Lecture 6

Time-Harmonic Fields, Complex

Power and Poynting’s Theorem

6.1 Time-Harmonic Fields—Linear Systems

The analysis of Maxwell’s equations can be greatly simplified by assuming the fields to be time harmonic, or sinusoidal (cosinusoidal). Electrical engineers use a method called phasor technique [31, 45], to simplify equations involving time-harmonic signals. This is also a poor-man’s Fourier transform [46]. That is one begets the benefits of Fourier transform technique without knowledge of Fourier transform. Since only one time-harmonic frequency is involved, this is also called frequency domain analysis.1

Figure 6.1: Courtesy of Wikipedia and Pinterest.

1 It is simple only for linear systems: for nonlinear systems, such analysis can be quite unwieldy. But rest assured, as we will not discuss nonlinear systems in this course.

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56 Electromagnetic Field Theory

To learn phasor techniques, one makes use the formula due to Euler (1707–1783) (Wikipedia)

e^{jα} = cos α + j sin α (6.1.1)

where j = √−1 is an imaginary number. But lo and behold, in other disciplines, √−1 is denoted by “i”, but “i” is too close to the symbol for current. So the preferred symbol for electrical engineering for an imaginary number is j: a quirkness of convention, just as positive charges do not carry current in a wire.

From Euler’s formula one gets

cos α = <e(e^{jα}) (6.1.2)

Hence, all time harmonic quantity can be written as

V(x, y, z, t) = V0(x, y, z) cos(ωt + α) (6.1.3)

= V0(r)<e(e^{j(ωt+α)}) (6.1.4)

= <e(V0(r)e^{jα}e^{jωt}) (6.1.5)

= <e(Ve(r)e^{jωt}) (6.1.6)

Now Ve(r) = V0(r)e^{jα} is a complex number called the phasor representation or phasor of V(r, t) a time-harmonic quantity.2 Here, the phase α = α(r) can also be a function of position r, or x, y, z. Consequently, any component of a field can be expressed as

Ex(x, y, z, t) = Ex(r, t) = <e[Exe(r)e^{jωt}] (6.1.7)

The above can be repeated for y and z components. Compactly, one can write

E(r, t) = <e[Ee(r)e^{jωt}] (6.1.8)

H(r, t) = <e[He(r)e^{jωt}] (6.1.9)

where Ee and He are complex vector fields. Such phasor representations of time-harmonic fields simplify Maxwell’s equations. For instance, if one writes

B(r, t) = <e(Be(r)e^{jωt}) (6.1.10)

then

∂/∂t B(r, t) = ∂/∂t <e[Be(r)e^{jωt}]

= <e((∂/∂t Be(r))jωe^{jωt})

= <e(Be(r)jωe^{jωt}) (6.1.11)

2 We will use under tilde to denote a complex number or a phasor here, but this notation will be dropped later. Whether a variable is complex or real is clear from the context.

Time-Harmonic Fields, Complex Power and Poynting’s Theorem 57

Therefore, a time derivative can be effected very simply for a time-harmonic field. One just needs to multiply jω to the phasor representation of a field or a signal. Therefore, given Faraday’s law that

∇ × E = −∂B/∂t − M (6.1.12)

assuming that all quantities are time harmonic, then

E(r, t) = <e[Ee(r)e^{jωt}] (6.1.13)

M(r, t) = <e[Mf(r)e^{jωt}] (6.1.14)

using (6.1.11), and (6.1.14), into (6.1.12), one gets

∇ × E(r, t) = <e[∇ × Ee(r)e^{jωt}] (6.1.15)

and that

<e[∇ × Ee(r)e^{jωt}] = −<e[Be(r)jωe^{jωt}] − <e[Mf(r)e^{jωt}] (6.1.16)

Since if

<e[Aejωt] = <e[B(r)e^{jωt}], ∀t (6.1.17)

then A = B, it must be true from (6.1.16) that

∇ × Ee(r) = −jωBe(r) − Mf(r) (6.1.18)

Hence, finding the phasor representation of an equation is clear: whenever we have ∂/∂t, we replace it by jω. Applying this methodically to the other Maxwell’s equations, we have

∇ × He(r) = jωDe(r) + Je(r) (6.1.19)

∇ · De(r) = ρe(r) (6.1.20)

∇ · Be(r) = ρm(r) (6.1.21)

In the above, the phasors are functions of frequency. For instance, He(r) should rightly be written as He(r, ω), but the ω dependence is implied.

6.2 Fourier Transform Technique

In the phasor representation, Maxwell’s equations has no time derivatives; hence the equations are simplified. We can also arrive at the above simplified equations using Fourier transform technique. To this end, we use Faraday’s law as an example. By letting

E(r, t) = 1/2π ∫∞−∞ E(r, ω)e^{jωt} dω (6.2.1)

B(r, t) = 1/2π ∫∞−∞ B(r, ω)e^{jωt} dω (6.2.2)

M(r, t) = 1/2π ∫∞−∞ M(r, ω)e^{jωt} dω (6.2.3)

Substituting the above into Faraday’s law given by (6.1.12), we get

∇ × ∫∞−∞ dω e^{jωt} E(r, ω) = −∂/∂t ∫∞−∞ dω e^{jωt} B(r, ω) − ∫∞−∞ dω e^{jωt} M(r, ω) (6.2.4)

Using the fact that

∂/∂t ∫∞−∞ dω e^{jωt} B(r, ω) = ∫∞−∞ dω ∂/∂t e^{jωt} B(r, ω) = ∫∞−∞ dω e^{jωt} jω B(r, ω) (6.2.5)

and that

∇ × ∫∞−∞ dω e^{jωt} E(r, ω) = ∫∞−∞ dω e^{jωt} ∇ × E(r, ω) (6.2.6)

Furthermore, using the fact that

∫∞−∞ dω e^{jωt} A(ω) = ∫∞−∞ dω e^{jωt} B(ω), ∀t (6.2.7)

implies that A(ω) = B(ω), and using (6.2.5) and (6.2.6) in (6.2.4), and the property (6.2.7), one gets

∇ × E(r, ω) = −jωB(r, ω) − M(r, ω) (6.2.8)

These equations look exactly like the phasor equations we have derived previously, save that the field E(r, ω), B(r, ω), and M(r, ω) are now the Fourier transforms of the field E(r, t), B(r, t), and M(r, t). Moreover, the Fourier transform variables can be complex just like phasors. Repeating the exercise above for the other Maxwell’s equations, we obtain equations that look similar to those for their phasor representations. Hence, Maxwell’s equations can be simplified either by using phasor technique or Fourier transform technique. However, the dimensions of the phasors are different from the dimensions of the Fourier-transformed fields: Ee(r) and E(r, ω) do not have the same dimension on closer examination.

Time-Harmonic Fields, Complex Power and Poynting’s Theorem 59

6.3 Complex Power

Consider now that in the phasor representations, Ee(r) and He(r) are complex vectors, and their cross product, Ee(r) × He*(r), which still has the unit of power density, has a different physical meaning. First, consider the instantaneous Poynting’s vector

S(r, t) = E(r, t) × H(r, t) (6.3.1)

where all the quantities are real valued. Now, we can use phasor technique to analyze the above. Assuming time-harmonic fields, the above can be rewritten as

S(r, t) = <e[Ee(r)e^{jωt}] × <e[He(r)e^{jωt}]

= 1/2 [Ee e^{jωt} + (Ee e^{jωt})*] × 1/2 [He e^{jωt} + (He e^{jωt})*] (6.3.2)

where we have made use of the formula that

<e(Z) = 1/2 (Z + Z*) (6.3.3)

Then more elaborately, on expanding (6.3.2), we get

S(r, t) = 1/4 Ee × He e^{2jωt} + 1/4 Ee × He* + 1/4 Ee* × He + 1/4 Ee* × He* e^{-2jωt} (6.3.4)

Then rearranging terms and using (6.3.3) yield

S(r, t) = 1/2 <e[Ee × He*] + 1/2 <e[Ee × He e^{2jωt}] (6.3.5)

where the first term is independent of time, while the second term is sinusoidal in time. If we define a time-average quantity such that

Sav = hS(r, t)i = lim T→∞ 1/T ∫T0 S(r, t) dt (6.3.6)

then it is quite clear that the second term of (6.3.5) time averages to zero, and

Sav = hS(r, t)i = 1/2 <e[Ee × He*] (6.3.7)

Hence, in the phasor representation, the quantity

Se = Ee × He (6.3.8)

is termed the complex Poynting’s vector. The power flow associated with it is termed complex power.

60 Electromagnetic Field Theory

Figure 6.2:

To understand what complex power is , it is fruitful if we revisit complex power [47, 48] in our circuit theory course. The circuit in Figure 6.2 can be easily solved by using phasor technique. The impedance of the circuit is Z = R + jωL. Hence,

Ve = (R + jωL)Ie (6.3.9)

where Ve and Ie are the phasors of the voltage and current for time-harmonic signals. Just as in the electromagnetic case, the complex power is taken to be

Pe = Ve Ie* (6.3.10)

But the instantaneous power is given by

Pinst(t) = V(t)I(t) (6.3.11)

where V(t) = <e{Ve e^{jωt}} and I(t) = <e{Ie e^{jωt}. As shall be shown below,

Pav = hPinst(t)i = 1/2 <e[Pe] (6.3.12)

It is clear that if V(t) is sinusoidal, it can be written as

V(t) = V0 cos(ωt) = <e[Ve e^{jωt}] (6.3.13)

where, without loss of generality, we assume that Ve = V0. Then from (6.3.9), it is clear that V(t) and I(t) are not in phase. Namely that

I(t) = I0 cos(ωt + α) = <e[Ie e^{jωt}] (6.3.14)

where Ie = I0e^{jα}. Then

Pinst(t) = V0I0 cos(ωt) cos(ωt + α)

= V0I0 cos(ωt)[cos(ωt) cos(α) − sin(ωt) sin α]

= V0I0 cos2(ωt) cos α − V0I0 cos(ωt) sin(ωt) sin α (6.3.15)

61 Time-Harmonic Fields, Complex Power and Poynting’s Theorem

It can be seen that the first term does not time-average to zero, but the second term does. Now taking the time average of (6.3.15), we get

Pav = hPinsti = 1/2 V0I0 cos α = 1/2 <e[VI*] (6.3.16)

= 1/2 <e[P] (6.3.17)

On the other hand, the reactive power

Preactive = 1/2 =m[P] = 1/2 =m[VI*] = 1/2 =m[V0I0e^{-jα}] = −1/2 V0I0 sin α (6.3.18)

One sees that amplitude of the time-varying term in (6.3.15) is precisely proportional to =m[P].3

The reason for the existence of imaginary part of P is because V(t) and I(t) are out of phase or V = V0, but I = I0e^{jα}. The reason why they are out of phase is because the circuit has a reactive part to it. Hence the imaginary part of complex power is also called the reactive power [34,47,48]. In a reactive circuit, the plot of the instantaneous power is shown in Figure 6.3. The reactive power corresponds to part of the instantaneous power that time averages to zero. This part is there when α 6= 0 or when a reactive component like an inductor or capacitor exists in the circuit. When a power company delivers power to our home, the power is complex because the current and voltage are not in phase. Even though the reactive power time averages to zero, the power company still needs to deliver it to our home to run our washing machine, dish washer, fans, and air conditioner etc, and hence, charges us for it.

Figure 6.3:

3 Because that complex power is proportional to VI*, it is the relative phase between V and I that matters. Therefore, α above is the relative phase between the phasor current and phasor voltage.

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