# Signals and Systems/LaPlace Transform

## The Laplace Transform

Whilst the Fourier Series and the Fourier Transform are well suited for analysing the frequency content of a signal, be it periodic or aperiodic, the Laplace transform is the tool of choice for analysing and developing circuits such as filters.

The Fourier Transform can be considered as an extension of the Fourier Series for aperiodic signals. The Laplace Transform can be considered as an extension of the Fourier Transform to the complex plane.

### Unilateral Laplace Transform

The Laplace Transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by:

$F(s)={\mathcal {L}}\left\{f(t)\right\}=\int _{0}^{\infty }e^{-st}f(t)\,dt.$ The parameter s is the complex number:

$s=\sigma +j\omega ,\,$ with a real part σ and an imaginary part ω.

### Bilateral Laplace Transform

The Bilateral Laplace Transform is defined as follows:

$F(s)={\mathcal {L}}\left\{f(t)\right\}=\int _{-\infty }^{\infty }e^{-st}f(t)\,dt.$ Comparing this definition to the one of the Fourier Transform, one sees that the latter is a special case of the Laplace Transform for $s=j\omega$ .

In the field of electrical engineering, the Bilateral Laplace Transform is simply referred as the Laplace Transform.

### Inverse Laplace Transform

The Inverse Laplace Transform allows to find the original time function on which a Laplace Transform has been made.:

:$>f(t)={\mathcal {L}}^{-1}\{F(s)\}={\frac {1}{2\pi i}}\lim _{T\to \infty }\int _{\gamma -iT}^{\gamma +iT}e^{st}F(s)\,ds,$ ## Laplace Transform Examples

• Unit impulse function (dirac-delta function) $\delta (t)$ ${\mathcal {L}}\left\{\delta (t)\right\}=\int _{-\infty }^{\infty }e^{-st}\delta (t)\,dt=1.$ • Unit step function, $u(t)$ ${\mathcal {L}}\left\{u(t)\right\}=\int _{-\infty }^{\infty }e^{-st}u(t)\,dt=\int _{0}^{\infty }e^{-st}\,dt=-{\frac {e^{-st}}{s}}{\Biggr |}_{0}^{\infty }.$ The above integral converges only when $\Re (s)>0.$ For $\Re (s)>0$ ${\mathcal {L}}\left\{u(t)\right\}={\frac {1}{s}}.$ • Exponential function, $e^{-at}u(t)$ ${\mathcal {L}}\left\{e^{-at}u(t)\right\}=\int _{-\infty }^{\infty }e^{-st}e^{-at}u(t)\,dt=\int _{0}^{\infty }e^{-(s+a)t}\,dt=-{\frac {e^{-(s+a)t}}{s+a}}{\Biggr |}_{0}^{\infty }$ The above integral converges only when $\Re (s+a)>0.$ ${\mathcal {L}}\left\{e^{-at}u(t)\right\}={\frac {1}{s+a}}$ ## Differential Equations

### Integral and Derivative

The properties of the Laplace transform show that:

• the transform of a derivative corresponds to a multiplication with $s$ • the transform of an integral corresponds to a division with $s$ This is summarized in the following table:

Time Domain Laplace Domain
$x(t)$ $X(s)={\mathcal {L}}\left\{x(t)\right\}$ ${\dot {x}}(t)$ $s\cdot X(s)$ $\int x(t)dt$ ${\frac {1}{s}}\cdot X(s)$ With this, a set of differential equations is transformed into a set of linear equations which can be solved with the usual techniques of linear algebra.

### Lumped Element Circuits

Lumped elements circuits typically show this kind of integral or differential relations between current and voltage:

$U_{C}={\frac {1}{sC}}\cdot I_{C}$ $U_{L}=sL\cdot I_{L}$ This is why the analysis of a lumped elements circuit is usually done with the help of the Laplace transform.

## Example

### Sallen-Key Lowpass Filter

The image on the side shows the circuit for an all-pole second order function.

Writing $v_{1}$ the potential between both resistances and $v_{2}$ the input of the op-amp follower circuit, gives the following relations:

${\begin{cases}R_{1}I_{R1}=V_{in}-V_{1}\\R_{2}I_{R2}=V_{1}-V_{2}\\I_{C2}=sC_{2}V_{2}\\I_{C1}=sC_{1}(V_{1}-V_{out})\\I_{R1}=I_{R2}+I_{C1}\\I_{R2}=I_{C2}\\V_{2}=V_{out}\end{cases}}$ Rewriting the current node relations gives:

${\begin{cases}R_{1}R_{2}I_{R1}=R_{1}R_{2}I_{R2}+R_{1}R_{2}I_{C1}\\R_{2}I_{R2}=R_{2}I_{C2}\end{cases}}$ ${\begin{cases}R_{2}(V_{in}-V_{1})=R_{1}(V_{1}-V_{out})+R_{1}R_{2}sC_{1}(V_{1}-V_{out})\\V_{1}-V_{out}=R_{2}sC_{2}V_{out}\end{cases}}$ ${\begin{cases}R_{2}V_{in}=(R_{1}+R_{2}+R_{1}R_{2}sC_{1})V_{1}-(R_{1}+R_{1}R_{2}sC_{1})V_{out}\\V_{1}=(1+sR_{2}C_{2})V_{out}\end{cases}}$ $R_{2}V_{in}={\begin{bmatrix}(1+sR_{2}C_{2})(R_{1}+R_{2}+R_{1}R_{2}sC_{1})-R_{1}-R_{1}R_{2}sC_{1}\end{bmatrix}}V_{out}$ $V_{in}={\begin{bmatrix}(1+sR_{2}C_{2})(R_{1}/R_{2}+1+sR_{1}C_{1}-R_{1}/R_{2}-sR_{1}C_{1}\end{bmatrix}}V_{out}$ ${\frac {V_{in}}{V_{out}}}=R_{1}/R_{2}+1+sR_{1}C_{1}+sR_{1}C_{2}+sR_{2}C_{2}+s^{2}R_{1}R_{2}C_{1}C_{2}-R_{1}/R_{2}-sR_{1}C_{1}$ and finally:

${\frac {V_{out}}{V_{in}}}={\frac {1}{1+s(R_{1}+R_{2})C_{2}+s^{2}R_{1}R_{2}C_{1}C_{2}}}$ Thus, the transfer function is:

$H(s)={\frac {\frac {1}{R_{1}R_{2}C_{1}C_{2}}}{s^{2}+{\frac {R_{1}+R_{2}}{R_{1}R_{2}}}{\frac {1}{C_{1}}}s+{\frac {1}{R_{1}R_{2}C_{1}C_{2}}}}}$ 