When working with phasors we have $$\bar{Z} = A\cos(\omega t + \phi) = Re(Ae^{j(\omega t + \phi)}) = Re(Ae^{j\phi}e^{j\omega t})$$ where $$\bar{P}_1 = Ae^{j\phi}$$ is the phasor.

But when we multiply we should have $$\bar Z_{1, \phi} \times \bar Z_{2, \alpha} = Re(A_1e^{j\phi}e^{j\omega t}) \times Re(A_2e^{j\alpha}e^{j\omega t}) = Re(A_1A_2e^{j(\phi + \alpha)}e^{j2\omega t})$$ where \$\small\bar{S} = A_1A_2e^{j(\phi + \alpha)}\$ is the phasor of the sum, but we now have the argument \$e^{j2\omega t}\$ instead of \$e^{j\omega t}\$ which would mean the frequency of the signal is doubled - which is not the case for an electric circuit.

The resultant sinusoid should be $$\bar{S} = A_1A_2\cos(2\omega t + \phi + \alpha)$$ which is not the correct result we have on a circuit. So what am I misinterpreting here?

  • \$\begingroup\$ Then you circuit either can't be multiplying or the signals you are multiplying have large enough DC offsets on them to prevent double frequencies appearing. \$\endgroup\$
    – Andy aka
    Aug 27, 2022 at 10:26
  • \$\begingroup\$ When phasors multiply, yes the frequency doubles so V(t)*I(t) results in a spectrum of 2f. Otherwise, in the same domain, of say voltage, phasors are added. wikiwand.com/en/Phasor Phasor notation can only represent systems with one frequency, such as a linear system stimulated by a sinusoid. \$\endgroup\$ Aug 27, 2022 at 10:55
  • 1
    \$\begingroup\$ @TonyStewartEE75 But isn't the premisse of phasor that the frequency must be the same for every component of the circuit? \$\endgroup\$
    – ludicrous
    Aug 27, 2022 at 11:19
  • 2
    \$\begingroup\$ Phasors are only useful in linear circuits and with operations like addition, subtraction, differentiation and integration. All of these operations, when applied to a single frequency signal results in another signal with altered amplitude and phase but the same frequency. Non-linear operations, such as multiplication, are not used with phasors because they result in frequencies other than the original signal frequency. \$\endgroup\$
    – Barry
    Aug 27, 2022 at 12:47
  • \$\begingroup\$ if P=V*I for a single frequency, is plotted as phasors in either V or I but not the product of both since multiplying is not a linear property. ... even though there are "linear multipliers" (modulators) per se. e.g. 100 Hz power from 50 Hz voltage. tinyurl.com/2e6tpv23 \$\endgroup\$ Aug 27, 2022 at 13:04

1 Answer 1


Passive components (\$Z\$ stuff) don't have a rotation that varies with time. So that's the issue you are struggling over, I guess.

Let's look at a power source. There are four parts:

$$\begin{align*} V &=\underbrace{\large{e^{^a}}}_{\text{magnitude}}\quad\underbrace{\large{e^{^{\,\sigma\, t}}}}_{\substack{\text{magnitude}\\\text{spiraling}\\\text{with time}}}\quad\underbrace{\large{e^{^{j\,\theta}}}}_{\text{phase}}\quad\underbrace{\large{e^{^{j\,\omega\, t}}}}_{\substack{\text{phase}\\\text{rotating}\\\text{with time}}}\\ \end{align*}$$

First factor: Using your notation and comparing it with the above, just recognize that \$A=e^{^a}\$. The first factor above is just another way of writing the magnitude, but where \$a=\ln\left(A\right)\$.

Second factor: This is the part of the magnitude that changes with time. I'm sure you've seen \$s=\sigma+j\omega\$ enough times, already. Especially where it is assumed \$\sigma=0\$ (or, perhaps \$\sigma\lt 0\$ to keep things on the left-hand side of the complex plane.) All multiplication of complex numbers (without time, for now) involves two things: scaling and rotation. (All multiplication with real numbers is just scaling, without rotation.) The second factor here usually isn't used. I'm just including it out of being pedantic. If, for some reason, \$\sigma\ne 0\$ then there will be some scaling either inward (negative value) or outward (positive value) without limit as time progresses. It's not normally considered in analysis. So ignore this factor, for now.

Third factor: This is the phase angle. Note that it isn't a function of time.

Fourth factor: This is the part of the phase angle that rotates as a function of time.

In the usual case, we can write:

$$\begin{align*} V &= A\:\large{e^{^{j\left(\omega\,t+\theta\right)}}} \\\\ &=\underbrace{\large{e^{^{\ln A}}}}_{\text{magnitude}}\quad\underbrace{\large{1}}_{\substack{\text{magnitude}\\\text{doesn't change}\\\text{with time}}}\quad\underbrace{\large{e^{^{j\theta}}}}_{\text{phase}}\quad\underbrace{\large{e^{^{j\omega t}}}}_{\substack{\text{phase}\\\text{rotating}\\\text{with time}}}\\ \end{align*}$$

But in the case of an impedance, it's more like this:

$$\begin{align*} Z &= X\:\large{e^{^{j\,\phi}}} \\\\ &=\underbrace{\large{e^{^{\ln X}}}}_{\text{magnitude}}\quad\underbrace{\large{1}}_{\substack{\text{magnitude}\\\text{doesn't change}\\\text{with time}}}\quad\underbrace{\large{e^{^{j\phi}}}}_{\text{phase}}\quad\underbrace{\large{1}}_{\substack{\text{phase doesn't}\\\text{rotate}\\\text{with time}}}\\ \end{align*}$$

There nothing about \$Z\$ where its phasor (phase) rotates with time.

This polar notation I'm showing above for \$Z\$ is unusual as it is normally couched instead in cartesian form with \$R\$ (on the x-axis) and (for example) \$X_L=2\pi\,f\,L\$ (on the y-axis) specified, instead. You can then work out the polar form from that, if you want. But it's not commonly done that way.

So, bottom line? You never get the case you discuss when multiplying \$I\$, which does carry a phase that rotates with time, by \$Z\$, which does not. So there's no frequency-doubling going on with a linear \$Z\$.

Make sense?

(You may very well see a frequency-doubling in the case of instantaneous real power in a resistor, as power is the product of voltage and current and each of those does have its own time-rotating phase.)


You can also re-organize \$V\$ in these two added ways:

$$\begin{align*} V &=\underbrace{\large{e^{^{\left(\sigma+j\,\omega\right)t}}}}_{\text{time-varying}\\\text{ scaling and}\\\:\:\text{rotation}}\quad\underbrace{\large{e^{^{a+j\,\theta}}}}_{\text{time-invariant}\\\:\:\text{ scaling and}\\\quad\text{rotation}} \\\\\text{or},\\\\ V &=\underbrace{\large{e^{^{\sigma\,t+a}}}}_{\text{scaling as a}\\\text{function of t}}\quad\underbrace{\large{e^{^{j\left(\omega\,t+\theta\right)}}}}_{\text{rotation as a}\\\text{function of t}} \end{align*}$$

Steinmetz, in his presentation in Chicago in 1893, carefully pointed out that one can simply drop the part of the voltage or current where its phase rotates with time, the \$e^{j\omega t}\$ part, in order to simplify things in cases where there isn't a need for the instantaneous values. (Just measuring RMS, instead, he said.) This is the beginning of the idea for using phasors (where that part is dropped for AC power sources.)

As an interesting footnote, the 1893 Chicago presentation that preceeds Steinmetz's, came from MacFarlane. MacFarlane chose to use:

$$a^\theta=\cos\,\theta +a^{^\frac{\pi}2}\cdot \sin\,\theta $$

where \$a\$ designates an axis (vertical) to some plane created by two vectors, \$\theta\$ a rotation about the axis, and \$a^{^\frac{\pi}2}\$ then represents the obvious \$90^\circ\$ rotation.

MacFarlane called \$a^\theta\$ a general rotator and \$a^{^\frac{\pi}2}\$ a rectangular rotator.

He preferred this kind of writing rather than using complex numbers for reasons which I suspect are more personal. (I've read his reasons but he fails to provide any workable examples to make his point more concrete.) I'm only noting it here for historical reasons and to point out that while many agreed with the geometric aspects they were also at the time were still struggling over notation conventions. (And I can readily say that I'm glad that complex notation won out.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.