ve216_l6.md

VE216 Lecture 6

Laplace Transform

Concept Map

Laplace Transform Definition

$X(s) = \int x(t)e^{-st}dt$

Two versions

• Unilateral: $X(s) = \int^\infty_0 x(t)e^{-st}dt$
• Bilateral: $X(s) = \int_{-\infty}^\infty x(t)e^{-st}dt$

(We focus on bilateral version.)

Region of Convergence (ROC)

This is based on the $X(s) = \int_{-\infty}^\infty x(t)e^{-st}dt$, we can see from the examples.

Example

Exercise 1

$x_1(t) = \begin{cases} e^{-t} &t \geq 0\0 &t< 0 \end{cases}$ Laplace transform.

$X_1(s) = L[x_1(t)] = \int_0^\infty e^{-t}e^{-st}dt = -\frac{1}{1+s}\int^\infty_0 -(1+s)e^{-(1+s)t}dt = -\frac{1}{1+s} e^{-(1+s)t}\Big|^\infty_0 = -\frac{1}{1+s}$

Since $e^{-(1+s)t}$ is convergent when $1+s &lt; 0$, or $Re(s) &gt; -1$, so ROC shows

Exercise 2

$x_2(t) = \begin{cases}-e^{-t} &t\leq 0\0 &t>0 \end{cases}$ Laplace transform.

$X_2(s) = \int_{-\infty}^0-e^{-t} e^{-st}dt = \frac{1}{1+s}\int^0_{-\infty}-(1+s)e^{-(1+s)t}dt = \frac{1}{1+s}$

Since $e^{-(1+s)t}$ is convergent for $x\in (-\infty,0)$, so $-(1+s) &gt; 0$ or $Re(s) &lt; -1$ with ROC shows

Exercise 3

$x_3(t) = e^{-|t|}$ Laplace transform.

$X_3(s) = \int_{-\infty}^\infty e^{-|t|-st}dt = \int_0^\infty e^{-(1+s)t}dt + \int_{-\infty}^0 e^{(1-s)t}dt = \frac{e^{(1-s)t}}{1-s}\Big|^0_{-\infty}-\frac{e^{-(1+s)t}}{1+s}\Big|^\infty_0 = \frac{1}{1-s}+\frac{1}{1+s}$

So $1-s &gt; 0$ and $1+s &gt; 0$, $1 &gt; s &gt; -1$, with ROC shows

Exercise 4

$\frac{2s}{s^2-4}$ Laplace inverse transform, how many possible solution and ROCs?

$\frac{2s}{s^2-4} = \frac{1}{s+2} + \frac{1}{s-2}$, with $\pm 2$ as poles.

• $\frac{1}{s+2}$ has 2 forms with 2 ROCs.
  • $\frac{1}{s+2} = \int^\infty_{-\infty}e^{-st}e^{-2t}u(t)dt = \int^\infty_0e^{-(2+s)t}dt = -\frac{1}{s+2}e^{-(2+s)t}\Big|^\infty_0$, so $2+s &gt; 0$ with $Re(s) &gt; -2$.
  • $\frac{1}{s+2} = \int^\infty_{-\infty} -e^{-st}e^{-2t}u(-t)dt = \int^0_{-\infty}-e^{-(s+2)t} = \frac{1}{s+2}e^{-(s+2)t}\Big|^0_{-\infty}$, so $2+s &lt; 0$ with $Re(s) &lt; -2$.
• $\frac{1}{s-2}$ has 2 forms with 2 ROCs
  • $\frac{1}{s-2} = \int^\infty_{-\infty}e^{-st}e^{2t}u(t)dt = \int^\infty_0e^{(2-s)t}dt = \frac{1}{2-s}e^{(2-s)t}\Big|^\infty_0$, so $2-s &lt; 0$ with $Re(s) &gt; 2$.
  • $\frac{1}{s-2} = \int^\infty_{-\infty} -e^{-st}e^{2t}u(-t)dt = \int^0_{-\infty}-e^{(2-s)t}dt = \frac{1}{s-2}e^{(2-s)t}\Big|^0_{-\infty}$, so $2-s &gt; 0$ with $Re(s) &lt; 2$.

So there are totally 3 solutions:

• $x(t) = e^{-2t}u(t) + e^{2t}u(t)$ with $Re(s) &gt; 2$
• $x(t) = -e^{-2t}u(-t)-e^{2t}u(-t)$ with $Re(s) &lt; -2$
• $x(t) = e^{-2t}u(-t)-e^{2t}u(-t)$ with $2 &gt; Re(s) &gt; -2$

Laplace Transform of a Derivative

$X_d(s) = \int^\infty_{-\infty}x'(t)e^{-st}dt = x(t)e^{-st}\Big|^\infty_{-\infty}-\int^\infty_{-\infty}x(t)(-s)e^{-st}dt$

Since $X(s)$ is convergent, so $x(t)e^{-st}\Big|^\infty_{-\infty}=0$, thus $X_d(s) = sX(s)$.

Laplace Transform Properties

Initial Value Theorem

If $x(t) = 0$ $\forall t &lt; 0$, $x(t)$ contains no impulses at $t=0$, then

$x(0^+) =\lim_{s\to\infty}sX(s) = \lim_{s\to\infty}\int^\infty_0x(t)se^{-st}dt = \lim_{s\to\infty}\int^\infty_0x(t)\delta(t)dt = x(0^+)$

(As $s \to \infty$ $e^{-st}$ shrink to 0 very fast.)

Final Value Theorem

If $x(t) = 0$ $\forall t &lt; 0$, $x(t)$ contains no impulses at $t=0$, then

$x(\infty) = \lim_{s\to0}sX(s) = \lim_{s\to0}\int^\infty_0x(t)se^{-st}dt$

Since $se^{-st}$ is flattened when $s\to 0$, $se^{-st}$ covers area of 1 whatever $s$.

So $x(\infty) = \frac1\infty\int^\infty_0x(t)dt = x(\infty)$.