# Chapter 3 - The Laplace Transform

$f(t)$ is causal, $f(t) = 0$ for $t < 0$.

$$F(s) = L[f(t)] = \int\limits_0^\infty f(t) e^{-st}dt$$

$s$ is a complex variable of the form:
$$s = \sigma + j\omega$$

### The Region of Convergence
The ROC is the set of all values of $s$ for which the one-sided Laplace transform exists.

If the causal $f(t)$ has finite support in the temporal region:
$$0 \le t_1 \le t_2 < \infty$$
$$F(s) = L[f(t)] = \int\limits_{t_1}^{t_2} f(t)e^{-st}dt$$

If $f(t)$ is causal, the ROC includes $s = \infty$.

If $f(t)$ has infinite support, $F(s)$ can be written as the ratio of two functions $N(s)$ and $D(s)$.
$$F(s) = {N(s) \over D(s)}$$

### Example (Finite Support)
$$F(s) = {s \over s^2 + 2s + 2}$$
Zero at $s=0$, poles at $s = -1 \pm j$.

ROC does not contain any poles, and is not influenced by zeros.

### Example (Infinite Support)
$$F(s) = {s+1 \over (s+2)(s+5)}$$

ROC is $\Re\{s\} > -2$

### Laplace Transform of Impulse
$$L[\delta(t)] = \int\limits_0^\infty \delta(t)e^{-st}dt = 1$$
$$L[\delta(t - \alpha)] = e^{-\alpha s}$$

### Example
Find the Laplace transform of a causal version of the complex exponential
$$L[e^{ja t}u(t)] = \int\limits_0^\infty e^{ja t}e^{-st}dt$$
$$F(s) = \lim_{v \to \infty} \int\limits_0^v e^{(ja-s)t}dt$$
$$F(s) = -{1\over s-ja} \lim_{v\to\infty}[e^{(ja-s)v}-1]={1\over s- ja}$$

Since $a$ is generally a complex number, $s = \sigma + j\omega$ and $a = \alpha + j\omega$.

## The Inverse Laplace Transform

$$F(s) = {3s + 5 \over (s+1)(s+2)} = {A \over s+1} + {B \over s+2}$$

Region of convergence:
$$\Re\{s\} > -1$$

Solve for $A$ and $B$:
$$3s + 5 = A(s+2) + B(s+1)$$
Eliminate $B$ by setting $s$ to -1:
$$3(-1) + 5 = A(-1 + 2) + B(-1 + 1)$$
$$2 = A$$
Eliminate $A$ by setting $s$ to -2:
$$3(-2) + 5 = A(-2 + 2) + B(-2 + 1)$$
$$-1  = -B \therefore B=1$$

Plug back in for $F(s)$:
$$F(s) = {2 \over s+1} + {1 \over s+2}$$
$$\therefore f(t)=\left[2e^{-t} + e^{-2t}\right]u(t)$$
**Note:** multiplying by $u(t)$ is required to make $f(t)$ causal.

### Example of the Delay Property
$$F(s) = {4e^{-10s} \over s(s+2)^2}$$
Take out the $e^{-10s}$ term and evaluate.
$$G(s) = {4 \over s(s+2)^2} = {A \over s} + {B \over s+2} + {C \over (s+2)^2}$$
$$4 = A(s+2)^2 + Bs(s+2) + Cs$$
Setting $s=0$ eliminates $B$ and $C$
$$4 = A(0+2)^2 + B(0)(0+2) + C(0)$$
$$4 = 4A \therefore A = 1$$
Setting $s=-2$ eliminates $A$ and $B$:
$$4 = A(-2 + 2)^2 + B(-2)(-2+2) + C(-2)$$
$$4 = -2C \therefore C=-2$$
Set an easy $s$ value to solve for $B$, $s=1$:
$$4 = 1(1+2)^2 + B(1)(1+2) + -2(1)$$
$$4 = 9 + 3B-2 \therefore B=-1$$
Plug back in:
$$G(s) = {4 \over s(s+2)^2} = {1 \over s} - {1 \over s+2} - {2 \over (s+2)^2}$$
Solve for $g(t)$
$$g(t) = \left[1 -e^{-2t} -2te^{-2t}\right]u(t)$$
By taking out the delay term, $e^{-10t}$, we offset $g(t)$ with respect to $f(t)$. To solve for $f(t)$, we must take this offset into account. The delay is by $10$, so for each $t$ in $g(t)$, there is a $(t-10)$ in $f(t)$. Plug in and solve:
$$f(t) = \left[1-e^{-2(t-10)}-2(t-10)e^{-2(t-10)}\right] u(t-10)$$

## Differential Equations using Laplace Transforms
$${df \over dt} + 6f(t) = u(t), f(0^-) = 1$$
$$s F(s) - f(0^-) + 6F(s) = {1 \over s}$$
$$(s + 6) F(s) = {1\over s} + 1$$
$$F(s) = {1 + s \over s(s + 6)}$$
$$F(s) = {A \over s} + {B \over s + 6}$$
$$1 + s = (s +6)A + sB$$
$$A = {1\over6}, B = {5\over6}$$
$$F(s) = {1 \over 6s} + {5 \over 6(s + 6)}$$
$$f(t) = \left[{1 \over 6} + {5 \over 6} e^{-6t}\right]u(t)$$