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I am trying to understand why impedance is not represented using vectors.

I assume it is due to complex numbers having the property that $$j = \sqrt {-1}$$ but with my limited knowledge I can't see how this relates to impedance or why this property would be desired. I'm not sure what reactance has to do with the square root of \$-1.\$

Could someone explain to me why complex numbers are used rather than vectors?
An intuitive answer is fine; I don't need a complex proof.

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  • \$\begingroup\$ en.wikipedia.org/wiki/Complex_number \$\endgroup\$ – Bruce Abbott Jul 9 at 21:37
  • \$\begingroup\$ @Sanmvegsaini this makes sense, thanks. Why is the fact that j = sqrt(-1) used though? To me it seems like complex numbers were invented to solve a completely different problem (solving equations with sqrts of negative numbers) where this property makes sense, but I don't understand at all why this property is also applied to impedance. There must be more to it than just "complex numbers allow more operations than vectors", the fact that j = sqrt(-1) must also apply to impedance too but I can't see how. \$\endgroup\$ – JShorthouse Jul 9 at 22:22
  • \$\begingroup\$ Quite similar to electronics.stackexchange.com/questions/28285/… \$\endgroup\$ – clabacchio Jul 10 at 6:21
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    \$\begingroup\$ "I'm not sure what reactance has to do with the square root of -1" - to put it simply, reactance is derivative, derivative is 90-degree phase lead, and by representing sinusoids as phasors (complex numbers) \$ j \cdot \$ becomes a 90-degree phase lead. \$\endgroup\$ – fghzxm Jul 10 at 13:14
  • \$\begingroup\$ @JShorthouse Consider this, maybe it'll help. \$j\$ can be seen as a vector having a direction unobtainable via linear combinations of any set of vectors in \$\mathbb{R^n}\$. You have a minimal basis for, for example, \$\mathbb{R^2}\$, mainly \$(0,1)\$ and \$(1,0)\$ which point in the direction of the y- and x-axis. We simply poof out \$j\$ out of thin air and give it a value unobtainable from composition of the previous basis. On top of this, as others have expressed, the way we chose the rules for complex numbers make them useful for calculating derivatives and integrals. \$\endgroup\$ – Novicegrammer Jul 11 at 15:46
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Complex numbers are similar to vectors, but have some extra mathematical properties that make them useful. Most notably, using the complex exponential \$e^{j\omega t}\$ instead of sines and cosines makes differential equations much easier to deal with. That's how you get to complex impedance in the first place:

$$v(t) = A\mathrm e^{\mathrm{j} \omega t + \theta}$$ $$i(t) = B \mathrm e^{\mathrm j \omega t + \phi}$$ $$\frac {v(t)} {i(t)} = Z = \frac A B \mathrm e ^ {\mathrm j (\theta - \phi)}$$

Or, in phasor notation:

$$\hat V = A\angle \theta$$ $$\hat I = B\angle \phi$$ $$\frac {\hat V} {\hat I} = Z = \frac A B \angle (\theta - \phi)$$

You could use something like vector notation for the magnitude and phase, but vectors don't multiply and divide like complex numbers do, so it wouldn't improve anything.

EDIT: Complex numbers developed to solve certain algebra problems. If you want to know more about the history, check out the first chapter of Visual Complex Analysis by Tristan Needham. (You can read the preview on Amazon if you don't have a good library handy.)

The second chapter of the book can probably answer your question by itself, but I'll give it a shot too. Complex numbers are, in a sense, two-dimensional quantities, but what makes them useful here is that they also include the concept of rotation. Multiplication by \$\sqrt{-1}\$ is equivalent to a 90° rotation in a 2D plane:

$$\mathrm i ^ 0 = 1$$ $$\mathrm i ^ 1 = \mathrm i$$ $$\mathrm i ^ 2 = -1$$ $$\mathrm i ^ 3 = -\mathrm i$$ $$\mathrm i ^ 4 = 1$$

We can expand on this with complex exponentials, with let us represent a rotation by any amount:

$$\mathrm e^{j\pi/4} \cdot\mathrm e^{j\pi/4} = \mathrm e^{j(\pi/4 + \pi/4)} = \mathrm e ^ {j\pi/2} = \mathrm i$$ $$45^\circ + 45^\circ = 90^\circ$$

Notice that we get this by doing normal arithmetic -- multiplying real-valued exponentials works the same way.

Why does that matter? We can already represent rotations with sines and cosines, right? But that gets nasty in differential equations, mainly because you can't combine sines and cosines by adding them. On the other hand, the derivative of \$\mathrm e^x\$ is... itself. No trouble there!

So where does impedance come in? Well, think about the difference between DC and the sinusoidal steady state. At DC, node voltages are constant values with different magnitudes. At AC, node voltages are sinusoidal with the same frequency but different magnitudes and phase angles. The voltage/current relationships change too. With a resistor, voltage and current are in phase. In an inductor or a capacitor, there's a 90° phase difference between them.

So now the concept of rotation (phase "angle") has crept into our circuit analysis. We could stay in the time domain and do stuff like this:

$$v = L \frac {\mathrm d i} {\mathrm d t}$$ $$V\cos(\omega t) = \omega L\cdot I\cos(\omega t - 90^\circ)$$

Or we use could complex numbers, where a \$90^\circ\$ rotation just means multiplying by i (well, \$j\$ in our case -- this is EE):

$$V\mathrm e^{\mathrm j \omega t} = \mathrm j\omega L \cdot I \mathrm e^{\mathrm j \omega t}$$

The key benefit here is that all of the \$\mathrm e^{\mathrm j \omega t}\$ terms cancel out of equations, so now our voltage/current relationship is just Ohm's Law with complex numbers:

$$\hat V = \mathrm j \omega L \hat I$$

If I had to sum all this up in one sentence, I would say that complex numbers let you represent rotation by grouping the magnitude and phase together separate from the frequency, while sinusoids group the frequency and phase together.

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    \$\begingroup\$ I liked this answer better than the other. It's a start. But there is so, so much more of beauty, too. Complex numbers are in unitary group U(1) and is a good segue into the study of Lie groups and algebras. \$\endgroup\$ – jonk Jul 9 at 23:50
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    \$\begingroup\$ @JShorthouse I added some more to my answer. Please let me know if there's anything you still don't understand. \$\endgroup\$ – Adam Haun Jul 10 at 1:27
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    \$\begingroup\$ Great edit, thank you. I also had a quick look at the book you recommended and everything is starting to become clearer now. Your explanation of how sqrt(-1) can be used to perform rotation is what made things really click - you've made me realise the complex numbers are quite amazing and now I want to learn more about them. \$\endgroup\$ – JShorthouse Jul 10 at 2:34
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    \$\begingroup\$ @AdamHaun Coxeter's "Polytropes" book and Robert Gilmore's "Lie Groups and Algebras" are really good starts. I used to go, once a month, to Dr Sirag's home near the University of Oregon to study and argue about string theory ideas. (In fact, chapter 3 of his new book comes out from those early discussions we had.) \$\endgroup\$ – jonk Jul 10 at 2:52
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    \$\begingroup\$ @jonk No worries. Best wishes for you and your daughter. \$\endgroup\$ – Adam Haun Jul 10 at 6:37
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Why are complex numbers used and not Vectors?

simply because there is no vector division defined in vector algebra, so simply you cannot use Ohm's law in division form, thereby making calculations more complicated. On the other hand the domain of complex number athematic has more progressed over time than vector counterpart, so you have many theorems at your disposal to simply your expression and (easily) carry out analysis. So, even though you could work around with vector algebra, it is easier to work with complex number.

read more: https://math.stackexchange.com/questions/246594/what-is-vector-division

why impedance are represented as complex number?

consider the following circuit: enter image description here

if Q is the charge on the capacitor, and i is the current, then using KVL we will have

$$R\times i + \frac QC + L\times \frac{di}{dt} = V \dots(1)$$ $$ \implies \frac{d^2i}{dt^2} + \frac RL\times \frac{dQ}{dt} + \frac 1{LC}\times i = 0\dots (2)$$ $$\implies i = Ae^{a_1t}+Be^{a^2t}$$ where $$a_1, a_2 \in C$$ and general solutions of 2nd order Differential equation are always complex in nature.

hence, your i is complex expression and putting this value in eq 1 will gives V which will also be a complex expression. On Dividing V by i, you will get another complex expression which we call impedance of this circuit. So you see, the reason why an impedance is complex is because of the mathematics involved.

Now, if you want to have a "feel" of complex impedance, you should learn about phasors and have an analogy with that.

Read More: https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-007-electromagnetic-energy-from-motors-to-lasers-spring-2011/lecture-notes/MIT6_007S11_lec19.pdf

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Just to remark on that you can represent impedance as a matrix:

$$ R + \mathrm j X \leftrightarrow \begin{bmatrix} R & X \\ -X & R \end{bmatrix} $$

This is in fact the matrix representation of complex numbers. On the other hand you can represent sinusoidal signals (but not impedance) using vectors:

$$ x_{\cos} + \mathrm j x_{\sin} \leftrightarrow \begin{bmatrix} x_{\cos} \\ x_{\sin} \end{bmatrix} $$

Addition/subtraction/scaling of impedance and sinusoids are obviously just the homonymous operations on matrices and vectors. Admittance is the matrix inverse of impedance:

$$ (R + \mathrm j X)^{-1} \leftrightarrow \begin{bmatrix} R & X \\ -X & R \end{bmatrix}^{-1} = \frac 1 {(R^2 + X^2)} \begin{bmatrix} R & -X \\ X & R \end{bmatrix} $$

You can matrix-multiply impedance with current, or admittance with voltage:

\begin{align} \begin{bmatrix} R & X \\ -X & R \end{bmatrix} \begin{bmatrix} i_{\cos} \\ i_{\sin} \end{bmatrix} &= \begin{bmatrix} R i_{\cos} + X i_{\sin} \\ R i_{\sin} - X i_{\cos} \end{bmatrix} \\ \begin{bmatrix} G & B \\ -B & G \end{bmatrix} \begin{bmatrix} u_{\cos} \\ u_{\sin} \end{bmatrix} &= \begin{bmatrix} G u_{\cos} + B u_{\sin} \\ G u_{\sin} - B u_{\cos} \end{bmatrix} \end{align}

Phase difference is also a matrix:

$$ {\mathrm e}^{\mathrm j \varphi} = \cos \varphi + \mathrm j \sin \varphi \leftrightarrow \begin{bmatrix} \cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{bmatrix} $$

Derivative is simply \$ \omega \$ times a 90-degree phase lead:

$$ \mathrm j \omega \leftrightarrow \begin{bmatrix} 0 & \omega \\ -\omega & 0 \end{bmatrix} $$

With what we have got ourselves so far we can write differential equations as matrix equations

\begin{align} U_0 \cos {\omega t} = u + R C \frac {\mathrm d u} {\mathrm d t} \leftrightarrow \begin{bmatrix} U_0 \\ 0 \end{bmatrix} = (\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + R C \begin{bmatrix} 0 & \omega \\ -\omega & 0 \end{bmatrix}) \mathbf u = \begin{bmatrix} 1 & R C \omega \\ -R C \omega & 1 \end{bmatrix} \mathbf u \end{align}

... and solve it by calculating the inverse matrix of \$ \begin{bmatrix} 1 & R C \omega \\ -R C \omega & 1 \end{bmatrix} \$ and then multiply it onto the \$ U_0 \$ vector.


As you can see though, this system of notation is quite verbose, and doesn't provide an intuitive representation of phase and amplitude (everything is in Cartesian coordinates essentially).

BTW, power has a neat representation as the vector dot product:

$$ \frac 1 2 (u_{\cos} i_{\cos} + u_{\sin} i_{\sin}) = \frac 1 2 {\mathbf i}^{\mathrm T} \mathbf u = \frac 1 2 \begin{bmatrix} i_{\cos} & i_{\sin} \end{bmatrix} \begin{bmatrix} u_{\cos} \\ u_{\sin} \end{bmatrix} $$

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  • \$\begingroup\$ He didn't want a mathematical proof and this doesn't intuitively answer why we use Cartesian Coordinates instead of Polar used by Phasors or Vectors at one instant in time. -1 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 10 at 22:28
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    \$\begingroup\$ +1, I don't consider this a proof, really more of a demonstration. This answer is good because it answers my question of "could vectors be used instead?" and also gives a good argument for why complex numbers are used by showing how messy and verbose these calculations look with vectors. \$\endgroup\$ – JShorthouse Jul 11 at 0:11
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In short: you can visualize an impedance as a type of vector, but vector math does not capture the behavior of the impedance. Complex numbers are not as visually appealing, initially, but mathematically they operate in a similar manner to the impedance's function within a circuit.

This combines two concepts that I will address separately: how does a complex impedance behave, and how a complex number represents that.

While a resistance only changes the magnitude of a signal by absorbing energy, a complex impedance can change both the magnitude and phase of the signal. This means that the impedance may store energy from the signal and later return that energy to the system; this causes a delayed response, which for periodic signals may appear as a rotation in either direction.

So the combined effect on magnitude and direction brings us back to your question: why don't we use a vector? In a general sense, we do! Power systems use a similar concept called a phasor.

Impedance analogue of V=IR

This represents what happens when a signal (current I) of a certain frequency is pushed through an impedance Z. The current starts with a magnitude and phase (angle), which the impedance modifies by its own magnitude and phase (rotation). The resulting voltage V is the product of the magnitudes, rotated by the sum of the angles.

Phasors are critical when working with multiple phases of power; where each phasor is tracking the difference between complex values. For most audio or RF signals, where a common reference is apparent, the V,I,Z phasors collapse into single (complex) values.

This leads to the final part of the answer. Complex scalars capture the same information as vectors - magnitude and angle - but they do not operate the same way mathematically. If an RF frequency were described as a vector value, then modeling an impedance would require matrix multiplication to capture the effects on both magnitude and phase; no sort of vector multiplication would do. Complex numbers operate in the same way as the impedance, providing the perfect tool to represent both the value and function of an impedance.

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  • \$\begingroup\$ Some technical errors using the word Vector with frequency instead of Phasor. -1 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 10 at 22:26
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    \$\begingroup\$ @TonyStewartSunnyskyguyEE75 Why not kindly point out the mistake or edit the answer to be correct instead of being hostile to new contributors? \$\endgroup\$ – JShorthouse Jul 11 at 0:13
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    \$\begingroup\$ @JShorthouse I could but was not trying be hostile like some admins who downvote without comments, but I just wanted to point out that "Vectors" are used for DC and Phasors for AC as rotating phase vectors or phasors. So if the word Vector is replaced with Phasor, I'll upvote it. Complex cartesian coordinates also have sinusoidal qualities but for impedance the phase is defined by real loss and +/- 90 deg reactive impedance amplitudes for some frequency. We don't use Vectors for DC in RLC components but we can use it for Force or Current. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 11 at 0:35
  • \$\begingroup\$ Although they use the term Vector Controlled AC drives I believe because they are variable frequency PWM voltages not for the impedance. So it can be confusing sciencedirect.com/topics/engineering/vector-control \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 11 at 0:41
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    \$\begingroup\$ Thank you for the constructive criticism. I initially tried to distinguish vector/phasor while generally using familiar language. I interpret the heart of the question to be functional and not deeply theoretical, but I will try to better address both audiences. \$\endgroup\$ – mbedded Jul 13 at 14:19
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The imaginary part represents the phase or delay of a sine wave. It can be represented by units of pi, degrees, or a complex number.

enter image description here
Source: https://www.mathsisfun.com/algebra/amplitude-period-frequency-phase-shift.html

An electrical component can cause a phase shift in a sine wave (inductors and capacitors do this). We can represent how much a capacitor or inductor shifts the phase as an imaginary component, and treat them as resistors. This simplifies circuit analysis

The property is desired because we can use imaginary math to carry around the phase information, which is much easier than adding sin functions with the phase together.

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  • \$\begingroup\$ Yes, I understand what the imaginary part represents but I don't understand why a complex number is used. Why could a vector with two dimensions not instead be used to represent this? \$\endgroup\$ – JShorthouse Jul 9 at 21:27
  • \$\begingroup\$ Vectors are used, it depends on what system is used to represent the phase. Either the phase can be represented as a vector, and actually some forms of AC analisys only use vectors. allaboutcircuits.com/textbook/alternating-current/chpt-2/… \$\endgroup\$ – Voltage Spike Jul 9 at 21:31
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    \$\begingroup\$ I am asking specifically why the property of j = sqrt(-1) is needed, meaning that j * j = -1 (which wouldn't happen with a vector representation). There must be some reason why this property is needed and therefore why complex numbers are used, I just can't figure out how this property is needed for impedance calculations. \$\endgroup\$ – JShorthouse Jul 9 at 21:35
  • \$\begingroup\$ The vectors are in the complex space, so the y axis is the imaginary part, the x axis is the real part. hackmath.net/en/calculator/complex-number \$\endgroup\$ – Voltage Spike Jul 9 at 21:42
  • \$\begingroup\$ @JShorthouse I guess mainly because complex numbers make the math much easier than other ways to compute the same result. \$\endgroup\$ – dronir Jul 10 at 8:25
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Complex Impedance may be expressed in either Phasor (Polar domain) or Orthogonal (Cartesian domain)

Polar coordinates are more useful for single-frequency phase shift in power system analysis.

The orthogonal domain is more useful for Electronics where explicit parameters for DCR, ESR and loss vs stored reactive measures are available and commonly specified in datasheets.

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Math: complex number is used to change the domain from t to frequency. In the t domain the equations will be differential and integral, in the frequency domain the equations will be simple. See Laplace transformation. This is a maths solution and it creates the idea about phasor . The physical effect that you see in the original time domain due current or voltage changes in the time by di/dt or integral of i.dt for sample you can se in the frequency domain to use the imaginary componente of the complex number. Z=r+jx coontains a real part R and a part X that means the effects of the changes due the alternating current in the inductance as Faraday law and in the capacitance. Physical idea about phasor is different of vector, it means an alternating changes in time as a senoidal curve but it is writed without use time.

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Actually, impedance is the sun of a real value (resistance) and a vector. Your j = sqrt(-1) is actually a unit vectors. Please keep this top secret, but there are two other unit vectors orthogonal to j. We call them i and k. i, j, and k are the standard unit vectors in 3 dimensional space, and each is a square root of -1. Further, the cross product i X j = k. So complex numbers are just a subset of this weird space of vectors plus real numbers. Think of adding apples and monkeys.🥴

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