Alternating Current Impedance
Lynn C. Kurtz, Ph.D.
Arizona State University
Department of Mathematics and Statistics, Retired
Abstract
This is a pedagogical article written with the intention of explaining electrical impedance and why it can be modeled with complex arithmetic.
1
Introduction
AC circuits involve voltages and currents that are sine or cosine functions. You
generally drive a simple RLC electric AC circuit with a voltage like sin(ωt) or
cos(ωt) and get resulting currents in the same form. To a neophyte, it isn’t
obvious what this has to do with complex arithmetic.
In this article we will develop the material leading to the notion of impedance.
There is nothing in this paper that isn’t already well known to mathematicians
and electrical engineers, but our approach builds on the previous pedagogical
article linked as Complex Numbers on the same web page as this article.
2
Cartesian Plane, Polar Coordinates, Complex
Numbers, Phasors
Y
R = [ r, θ]
R = (a,b) = (r cos(θ), r sin(θ))
R = a + bi
R
b
r = |R|
θ
a
X
Figure 1: The Cartesian Plane
The plane can be used to represent various things. When we express a point
in the plane in terms of its x and y coordinates (a, b), it is called the Cartesian
plane. We can also represent the point in the Cartesian plane with its standard
polar radius r and angle θ. In order to not mix up the two notations we will
use square brackets for polar notation. So the point (a, b) is the same point as
[r, θ] where a = r cos(θ) and b = r sin(θ).
1
Another interpretation is to write the point as a + bi and interpret it as a
complex number with the usual complex arithmetic operations. In that setting
we call it the Complex plane.
Finally, we can use the notion that the point is represented by the position
vector R. We will use capital letters to represent such vectors and, following
the tradition in Electrical Engineering, call them Phasors. So, as in the figure,
we will generally use the polar representation to write the phasor R = [r, θ] and
we will call r the magnitude of the phasor, and θ its angle.
We can define phasor addition ⊕ and phasor multiplication ⊙ to do arithmetic with phasors. For phasors expressed in rectagular form addition is defined
as componentwise addition
def
(a, b) ⊕ (c, d) = (a + c, b + d)
Figure 2 illustrates addition for:
V = (a, b), W = (c, d), Z = V ⊕ W = (a + c, b + d)
(a+c,b+d)
Y
(c,d)
Z
W
(a,b)
V
X
Figure 2: Phasor addition
Multiplication of phasors is more conveniently expressed in polar form, so if
V = (a, b) = [r, θ] and W = (c, d) = [s, α], multiplication is defined as
def
Z = V ⊙ W = [rs, θ + α]
You simply multiply the magnitudes and add the angles.
This is a good point at which to observe a couple of properties about the
phasor multiplication operation ⊙. The first is that [1, 0] is the multiplicative
identity because [A, θ] ⊙ [1, 0] = [A · 1, θ + 0] = [A, θ]. Also, if A ̸= 0, the
multiplicative inverse of [A, θ] is [ A1 , −θ] because if you ⊙ them you get the
multiplicative identity [1, 0]. If you call P = [A, θ], the inverse [ A1 , −θ] is denoted
by P −1 or P1 . So you have P ⊙ P −1 = [1, 0] or P ⊙ P1 = [1, 0], the multiplicative
identity.
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Also, if we multiply a phasor [A, θ] by a scalar c, it just multiplies the length
without changing the angle: c[A, θ] = [cA, θ]. This has the same effect as
multiplying by [c, 0] since [A, θ] ⊙ [c, 0] = [cA, θ]. Because of this we may use c
and [c, 0] interchangeably depending on context.
Readers familiar with the complex plane will recognize that the ⊕ and ⊙
are the same operations as complex addition and multiplication in the complex
plane.
In this article, we will use the phasor plane because its geometric properties
will be very useful in helping to give a visual representation for our equations.
3
The Phasor Map
So, you may wonder, what does the phasor plane have to do with AC circuits?
Well, a typical voltage or current will be a linear combination of sines and cosines
such as
A cos θ + B sin θ
where often θ = ωt. Any such expression can always be written as a cosine with
a phase angle.
(
)
√
A
B
A cos θ + B sin θ = A2 + B 2 √
cos θ + √
sin θ
A2 + B 2
A2 + B 2
This can be written as
√
√
A2 + B 2 (cos ϕ cos θ + sin ϕ sin θ) = A2 + B 2 cos(θ − ϕ)
one such expression from
where tan ϕ = B
A . The only thing that distinguishes
√
2
2
another similar one is the amplitude D = A + B and the angle θ − ϕ. To
put it simply and informally, all sums of sines and cosines look alike. They
are all cosines and just have different amplitudes and angles. The idea behind
the phasor map is that if the only distinguishing features of these trigonometry
functions are their amplitudes and angles, lets represent them as phasors, which
give an easy visualization of the amplitudes and angles.
To be specific, lets call F the span of {sin θ, cos θ} which includes all linear
combinations A cos θ + B sin θ with the usual arithmetic. For brevity and convenience I will call this space the trig functions. And lets denote by P the phasor
plane discussed above with its ⊙ and ⊕ operations.
The phasor map is the function p : F → P which maps each trig function
f ∈ F to a phasor F ∈ P. The definition of p is given by p(A cos θ) = [A, θ]. In
other words, if f = A cos θ, then F = p(f ) = [A, θ]. This can be written in a
shorthand notation as
p
A cos θ → [A, θ]
As illustrated here, we will use upper case F to represent the phasor which is
the image of f . This sets up a correspondence with each of our sine and cosine
functions with a corresponding phasor. Another example is
π
π
s = B sin θ, S = p(s) = p(Bcos(θ − )) = [B, θ − ]
2
2
3
or, in shorthand notation:
p
B sin θ → [B, θ −
π
]
2
Note that from its phasor, it is easy to see that the sine lags the cosine by π/2.
There is more to it, though. It turns out that the phasor map preserves
addition. In other words, adding two trig functions in F corresponds to adding
their phasors in P. That is shown next.
Theorem 1: p(A cos θ + B sin θ) = p(A cos θ) ⊕ p(B sin θ)
To help visualize the proof of this theorem the figure below illustrates the
case where θ is a second quadrant angle so the phasor for cos θ is in the second
quadrant and the phasor for sin θ lags it by π/2. To simplify the argument, I
will assume A and B are nonnegative.
Y
2 1/2
2 B )
(A +
π/2 − α
θ
A
α
B
θ − π/2
X
Figure 3: Addition of trig functions.
Proof: Beginning with the right side
p(A cos θ) ⊕ p(B sin θ) =
=
=
=
=
=
π
[A,
√ θ] ⊕ [B, θ − 2π]
2
2
[√A + B , θ − 2 + α]
[√A2 + B 2 , θ − ( π2 − α)]
[ √
A2 + B 2 , θ − ϕ]
p( A2 + B 2 cos(θ − ϕ))
p(A cos θ + B sin θ)
where the last line follows from the previous as shown at the beginning of this
section. Note that ϕ = π2 − α. From the upper left triangle you can see that its
tangent is B
A.
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This theorem implies that for f, g ∈ F with corresponding phasors F, G ∈ P,
the phasor for f + g is F ⊕ G, that is,
p
f +g →F ⊕G
4
Differentiation and the phasor map
In Electrical Engineering applications, the trig functions will usually vary with
time t and have an angular frequency ω, so typically θ = ωt. So let’s consider
f = A cos(ωt) and its phasor F = [A, ωt]. If we differentiate f with respect to
t, we get
π
f˙(t) = −Aω sin(ωt) = Aω cos(ωt + )
2
Now let’s look at its phasor F = [A, ωt] and differentiate it with respect to t:
d
d
[A, ωt] = (A cos ωt, A sin ωt) = (−Aω sin ωt, Aω cos ωt)
dt
dt
π
π
π
= (Aω cos(ωt + ), Aω sin(ωt + )) = [ωA, ωt + ]
2
2
2
Notice that we switched between polar and rectangular coordinates in that
argument. Looking at the derivatives of f and F , you can see that Ḟ is the
p
phasor of f˙. So we have f˙ → Ḟ . Also notice that the final result is equal to
π
[A, ωt] ⊙ [ω, 2 ]. If we call P = [A, ωt] this says
Ḟ =
Ṗ = P ⊙ [ω,
π
]
2
This will be our standard formula for differenting a phasor.This says that differentiating the phasor [A, ωt] with respect to t is the same as multiplying it by
the phasor [ω, π2 ] and the result simply rotates the phasor π2 and modifies its
length depending on ω. So, putting it all together, we see that differentiating
a trig function corresponds to rotating its phasor and multiplying its length by
the angular frequency ω.
5
Inductors and Capacitors
di
The voltage v(t) across a coil with inductance L is given by v(t) = L dt
. If
˙
we translate this into the phasor space we get V = LI. Lets say a current,
represented by its phasor of I = [A, ωt] flows in the inductor. Then
V = LI˙ = LI ⊙ [ω,
π
π
] = I ⊙ [ωL, ]
2
2
This last expression looks similar to Ohm’s law for DC circuits: V = IR.
For that reason the quantity on the right is called the inductive impedance:
ZI = [ωL, π2 ]. With that definition we have
V = I ⊙ ZI
5
We have a similar situation for a capacitor. If the charge in a capacitor of
capacitance C is Q, the voltage v across it is given by the equation Q = Cv and
the current passing through it is given by
i(t) =
dQ
dv
=C
dt
dt
Translating this equation into the phasor space gives I = C V̇ = V ⊙ [Cω, π2 ]. If
1
we phasor multiply both sides of this by [ Cω
, − π2 ], which is the multiplicative
π
inverse of [Cω, 2 ] we get
I ⊙[
1
π
π
1
π
, − ] = V ⊙ [Cω, ] ⊙ [
, − ] = V ⊙ [1, 0] = V
Cω
2
2
Cω
2
Again, this looks similar to Ohm’s law, so we define the capacitive impedance
1
, − π2 ]. With this definition, we have
as ZC = [ ωC
V = I ⊙ ZC
Finally, for a resistor with resistance R, we know the current and voltage are
related by v(t) = i(t)R which translates into the phasor equation V = I ⊙ [R, 0]
so the impedance phasor is ZR = [R, 0] as we would expect, giving
V = I ⊙ ZR
Note that the capacitive and inductive impedances depend on the angular frequency ω. Figure 4 illustrates these three impedances.
Y
[ωL,π/2]
[R,0]
X
[1/(ωC),−π/2]
Figure 4: Impedance Phasors
6
In each case, whether a circuit element is a resistor, inductor, or capacitor,
we have that the voltage and current phasors are related by
V =I ⊙Z
where Z is the impedance of the circuit element. If we wish to solve this for I
we can just multiply both sides by Z −1 giving
V ⊙ Z −1 = I ⊙ Z ⊙ Z −1 = I
which can also be written
I=
6
V
Z
Combining impedances
Now let’s suppose we have two circuit elements with impedances Z1 and Z2 in
series, and we want to calculate the impedance of the circuit.
We know the total voltage drop vT across the two components is the sum of
the separate voltage drops. So vT = v1 +v2 . Applying the phasor transformation
p to this equation gives VT = V1 ⊕ V2 = (I ⊙ Z1 ) ⊕ (I ⊙ Z2 ). If we call the total
impedance across the two components ZT this gives us
I ⊙ ZT = (I ⊙ Z1 ) ⊕ (I ⊙ Z2 ) = I ⊙ (Z1 ⊕ Z2 )
Multiplying through by the inverse of the I phasor gives
ZT = Z1 ⊕ Z2
so the impedance of two series elements is the sum of their impedances.
Now let’s consider the same circuit elements in parallel. In this case the
total current iT through the circuit is the sum of the individual currents so
iT = i1 + i2 . Applying the phasor map to this gives IT = I1 ⊕ I2 . If we call the
total impedance of this circuit ZT we have
V
V
V
=
⊕
ZT
Z1
Z2
If we multiply both sides by V −1 we get
[1, 0] [1, 0]
[1, 0]
=
⊕
ZT
Z1
Z2
That may look confusing, but remember that [1, 0] is the multiplicative identity.
Since there is little chance of confusion, sometimes the notation is slightly abused
by identifying 1 and [1, 0] and you will see the above equation written as
1
1
1
=
⊕
ZT
Z1
Z2
In any case the formula for the total impedance for parallel elements can be
written
Z1 ⊙ Z2
ZT =
Z1 ⊕ Z2
7
7
Summary
We have seen that the phasor map transforms circuit equations to phasor equations. The map takes sums of trig functions to sums of phasors and takes
derivatives of trig functions to rotations of their phasors. Using the phasor multiplication and addition operators ⊙ and ⊕ we have a definition of impedance
Z and we also have an Ohm’s law for phasors
V =I ⊙Z
The upshot of all this is that instead of working with the AC circuit equations
themselves, we can work with the algebra of phasors. This will be illustrated in
the final section.
8
An Example
R = 12 Ohms
V = 26cos(50t)
L = 1/10 H
Figure 5: RL Circuit
Consider the circuit given in Figure 5. We have a source voltage v(t) =
26 cos(50t) and we would like to calculate the amplitude and phase of steady
state current i(t). The differential equation for this circuit is
L
di
+ iR = A cos(ωt)
dt
which, for this circuit is
1 di
+ 12i = 26 cos(50t)
10 dt
The homogeneous solution gives only a transient current, so we look for a particular solution for the non-homogeneous equation of the form
i(t) = C cos(50t) + D sin(50t)
8
i′ (t) = −50C sin(50t) + 50D cos(50t)
Substituting these values in our equation gives
1
(−50C sin(50t) + 50D cos(50t)) + 12(C cos(50t) + D sin(50t)) = 26 cos(50t)
10
Simplifying and collecting terms on the sine and cosine gives
(−5C + 12D) sin(50t) + (12C + 5D) cos(50t) = 26 cos(50t)
Setting coefficients equal gives the system of equations for C and D
−5C + 12D
12C + 5D
Solving this system gives C =
i(t) =
24
13 ,
D=
10
13 .
= 0
= 26
This gives our solution
24
10
cos(50t) +
sin(50t)
13
13
We can factor out a 2, leaving us with
(
)
12
5
i(t) = 2
cos(50t) +
sin(50t)
13
13
The numbers in this example have worked out nicely so that the sum of the
5
squares of the coefficients is 1. Now if we set cos ϕ = 12
13 and sin(ϕ) = 13 the
cosine addition formula gives us the expression
i(t) = 2 cos(50t − ϕ)
5
where tan ϕ = 12
. So the amplitude is 2 and the phase angle is 22.62◦ or .395
radians.
This has taken quite a bit of work, especially noting that I left out the steps
solving for the constants and configuring the sum as an addition formula to get
the phase angle. Now let’s look at solving the same problem with phasors.
Using capital letters for the phasors, we know since the impedances of the
series components add, that
I=
V
V
=
Z
ZR ⊕ ZL
where
V = [A, ωt] = [26, 50t], ZR = [12, 0], ZL = [ωL,
π
π
] = [5, ]
2
2
To add ZR and ZL it is best to change them to rectangular coordinates
ZR ⊕ ZL = [12, 0] ⊕ [5,
π
] = (12, 0) ⊕ (0, 5) = (12, 5) = [13, ϕ]
2
9
where ϕ is the same angle we have above. So now we are ready to calculate I:
I=
[26, 50t]
= [2, 50t − ϕ] = [2, 50t − 22.62◦ ]
[13, ϕ]
giving the amplitude and phase of the current.
This document is “copyleft” i.e., not copyrighted 2014. It is for educational
purposes and may be freely distributed either electronically or in printed form
as long as it is distributed in its entirety. Comments, corrections, and suggestions are welcome. Let me know if you find it useful. My email address is
kurtz@asu.edu.
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