Phasors as Complex Numbers

Section 1.3 Phasors as Complex Numbers

You may have encountered complex numbers in your math classes. Recall that a complex number

z

can be written as

\begin{matrix} (1.8) & z = a + i b \end{matrix}

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with

a

and

b

real numbers, and that

i

is the imaginary unit,

i = \sqrt{- 1} .

Here

a

is called the real part of

z

[written

Re (z)

], while

b

is called the imaginary part of

z

[written

Im (z)

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It's easy to represent a complex number graphically. Just plot the real part complex number along the horizontal, or real axis, and the imaginary part along the vertical, or imaginary axis, as shown in Figure 1.10.

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Figure 1.10. A complex number can be represented graphically as a point in the complex plane.

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The length of the diagonal line from the origin to the point representing the complex number

z

is the magnitude of

z,

and corresponds the amplitude of the oscillation

A,

and the angle with respect to the positive real axis corresponds to the phase angle

ϕ,

as shown in Figure 1.11, which is really just a phasor diagram!

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Figure 1.11. Phasors correspond to complex numbers.

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Applying simple trig to the diagram gives

\begin{matrix} (1.9) & a = Re (z) = A \cos ϕ b = Im (z) = A \sin ϕ . \end{matrix}

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It is the real part of the complex number that corresponds to the physical oscillation, but using the complex representation actually simplifies many calculations.

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The connection to phasors gets even stronger when we make use of the Euler identity

\begin{matrix} (1.10) & e^{i ϕ} = \cos ϕ + i \sin ϕ, \end{matrix}

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so that

\begin{aligned} z = a + i b = & A \cos ϕ + i A \sin ϕ \\ = & A (\cos ϕ + i \sin ϕ) \\ = & A e^{i ϕ} . \end{aligned}

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Thus, if we have an oscillation represented in complex exponential form, we can immediately draw its phasor picture: Make an arrow of length

A

at angle

ϕ

from the real (horizontal) axis.

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Example 1.12. Another Day at the Beach.

The change in water level at your location is now represented with the following complex function (whose real part is the actual water height):

\begin{matrix} (1.11) & y (t) = 25 e^{i (π / 4 t + π / 2)}, \end{matrix}

with $y$ in cm and $t$ in sec. Draw a phasor diagram to depict the level of the water at $t = 2,$ 3 and $4 s .$

Solution.

Draw the real and imaginary axes. Starting at the real axis, go counterclockwise for $π / 2$ radians (the phase constant $ϕ_{0}$ ). Then keep going counterclockwise $π / 4$ radians for each second of elapsed time. The phasors are shown in Figure 1.13.

Figure 1.13. Phasors for the water level in Example 1.12 at times $t = 2 s,$ $t = 3 s,$ and $t = 4 s .$

To find actual water levels, take the real part (the horizontal projection). The water levels are thus

\begin{aligned} at t = 2 s : - 25 cm \\ at t = 3 s : - 25 \cos (π / 4) = - \frac{25}{\sqrt{2}} ≃ - 17.7 cm \\ at t = 4 s : 0 \end{aligned}