EIS Mathematics #2 – (The simplicity of Laplace transform) Battery – Application Note 50-2

Introduction

Several electrochemical techniques use results obtained using the Laplace transform. Examples can be found in Application notes #21, #28, #38 and #41b [1-4].

Definitions

The Laplace¹ transform of a function f(t) is defined by the integral

$$F(s)= \text L \{f(t)\} = \int^{\infty}_0 f(t)\text{exp}(-st)\text d t\tag{1}$$

Where s is the Laplace complex variable [5]. For example the Laplace transform of the sine function sin(t) is given by

$$\int^{\infty}_0 \text{sin}(t)\text{exp}(-st)\text d t\tag{2}$$

On the other hand, the time function is given by a more complicated expression from the Laplace transform in the s-domain by

$$f(t)= \text L^{-1} \{f(s)\} = \frac{1}{2 \pi i} \lim\limits_{T \to \infty} \int^{\gamma + iT}_{\gamma – iT}\text{exp}(st)F(s)\text{d}s\tag{3}$$

Table Of Laplace Transforms

Calculation of the integral of a function of a real variable is sometimes easy, e.g. it is well known that $\int xdx = x^2/2$. Calculation of the integral of a function of a complex variable is much more difficult. Fortunately it is very rare to have to use Eqs. (1) or (3). Laplace transforms and inverse Laplace transforms usually used in electrochemistry have already been calculated. The result is obtained by consulting a table of Laplace transform [5] (Tabs. I and II), or better using a software such as Wolfram Alpha [6].

Transfer Function of a Linear and Time Invariant (LTI) System

Definition

Laplace transform is the tool of choice for the study of linear and time invariant systems (Fig. 1) characterized by their transfer function H(s) defined by [7, 8]

$$H(s)= \frac{Y(s)}{X(s)} = \frac{\text{L}\{y(t)\}}{\text{L}\{x(t)\}}\tag{4}$$

Figure 1: Sketch of a scalar dynamic system.

For example, the impedance of a dipole is the transfer function for a current input

$$Z(s)= \frac{\text{L}\{\Delta E(t)\}}{\text{L}\{\Delta I(t)\}}= \frac{\Delta E(s)}{\Delta I(s)} \tag{5}$$

With $\Delta E(t)= E(t)-E_{t=0}$ and $\Delta I(t)= I(t)-I_{t=0}$.

Response of a LTI System

Introduction

Knowing the transfer function of a system allows us to calculate its response to an input. For example, if Z(s) is known, it is possible to determine the response of the dipole in the Laplace domain

$$\Delta E(s)= \Delta I(s)Z(s)\tag{6}$$

Then in the time domain using inverse Laplace transform

$$\Delta E(t)= \text L^{-1}\{\Delta E(s)\}\tag{7}$$

Response of a Parallel RC Circuit to a Current Heaviside Step Function

a. Impedance of a parallel RC circuit

The impedance of a parallel RC circuit is directly written from the admittance Y(s) = 1/Z(s) [9]:

$$Y(s)= \frac{1}{R}+sC \implies Z(s)= \frac{R}{1+RCs} \tag{8}$$

b. Response to a Heaviside step function

The Heaviside step function is equal to 0 for t<0 and 1 for t>0 (Fig. 2). The Laplace transform of the Heaviside step function is given by (Tab. I, Eqs. (1) and (2)).

Figure 2: Potential response of the parallel RC circuit to a Heaviside step current using Eq. (11).

$$\Delta I(s)= \text L \{\delta I(t)\} = \frac{\delta I}{s} \tag{9}$$

The Laplace transform of the output is written

$$\Delta E(s)= \frac{\delta I}{s} \frac{R}{1+RCs} \tag{10}$$

And the time domain response is given by (Tab. II, Eqs. (1) and (2)):

$$\Delta E(s)= \text L^{-1} \{\Delta E(s)\} = \text L^{-1} \left\{\frac{\delta I}{s}\frac{R}{1+RCs}\right\}= \delta IR \left( 1- \text{exp}\left(\frac{-t}{CR}\right) \right) \tag{11}$$

c. Response to a sinusoidal input

The Laplace transform of a sinusoidal signal is given by (Tab. I, Eqs (1) and (4))

$$I(t) = \delta I \text{sin}(\omega t) \implies I(s) = \text L\{\delta I(t)\}= \frac{\delta I \omega}{s^2 + \omega^2} \tag{12}$$

Therefore the output potential in the Laplace domain is

$$\Delta E(s) = \frac{\delta I \omega}{s^2 + \omega^2}\frac{R}{1+RCs}$$

And the time domain response is given by (Tab. II, Eqs. (1) and (4))

$$\Delta E(t) = \text L^{-1}\{\Delta E(s)\}= \text L^{-1}\left\{\frac{\delta I \omega}{s^2 + \omega^2}\frac{R}{1+\tau s}\right\} = \frac{\delta IR \left(\tau\omega\text{exp}\left(\frac{-t}{\tau}\right)\right)-\tau\omega\text{cos}(\omega t) + \text{sin}(\omega t)}{1+\tau^2\omega^2}\tag{13}$$

With τ = RC. The establishment of the sinusoidal steady-state of potential is shown in Fig. 3.

Figure 3: Potential response of the parallel RC circuit to a sinusoidal current using Eq. (13).

Ramp Response of a Series RC Circuit

a. Impedance of a series RC circuit

The impedance of a series RC circuit is given by

$$Z(s)= R+\frac{1}{C_s}\tag{14}$$

b. Response to a potential ramp

The signal used in linear sweep voltammetry is a potential ramp defined as:

$$E(t)=E_i+\upsilon t \implies \Delta E(t) -E_i= \upsilon t \tag{15}$$

The Laplace transform of a ramp is given by (Tab I., Eqs. (1) and (3))

$$\Delta E(s) =\text L \{\Delta E(t)\} =\text L \{\upsilon t\} = \frac{\upsilon}{s^2}\tag{16}$$

With

$$Z(s)= \frac{\Delta E(s)}{\Delta I(s)} \implies \Delta I(s) =\frac{\Delta E(s)}{Z(s)}$$

It is obtained

$$\Delta I(s) =\text L^{-1}\{\delta I(s)\} =\text L^{-1} \left\{\frac{C\upsilon}{s(1+CRs)}\right\}$$

And using Eqs. (1) and (2) (Tab. II),

$$\Delta I(t)= I(t)=C\upsilon\left( 1- \text{exp}\left(\frac{-t}{CR}\right) \right) \tag{17}$$

Figure 4. Current response of the series RC circuit to a ramp of potential calculated using Eq. (17).

Laplace and Inverse Laplace Transforms Tables

Laplace Transforms

Laplace transforms used in this note are shown in Tab. I.

Table I: Table of Laplace transforms.

Inverse Laplace Transforms

Inverse Laplace transforms used in this note are shown in Tab. II.
Table II: Table of inverse Laplace transforms.