Laplace Transforms

Laplace Transform

Extremely useful for solving differential equations, but also helps transforms functions/waveforms from the time domain to the frequency domain.

The notation for Laplace transforms is:

$$\mathcal{L}\{f(t)\}\to \mathcal{F}(s)$$

What is a Transform?

Sort of a function of functions. Similar to how some functions take us from one set of numbers to another set. Transforms take us from one set of functions to another set of functions.

Definition of Laplace Transform

$$\mathcal{L}\{f(t)\}={\int }_0^{\infty }{e}^{-st}f(t)dt$$

This is an improper integral.

Laplace Transform Table

These are well known Laplace transforms. Some of the derivations are shown below.

$$\mathcal{L}\{1\}=\frac{1}{s}$$

Derivation of Laplace Transform on 1

$$\begin{array}{r}\mathcal{L}\{1\}={\int }_0^{\infty }{e}^{-st}dt=\underset{\mathcal{A}\to \infty }{\lim }{\int }_0^{\mathcal{A}}{e}^{-st}dt\\ =\underset{A\to \infty }{\lim }{\left[-\frac{1}{s}{e}^{-st}\right]}_0^A=\underset{A\to \infty }{\lim }\left(-\frac{1}{s}{e}^{-sA}-\left(-\frac{1}{s}\right)\right)\\ \underset{A\to \infty }{\lim }\left[-\frac{1}{s}{e}^{-sA}+\frac{1}{s}\right]\text{for }s>0=\frac{1}{s}\end{array}$$

Derivation of Laplace Transform on $e^{at}$

$$\begin{array}{r}\mathcal{L}\{e^{at}\}={\int }_0^{\infty }{e}^{-st}{e}^{at}dt={\int }_0^{\infty }{e}^{(a-s)t}dt\\ =\frac{1}{a-s}[e^{(a-s)t}{]}_0^{\infty }=\frac{1}{a-s}[]\end{array}$$

System Response

Imagine we have $x(t)\to H\to y(t)$. Where $H$ is some black box that performs a function on input signal $x(t)$. We want to find output signal $y(t)$.

The system $H$ can be described as:

• Differential Equation (ODE) $a_2{y}^{\prime \prime }(t)+a_1{y}^{\prime }(t)+a_{0}y(t)=b_{0}x(t)$
• Difference Equation $a_{0}y[n]+a_{1}y[n-1]+a_{2}y[n-2]={\beta }_{0}x[n]$
• Impulse Response $h(t)$
• Step Response $h_1(t)$
• Frequency Response $H(jw)$
• Transfer Function $H(S)$
• Block Diagrams

Input Signals

The input signal $x(t)$ was generally limited to either

• Impulse $x(t)=S(t)$
• Step $x(t)=u(t)$
• Ramp $x(t)=t(wt)$

Transforms

Time domain:

$$1.\text{Convolution: }y(t)=x(t)\times h(t)={\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )dt$$

By transforming the time domain problem into the frequency domain, we changed from convolution to multiplication. That means that the ODE is solved by algebraic techniques.

Eventually, we get to to $Y(s)$, but we want to get $Y(t)$.

We look to the Inverse Laplace Transform.