Problem 5.6.6: A modulated wave on a loop, and a mixer

Problem 5.6.6

A modulated wave on a loop, and a mixer

An information signal of frequency $f_{m}$ is modulated with a carrier wave of frequency $f_{c}$ . The resulting wave which is propagating in air has the magnetic field component

H (z; t) = H_{0} sin [2 π (f_{c} + f_{m}) t - β z] a_{x} [A/m] .

A circular antenna with a radius of $b$ is on the $y z$ -plane with its centre at the origin.

Find the induced emf.
There is a local oscillator with a frequency of $f_{c}$ . The induced emf and the output of the local oscillator are put into a mixer. Determine the output of the mixer and comment on the result.

Note: a basic oscillator could be an $L C$ circuit with a resonance frequency of $f_{c} = 1/2 π L C$ . A Colpitts oscillator is another choice.

Adapted from: B. M. Notaroš, Electromagnetics, Example 9.9.

Try it yourself first.

Strategy

The loop does not move, so only the transformer emf $- d Φ/ d t$ of Faraday's law appears. A loop much smaller than the wavelength sees the same field all over its surface. An ideal mixer multiplies its two inputs, and $cos A cos B = \frac{1}{2} [cos (A + B) + cos (A - B)]$ shows which frequencies come out.

Bookkeeping

There is one frequency, $ω^{'} = 2 π (f_{c} + f_{m}) rad/s$ . In air $B = μ_{0} H = μ_{0} H_{0} sin (ω^{'} t - β z) a_{x} T$ . The loop lies in the $y z$ -plane, so $d S = d y d z a_{x}$ , with the emf counted anticlockwise seen from $+ x$ . A stationary loop picks up only the transformer emf, $emf = - d Φ/ d t$ . An antenna for a wave of wavelength $λ$ is "small" when $b ≪ λ$ : then $∣ β z ∣ \leq β b ≪ 1$ on the loop.

The flux through the disc $S$ of radius $b$ is

Φ (t) = \int_{S} B \cdot d S = μ_{0} H_{0} \iint_{S} sin (ω^{'} t - β z) d y d z .

For a small loop, $sin (ω^{'} t - β z) ≃ sin ω^{'} t$ all over $S$ (the origin is at the centre):

Φ (t) ≃ μ_{0} H_{0} sin (ω^{'} t) \iint_{S} d y d z = π b^{2} μ_{0} H_{0} sin ω^{'} t,

emf = - \frac{d Φ}{d t} = - 2 π (f_{c} + f_{m}) π b^{2} μ_{0} H_{0} \times cos [2 π (f_{c} + f_{m}) t] .

Answer

emf (t) V_{emf} = V_{emf} cos [2 π (f_{c} + f_{m}) t], = - 2 π (f_{c} + f_{m}) π b^{2} μ_{0} H_{0} [V]

The minus sign only fixes the phase for the chosen sense of the loop: reversing the terminals reverses it. The emf is a property of the whole loop, so it cannot depend on $z$ : the phase $β z$ is integrated out, and what survives is the amplitude, proportional to $ω^{'} b^{2} H_{0}$ , and the frequency $f_{c} + f_{m}$ of the wave.

The local oscillator gives $V_{L O} (t) = V_{0} cos (2 π f_{c} t)$ . An ideal mixer multiplies its two inputs:

V_{mix} = V_{emf} cos [2 π (f_{c} + f_{m}) t] V_{0} cos (2 π f_{c} t) = \frac{1}{2} V_{emf} V_{0} {cos [2 π (2 f_{c} + f_{m}) t] + cos (2 π f_{m} t)} .

A low-pass filter with its cut-off between $f_{m}$ and $2 f_{c}$ keeps the second term:

Answer

V_{out} (t) = \frac{1}{2} V_{emf} V_{0} cos (2 π f_{m} t)

Comment

The mixer shifts the received line down by $f_{c}$ : the carrier is removed and the message at $f_{m}$ is recovered. This is the down-conversion in every radio receiver (Figure 5.6.18). An oscillator off by $Δ f$ would deliver the tone at $f_{m} - Δ f$ , so the local oscillator must be accurate.

Check

Units: $ω^{'} [s^{- 1}] \times b^{2} [m^{2}] \times μ_{0} [H/m] \times H_{0} [A/m] = H \cdot A/s = V$ .

Problem 5.6.6. The flux factor F=2J_1(\beta b)/\beta b of a circular loop of radius b in a wave that travels in the plane of the loop (computed from Bessel's integral). Shaded: the small-loop range b\le\lambda/20, where F\ge0.988. The first zero is at b=0.61\lambda. — **Figure 5.6.17.** Problem 5.6.6. The flux factor $F = 2 J_{1} (β b) / β b$ of a circular loop of radius $b$ in a wave that travels in the plane of the loop (computed from Bessel's integral). Shaded: the small-loop range $b \leq λ /20$ , where $F \geq 0.988$ . The first zero is at $b = 0.61 λ$ .