Phasor

Multiplication of the phasor ${\displaystyle Ae^{i\theta }e^{i\omega t}}$ by a complex constant, ${\displaystyle Be^{i\phi }}$, produces another phasor. That means its only effect is to change the amplitude and phase of the underlying sinusoid:

$${\displaystyle {\begin{aligned}&\operatorname {Re} \left(\left(Ae^{i\theta }\cdot Be^{i\phi }\right)\cdot e^{i\omega t}\right)\\={}&\operatorname {Re} \left(\left(ABe^{i(\theta +\phi )}\right)\cdot e^{i\omega t}\right)\\={}&AB\cos(\omega t+(\theta +\phi )).\end{aligned}}}$$

In electronics, ${\displaystyle Be^{i\phi }}$ would represent an impedance, which is independent of time. In particular it is not the shorthand notation for another phasor. Multiplying a phasor current by an impedance produces a phasor voltage. But the product of two phasors (or squaring a phasor) would represent the product of two sinusoids, which is a non-linear operation that produces new frequency components. Phasor notation can only represent systems with one frequency, such as a linear system stimulated by a sinusoid.

The sum of phasors as addition of rotating vectors

The sum of multiple phasors produces another phasor. That is because the sum of sinusoids with the same frequency is also a sinusoid with that frequency:

$${\displaystyle {\begin{aligned}&A_{1}\cos(\omega t+\theta _{1})+A_{2}\cos(\omega t+\theta _{2})\\[3pt]={}&\operatorname {Re} \left(A_{1}e^{i\theta _{1}}e^{i\omega t}\right)+\operatorname {Re} \left(A_{2}e^{i\theta _{2}}e^{i\omega t}\right)\\[3pt]={}&\operatorname {Re} \left(A_{1}e^{i\theta _{1}}e^{i\omega t}+A_{2}e^{i\theta _{2}}e^{i\omega t}\right)\\[3pt]={}&\operatorname {Re} \left(\left(A_{1}e^{i\theta _{1}}+A_{2}e^{i\theta _{2}}\right)e^{i\omega t}\right)\\[3pt]={}&\operatorname {Re} \left(\left(A_{3}e^{i\theta _{3}}\right)e^{i\omega t}\right)\\[3pt]={}&A_{3}\cos(\omega t+\theta _{3}),\end{aligned}}}$$

where:

$${\displaystyle A_{3}^{2}=(A_{1}\cos \theta _{1}+A_{2}\cos \theta _{2})^{2}+(A_{1}\sin \theta _{1}+A_{2}\sin \theta _{2})^{2},}$$

and, if we take ${\textstyle \theta _{3}\in \left[-{\frac {\pi }{2}},{\frac {3\pi }{2}}\right]}$, then ${\displaystyle \theta _{3}}$ is:

• ${\textstyle \operatorname {sgn}(A_{1}\sin(\theta _{1})+A_{2}\sin(\theta _{2}))\cdot {\frac {\pi }{2}},}$ if ${\displaystyle A_{1}\cos \theta _{1}+A_{2}\cos \theta _{2}=0,}$ with ${\displaystyle \operatorname {sgn} }$ the signum function;
• ${\displaystyle \arctan \left({\frac {A_{1}\sin \theta _{1}+A_{2}\sin \theta _{2}}{A_{1}\cos \theta _{1}+A_{2}\cos \theta _{2}}}\right),}$ if ${\displaystyle A_{1}\cos \theta _{1}+A_{2}\cos \theta _{2}>0}$;
• ${\displaystyle \pi +\arctan \left({\frac {A_{1}\sin \theta _{1}+A_{2}\sin \theta _{2}}{A_{1}\cos \theta _{1}+A_{2}\cos \theta _{2}}}\right),}$ if ${\displaystyle A_{1}\cos \theta _{1}+A_{2}\cos \theta _{2}<0}$.

or, via the law of cosines on the complex plane (or the trigonometric identity for angle differences):

$${\displaystyle A_{3}^{2}=A_{1}^{2}+A_{2}^{2}-2A_{1}A_{2}\cos(180^{\circ }-\Delta \theta )=A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2}\cos(\Delta \theta ),}$$

where ${\displaystyle \Delta \theta =\theta _{1}-\theta _{2}.}$

A key point is that A₃ and θ₃ do not depend on ω or t, which is what makes phasor notation possible. The time and frequency dependence can be suppressed and re-inserted into the outcome as long as the only operations used in between are ones that produce another phasor. In angle notation, the operation shown above is written:

$${\displaystyle A_{1}\angle \theta _{1}+A_{2}\angle \theta _{2}=A_{3}\angle \theta _{3}.}$$

Another way to view addition is that two vectors with coordinates [A₁ cos(ωt + θ₁), A₁ sin(ωt + θ₁)] and [A₂ cos(ωt + θ₂), A₂ sin(ωt + θ₂)] are added vectorially to produce a resultant vector with coordinates [A₃ cos(ωt + θ₃), A₃ sin(ωt + θ₃)] (see animation).

Phasor diagram of three waves in perfect destructive interference

In physics, this sort of addition occurs when sinusoids interfere with each other, constructively or destructively. The static vector concept provides useful insight into questions like this: "What phase difference would be required between three identical sinusoids for perfect cancellation?" In this case, simply imagine taking three vectors of equal length and placing them head to tail such that the last head matches up with the first tail. Clearly, the shape which satisfies these conditions is an equilateral triangle, so the angle between each phasor to the next is 120° (2π⁄3 radians), or one third of a wavelength λ⁄3. So the phase difference between each wave must also be 120°, as is the case in three-phase power.

In other words, what this shows is that:

$${\displaystyle \cos(\omega t)+\cos \left(\omega t+{\frac {2\pi }{3}}\right)+\cos \left(\omega t-{\frac {2\pi }{3}}\right)=0.}$$

In the example of three waves, the phase difference between the first and the last wave was 240°, while for two waves destructive interference happens at 180°. In the limit of many waves, the phasors must form a circle for destructive interference, so that the first phasor is nearly parallel with the last. This means that for many sources, destructive interference happens when the first and last wave differ by 360 degrees, a full wavelength ${\displaystyle \lambda }$. This is why in single slit diffraction, the minima occur when light from the far edge travels a full wavelength further than the light from the near edge.

As the single vector rotates in an anti-clockwise direction, its tip at point A will rotate one complete revolution of 360° or 2π radians representing one complete cycle. If the length of its moving tip is transferred at different angular intervals in time to a graph as shown above, a sinusoidal waveform would be drawn starting at the left with zero time. Each position along the horizontal axis indicates the time that has elapsed since zero time, t = 0. When the vector is horizontal the tip of the vector represents the angles at 0°, 180°, and at 360°.

Likewise, when the tip of the vector is vertical it represents the positive peak value, (+A_max) at 90° or π⁄2 and the negative peak value, (−A_max) at 270° or 3π⁄2. Then the time axis of the waveform represents the angle either in degrees or radians through which the phasor has moved. So we can say that a phasor represents a scaled voltage or current value of a rotating vector which is "frozen" at some point in time, (t) and in our example above, this is at an angle of 30°.

Sometimes when we are analysing alternating waveforms we may need to know the position of the phasor, representing the alternating quantity at some particular instant in time especially when we want to compare two different waveforms on the same axis. For example, voltage and current. We have assumed in the waveform above that the waveform starts at time t = 0 with a corresponding phase angle in either degrees or radians.

But if a second waveform starts to the left or to the right of this zero point, or if we want to represent in phasor notation the relationship between the two waveforms, then we will need to take into account this phase difference, Φ of the waveform. Consider the diagram below from the previous Phase Difference tutorial.

The time derivative or integral of a phasor produces another phasor.[lower-alpha 2] For example:

$${\displaystyle {\begin{aligned}&\operatorname {Re} \left({\frac {\mathrm {d} }{\mathrm {d} t}}{\mathord {\left(Ae^{i\theta }\cdot e^{i\omega t}\right)}}\right)\\={}&\operatorname {Re} \left(Ae^{i\theta }\cdot i\omega e^{i\omega t}\right)\\={}&\operatorname {Re} \left(Ae^{i\theta }\cdot e^{i\pi /2}\omega e^{i\omega t}\right)\\={}&\operatorname {Re} \left(\omega Ae^{i(\theta +\pi /2)}\cdot e^{i\omega t}\right)\\={}&\omega A\cdot \cos \left(\omega t+\theta +{\frac {\pi }{2}}\right).\end{aligned}}}$$

Therefore, in phasor representation, the time derivative of a sinusoid becomes just multiplication by the constant ${\textstyle i\omega =e^{i\pi /2}\cdot \omega }$.

Similarly, integrating a phasor corresponds to multiplication by ${\textstyle {\frac {1}{i\omega }}={\frac {e^{-i\pi /2}}{\omega }}.}$ The time-dependent factor, ${\displaystyle e^{i\omega t},}$ is unaffected.

When we solve a linear differential equation with phasor arithmetic, we are merely factoring ${\displaystyle e^{i\omega t}}$ out of all terms of the equation, and reinserting it into the answer. For example, consider the following differential equation for the voltage across the capacitor in an RC circuit:

$${\displaystyle {\frac {\mathrm {d} \,v_{\text{C}}(t)}{\mathrm {d} t}}+{\frac {1}{RC}}v_{\text{C}}(t)={\frac {1}{RC}}v_{\text{S}}(t).}$$

When the voltage source in this circuit is sinusoidal:

$${\displaystyle v_{\text{S}}(t)=V_{\text{P}}\cdot \cos(\omega t+\theta ),}$$

we may substitute ${\displaystyle v_{\text{S}}(t)=\operatorname {Re} \left(V_{\text{s}}\cdot e^{i\omega t}\right).}$

$${\displaystyle v_{\text{C}}(t)=\operatorname {Re} \left(V_{\text{c}}\cdot e^{i\omega t}\right),}$$

where phasor ${\displaystyle V_{\text{s}}=V_{\text{P}}e^{i\theta },}$ and phasor ${\displaystyle V_{\text{c}}}$ is the unknown quantity to be determined.

In the phasor shorthand notation, the differential equation reduces to:

$${\displaystyle i\omega V_{\text{c}}+{\frac {1}{RC}}V_{\text{c}}={\frac {1}{RC}}V_{\text{s}}.}$$

Derivation

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\operatorname {Re} \left(V_{\text{c}}\cdot e^{i\omega t}\right)+{\frac {1}{RC}}\operatorname {Re} (V_{\text{c}}\cdot e^{i\omega t})={\frac {1}{RC}}\operatorname {Re} \left(V_{\text{s}}\cdot e^{i\omega t}\right)}$

(Eq.1)

Since this must hold for all ${\displaystyle t}$, specifically: ${\textstyle t-{\frac {\pi }{2\omega }},}$ it follows that:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\operatorname {Im} \left(V_{\text{c}}\cdot e^{i\omega t}\right)+{\frac {1}{RC}}\operatorname {Im} \left(V_{\text{c}}\cdot e^{i\omega t}\right)={\frac {1}{RC}}\operatorname {Im} \left(V_{\text{s}}\cdot e^{i\omega t}\right).}$

(Eq.2)

It is also readily seen that:

$${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}\operatorname {Re} \left(V_{\text{c}}\cdot e^{i\omega t}\right)&=\operatorname {Re} \left({\frac {\mathrm {d} }{\mathrm {d} t}}{\mathord {\left(V_{\text{c}}\cdot e^{i\omega t}\right)}}\right)=\operatorname {Re} \left(i\omega V_{\text{c}}\cdot e^{i\omega t}\right)\\{\frac {\mathrm {d} }{\mathrm {d} t}}\operatorname {Im} \left(V_{\text{c}}\cdot e^{i\omega t}\right)&=\operatorname {Im} \left({\frac {\mathrm {d} }{\mathrm {d} t}}{\mathord {\left(V_{\text{c}}\cdot e^{i\omega t}\right)}}\right)=\operatorname {Im} \left(i\omega V_{\text{c}}\cdot e^{i\omega t}\right).\end{aligned}}}$$

Substituting these into Eq.1 and Eq.2, multiplying Eq.2 by ${\displaystyle i,}$ and adding both equations gives:

$${\displaystyle {\begin{aligned}i\omega V_{\text{c}}\cdot e^{i\omega t}+{\frac {1}{RC}}V_{\text{c}}\cdot e^{i\omega t}&={\frac {1}{RC}}V_{\text{s}}\cdot e^{i\omega t}\\\left(i\omega V_{\text{c}}+{\frac {1}{RC}}V_{\text{c}}\right)\!\cdot e^{i\omega t}&=\left({\frac {1}{RC}}V_{\text{s}}\right)\cdot e^{i\omega t}\\i\omega V_{\text{c}}+{\frac {1}{RC}}V_{\text{c}}&={\frac {1}{RC}}V_{\text{s}}.\end{aligned}}}$$

Solving for the phasor capacitor voltage gives:

$${\displaystyle V_{\text{c}}={\frac {1}{1+i\omega RC}}\cdot V_{\text{s}}={\frac {1-i\omega RC}{1+(\omega RC)^{2}}}\cdot V_{\text{P}}e^{i\theta }.}$$

As we have seen, the factor multiplying ${\displaystyle V_{\text{s}}}$ represents differences of the amplitude and phase of ${\displaystyle v_{\text{C}}(t)}$ relative to ${\displaystyle V_{\text{P}}}$ and ${\displaystyle \theta .}$

In polar coordinate form, the first term of the last expression is:

$${\displaystyle {\frac {1-i\omega RC}{1+(\omega RC)^{2}}}={\frac {1}{\sqrt {1+(\omega RC)^{2}}}}\cdot e^{-i\phi (\omega )},}$$

where ${\displaystyle \phi (\omega )=\arctan(\omega RC)}$.

Therefore:

$${\displaystyle v_{\text{C}}(t)=\operatorname {Re} \left(V_{\text{c}}\cdot e^{i\omega t}\right)={\frac {1}{\sqrt {1+(\omega RC)^{2}}}}\cdot V_{\text{P}}\cos(\omega t+\theta -\phi (\omega )).}$$

A quantity called complex impedance is the ratio of two phasors, which is not a phasor, because it does not correspond to a sinusoidally varying function.

With phasors, the techniques for solving DC circuits can be applied to solve linear AC circuits.[lower-alpha 1]

Ohm's law for resistors

A resistor has no time delays and therefore doesn't change the phase of a signal therefore V = IR remains valid.

Ohm's law for resistors, inductors, and capacitors

V = IZ where Z is the complex impedance.

Kirchhoff's circuit laws

Work with voltages and current as complex phasors.

In an AC circuit we have real power (P) which is a representation of the average power into the circuit and reactive power (Q) which indicates power flowing back and forth. We can also define the complex power S = P + jQ and the apparent power which is the magnitude of S. The power law for an AC circuit expressed in phasors is then S = VI^* (where I^* is the complex conjugate of I, and the magnitudes of the voltage and current phasors V and of I are the RMS values of the voltage and current, respectively).

Given this we can apply the techniques of analysis of resistive circuits with phasors to analyze single frequency linear AC circuits containing resistors, capacitors, and inductors. Multiple frequency linear AC circuits and AC circuits with different waveforms can be analyzed to find voltages and currents by transforming all waveforms to sine wave components (using Fourier series) with magnitude and phase then analyzing each frequency separately, as allowed by the superposition theorem. This solution method applies only to inputs that are sinusoidal and for solutions that are in steady state, i.e., after all transients have died out.[15]

The concept is frequently involved in representing an electrical impedance. In this case, the phase angle is the phase difference between the voltage applied to the impedance and the current driven through it.

In analysis of three phase AC power systems, usually a set of phasors is defined as the three complex cube roots of unity, graphically represented as unit magnitudes at angles of 0, 120 and 240 degrees. By treating polyphase AC circuit quantities as phasors, balanced circuits can be simplified and unbalanced circuits can be treated as an algebraic combination of symmetrical components. This approach greatly simplifies the work required in electrical calculations of voltage drop, power flow, and short-circuit currents. In the context of power systems analysis, the phase angle is often given in degrees, and the magnitude in RMS value rather than the peak amplitude of the sinusoid.

The technique of synchrophasors uses digital instruments to measure the phasors representing transmission system voltages at widespread points in a transmission network. Differences among the phasors indicate power flow and system stability.

A: phasor representation of amplitude modulation, B: alternate representation of amplitude modulation, C: phasor representation of frequency modulation, D: alternate representation of frequency modulation

The rotating frame picture using phasor can be a powerful tool to understand analog modulations such as amplitude modulation (and its variants[16]) and frequency modulation.

$${\displaystyle x(t)=\operatorname {Re} \left(Ae^{i\theta }\cdot e^{i2\pi f_{0}t}\right),}$$

where the term in brackets is viewed as a rotating vector in the complex plane.

The phasor has length ${\displaystyle A}$, rotates anti-clockwise at a rate of ${\displaystyle f_{0}}$ revolutions per second, and at time ${\displaystyle t=0}$ makes an angle of ${\displaystyle \theta }$ with respect to the positive real axis.

The waveform ${\displaystyle x(t)}$ can then be viewed as a projection of this vector onto the real axis. A modulated waveform is represented by this phasor (the carrier) and two additional phasors (the modulation phasors). If the modulating signal is a single tone of the form ${\displaystyle Am\cos {2\pi f_{m}t}}$, where ${\displaystyle m}$ is the modulation depth and ${\displaystyle f_{m}}$ is the frequency of the modulating signal, then for amplitude modulation the two modulation phasors are given by,

$${\displaystyle {1 \over 2}Ame^{i\theta }\cdot e^{i2\pi (f_{0}+f_{m})t},}$$

$${\displaystyle {1 \over 2}Ame^{i\theta }\cdot e^{i2\pi (f_{0}-f_{m})t}.}$$

The two modulation phasors are phased such that their vector sum is always in phase with the carrier phasor. An alternative representation is two phasors counter rotating around the end of the carrier phasor at a rate ${\displaystyle f_{m}}$ relative to the carrier phasor. That is,

$${\displaystyle {1 \over 2}Ame^{i\theta }\cdot e^{i2\pi f_{m}t},}$$

$${\displaystyle {1 \over 2}Ame^{i\theta }\cdot e^{-i2\pi f_{m}t}.}$$

Frequency modulation is a similar representation except that the modulating phasors are not in phase with the carrier. In this case the vector sum of the modulating phasors is shifted 90° from the carrier phase. Strictly, frequency modulation representation requires additional small modulation phasors at ${\displaystyle 2f_{m},3f_{m}}$ etc, but for most practical purposes these are ignored because their effect is very small.