Lecture 10
Inverse Laplace Transform

In today’s lecture we introduce the Inverse Laplace Transform for CT signals and systems.

Inverse Unilateral Laplace Transform

Consider a causal function \(f(t)\) that may or may not be absolutely integrable. Let \(g(t)=f(t) e^{-c t}\) for some \(c>0\) such that

\[\int_{0}^{\infty}|g(t)| dt = \int_{0}^{\infty}\left|f(t) e^{-c t}\right| dt < \infty\]

Then the Fourier Transform of \(g(t)\) exists

\[\begin{aligned} G(\omega) & =\int_{-\infty}^{\infty} g(t) e^{-j \omega t} dt = \int_{0}^{\infty} f(t) e^{-c t} e^{-j \omega t} dt \\ & =\int_{0}^{\infty} f(t) e^{-(c+j \omega) t} dt \\ & =\left.\mathcal{L}_{1}\{f(t)\}\right|_{s = c+j\omega} \quad c \text{ fixed s.t. } c+j\omega \in \text{ROC} \end{aligned}\]

The inverse Fourier Transform is

\[g(t)=f(t) e^{-c t}=\frac{1}{2 \pi} \int_{-\infty}^{\infty} G(\omega) e^{j \omega t} d\omega\]

multiply through by \(e^{c t}\)

\[\begin{aligned} e^{c t} g(t)=f(t) & =\frac{1}{2 \pi} \int_{-\infty}^{\infty} G(\omega) e^{c t} e^{j \omega t} d\omega \\ & =\frac{1}{2 \pi} \int_{-\infty}^{\infty} G(\omega) e^{(c+j \omega) t} d\omega \end{aligned}\]

Substitute \(G(\omega)\)
\[f(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left[\underbrace{\int_{0}^{\infty} f(\tau) e^{-(c+j \omega) \tau}}_{\left.F(s)\right|_{s=c+j \omega}} d\tau\right] e^{(c+j \omega) t} d \omega\] \[f(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(c+j \omega) e^{(c+j \omega) t} d \omega\]

Let \(s=c+j \omega\). Since \(c\) is a constant \(ds = 0+j d\omega\) implying \(d \omega=\frac{1}{j} d s\) and \[f(t)=\frac{1}{2 \pi j} \int\limits_{c-j\infty}^{c+j\infty} F(s) e^{st}\; ds\] a complex integral.

Note: we can think of \(c\) as the real values such that \(f(t) e^{-c t}\) has a Fourier Transform. This is the ROC.

Does the above integral look like our contar Integrals from previous lectures?

Yes, if we use the Bromwitch contour.

\[f(t)=\frac{1}{2 \pi j} \int\limits_{c-j \infty}^{c+j \infty} F(s) e^{s t} d s = \lim_{R \rightarrow \infty} \oint_{C(R)} F(s) e^{s t} ds\]

where \(C(R)\) is the Bromwitch contour.
This allows us to use the method of residues since \(C \in \text{ROC}\) implies \(C(R)\) encloses all singulartes of \(F(s)\).

Example: Recall that \(\mathcal{L}\{x(t)\}=\mathcal{L}\left\{e^{-a t} u(t)\right\}\) for \(a\in \mathbb{R}\) was \(\frac{1}{s+a} \text{Re}(s) > -a\)

Then

\[x(t)=\mathcal{L}_1 \left\{\frac{1}{s+a}\right\}=\frac{1}{2 \pi j} \int_{c-j \infty}^{c+ j\infty} \frac{1}{s+a} e^{s t} d s \quad \text{for} c > -a\] or using the Bromwitch contour \[x(t) =\lim_{R \rightarrow \infty} \frac{1}{2 \pi j} \oint_{C(R)} \frac{e^{s t}}{s+a} ds \qquad c>-a\]

Now from last time, since \(e^{s t}\) is analytic and \(C(R)\) encloses the singulary at \(-a\), from the Residue Theorem

\[\oint \frac{e^{st}}{s+a}\; ds = 2\pi j k_1\] where the residual is

\[k_1 = \left. (s+a) \frac{e^{st}}{s+a} \right|_{s = -a} = \left. e^{st} \right|_{s = -a} = e^{-a t} \qquad t > 0\]

Thus

\[x(t) = \frac{1}{2\pi j} \cdot 2\pi j e^{-at} \qquad t > 0 = e^{-at}u(t)\]

Inverse Bilateral Laplace Transform

Given a non-causal signal \(x(t)\) we can write it as

\[x(t) =\underbrace{x_{1}(t) u(-t)}_{\text{anticausal part}} +\underbrace{x_{2}(t) u(t)}_{\text{causal part}}\]

\[\begin{aligned} L_2\left\{ x(t) \right\} &= \int_{-\infty}^{0} x_{1}(t) e^{-s t} \; dt + \int_{0}^{\infty} x_{2}(t) e^{-s t} \; dt\\ &= \int_{0}^{\infty} x_{1}(-t) e^{s t} \; dt + \mathcal{L}_1 \left\{x_{2}(t)\right\}\\ &= \left. \mathcal{L}_1 \left\{x_{1}(-t)\right\} \right|_{s = -s} + \mathcal{L}_1 \left\{x_{2}(t)\right\}\\ X(s) &= \underbrace{X_1(s)}_{\text{Re}(s) < U} + \underbrace{X_2(s)}_{\text{Re}(s) > L} \end{aligned}\] where the ROC of \(X(s)\) is the intersection of the ROC for \(X_1\) and \(X_2\), a strip in the complex plane \(L < \text{Re}(s) < U\).

To use this for the inverse we apply this in reverse.

\[x(t) = \left. \mathcal{L}_1^{-1}\left\{X_1(-s)\right\}\right|_{t\rightarrow -t} + \mathcal{L}_1^{-1}\left\{X_2(s)\right\}\]

Example: Recall an example of a forward bilateral Laplace transform from lecture 9 \[\mathcal{L}_{2}\left\{e^{-|t|}\right\}=\underbrace{\frac{-1}{s-1}}_{\text{Re}(s) < 1}+\underbrace{\frac{1}{s+1}}_{\text{Re}(s) > -1} = X(s)\]

Then \[x(t)=\mathcal{L}_{2}^{-1}\{X(s)\}=\left.\mathcal{L}_{1}^{-1}\left\{\frac{1}{s+1}\right\}\right|_{t\rightarrow -t} \mathcal{L}_1^{-1}\left\{\frac{1}{s+1}\right\}\]

From our previous result when \(a=1\)

\[\begin{aligned} x(t) & =\left.e^{-t} u(t)\right|_{t \rightarrow -t}+e^{-t} u(t) \\ & =e^{t} u(-t)+e^{-t} u(t)=e^{-|t|} \end{aligned}\]

Example: Find the inverse Laplace transform of \[X(s) = \frac{1}{s} \quad \text{Re}(s) > 0\; .\]

Solution: Since the ROC corresponds to a causal signal, for \(t > 0\) \[\begin{aligned} x(t) &= \frac{1}{2 \pi j} \int\limits_{c-j \infty}^{c+j \infty} X(s) e^{s t} \; ds \quad c > 0 \\ &= \frac{1}{2 \pi j} \oint \frac{e^{s t}}{s} \; ds\\ &= \frac{1}{2 \pi j} 2 \pi j \; k \\ &= \left.s \frac{e^{st}}{s}\right|_{s=0}\\ &=1 \quad t > 0 \\ \end{aligned}\] which can be written as \(x(t) = u(t)\) for all times \(t\).

Example: Find the inverse Laplace transform of \[X(s) = \frac{10}{s^{2} + 5s + 6} \quad \text{Re}(s) >-2 \; .\]

First we use the partial fraction expansion.

\[X(s) = \frac{10}{(s+2)(s+3)} =\frac{A}{s+2}+\frac{B}{s+3}\]

\[A=\left.\frac{10}{s+3}\right|_{s=-2}=\frac{10}{1}=10\]

\[B=\left.\frac{10}{s+2}\right|_{s=-3}=\frac{10}{-1}=-10\]

Then \[x(t)=\mathcal{L}_{1}^{-1}\left\{\frac{10}{s+2}\right\}+\mathcal{L}_{1}^{-1}\left\{\frac{-10}{s+3}\right\}\]

Using the Residue theorem,

\[\begin{aligned} x(t)&=\frac{10}{2 \pi j} \int\limits_{c-j\infty}^{c+j \infty} \frac{e^{s t}}{s+2} \; ds - \frac{10}{2 \pi j} \int\limits_{c-j \infty}^{c+j \infty} \frac{e^{s t}}{s+3} \; ds \\ & c>-2 \quad c>-2 \\ & =\frac{10}{2 \pi j} \oint \frac{e^{s t}}{s+2} \; d s - \frac{10}{2 \pi j} \oint \frac{e^{s t}}{s+3} \; ds \\ & =\frac{10}{2 \pi j} 2 \pi j e^{-2 t} - \frac{10}{2 \pi j} 2 \pi j e^{-3 t} \\ &=10 e^{-2 t}-10 e^{-3 t} \quad t>0 \end{aligned}\]

Writing the expression for all \(t\) gives:

\[x(t)=10\left(e^{-2 t}-e^{-3 t}\right) u(t) .\]

Example: Here is an example with complex singularties. Find the inverse Laplace transform of

\[X(s)=\frac{s}{s^{2}+2 s+5} \quad \text{Re}(s)>-1\]

Using the partial fraction expansion

\[X(s) = \frac{s}{(s + 1 + j2)(s + 1 - j2)} = \frac{A}{s + 1 + j2} + \frac{B}{s + 1 - j2}\] where \[A=\frac{-1-j2}{-1-j2+1-j2}=\frac{-1-j2}{-j4}=\frac{1}{2}-\frac{1}{4} j\]

\[B=\frac{-1+j2}{-1+j2+1+j2}=\frac{-1+j2}{j4}=\frac{1}{2}+\frac{1}{4} j\]

Then

\[x(t) = \mathcal{L}_{1}^{-1}\left\{\frac{\frac{1}{2}-\frac{1}{4} j}{s+1+j2}\right\}+\mathcal{L}_{1}^{-1}\left\{\frac{\frac{1}{2}+\frac{1}{4} j}{s+1-j2}\right\}\]

Using residues, let \(c = 0 > -1\)

\[x(t) = \frac{1}{2 \pi j} \int\limits_{-j \infty}^{j\infty} \frac{\frac{1}{2}-\frac{1}{4} j}{s+1+j2} e^{st}\; ds + \frac{1}{2 \pi j} \int\limits_{-j \infty}^{j\infty} \frac{\frac{1}{2}+\frac{1}{4} j}{s+1-j2} e^{st}\; ds\]

\[x(t)=\frac{1}{2 \pi j} 2 \pi j \, k_{1}+\frac{1}{2 \pi j} 2 \pi j\, k_{2}\]

\[\begin{aligned} & k_{1}=\left(\frac{1}{2}-\frac{1}{4} j\right) e^{(-1-j 2) t} \quad k_{2}=\left(\frac{1}{2}+\frac{1}{4} j\right) e^{(-1+j 2) t} \\ & x(t)=k_{1}+k_{2}=\left(\frac{1}{2}-\frac{1}{4} j\right) e^{(-1-j 2) t}+\left(\frac{1}{2}+\frac{1}{4} j\right) e^{(-1+j 2) t} \\ & \text { } \quad \text{for } t > 0 \end{aligned}\]

with some effort you can write this as

\[x(t)=e^{-t}\left(\cos (2 t)-\frac{1}{2} \sin (2 t)\right)\; u(t)\]

for all \(t\).