The Laplace transform of a continuous time signal $\text{x}(t)$ can be expressed as: $$\text{X}(s) = \int_{-\infty}^{\infty} \text{x}(t) e^{-st} dt$$

Where $s = \sigma + j \omega$ represent a complex frequency on a complex number plane.

$$e^{-st} = e^{-(\sigma+j\omega)t} = e^{-\sigma t} e^{-j\omega t}$$

We can clearly see that

$Re\{e^{-s t}\} = e^{-\sigma t} \cos(\omega t)$

$Im\{e^{-s t}\} = - e^{-\sigma t} \sin(\omega t)$

Eigenfunctions of LTI systems

Let the system $h(t)$ be a linear and time invariant (LTI) system and the input $\text{x}(t) = e^{st}$. The output $\text{y}(t)$ of this system can be written as

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$~~~~~~~ = \int\_{-\infty}^{\infty} h(\tau) e^{s(t-\tau)} d\tau$

$~~~~~~~ = \int\_{-\infty}^{\infty}~ h(\tau) ~e^{st}~ e^{-s\tau} ~d\tau$

$~~~~~~~= e^{st} \int\_{-\infty}^{\infty}~ h(\tau)~ e^{-s\tau} d\tau$

$~~~~~~~ = H(s) e^{st}$

$e^{st}$ is known as eigen function for LTI system and it preserves the shape of the input signal.

Example - All RLC circuits are LTI systems. Recall that for any AC source, we always get sinusoidal voltage/current across any part of the circuit. The amplitude and phase might change but the shape remains the same

The Laplace transform has always two parts:

1. Mathematical expression

2. Region of convergence: the region in the complex s-plane, where the mathematical expression is valid.

Example

1. $\text{x}(t) = e^{-t} u(t)$

   Solution: The Laplace transform is given as:

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

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \int\_{0}^{\infty} e^{-t} e^{-st} dt$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \int\_{0}^{\infty} e^{-(s+1)t} dt$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{1}{s+1}$

• Region of convergence (ROC) is region in the s-plane where the above expression is valid.

• The expression is valid if $Re{\{s+1\}} > 0$ i.e. $Re{\{s\}} > -1$. $s = -1$ is a point of singularity.

Let $\text{X}(s)$ be the polynomials in $s$ variable as below:

$$\text{X}(s) = \frac{\text{A}(s)}{\text{B}(s)}$$

Roots of $\text{A}(s)$ i.e. $\text{A}(s) = 0 \rightarrow$ zeros

Roots of $\text{B}(s)$ i.e. $\text{A}(s) = 0 \rightarrow$ poles

Note: Poles play important role to decide ROC. ROC cannot have poles within them.

Exercise

What is the ROC for the pole-zero plot shown below?

Linear constant coefficient differential equations (LCCDE)

Any linear constant coefficient differential equations (LCCDE) in Laplace domain can be expressed as ratio of polynomials.

Example: Let $\text{x}(t) = \frac{d\text{y}}{dt} + a~\text{y}(t)$

Taking Laplace transform of the above equation:

$~~~~~~~~~~~\text{X}(s) = s\text{Y}(s) + a \text{Y}(s)$

$~~~~~~~~~~~ \text{X}(s) = (s + a)~ \text{Y}(s)$

$~~~~~~~~~~~\text{Y}(s) = \frac{\text{X}(s)}{s + a}$

$~~~~~~~~~~~ \text{H}(s) = \frac{\text{Y}(s)}{\text{X}(s)} = \frac{1}{s + a}$

Here $\text{H}(s)$ is known as transfer function or system function.

Application of Laplace transform for system analysis

Let $h(t)$ be the system with input $\text{x}(t)$ and output being $\text{y}(t)$.

$~~~~~\text{y}(t) = \text{x}(t) * h(t)$

The system input-output relationship in Laplace transform can be expressed as:

$~~~\text{Y}(s) = \text{X}(s) \text{H}(s)$

We are interested in additional system properties and their effect on impulse response and system function

Causality of LTI system

For LTI system to be causal, we should have

$~~~~h(t) =0, ~~\forall~ t<0$

$~~~~ \text{y}(t) = \text{x}(t)*h(t) = \int_{-\infty}^{\infty} \text{x}(\tau) h(t-\tau) d\tau$

For Causality: we should only use $\text{x}(\tau)$ for $\tau \leq t$

$\tau > t,~ h(t-\tau) = 0$

$h(t) = 0~~ \forall~ t<0$

• For a causal system (i.e.,$h(t)$ is a right-sided signal) , what is nature of $\text{H}(s)$ ?

• ROC of $\text{H}(s)$ will be right-sided plane. (Converse is not true in general)

• For a system with rational system function, causality is equivalent to the ROC being right-sided plane to the right of right most pole.

Example:

1. $\text{H}(s) = \frac{1}{s+1}~\text{and}~ Re\{s\} > -1$

$~~~~~~~~~~h(t) = e^{-t} u(t)$ (Causal)

2. $\text{H}(s) = \frac{e^s}{s+1}$ and $Re\{s\} > -1$

$~~~~~~~~~~h(t) = e^{-(t+1)} u(t+1)$ (Non-causal)

Stability of LTI system (bounded input bounded output)

An LTI system is said to be stable if

$$\int_{\infty}^{\infty} |h(t)| dt < \infty$$

i.e. $h(t)$ is absolutely integrable

We know Laplace transform is given as:

$$~~~~\text{H}(s) = \int_{\infty}^{\infty} h(t) e^{-st} dt$$

$$\text{H}(s)|_{s=0} = \int_{\infty}^{\infty} h(t) dt < \infty$$

Laplace transform converges at $s=0$, In fact, for any $s=j\omega = 0+j\omega, ~\text{H}(s)|\_{s=j\omega} < \infty \rightarrow$ system is stable

Note: $j\omega$-axis i.e. $Re\{s\} = 0$ is part of the ROC

Example:

1. Shifting operator: $h(t) = \delta(t-t_0)$

   Taking Laplace transform: $\text{H}(s) = e^{st_0}$

   ROC: Full $s$-plane (stable)

2. Integrator: $h(t) = u(t)$
   Taking Laplace transform: $\text{H}(s) = \frac{1}{s}$

   ROC: $Re\{s\} > 0$ (Not stable)

Note: A causal system with rational system function is stable if and only if: all the poles have negative real part

Frequency analysis and geometric interpretation

Let $\text{H}(s) = \frac{1}{s+1}$

The pole-zero plot on $s$-plane is

Evaluating $\text{H}(s)$ at any point at $s = s_0$ in the $s$-plane

$\text{H}(s)|_{s=s_0} = \frac{1}{s_0 +1}$

$\text{H}(s)|_{s=s_0} = |\text{H}(s_0)| e^{j\angle \text{H}(s_0)}$

$|\text{H}(s_0)| = \frac{1}{|s_0 + 1|}$ and $\angle \text{H}(s_0) = \theta$

Given a rational system function:

$\text{H}(s) = \frac{\prod*{i=1}^{M} (s-z_i)}{\prod*{i=1}^{N } (s-p_i)} \quad \text{and ROC~} N>M$

$|\text{H}(s)|_{s=s_0} = \frac{\prod_{i=1}^{M} (s-z_i)}{\prod_{i=1}^{N }|s_0 - P_i|}$

$\angle \text{H}(s)|_{s=s_0} = \sum_{i=1}^{M} \angle(s*0 – \text{z}\_i) - \sum*{i=1}^{N} \angle (s_0 – P_i)$

For any LTI system $\cos({\omega_o}) \rightarrow |\text{H}(j\omega_o)|\cos({\omega_ot-\theta})$

and $\theta = \angle ~\text{H}{(j\omega_0})$

where $|\text{H}(j\omega_o)| \rightarrow \text{magnitude response}$ and $\angle \text{H}(j\omega_o) \rightarrow \text{phase response}$

Example:

Consider the input $\text{x}(t) = \cos(\omega_0 t)$ for a system with transfer function $\text{H}(s) = \frac{1}{s+1}$ and ROC $Re\{s\} > -1$

$\cos(\omega_0 t) \rightarrow \frac{1}{\sqrt{1+\omega_0^2}} \cos({\omega_ot-\theta})$

where $\theta = tan^{-1}({\omega_0})$

Output signal magnitude is $|\text{H}(j\omega_o)| = \frac{1}{\sqrt{1+\omega_0^2}} \rightarrow$ Low pass filter