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.:



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 $$\Re(s+a) > 0.$$

$$\mathcal{L}\left\{e^{-at} u(t)\right\} = \frac{1}{s+a}$$

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:

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.

Sallen-Key Lowpass Filter
 The Sallen-Key circuit is widely used for the implementation of analog second order sections.

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} = s C_2 V_2 \\ I_{C1} = s C_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 s C_1 (V_1 - V_{out}) \\ V_1 - V_{out} = R_2 s C_2 V_{out} \end{cases} $$


 * $$ \begin{cases}

R_2 V_{in} = (R_1 + R_2 + R_1 R_2 s C_1) V_1 - (R_1 + R_1 R_2 s C_1) V_{out} \\ V_1 = (1 + s R_2 C_2) V_{out} \end{cases} $$


 * $$ R_2 V_{in} = \begin{bmatrix} (1 + s R_2 C_2) (R_1 + R_2 + R_1 R_2 s C_1) - R_1 - R_1 R_2 s C_1 \end{bmatrix} V_{out} $$


 * $$ V_{in} = \begin{bmatrix} (1 + s R_2 C_2) (R_1/R_2 + 1 + s R_1 C_1 - R_1/R_2 - s R_1 C_1 \end{bmatrix} V_{out} $$


 * $$ \frac{V_{in}}{ V_{out}} = R_1/R_2 + 1 + s R_1 C_1 + s R_1 C_2 + s R_2 C_2 + s^2 R_1 R_2 C_1 C_2 - R_1/R_2 - s R_1 C_1 $$

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:
 * $$ H(s) = \frac{\frac{1}{R_1 R_2 C_1 C_2}}{s^2 + \frac{R_1 + R_2}{R_1 R_2} \frac{1}{C_1} s + \frac{1}{R_1 R_2 C_1 C_2}} $$