Phasor Introduction and Demo

Contents

What is a phasor?

A phasor is simply a shorthand way of representing a signal that is sinusoidal in time. Though it may seem difficult at first, it makes the mathematics involved in tha analysis of systems with sinusoidal inputs much simpler. To start, we take a sinusoidal signal in time defined by magnitude, phase and frequency (A, θ and ω)

\[f\left( t \right) = A\cos \left( {\omega t + \theta } \right)\]

and represent it in phasor form as a complex number with a magnitude and phase (A, and θ). Note that frequency (ω) is not included, but is implicit in the concept of a phasor.

\[{\mathbf{F}} = A{e^{j\theta }}=A\left(cos(\theta)+j sin(\theta)\right)\]

We frequently write the phasor in terms of its magnitude and phase.

\[\mathbf{F} = A \angle \theta\]

If we multiply the phasor, F, by ejωt, we are simply rotating F by an angle ωt (i.e., the resulting vector has an angle of (ωt+θ).

\[{\mathbf{F}}{e^{j\omega t}} = A{e^{j\theta }}{e^{j\omega t}} = A{e^{j\left( {\omega t + \theta } \right)}} = A \angle \left( {\omega t + \theta } \right) \]

We can recover the time domain function, f(t), by taking the real part of this rotating vector.

\[f\left( t \right) = \operatorname{Re} \left\{ {{\mathbf{F}}{e^{j\omega t}}} \right\} = \operatorname{Re} \left\{ {A{e^{j\left( {\omega t + \theta } \right)}}} \right\} = \operatorname{Re} \left\{ {A\left( {\cos (\omega t + \theta ) + j\sin (\omega t + \theta )} \right)} \right\} = A\cos \left( {\omega t + \theta } \right)\]

The demo below illustrate the relationship between f(t) and F. There are sliders for changing A, θ, ω and t. You can also change t by clicking and dragging in the graph on the right, or by hitting the "Animate" button.

Simple Phasor Demo

f(t) = ___


Important things to note

Key Concept: A sinusoidal signal can be represented by a vector in the complex plane called a phasor

A sinusoidal signal f(t)=A·cos(ωt+θ) can be represented by a phasor F=Ae, which is a vector in the complex plane with length A, and an angle θ measured in the counterclockwise direction. If we multiply F by ejωt, we get a vector that rotates counterclockwise with a rotational velocity of ω. The real part of this vector is equal to f(t).

\[f\left( t \right) = \operatorname{Re} \left\{ {{\mathbf{F}}{e^{j\omega t}}} \right\} = \operatorname{Re} \left\{ {A{e^{j\left( {\omega t + \theta } \right)}}} \right\} = \operatorname{Re} \left\{ {A\left( {\cos (\omega t + \theta ) + j\sin (\omega t + \theta )} \right)} \right\} = f\left( t \right) = A\cos \left( {\omega t + \theta } \right)\]

The phasor, F, is a constant. We need to know the frequency, ω, in order to make a rotating vector and to determine the time domain function f(t).

Why use phasors?

To understand why phasors are useful, consider the circuit shown which shows a current source driving a series combination of a resistor and inductor. The input is a sinusoidal current, Is(t) at 6.28 kHz = 1000 rad/sec, and the output is the voltage across the elements, Vout(t). We'll do it first with brute force, and then show how phasors can make this easier. Note: this won't be a full explanation of how phasors work - if you need that you should check a circuits textbook or do a search for material on line.

The hard way

Let us assume the input current is is(t)=2·cos(1000·t).

This technique of taking derivatives and recombining sine and cosine terms using trigonometric identities quickly becomes tedious and difficult. It turns out the problem is greatly simplified by using phasors.

Also note that adding a phase angle to the input presents no particular difficult - whatever phase (i.e., a time delay) we add to the input, we also add to the output. In the example above if the input had been 2·cos(1000·t+30°) the output would be 23.6·cos(1000·t+30°+32.2°) = 23.6·cos(1000·t+62.2°)

The easier way: Using phasors

We will make use of the fact that the derivative of a rotating phasor is just a rotating phasor whose magnitude is multiplied by ω and whose phase is shifted by 90° Consider:

\[f(t) = A\cos \left( {\omega t + \theta } \right),\quad so\;{\mathbf{F}} = A\angle \theta \]

The rotating phasor is Fejωt, so the derivative of the rotating phasor is jωFejωt. Represented as a phasor this is jωF.

The calculations above can be simplified using phasors. Start by assuming the input current is a phasor of magnitude 2, with phase of 0. Is=2∠0.

Key Concept: Multiplying a phasor by a complex number

Multiplication of a phasor by a complex number yields a scaled and phase shifted phasor at the same frequency.

Key Concept: Integration and differentiation of signals represented by phasors.

Though we won't dwell on it here, one of the important reasons we use phasors is because it turns the problem of differentiation and integration to an algebraic problem. If we have function f(t), the phasor representation is F, and the rotating vector is Fejωt.

Differention: If we differentiate f(t), we differentiate the rotating vector and get jωFejωt. So the differentiated phasor is jωF.

Integration: If we integrate f(t), we differentiate the rotating vector and get (1/(jω))Fejωt. So the differentiated phasor is (1/(jω))F.

This is summarized below, where the arrow denotes that the time domain function is represented by a phasor.

\[\begin{gathered} f(t)\; \Rightarrow \;{\mathbf{F}} \\ \frac{{df(t)}}{{dt}}\; \Rightarrow \;j\omega {\mathbf{F}} \\ \int {f(t)dt \Rightarrow \;\frac{1}{{j\omega }}{\mathbf{F}}} \\ \end{gathered} \]

Phasor Demo

The diagrams below illustrate the relationship between the phasor I, the complex number Z, the phasor V=I·Z, as well as i(t) and v(t). There are sliders for changing the magnitude and phase of I and Z as well as ω and t. Z can also be defined in terms of its real and imaginary parts and the value of t can also be changed by clicking and dragging in the graph on the right, or by hitting the "Animate" button. There is a lot going on in the Phasor Diagram, so a key is given to the right.

Phasor Demo with Phasor Multiplication

Phasor I, A∠θ



Complex Number Z, either M∠φ   or   a+jb
M∠φ

a+jb


Phasor V = I·Z = A·M∠(θ+φ) =
v(t) = ___
i(t) = ___

Important things to note

Refer to the diagram at right

Test your understanding

Try each of the things below and predict what happens (none of the answers are difficult once you understand the concepts). When you can predict all of the changes, you have a good understanding of the material.

Though this example used voltage, current and impedance, there are many applications of phasor analysis in the study of physical systems.


References

© Copyright 2005 to 2019 Erik Cheever    This page may be freely used for educational purposes.

Erik Cheever       Department of Engineering         Swarthmore College