Laplace Transform

Given a function $f(t)$, the Laplace Transform of $f(t)$ denoted by $F(s)$ (given that the integral exists) is defined by $F(s)=\mathcal{L}\{f(t)\}={\int }_{-\infty }^{\infty }f(t)e^{-st}dt$

• Here $s$ is a complex parameter called the “frequency”.

Why is s called the frequency?

In $s=\sigma +j\omega$, the term ω relates to angular frequency in radians per second.

Prof gave these two examples:

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Properties of Laplace Transform

Theorem 1

For any function $f(t)$, the one-sided Laplace transform will always converge given that there is some sufficiently large $s\in \mathbb{R}$ such that ${\int }_0^{\infty }f(t)e^{-st}dt$ exists

Theorem 2: Laplace Transform is Linear

Suppose that $f(t)$ and $g(t)$ have Laplace transform $F(s)$ and $G(s)$.

Then for all $\alpha ,\beta \in C$ $\mathcal{L}${$\alpha f(t)+\beta g(t)$} $=\alpha F(s)+\beta G(s)$ the ROC is the intersection of the ROCs for $F(s)$ and $G(s)$.

Theorem 3: Time-Scaling

If $\mathcal{L}${$f(t)$}$=F(s)$ then for $c>0,\mathcal{L}${$f(ct)$}$=\frac{1}{c}F(\frac{s}{c})$

Theorem 4: Exponential Modulation

$\mathcal{L}\{e^{\alpha t}f(t)\}=F(s-\alpha )$

Theorem 5: Time-Shifting

If $F(s)=\mathcal{L}\{f(t)u(t)\}$ and $g(t)=f(t-T)u(t-T)$ then $G(s)=e^{-sT}F(s)$

Theorem 6: Multiplication by t

If $\mathcal{L}\{f(t)\}=F(s)$ then $\mathcal{L}\{tf(t)\}=-\frac{d}{ds}F(s)$.

Theorem 7: Laplace Transform of a Derivative / Integral

Let $f(t)$ be such