Time-Harmonic Fields
Almost every field an engineer meets oscillates — at the power-line hum, a radio carrier, or a microwave tone — and any signal is a sum of such tones. The phasor is the trick that strips the time out of the problem: it turns Maxwell's coupled differential equations into ordinary algebra, and packs amplitude and phase into a single complex number. This is the language the rest of the course is written in.
- How to represent a sinusoidal field as a complex phasor \(\vec{E}_s\).
- The \(j\omega\) substitution that turns \(\partial/\partial t\) into multiplication.
- Maxwell's four equations in compact phasor form.
- Complex permittivity, the loss tangent, and the conductor-vs-dielectric test.
- Time-average power and the \(\tfrac12\mathrm{Re}\{\cdot\}\) rule.
- The Helmholtz equation and the propagation constant \(\gamma=\alpha+j\beta\).
Phasors
When everything varies as \(\cos\omega t\), a field carries only two pieces of information at each point: an amplitude and a phase. Both fit inside one complex number — the phasor \(\vec{E}_s(\vec{r})\) — with the time dependence \(e^{j\omega t}\) understood and factored out. The real, physical field is recovered at the end:
The bridge between the two pictures is one rule: a cosine that lags in space or time becomes a complex exponential. For instance \(\vec{E}=E_0\cos(\omega t-\beta z)\,\hat{a}_x\) has the phasor \(\vec{E}_s=E_0\,e^{-j\beta z}\,\hat{a}_x\). All the dynamics now live in the phase angle, not in a time derivative.
The jω Substitution
Here is the single fact that makes phasors worth the trouble. Because \(\partial/\partial t\) acting on \(e^{j\omega t}\) just pulls down a factor \(j\omega\), every time derivative becomes a multiplication, and every time integral becomes a division:
A partial differential equation in space and time collapses to an ordinary (often algebraic) equation in the space-dependent phasors alone. This is exactly the move you already use for AC circuits — impedances replace derivatives — extended to fields.
Maxwell in Phasor Form
Apply \(\partial/\partial t\to j\omega\) to the four equations of Chapter 16, with \(\vec{D}_s=\varepsilon\vec{E}_s\) and \(\vec{B}_s=\mu\vec{H}_s\). The time derivatives vanish, leaving a clean set in the phasors:
Notice how the Ampère–Maxwell law gathers the conduction current \(\sigma\vec{E}_s\) and the displacement current \(j\omega\varepsilon\vec{E}_s\) into one factor \((\sigma+j\omega\varepsilon)\). That combination — real part loss, imaginary part storage — is the seed of everything that follows about waves in matter.
Complex Permittivity
Factor \(j\omega\) out of \((\sigma+j\omega\varepsilon)\) and the medium is described by a single complex permittivity whose imaginary part carries the loss:
The loss tangent is the single number that classifies a medium at a given frequency — and because it carries \(\omega\) in the denominator, the same material changes character with frequency:
| Regime | Loss tangent | Behaviour |
|---|---|---|
| Good (low-loss) dielectric | \(\tan\delta\ll 1\) | Displacement dominates; wave passes with little loss |
| Quasi-conductor | \(\tan\delta\approx 1\) | Both currents matter; use the full \(\gamma\) |
| Good conductor | \(\tan\delta\gg 1\) | Conduction dominates; wave dies within a skin depth |
Time-Average Power
Power is a product of two fields, so it is not simply the product of phasors — the cross terms average out over a cycle. The time-average of the product of two sinusoids is given by the half-real-part rule, with the conjugate capturing their phase difference:
The factor \(\tfrac12\) is the average of \(\cos^2\) over a cycle; the conjugate \({}^{*}\) is what makes a \(90^\circ\) phase difference carry zero average power — pure reactive flow, just like a lossless reactance in circuit theory. The quantity \(\tfrac12\vec{E}_s\times\vec{H}_s^{*}\) is the complex Poynting vector, whose full meaning waits for Chapter 20. The average power dissipated per unit volume in a conductor is \(\tfrac12\sigma|\vec{E}_s|^2\).
The Helmholtz Equation
Take the curl of the phasor Faraday law and substitute Ampère–Maxwell, exactly as in Chapter 16 but now with no time derivatives to chase. In a source-free medium the wave equation becomes the algebraic Helmholtz equation:
The real part \(\alpha\) (nepers/m) is the attenuation constant — how fast the wave dies; the imaginary part \(\beta\) (rad/m) is the phase constant — how fast it oscillates in space. In a lossless medium \(\sigma=0\), so \(\gamma=j\beta\) with \(\beta=\omega\sqrt{\mu\varepsilon}=\omega/v\), and the wave propagates without decay. Solving this equation for each geometry is the entire programme of Parts 5 through 7.
Worked Examples
Problem. Write the phasor of \(\vec{E}=50\cos(10^{8}t-2z)\,\hat{a}_y\ \text{V/m}\).
Solution. Drop \(e^{j\omega t}\) and turn the space lag into a complex exponential:
Problem. Given \(\vec{H}_s=10\,e^{-j\beta z}\,e^{j\pi/4}\,\hat{a}_x\ \text{A/m}\) at \(\omega\), write the real field.
Solution. Multiply by \(e^{j\omega t}\) and take the real part:
Problem. Wet soil has \(\sigma=10^{-2}\ \text{S/m}\), \(\varepsilon_r=15\). Find \(\tan\delta\) at \(100\ \text{MHz}\) and classify it.
Solution. Use \(\tan\delta=\sigma/\omega\varepsilon\):
Problem. At what frequency does the same wet soil have \(\tan\delta=1\)?
Solution. Set \(\sigma=\omega\varepsilon\), so \(f=\sigma/2\pi\varepsilon\):
Problem. In free space \(\vec{E}_s=100\,\hat{a}_x\ \text{V/m}\) and \(\vec{H}_s=0.5\,\hat{a}_y\ \text{A/m}\) (in phase). Find the average power density.
Solution. Use \(\vec{P}_{\text{avg}}=\tfrac12\mathrm{Re}\{\vec{E}_s\times\vec{H}_s^{*}\}\):
Problem. A lossless dielectric has \(\varepsilon_r=4\), \(\mu_r=1\). Find \(\beta\), \(\lambda\), and \(v\) at \(300\ \text{MHz}\).
Solution. Use \(\beta=\omega\sqrt{\mu\varepsilon}=(\omega/c)\sqrt{\varepsilon_r}\):
Chapter Summary
\(\vec{E}=\mathrm{Re}\{\vec{E}_s e^{j\omega t}\}\); amplitude and phase in one complex number.
\(\partial/\partial t\to j\omega\); \(\partial^2/\partial t^2\to-\omega^2\); time calculus becomes algebra.
\(\nabla\times\vec{E}_s=-j\omega\mu\vec{H}_s\); \(\nabla\times\vec{H}_s=(\sigma+j\omega\varepsilon)\vec{E}_s\).
\(\varepsilon_c=\varepsilon-j\sigma/\omega\); \(\tan\delta=\sigma/\omega\varepsilon\) sorts conductor vs dielectric.
\(\langle AB\rangle=\tfrac12\mathrm{Re}\{A_sB_s^{*}\}\); \(\vec{P}_{\text{avg}}=\tfrac12\mathrm{Re}\{\vec{E}_s\times\vec{H}_s^{*}\}\).
\(\nabla^2\vec{E}_s=\gamma^2\vec{E}_s\); \(\gamma=\alpha+j\beta\); lossless \(\beta=\omega\sqrt{\mu\varepsilon}\).
Problems
For each item, decide whether you are converting between domains, classifying a medium, computing power, or extracting a propagation constant — then apply the matching rule. Difficulty rises down the list.
- Write the phasor of \(\vec{E}=20\sin(\omega t-\beta z)\,\hat{a}_x\ \text{V/m}\). (Mind the sine.)
- Given \(\vec{E}_s=(3+j4)\,e^{-j\beta z}\,\hat{a}_y\ \text{V/m}\), find the amplitude and phase of the real field.
- Use the \(j\omega\) rule to find \(\vec{H}_s\) from \(\vec{E}_s=E_0 e^{-j\beta z}\hat{a}_x\) via the phasor Faraday law.
- Sea water has \(\sigma=4\ \text{S/m}\), \(\varepsilon_r=81\). Find \(\tan\delta\) at \(1\ \text{kHz}\) and at \(10\ \text{GHz}\); classify each.
- Find the frequency at which copper (\(\sigma=5.8\times10^{7}\)) ceases to be a "good conductor" (\(\tan\delta=1\)), taking \(\varepsilon\approx\varepsilon_0\).
- A medium has \(\varepsilon_c=(8-j2)\varepsilon_0\). Find its effective \(\sigma\) at \(1\ \text{GHz}\).
- Compute the average power density for \(\vec{E}_s=80\,\hat{a}_x\), \(\vec{H}_s=0.3\,e^{j30^\circ}\,\hat{a}_y\).
- Show that two fields \(90^\circ\) out of phase carry zero average power.
- For a lossless medium with \(\varepsilon_r=2.25\), find \(\beta\), \(\lambda\) and \(v\) at \(1\ \text{GHz}\).
- Evaluate \(\gamma=\sqrt{j\omega\mu(\sigma+j\omega\varepsilon)}\) for a good conductor and show \(\alpha=\beta=\sqrt{\pi f\mu\sigma}\).
- Find the average dissipated power per cubic metre in a medium with \(\sigma=2\times10^{-3}\ \text{S/m}\) and \(|\vec{E}_s|=40\ \text{V/m}\).
- Explain why the phasor method fails for a non-sinusoidal pulse, and what must be done (Fourier) to use it anyway.