Laplace transform
Laplace transform
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
$ \text{y}(t) = e^{st} * h(t) $
$~~~~~~~ = \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 preserved the shape of the input signal.
Note:
All RLC circuits are LTI systems
Any AC source always gives 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:
Mathematical expression
Region of convergence: the region in s-plane, where the mathematical expression is valid.
Example:
$\text{x}(t) = e^{-t} u(t) $
Solution: We know that Laplace tranform 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}^{\ infity} 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 nit ROC cannot have poles within them.
Exercise:
What is the ROC in above example?
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 – Causality and stability
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 ($h(t)$ 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:
- $\text{H}(s) = \frac{1}{s+1}
\text{and}Re{s} > -1$
$~~~~~~~~~~h(t) = e^{-t} u(t)$ (Causal)
- $\text{H}(s) = \frac{e^s}{s+1}$ 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:
- Shifting operator: $h(t) = \delta(t-t_0)$Taking Laplace transform: $ \text{H}(s) = e^{st_0}$ ROC: Full $s$-plane (stable)
- 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) $
Example: Let $\text{x}(t) = \cos(\omega_0 t)$ for given $\text{H}(s) = \frac{1}{s+1}$ and $Re{s} > -1$
$\cos(\omega_0 t) \rightarrow \frac{1}{\sqrt{1+\omega_0^2}} \cos({\omega_ot-\theta})$
where $\theta = tan^{-1}({\omega_0})$
$|\text{H}(j\omega_o)| = \frac{1}{\sqrt{1+\omega_0^2}} \rightarrow$ Low pass filter
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 $\angle \text{H}(j\omega_o) \rightarrow \text{phase response}$
The $\text{H}(s) = \frac{s}{s+1} \rightarrow$ High Pass filter