also has a shift of $a = -1$.
$$\mathcal{L}^{-1} \left\{ \frac{2!}{(s - (-1))^3} \right\} = e^{-1t} \cdot t^2 = t^2 e^{-t}$$Problem: Find $\mathcal{L}^{-1} \left\{ \frac{s}{s^2 - 2s + 5} \right\}$
Whenever the denominator is a quadratic that cannot be easily factored into real roots, complete the square to prepare it for a shifted sine/cosine transform.
Step 1: Complete the square in the denominator.
$s^2 - 2s + 5 = (s^2 - 2s + 1) + 4 = (s-1)^2 + 2^2$ So our fraction is now: $\frac{s}{(s-1)^2 + 2^2}$.
Step 2: Massage the numerator.
Because our denominator is shifted by $1$ (i.e., $(s-1)$), we must have an $(s-1)$ in the numerator to use the shifted cosine rule ($\frac{s-a}{(s-a)^2 + \omega^2}$).
We force this by subtracting and adding $1$ in the numerator: $\frac{s-1+1}{(s-1)^2 + 2^2} = \frac{s-1}{(s-1)^2 + 2^2} + \frac{1}{(s-1)^2 + 2^2}$
Step 3: Apply the inverse transform.
We use the shifted cosine and sine rules: $\mathcal{L}\{e^{at}\cos(\omega t)\} = \frac{s-a}{(s-a)^2 + \omega^2}$
$\mathcal{L}\{e^{at}\sin(\omega t)\} = \frac{\omega}{(s-a)^2 + \omega^2}$
For our expression, $a = 1$ and $\omega = 2$.
The first term perfectly matches the cosine rule.
The second term is missing the $\omega = 2$ in the numerator, so we multiply by $\frac{2}{2}$: $\mathcal{L}^{-1} \left\{ \frac{s-1}{(s-1)^2 + 2^2} + \frac{1}{2} \cdot \frac{2}{(s-1)^2 + 2^2} \right\} = e^t \cos(2t) + \frac{1}{2} e^t \sin(2t)$
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