Block 03 — Complex Calculus in EE

Learning objectives

Preparation at Home

Well, again

• read through the present chapter and write down anything you did not understand.
• Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting).

For checking your understanding please do the following exercises:

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90-minute plan

1. Warm-up (x min):
   1. ….
2. Core concepts & derivations (x min):
   1. …
3. Practice (x min): …
4. Wrap-up (x min): Summary box; common pitfalls checklist.

Conceptual overview

Core content

Representation and Interpretation

Up to now, we used for the AC signals the formula $x(t)= \sqrt{2} X \cdot \sin (\omega t + \varphi_x)$ - which was quite obvious.

However, there is an alternative way to look at the alternating sinusoidal signals. For this, we look first at a different, but already a familiar problem (see Abbildung 1).

1. A mechanical, linear spring with the characteristic constant $D$ is displaced due to a mass $m$ in the Earth's gravitational field. The deflection only based on the gravitational field is $X_0$.
2. At the time $t_0=0$ , we deflect this spring a bit more to $X_0 + x(t_0)=X_0 + \hat{X}$ and therefore induce energy into the system.
3. When the mass is released, the mass will spring up and down for $t>0$. The signal can be shown as a shadow when the mass is illuminated sideways.
   For $t>0$, the energy is continuously shifted between potential energy (deflection $x(t)$ around $X_0$) and kinetic energy (${{\rm d}\over{{\rm d}t}}x(t)$)
4. When looking onto the course of time of $x(t)$, the signal will behave as: $x(t)= \hat{X} \cdot \sin (\omega t + \varphi_x)$
5. The movement of the shadow can also be created by the sideways shadow of a stick on a rotating disc.
   This means, that a two-dimensional rotation is reduced down to a single dimension.

The transformation of the two-dimensional rotation to a one-dimensional sinusoidal signal is also shown in Abbildung 2.

The two-dimensional rotation can be represented with a complex number in Euler's formula. It combines the exponential representation with real part $\Re$ and imaginary part $\Im$ of a complex value: $$ \underline{x}(t)=\hat{X}\cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} = \Re(\underline{x}) + {\rm j}\cdot \Im(\underline{x})$$

For the imaginary unit ${\rm i}$ the letter ${\rm j}$ is used in electrical engineering since the letter ${\rm i}$ is already taken for currents.

Complex Current and Voltage

The concepts of complex numbers shall now be applied to voltages and currents. Up to now, we used the following formula to represent alternating voltages:

$$u(t)= \sqrt{2} U \cdot \sin (\varphi)$$

This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$.

The parts on the complex plane are then given by:

1. The real part $\Re{(\underline{u}(t))} = \sqrt{2}U \cdot \cos (\omega t + \varphi_u)$
2. The imaginary part $\Im{(\underline{u}(t))} = \sqrt{2}U \cdot \sin (\omega t + \varphi_u)$

This is equivalent to the complex phasor $\underline{u}(t)=\sqrt{2}U \cdot {\rm e} ^{{\rm j} (\omega t + \varphi_u)}$

The complex phasor can be separated: \begin{align*} \underline{u}(t) &=\sqrt{2} U \cdot {\rm e}^{{\rm j} (\omega t + \varphi_u)} \\ &=\sqrt{2}\color{blue} {U \cdot {\rm e}^{{\rm j} \varphi_u}} \cdot {\rm e}^{{\rm j} \omega t} \\ &=\sqrt{2}\color{blue}{\underline{U}} \cdot {\rm e}^{{\rm j} \omega t} \\ \end{align*}

The fixed phasor (in German: komplexer Festzeiger) of the voltage is given by $\color{blue}{\underline{U}}= \color{blue}{U \cdot e ^{j \varphi_u}} $

Generally, from now on not only the voltage will be considered as a phasor, but also the current $\underline{I}$ and derived quantities like the impedance $\underline{X}$.
Therefore, the known properties of complex numbers from Mathematics 101 can be applied:

• A multiplication with $j$ equals a phase shift of $+90°$
• A multiplication with ${{1}\over{j}}$ equals a phase shift of $-90°$

Complex Impedance

Introduction to Complex Impedance

The complex impedance is „nearly“ similar calculated like the resistance. In the subchapters before, that impedance $Z$ was calculated by $Z=\frac{U}{I}$.
Now the complex impedance is:

\begin{align*} \underline{Z}&=\frac{\underline{U}}{\underline{I}} \\ &= \Re{(\underline{Z})} + {\rm j} \cdot \Im{(\underline{Z})} \\ &= R + {\rm j} \cdot X \\ &= Z \cdot {\rm e}^{{\rm j} \varphi} \\ &= Z \cdot (\cos \varphi + j \cdot \sin \varphi ) \end{align*}

With

• the resistance $R$ (in German: Widerstand) as the pure real part
• the reactance $X$ (in German: Blindwiderstand) as the pure imaginary part
• the impedance $Z$ (in German: Scheinwiderstand) as the complex number given by the complex addition of resistance and the reactance as a complex number

The impedance can be transformed from Cartesian to polar coordinates by:

• $Z=\sqrt{R^2 + X^2}$
• $\varphi = \arctan \frac{X}{R} $

The other way around it is possible to transform by:

• $R = Z \cos \varphi$
• $X = Z \sin \varphi$

Application on pure Loads

With the complex impedance in mind, the Tabelle ## can be expanded to:

The relationship between ${\rm j}$ and integral calculus should be clear:

1. The derivative of a sinusoidal value - and therefore a phasor - can simply be written as „$\cdot {\rm j}\omega$“,
   which also means a phase shift of $+90°$:
   \begin{align*}{{\rm d}\over{{\rm d}t}} {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}\end{align*}
2. The integral of a sinusoidal value - and therefore a phasor - can simply be written as „$\cdot (-{{1}\over{ {\rm j}\omega}})$“,
   which also means a phase shift of $-90°$.¹⁾

   \begin{align*} \int {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {{1}\over{\rm j\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} = -{{\rm j}\over{\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} \end{align*}

Once a fixed input voltage is given, the voltage phasor $\underline{U}$, the current phasor $\underline{I}$, and the impedance phasor $\underline{Z}$. In Abbildung 5 these phasors are shown.

Application on Impedance Networks

Simple Networks

In the chapter Kirchhoff's Circuit Laws we already had a look at simple networks like a series or parallel circuit of resistors.
These formulas not only apply to ohmic resistors but also to impedances:

Similarly, the voltage divider, the current divider, the star-delta transformation, and the Thevenin and Northon Theorem can be used, by substituting resistances with impedances. This means for example, every linear source can be represented by an output impedance $\underline{Z}_o$ and an ideal voltage source $\underline{U}$.

More "complex" Networks

For more complex problems having AC values in circuitries, the following approach is beneficial.
This concept will be used in the next chapter and in circuit design.

Common pitfalls

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Exercises

Worked examples

Exercise 6.5.1 Two voltage sources

Two ideal AC voltage sources $1$ and $2$ shall generate the RMS voltage drops $U_1 = 100~\rm V$ and $U_2 = 120~\rm V$.
The phase shift between the two sources shall be $+60°$. The phase of source $1$ shall be $\varphi_1=0°$.
The two sources shall be located in series.

1. Draw the phasor diagram for the two voltage phasors and the resulting phasor.

Solution 1

The phasor diagram looks roughly like this:

2. Calculate the resulting voltage and phase.

Solution 2

By the law of cosine, we get: \begin{align*} U&= \sqrt{{{U_1 }^2}+{{U_1 }^2}-2{U_1 } \cdot{U_2 }\cdot \cos(180°- {\varphi_2})} \\ &= \sqrt{{{100~V}^2}+{{120~V}^2}-2\cdot{100~V} \cdot{120~V}\cdot \cos(120°)} \end{align*}

The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. \begin{align*} \varphi&= \arctan 2 \left(\frac{Im\{\underline{U}\}} {Re\{\underline{U}\}} \right)\\ &= \arctan 2 \left(\frac{Im\{\underline{U}_1\}+Im\{\underline{U}_2\}}{Re\{\underline{U}_1\}+\{\underline{U}_2\}} \right)\\ &= \arctan 2 \left(\frac{{U}_2 \cdot\sin(\varphi_2)}{{U}_1 +{U}_2\cos(\varphi_2)} \right)\\ &= \arctan 2 \left(\frac{120~{\rm V}\cdot\sin(60°)} {100~{\rm V}+120~V\cos(60°)} \right) \end{align*}

Final value

\begin{align*} U &=190.79~{\rm V} \\ \varphi&=33° \end{align*}

3. Is the resulting voltage the RMS value or the amplitude?

Solution 3

The resulting voltage is the RMS value.

The source $2$ shall now be turned around (the previous plus pole is now the minus pole and vice versa).

4. Draw the phasor diagram for the two voltage phasors and the resulting phasor for the new circuit.

Solution 4

The phasor diagram looks roughly like this.
But have a look at the solution for question 5!

5. Calculate the resulting voltage and phase.

Solution 5

By the law of cosine, we get: \begin{align*} U&= \sqrt{{{U_1 }^2}+{{U_1 }^2}-2{U_1 } \cdot{U_2 }\cdot \cos(180°- {\varphi_1})} &= \sqrt{{{100~V}^2}+{{120~V}^2}-2\cdot{100~V} \cdot{120~V}\cdot \cos(60°)} \end{align*} The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. \begin{align*} \varphi&= \arctan 2 \left(\frac{Im\{\underline{U}\}} {Re\{\underline{U}\}} \right)\\ &= \arctan 2 \left(\frac{Im\{\underline{U}_1\}+Im\{\underline{U}_2\}}{Re\{\underline{U}_1\}+\{\underline{U}_2\}} \right)\\ &= \arctan 2 \left(\frac{-{U}_2 \cdot\sin(\varphi_2)}{{U}_1 -{U}_2\cos(\varphi_2)} \right)\\ &= \arctan 2 \left(\frac{-120~{\rm V}\cdot\sin(60°) }{100~{\rm V}-120~V\cos(60°)} \right) \\ &= \arctan 2 \left(\frac{-103.92...~{\rm V} }{+40~{\rm V}} \right) \end{align*} The calculated (positive) horizontal and (negative) vertical dimension for the voltage indicates a phasor in the fourth quadrant. Does it seem right?
The phasor diagram which was shown in answer 4. cannot be correct.
With the correct lengths and angles, the real phasor diagram looks like this: Here the phasor is in the fourth quadrant with a negative angle.

Final value

\begin{align*} U &=111.355~{\rm V}\\ \varphi&=-68.948° \end{align*}

Notice:

Be aware that some of the calculators only provide $\tan^{-1}$ or $\arctan$ and not $\arctan2$!
Therefore, you have always to check whether the solution lies in the correct quadrant.

Exercise 6.5.3 Series Circuit

The following circuit shall be given.

This circuit is used with different component values, which are given in the following.
Calculate the RMS value of the missing voltage and the phase shift $\varphi$ between $U$ and $I$.

1. $U_R = 10~\rm V$, $U_L = 10~\rm V$, $U_C = 20~\rm V$, $U=\rm ?$

Solution

The drawing of the voltage pointers is as follows: The voltage U is determined by the law of Pythagoras \begin{align*} U &= \sqrt{{{U_R } ^2}+{({U_L }-{U_C }})^2} \\ &= \sqrt{(10~{\rm V})^2+ {({10~{\rm V}}-{20~{\rm V}}})^2} \end{align*} The phase shift angle is calculated by simple geometry. \begin{align*} \tan(\varphi)&=\frac{{U_L }-{U_C }}{U_R}\\ &=\frac{{10~{\rm V}}-{20~{\rm V}}}{10~{\rm V}} \end{align*} Considering that the angle is in the fourth quadrant we get:

Final value

\begin{equation*} U=\sqrt{2}\cdot 10~{\rm V} = 14.1~{\rm V} \qquad \varphi=-45° \end{equation*}

2. $U_R = ?$, $U_L = 150~\rm V$, $U_C = 110~\rm V$, $U=50~\rm V$

Solution 2

The drawing of the voltage pointers is as follows: The voltage $U_R$ is determined by the law of Pythagoras \begin{align*} U_R&=\sqrt{{U ^2}+{({U_L} -{U_C}} )^2}\\ &=\sqrt{(50~\rm V)^2 +{(150~\rm V -110~\rm V})^2} \end{align*} The phase shift angle is calculated by simple geometry. \begin{align*} \tan(\varphi)&=\frac{{U_L} -{U_C} }{U_R}\\ &=\frac{{150~\rm V}-{110~\rm V}}{30~\rm V} \end{align*} Considering that the angle is in the fourth quadrant we get:

Final value

\begin{equation*} U_R= 30~{\rm V}\qquad \varphi=53.13° \end{equation*}

Embedded resources