22-Elec-A3 Signals and Communications · May 2016

Question 6 of 6: FM Modulator Output, Carson Bandwidth and Demodulation

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 6: FM Modulator Output, Carson Bandwidth and Demodulation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the modulator output $s(t)$ with every parameter named, (b) an approximate transmission bandwidth, and (c) a non-PLL demodulator.

Approach. Extract the frequency sensitivity from the dc calibration, integrate the message to obtain the instantaneous phase, then find the true peak of the two-tone message (which is not the sum of the tone amplitudes) before applying Carson's rule.

1. Frequency sensitivity from the dc calibration. The modulator is specified by a constant $k_f$ in Hz per volt such that the instantaneous frequency is $f_i(t)=f_c+k_f\,m(t)$. A steady 2 V shifts the frequency by 5 kHz, so $$k_f=\frac{\Delta f_{dc}}{V_{dc}}=\frac{5\ \text{kHz}}{2\ \text{V}} =\boxed{\;2.5\ \text{kHz/V}\;}$$
2. Integrate the message to get the phase. An FM signal is $s(t)=A_c\cos\!\left[2\pi f_ct+2\pi k_f\!\int_0^{t}m(\tau)\,d\tau\right]$. Integrating the two tones term by term, $$\int_0^{t}m\,d\tau=\frac{\sin(2\pi f_1t)}{2\pi f_1}-\frac{3\cos(4\pi f_1t)}{4\pi f_1} +\frac{3}{4\pi f_1},$$ so, absorbing the constant into an arbitrary phase reference, the phase deviation is $$\theta(t)=\frac{k_f}{f_1}\sin(2\pi f_1t)-\frac{3k_f}{2f_1}\cos(4\pi f_1t) =\beta_1\sin(2\pi f_1t)-\beta_2\cos(4\pi f_1t).$$
3. Evaluate the two modulation indices. Each tone contributes its own index, equal to its own peak frequency deviation divided by its own frequency: $$\beta_1=\frac{k_f\times 1\ \text{V}}{f_1}=\frac{2500}{2000}=1.25\ \text{rad},\qquad \beta_2=\frac{k_f\times 3\ \text{V}}{2f_1}=\frac{7500}{4000}=1.875\ \text{rad}.$$ Notice that the larger-amplitude tone sits at twice the frequency, so its index is not simply three times the first — the frequency in the denominator partly offsets the amplitude.
4. State the modulator output (part a). Collecting everything, $$\boxed{\;s(t)=A_c\cos\!\Bigl[2\pi(10^{6}\!\ \text{-class }f_c)t +1.25\sin\!\left(2\pi\,2000\,t\right)-1.875\cos\!\left(2\pi\,4000\,t\right)\Bigr]\;}$$ with the parameters: carrier amplitude $A_c$ (not specified by the question, so it is carried symbolically), nominal carrier frequency $f_c$ (the grounded-input frequency), $k_f=2.5$ kHz/V, $\beta_1=1.25$ rad at 2 kHz and $\beta_2=1.875$ rad at 4 kHz. The instantaneous frequency is $f_i(t)=f_c+2500\left[\cos(2\pi 2000t)+3\sin(2\pi 4000t)\right]$ Hz.
5. Find the true peak of the message. The peak frequency deviation is $\Delta f=k_f\max|m(t)|$, and because the two tones are harmonically related the peak is not $1+3=4$ V. Writing $x=2\pi f_1t$ and using $\sin 2x=2\sin x\cos x$, $$m=\cos x+3\sin 2x=\cos x+6\sin x\cos x.$$ Setting the derivative to zero, $-\sin x+6\cos 2x=0$, and substituting $\cos 2x=1-2\sin^{2}x$ gives the quadratic $12\sin^{2}x+\sin x-6=0$, whose physical root is $\sin x=2/3$ (hence $\cos x=\sqrt5/3$). Substituting back, $$\max|m|=\frac{\sqrt5}{3}+6\cdot\frac{2}{3}\cdot\frac{\sqrt5}{3}=\frac{5\sqrt5}{3} =\boxed{\;3.727\ \text{V}\;}$$ — some 7 % below the naive sum of the amplitudes, because the two tones never reach their peaks at the same instant.
6. Peak deviation and Carson bandwidth (part b). Hence $$\Delta f=k_f\max|m|=2500\times 3.727=\boxed{\;9.32\ \text{kHz}\;}$$ and with the highest message frequency $W=2f_1=4$ kHz, Carson's rule gives $$B_T=2\left(\Delta f+W\right)=2\left(9.32+4\right)\ \text{kHz} =\boxed{\;26.6\ \text{kHz}\;}$$ The deviation ratio is $D=\Delta f/W=2.33$, comfortably in the wide-band FM regime, so Carson's rule (rather than the narrow-band approximation $B_T\approx 2W$) is the right tool. The output spectrum consists of lines spaced 2 kHz apart — every combination $n f_1+m(2f_1)$ — whose significant members fall inside that 26.6 kHz band.

7. Non-PLL demodulator (part c). The classical answer is a frequency discriminator: convert frequency variations into amplitude variations, then detect the envelope. In cascade: a limiter removes any incidental amplitude modulation picked up in the channel (FM carries no information in its envelope, so hard limiting costs nothing and rejects noise); a band-pass filter removes the harmonics the limiter creates; a differentiator (or a slope circuit / detuned resonant network, as in a balanced Travis or ratio detector) multiplies the signal by its instantaneous frequency, producing $A_c\,\omega_i(t)$ as an amplitude; and an envelope detector with a DC block recovers $m(t)$ up to a scale factor. Differentiating $s(t)=A_c\cos[\omega_ct+\theta(t)]$ gives $$\dot s(t)=-A_c\left[\omega_c+2\pi k_f m(t)\right]\sin\left[\omega_ct+\theta(t)\right],$$ whose envelope $A_c\left[\omega_c+2\pi k_fm(t)\right]$ is an exact affine image of the message. This satisfies the “no phase-locked loop” restriction, and a quadrature detector or zero-crossing counter would be equally acceptable answers.