Rowan-Classes/5th-Semester-Fall-2023/Signals-and-Systems/Notes/Chapter3.md

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)