\(\)Electrical impedance is a complex-valued measure of the opposition that a circuit presents to a time-varying electric current when a voltage is applied. Combined with the Laplace transform it allows us to use algebra to calculate electrical networks.

\(u\)Instead of \(\Delta v\), we use the European symbol for voltage difference: \(u\). The letter ‘u’ stands for “Potential__u__nterschied”.

## Capacitor impedance

Capacitors store energy in an electric field. The amount of charge \(q\) stored in a capacitor is linearly proportional to the voltage \(u\) over the capacitor. [MIT] $$ q(t) = Cu(t)\label{eq:c_equiv} $$ where \(C\) is a constant called capacitance. The SI unit for capacitance is Farad with values typically range from from 2.2 pF to 470 μF.

The electrical charge \(q(t)\) is a function of the current accumulated over time, assuming that there is no initial charge. $$ q(t)=\int\limits_{0}^{t} \! i(\tau) \, \mathrm{d}\tau \label{eq:c_integral} $$

Combining equation \(\eqref{eq:c_equiv}\) and \(\eqref{eq:c_integral}\) and solving for the current and taking the derivative on both sides, gives the current as a function of the voltage $$ \begin{align} \int\limits_{0}^{t} \! i(\tau) \, \mathrm{d}\tau = C\,u(t) \\ \Rightarrow i(t) &= C\frac{\mathrm{d}u(t) }{\mathrm{d}t} \end{align} $$

The Laplace transform allows us to use algebra in complex frequency domain, instead of working with differential equations. Using the Laplace transform of the first derivative $$ \left. \begin{align} I(s) &= \mathfrak{L}i(t) = \mathfrak{L}\left\{ C\frac{\mathrm{d}u(t)}{\mathrm{d}t} \right\} = C\mathfrak{L}\left\{ \frac{\mathrm{d}}{\mathrm{d}t}u(t) \right\} \nonumber \\ \mathfrak{L}\left\{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\right\}&= -f(0^-)+s\mathfrak{L}\left\{f(t)\right\} \nonumber \\ f(0^-) &= 0 \nonumber \end{align} \right\} \Rightarrow I(s) = s\,C\,U(s) \label{eq:c_laplace} $$ where capital letters are used to indicate complex domain variables, such as \(I(s)=\mathfrak{L}i(t)\) and \(U(s)=\mathfrak{L}u(t)\).

Solving \(\eqref{eq:c_laplace}\) for the complex impedance \(Z_{C}(s) \equiv\frac{U(s)}{I(s)}\) we get $$ \shaded{ Z_{C}(s) = \frac{1}{sC} } \label{eq:c_impedance} $$

The plot below shows the magnitude of the capacitor impedance \(Z_C\) as a function of the frequency where \(s=\sigma+j\omega\).

## Inductor impedance

An inductor stores energy in a magnetic field. The magnetic flux \(\phi\) in the inductor is linearly proportional to the current \(i\) through the inductor. The role played by the inductor in this magnetic case is analogous to that of a capacitor in the electric case. $$ \phi(t) = L i(t) \label{eq:l_equiv} $$ where \(L\) is a constant called inductance. The SI unit for inductance is Henry with values typically range from from 0.1 µH to 1 mH.

According to Faraday’s law of induction [MIT], an inductor opposes changes in current by developing a voltage \(\varepsilon=-u\) proportional to the negative of the rate of change of magnetic flux \(\phi\). $$ \varepsilon(t) = -u(t) = -\frac{\mathrm{d}\phi(t) }{\mathrm{d}t} \label{eq:l_emf} $$

Combining equation \(\eqref{eq:l_equiv}\) and \(\eqref{eq:l_emf}\) yields $$ \begin{align} u(t) = -\varepsilon(t) = +\frac{\mathrm{d}\phi(t) }{\mathrm{d}t} = L \frac{\mathrm{d}i(t) }{\mathrm{d}t} \end{align} $$

Once more, the Laplace transform allows us to use algebra in complex frequency domain, instead of working with differential equations. Using the Laplace transform of the first derivative $$ \left. \begin{align} U(s) &=\mathfrak{L}u(t) =\mathfrak{L}\left\{ L\tfrac{\mathrm{d}i(t)}{\mathrm{d}t} \right\} =L\mathfrak{L}\left\{ \tfrac{\mathrm{d}}{\mathrm{d}t}i(t) \right\} \nonumber \\ \mathfrak{L}\left\{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\right\} &= -f(0^-)+s\mathfrak{L}\left\{f(t)\right\} \nonumber \\ f(0^-)&=0 \nonumber \end{align} \right\} \Rightarrow U(s) = s\,L\,I(s) \label{eq:l_laplace} $$ where capital letters are used to indicate complex domain variables, such as \(I(s)=\mathfrak{L}i(t)\) and \(U(s)=\mathfrak{L}u(t)\).

Solving \(\eqref{eq:l_laplace}\) for the complex impedance \(Z_L(s) \equiv\frac{U(s)}{I(s)}\) we get $$ \shaded{ Z_L(s) = sL } \label{eq:c_inductance} $$

The plot below shows the magnitude of the inductor impedance \(Z_L\) as a function of the frequency

## Resistor impedance

For completeness we show the resistor impedance \(Z_r\) as a function of the frequency $$ \shaded{ Z_r(s) = R } \label{eq:r_inductance} $$

The plot below shows the magnitude of the resistor impedance \(Z_r\) as a function of the frequency

## Overview

name | symbol | unit | impedance |
---|---|---|---|

Voltage | \(u(t)\) | Volt | |

Current | \(i(t)\) | Ampere | |

Resistance | \(R\) | \(u(t)=R\cdot i(t)\) | \(Z_r=R\) |

Capacitance | \(C\) | \(u(t)=\frac{1}{C}\int_0^t i(\tau)\,\mathrm{d}\tau\) | \(Z_c=\frac{1}{j\omega C}\) |

Inductance | \(L\) | \(u(t)=L\frac{\mathrm{d}i(t)}{\mathrm{d}t}\) | \(Z_l=j\omega L\) |

\(\sum u(t)=0\) | \(\sum i(t)=0\) |