Phasor algebra

This beginners-level introduction to phasors concludes with a derivation of the average power for an ac circuit element when there is a phase shift between the current and voltage. For a more sophisticated example of phasor algebra that uses calculus to understand the momentum of electromagnetic radiation, visit:

• MyOpenMath/Electromagnetic momentum
• MyOpenMath/Complex phasors

Forms, phase, and angular frequency

Phasors have two components, the magnitude, $M$, and the phase angle, $\Phi$. Phasors are related to sinusoids through our cosine convention:

$\mathbb {C} =M\angle \Phi =M\cos(\omega t+\phi )$

Here, $M$ is real, and,

$\Phi =\omega t+\phi$,

depends on time. Angular frequency $\omega$ and period $T$ are related by $\omega T=2\pi$. Also, frequency $f$ and period are related by $fT=1$.

There are 3 forms to phasors:

• $\mathbb {C} =M\angle \Phi$ "Phasor Form"
• $\mathbb {C} =A+jB$ "Rectangular Form"
• $\mathbb {C} =Me^{j\Phi }$ "Exponential Form"

Phasor and Exponential forms are identical and are also referred to as polar form.

Converting between Forms. When working with phasors it is often necessary to convert between rectangular and polar form. To convert from rectangular form to polar form:

$M={\sqrt {A^{2}+B^{2}}}$

$\Phi =\arctan \left({\frac {B}{A}}\right)$

To convert from polar to rectangular form:

$A=M\cos {\Phi }$, where A is the part of the phasor along the real axis

$B=M\sin {\Phi }$, where B is the part of the phasor along the imaginary axis

Operations

Addition and subtraction. These can be performed in rectangular form:

$\mathbb {C} _{1}+\mathbb {C} _{2}=(A_{1}+A_{2})+j(B_{1}+B_{2})$, and

$\mathbb {C} _{1}-\mathbb {C} _{2}=(A_{1}-A_{2})+j(B_{1}-B_{2})$,

where, $\mathbb {C} _{1}=A_{1}+jB_{1}$ and $\mathbb {C} _{2}=A_{2}+jB_{2}$.

Inversion. An important relationship that is worth understanding is the inversion property of phasors:

$\mathbb {C} =M_{C}\angle 0=-M_{C}\angle 180$

On the normal cartesian plane, for instance, the negative X axis is 180 degrees around from the positive X axis. By using that fact on an imaginary axis, we can see that the Negative Real axis is facing in the exact opposite direction from the Positive Real axis, and therefore is 180 degrees apart.

Complex Conjugation. Similar to the inversion property is the complex conjugation property of phasors. Complex conjugation is denoted with an asterisk above the phasor to be conjugated. Since phasors can be graphed on the Real-Imaginary plane, a 90 degree phasor is a purely imaginary number, and a -90 degree phasor is its complex conjugate:

$\mathbb {C} =M\angle 90^{\circ }$

$\mathbb {C} ^{*}=M\angle -90^{\circ }=M\angle 270^{\circ }$

Essentially, this holds true for phasors with all angles: the sign of the angle is reversed to produce the complex conjugate of the phasor in polar notation. In general, for polar notation, we have:

$\mathbb {C} =M\angle \phi$

$\mathbb {C} ^{*}=M\angle -\phi$

In rectangular form, we can express complex conjugation as:

$\mathbb {C} =A+jB$

$\mathbb {C} ^{*}=A-jB$

Notice the only difference in the complex conjugate of C is the sign change of the imaginary part.

Multiplication and division of phasors as complex numbers

To multiply two phasors, we should first convert them to polar form to make things simpler. The product in polar form is simply the product of their magnitudes, and the phase is the sum of their phases. Keep in mind that in polar form, phasors are exponential quantities with a magnitude (M), and an argument (φ). Multiplying two exponentials together forces us to multiply the magnitudes, and add the exponents.	Multiply	Divide
	$\mathbb {C} _{1}\cdot \mathbb {C} _{2}=M_{1}\cdot M_{2}\angle (\Phi _{1}+\Phi _{2})$	${\mathbb {C} _{1} \over \mathbb {C} _{2}}={M_{1} \over M_{2}}\angle (\Phi _{1}-\Phi _{2})$
	where $\mathbb {C} _{1}=M_{1}\angle \Phi _{1}$, and $\mathbb {C} _{2}=M_{2}\angle \Phi _{2}$

Two types of multiplication

Let $\Re e\mathbb {X}$ to the real part of the phasor $\mathbb {X}$ (written in exponential form.) In most cases, the real part of the product, $\Re e\left[\mathbb {C} _{1}\cdot \mathbb {C} _{2}\right]$, is not equal to the product of the real parts, $(\Re e[\mathbb {C} _{1}])\cdot (\Re e[\mathbb {C} _{2}])$. For this reason, multiplication must be done with care when physical meaning is involved. We can express this discrepancy with an inequality:

$\Re e\left\{\mathbb {C} _{1}\mathbb {C} _{2}\right\}=M_{1}M_{2}\cos(\Phi _{1}+\Phi _{2})$   $\neq$   $\left(\Re e\mathbb {C} _{1}\right)\cdot \left(\Re e\mathbb {C} _{2}\right)=M_{1}M_{2}\cos \Phi _{1}\cdot \cos \Phi _{2}$

The former is the multiplication of the complex numbers associated with phasors, while the latter expresses the result of two (real valued) signals, as functions of time, where the complex phasors merely represent the signals.

A useful way to multiply phasors

Since it is common to multiply two signals, the most useful way to multiply phasors is probably the latter. Using the identity, $2\cos \alpha \,\cos \beta$$\,=\,$$\cos(\alpha -\beta )+\cos(\alpha +\beta )$, it can be shown that:

$\left(\Re e\mathbb {C} _{1}\right)\cdot \left(\Re e\mathbb {C} _{2}\right)={\frac {M_{1}M_{2}}{2}}\cos(\Phi _{1}-\Phi _{2})\;+\;{\frac {M_{1}M_{2}}{2}}\cos(\Phi _{1}+\Phi _{2})$

As will be shown next, using current and voltage as phasors, if both phases advance in time with the same angular frequency $\omega$, the first term is a constant while the second term oscillates at an angular frequency of $2\omega$.

Multiplication of two sinusoidal functions

Phasors are often used to represent sinusoidal signals, usually by taking only the real part of the exponential form. For example voltage $V$ might be expressed as:

$V(t)=V_{0}\cos(\omega t+\varphi _{v})$.

This voltage can be represented as the real part of the phasor $\mathbb {V}$,

$\mathbb {V} (t)=V_{0}e^{j(\omega t+\varphi _{v})}$.

In other words:

$V=\Re e[\mathbb {V} ]={\tfrac {1}{2}}(\mathbb {V} +\mathbb {V} ^{*})$,

where $V_{0}$ is a real number known as the amplitude of the signal, and * denotes complex conjugate. Current is represented in the same way:

$I(t)=I_{0}\cos(\omega t+\varphi _{i})=\Re e\left[I_{0}e^{j(\omega t+\varphi _{i})}\right]=\Re e[\,\mathbb {I} \,]$.

Power, $P=IV$ cannot be represented as a phasor because it contains an average part and an oscillating part. Moreover, the oscillating oscillates at angular frequency, $2\omega$ (instead of $\omega$.) Euler's formula permits us to write power in phasor notation as a real-valued entity:

$\underbrace {P(t)=I(t)\cdot V(t)} _{\text{real numbers}}=\left({\frac {\mathbb {I} +\mathbb {I} ^{*}}{2}}\right)\left({\frac {\mathbb {V} +\mathbb {V} ^{*}}{2}}\right)={\frac {\mathbb {I} ^{*}\cdot \mathbb {V} +\mathbb {I} \cdot \mathbb {V} ^{*}}{4}}+{\frac {\mathbb {I} \cdot \mathbb {V} +(\mathbb {I} \cdot \mathbb {V} )^{*}}{4}}$

Here we have treated the phasors as complex entities, where we have used the fact that, $\mathbb {I} ^{*}\mathbb {V} ^{*}=(\mathbb {I} \mathbb {V} )^{*},$ for complex entities. Both terms on the RHS are real, and this grouping of four terms into two fractions highlights the peculiar time dependence associated with power. The first term on the RHS of the following does not depend on time, while the second term oscillates at twice the frequency:

$I(t)\cdot V(t)={\frac {1}{2}}\underbrace {\left({\frac {\mathbb {I} ^{*}\cdot \mathbb {\mathbb {V} } +\mathbb {I} \mathbb {\cdot } \mathbb {V} ^{*}}{2}}\right)} _{\Re e[\mathbb {I} ^{*}\cdot \mathbb {V} ]}+{\frac {1}{2}}\underbrace {\left({\frac {\mathbb {I} \cdot \mathbb {\mathbb {V} } +\mathbb {I} ^{*}\mathbb {\cdot } \mathbb {V} ^{*}}{2}}\right)} _{\Re e[\mathbb {I} \cdot \mathbb {V} ]}$

The number of complex conjugates in the product of two phasors dramatically affects the time dependence:

${\begin{array}{ccl}\mathbb {I} ^{*}\cdot \mathbb {V} &=&I_{0}V_{0}e^{j(\varphi _{v}-\varphi _{i})}\\\mathbb {I} \cdot \mathbb {V} &=&I_{0}V_{0}e^{j(2\omega t+\varphi _{v}+\varphi _{i})}\end{array}}$

Time average and root-mean-square

These equations simplify considerably if we are interested in time averages because the oscillating part of, $P(t)=IV$, vanishes if we either average over many cycles, or carefully average over exactly one cycle. Denote such a time average with the overbar (overline) symbol:

${\overline {I(t)V(t)}}={\tfrac {1}{2}}\Re e[\mathbb {I} ^{*}\cdot \mathbb {V} ]={\tfrac {1}{2}}\Re e[\mathbb {I} \cdot \mathbb {V} ^{*}]={\frac {1}{2}}\left({\frac {\mathbb {I} ^{*}\cdot \mathbb {V} +\mathbb {I} \cdot \mathbb {V} ^{*}}{2}}\right)$

The definition of root-mean-square applies to any quantity that is squared in this fashion:

$I_{\text{rms}}\equiv {\sqrt {\overline {I^{2}}}}={\tfrac {1}{\sqrt {2}}}I_{0}$ $\quad {\text{and}}\quad$ $V_{\text{rms}}\equiv {\sqrt {\overline {V^{2}}}}={\tfrac {1}{\sqrt {2}}}V_{0}.$

Graphs of current i, voltage v, and power p for an ac circuit with a phase shift between current and voltage.

$I(t)V(t)=I_{\text{rms}}V_{\text{rms}}\cos(\varphi _{v}-\varphi _{i})+I_{\text{rms}}V_{\text{rms}}\cos(2\omega t+\varphi _{v}+\varphi _{i})$

$\qquad {\text{ where: }}\quad I_{\text{rms}}={\tfrac {1}{\sqrt {2}}}I_{0}\quad {\text{ and }}\quad V_{\text{rms}}={\tfrac {1}{\sqrt {2}}}V_{0}.$

To render the formula more familiar to those accustomed to direct currents, we express the result in terms of root-mean-square values of current and voltage. Power, $P=IV,$ cannot be represented as a phasor because it is not a sinusoidal. Instead, it is the sum of a time-independent term, $I_{\text{rms}}V_{\text{rms}}\cos(\varphi _{v}-\varphi _{i})$, plus a sinusoidal them with a frequency twice that of $I(t)$ or $V(t)$.