Analog Filtering

Bilateral Laplace transform

Definition and properties

The Bilateral Laplace Transform is a kind of generalization of the Fourier Transform. Indeed, for $x\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, we have $$ \hat x(\nu) = \int_{-\infty}^{+\infty} x(t)e^{i2\pi\nu t}\mathrm{d}t $$ by the change of variable $p=i2\pi\nu$, $p$ being a pure imaginary numbers, we get $$ \hat x(p) = \int_{-\infty}^{+\infty} x(t)e^{p t}\mathrm{d}t $$ The bilateral Laplace transform extend $p$ to all the complex plan.

Bilateral Laplace Transform Let $x:\mathbb{R}\rightarrow \mathbb{C}$. When it exists, the bilateral Laplace transform of $x$ is given by $$ X(p) = \int_{-\infty}^{+\infty} x(t) e^{-pt}\mathrm{d} t\quad p = \sigma + i\omega\quad (\sigma,\omega)\in\mathbb{R}^2 $$ The region of convergence is under the form $\mathcal{R}(p)=]\sigma_1,\sigma_2[$, delimited by two lines parallel to the imaginary axis with abscises $\sigma_1$ and $\sigma_2$.

Why the bilateral Laplace transform ? The usual definition of the Laplace transform is the unilateral Laplace transform applied to Causal signal: if $x$ is causal in the previous definition, then the bilateral Laplace transform becomes: $$ \hat x(p) = \int_{0}^{+\infty} x(t)e^{p t}\mathrm{d}t $$ which is the usual unilateral Laplace transform. However, we can be interested by acausal signals or system, hence the used of the more general bilateral Laplace transform.

Two functions may have the same Laplace transform expression. The difference is given by the region of convergence.

examples Let $x(t) = \theta(t)e^{-at}$, $a\in\mathbb{R}$. $$ \begin{aligned} X(p) & = \int_{\mathbb{R}} \theta(t)e^{-at}e^{-pt}\mathrm{d} t = \int_0^{+\infty} e^{-(a+p) t} \mathrm{d} t\\ & = \frac{1}{a+p}\quad \mathcal{R}(p) = \sigma > -a \end{aligned} $$ Let $x(t) = \theta(-t)e^{-at}$, $a\in\mathbb{R}$. $$ \begin{aligned} X(p) & = \int_{\mathbb{R}} \theta(-t)e^{-at}e^{-pt}\mathrm{d} t = \int_{-\infty}^0 e^{-(a+p) t} \mathrm{d} t\\ & = \frac{1}{a+p}\quad \mathcal{R}(p) = \sigma < -a \end{aligned} $$

Inverse Laplace Transform Let $x$ a function which admits a Laplace transform $X(p)$ with region of convergence $\mathcal{R}(p)$. Then $$ x(t) = \frac{1}{i 2\pi} \int_{\sigma^+-i\infty}^{\sigma^- + i\infty} X(p) e^{pt}\mathrm{d} p $$

Causality and stabilty can be directly read on the Laplace transform.

Laplace transform and Causality Let $x$ a function which admits a Laplace transform $X(p)$ with region of convergence $\mathcal{R}(p)$. $x$ is causal iff the region of convergence is on the form $R(p) = \sigma > \sigma^+$. Moreover, $\lim\limits_{\sigma\rightarrow+\infty} |X(p)| = 0$ $x$ is anti-causal iff the region of convergence is on the form $R(p) = \sigma < \sigma^-$. Moreover $\lim\limits_{\sigma\rightarrow -\infty} |X(p)| = 0$\ .

Laplace transform and stability Let $x$ a function which admits a Laplace transform $X(p)$ with region of convergence $\mathcal{R}(p)$. $x$ is stable iff the region $1\in R(p)$.

Indeed, if $1\in R(p)$, then the Laplace transform is defined for $p=i2\pi \nu$ (with $\nu\in\RR), and is equal to the Fourier Transform: $$ \hat x(\nu) = X(i2\pi \nu) $$

Initial and final values of a signal $x$ can be deduced from its Laplace transform

Initial value and final value Let $x(t)$ a causal signal and $X(p)$its Laplace transform. Then, when the limits exists, $$ \lim_{t\rightarrow 0^+} x(t) = \lim_{\sigma\rightarrow +\infty} \sigma X(p) $$ and $$ \lim_{t\rightarrow +\infty} x(t) = \lim_{\sigma\rightarrow 0^+} \sigma X(p) $$

Propeties Let $x$ be a function and $X(p)$ its Laplace transform with $\mathcal{R}(p) = ]\sigma^+,\sigma^-[$. The Laplace transform is linear Translation: let $y(t) = x(t-t_0)$ $Y(p) = e^{-pt_0}X(p)$ Modulation: let $y(t) = e^{pt}x(t)$ $Y(p) = X(p-p_0)$ Scaling change let $y(t) = x(at)$ $Y(p) = \frac{1}{a} X\left(\frac{p}{a}\right)$ and $\mathcal{R}(p) = ]a \sigma^+, a\sigma^-[$ Derivation: let $y(t) = x’(t)$ $Y(p) = p X(p)$ Integration: let $y(t) = \int_{a}^t x(u) \mathrm{d} u$, $a\in\mathbb{R}$, $Y(p) = \frac{1}{p} X(p)$ et $\mathcal{R} = ]\sigma^+,\sigma^-[\cup ]0,+\infty[ $ Convolution: let $y(t) = (h*x)(t)$ with $h$ a function with Laplace transform $H(p)$ and $\mathcal{R}_h(p) = ]\tilde \sigma^+,\tilde \sigma^-[$. $Y(p) = H(p)X(p)$ et $R = ]\max(\tilde \sigma^+, \sigma^+), \min(\tilde \sigma^-,\sigma^-)[$

Thanks to the derivation and the integration properties, the Laplace transform will be very usefull to solve differential equations.

Usual Laplace transform

Transfert function

Let a filter with an impulse response $h$. Then, if $y$ is the response of the system with the input $x$, we have $$ y(t) = (h\star x) (t) = \int_{-\infty}^{+\infty} h(u) x(t-u) \mathrm{d} u $$ By taking the Laplace transform, we get $$ Y(p) = H(p)X(p) $$ H is called the transfert function of the system.

transfert function The transfer function of a filter is the Laplace transform of its impulse response. For a filter with impulse response $h$, we denote its transfer function by $H(p)$.

If the system and the input are stable , we can write: $$ Y(i2\pi\nu) = H(i2\pi\nu)X(i2\pi\nu) $$ i.e. $$ \hat y(\nu) = \hat h(\nu)\hat x(\nu) $$ $\hat h$, the Fourier transform of the impulse response, is called the complex gain of the system. The spectrum $|h(\nu)|^2 is called the frequency response of the system.

The complex gain of a filter is meaningfull for filtering. Indeed, by taking the Fourier transform of the filtering equation (suppsosing that the Fourier transform exists) $$ y(t) = h\star x(t) = \int_{-\infty}^{+\infty} h(u)x(t-u)\mathrm{d} u\ . $$ we get $$ \hat y(\nu) = \hat h(\nu)\hat x(\nu). $$ We clearly see that a filtering operation allows one to act direcly on the frequency content of a signal: some frequencies will be attenuated while some others can be amplified.

The “complex exponential” signals play a particular role for the filters. By taking $x(t) = e^{st},\ s\in\mathbb{C}$ as the input to the filter $\mathcal{S}$ with the impulse response $h$: $$ \begin{aligned} \mathcal{S}(x(t)) = \mathcal{S}(e^{st}) & = \int_\mathbb{R} h(t-\theta) e^{s\theta} \mathrm{d} \theta\\& = e^{st}\int_\mathbb{R} h(u) e^{-su} \mathrm{d} u\\& = H(s) e^{st} = H(s) x(t) \end{aligned} $$ with $H(s) = \int_\mathbb{R} h(u) e^{-su} \mathrm{d} u$ the eigen value associated to the eigen signal $t\mapsto e^{st}$.

An example: the RC filter

The tension $v(t)$ at the terminals of the capacitor is given by the differential equation $$ RC v’(t) + v(t) = u(t) $$ which admits as solution $$ v(t) = \frac{1}{RC} \int_{-\infty}^{t} e^{\frac{-(t-s)}{RC}} u(s) \mathrm{d} s $$ This system is clearly linear, being governed by a linear differential equation.

We can write the output $v(t)$ as a convolution between the input $u(t)$ and an impulse response $h(t)$ to be determinate. By tacking $$ h(t) = \frac{1}{RC}\Theta(t) e^{\frac{-t}{RC}} $$ where $\Theta$ is the Heaviside function. One can check that $$ v(t) = h\star u (t)\ . $$

This result can be found by using the Laplace transform and its properties. By taking the Laplace transform of the differential equation we get $$ p V (p) + \frac{1}{RC}V(p) = \frac{1}{RC}U (p) $$ with a region of convergence under the form $p\in ]\sigma,+\infty[$, $\sigma \in \mathbb{R}$, the system being clearly causal. We get $$ V(p) = \frac{\frac{1}{RC}}{\frac{1}{RC}+p} U(p) $$ We can then identified the Laplace transform $H$ of the impulse response of the filter $$ H(p) = \frac{\frac{1}{RC}}{\frac{1}{RC}+p} $$ By inverting $H$ we get the impulse response of the RC filter $$ h(t) = \frac{1}{RC}\Theta(t) e^{\frac{-t}{RC}} $$

Ideal filters

Ideal filters allow one to vanish some frequencies of the input signal.

Low pass filter

The complex gain of an ideal low pass filter is given by $$ \hat h_{\nu_0}^{Low}(\nu) = \begin{cases} 1 & \text{ if } |\nu| < \nu_0 \\ 0 & \text{ otherwise}\end{cases} $$ where $\nu_0$ is the cutting frequency of the filter: all the high frequencies above are vanished in the filtered signal by the ideal low pass filter.

The impulse response of this filter is given by taking the invert Fourier transform: $$ \begin{aligned} h_{\nu_0}^{Low}(t) & = \int_{-\infty}^{\infty} \hat h(\nu) e^{i2\pi\nu t} \mathrm{d}\nu\\ &= \int_{-\nu_0}^{\nu_0} e^{i2\pi\nu t} \mathrm{d}\nu \\& = \frac{\sin(2\pi\nu_0 t)}{\pi t}\\ & = 2\nu_0\text{sinc}(2\nu_0 t) \end{aligned} $$ This filter is then clearly not causal, and is then not realizable.

High pass filter

The complex gain of an ideal high pass filter is given by $$ \hat h_{\nu_0}^{High}(\nu) = \begin{cases} 1 & \text{ if } |\nu| > \nu_0 \\ 0 & \text{ otherwise}\end{cases} $$ where $\nu_0$ is the cutting frequency of the filter: all the low frequencies below $\nu_0$ are vanished in the filtered signal by the ideal high pass filter.

Here, $\hat h$ cannot be inverted !

The ideal high pass filter with the cutting frequency $\nu_0$ can be written in function of the ideal low pass filter with he cutting frequency $\nu_0$: $$ \hat h_{\nu_0}^{High}(\nu) = 1 - \hat h_{\nu_0}^{Low}(\nu) $$

Band pass filter

The complex gain of an ideal band pass filter is given by $$ \hat h_{[\nu_0,\nu_1]}^{Band}(\nu) = \begin{cases} 1 & \text{ if } \nu_0 < |\nu| < \nu_1 \\ 0 & \text{ otherwise}\end{cases} $$ where $\nu_0$ and $\nu_1$ are the two cutting frequencies of the filter: all the low frequencies below $\nu_0$ and above $\nu_1$ are vanished in the filtered signal by the ideal band pass filter.

The ideal band pass filter can be expressed as the difference of two ideal low pass filters with cutting frequencies $\nu_0$ and $\nu_1$: $$ \hat h_{[\nu_0,\nu_1]}^{Band}(\nu) = \hat h_{\nu_1}^{Low}(\nu) - \hat h_{\nu_0}^{Low}(\nu) $$

Band cut filter

The complex gain of an ideal band cut filter is given by $$ \hat h_{[\nu_0,\nu_1]}^{Cut}(\nu) = \begin{cases} 0 & \text{ if } \nu_0 < |\nu| < \nu_1 \\ 1 & \text{ otherwise}\end{cases} $$ where $\nu_0$ and $\nu_1$ are the two cutting frequencies of the filter: all the low frequencies between $\nu_0$ and $\nu_1$ are vanished in the filtered signal by the ideal band pass filter.

The ideal band cut filter can be expressed as the difference of two ideal low pass filters with cutting frequencies $\nu_0$ and $\nu_1$: $$ \hat h_{[\nu_0,\nu_1]}^{Cut}(\nu) = \hat h_{\nu_0}^{Low}(\nu) - \hat h_{\nu_1}^{Low}(\nu) $$