Fundamentals of FM Modulation — Quiz

Correct answer: Frequency..

By definition, FM varies the instantaneous frequency of the carrier about its resting value f_c in proportion to the message amplitude, while the carrier amplitude remains constant. Varying amplitude is AM; FM is an angle-modulation scheme.

Correct answer: Δf / f_m..

β = Δf/f_m, the ratio of peak frequency deviation to the modulating tone frequency. Option (d), A_m/A_c, is the AM modulation index — a common confusion to avoid.

Correct answer: 75 kHz..

FM broadcasting (88–108 MHz) permits a maximum deviation of 75 kHz with audio bandwidth 15 kHz, giving a deviation ratio D = 5. The 200 kHz figure is the channel spacing, and 15 kHz is the maximum audio (message) frequency.

Correct answer: 2(Δf + f_m)..

Carson's rule: BW ≈ 2(Δf + f_m) = 2(β + 1)f_m for a single tone, or 2(Δf_max + W) for a general message band W. It accounts for the significant sideband pairs extending to about f_c ± (β+1)f_m.

Correct answer: A_c J_n(β)..

Bessel expansion gives sidebands at f_c ± nf_m with amplitudes A_c J_n(β), where J_n is the Bessel function of the first kind of order n. Option (b) is the AM sideband amplitude formula.

Correct answer: 10.7 MHz..

FM broadcast receivers use a 10.7 MHz IF (455 kHz is the AM broadcast IF). The high IF places image frequencies far from the tuned station, and the wide IF bandwidth accommodates the wideband FM signal.

Correct answer: A limiter..

FM information lives only in the frequency/phase; amplitude carries nothing. A limiter clips amplitude variations (noise, fading) to a constant level before the discriminator or PLL, so they cannot reach the output. This is impossible in AM, where amplitude is the message.

Correct answer: the density of the zero crossings..

Because the carrier amplitude is constant, all the information is encoded in how the instantaneous frequency — equivalently the spacing of zero crossings — varies with time: crowded cycles = higher frequency, spread cycles = lower frequency.

Correct answer: the sidebands carry all the power there..

The carrier-line amplitude is A_c J_0(β), and J_0(β) has zeros at β ≈ 2.405, 5.520, 8.654, …; at those points the carrier line vanishes from the spectrum while the sidebands redistribute the power (Σ J_n²(β) = 1 keeps the total constant). This can never happen in AM, where the carrier line is fixed at A_c.

Correct answer: increased deviation and wider bandwidth..

Since Δf = k_f A_m, a larger message amplitude pushes the instantaneous frequency further from f_c, raising β = Δf/f_m. Higher β increases the number of significant Bessel sidebands, widening the occupied bandwidth — FM trades bandwidth for stronger modulation.

Correct answer: more bandwidth in exchange for better noise immunity..

WBFM needs roughly 2(Δf + W) of spectrum (e.g., 180 kHz for broadcasting versus 30 kHz for AM). In exchange, the constant-amplitude carrier passes through a limiter that strips amplitude noise, and wide deviation spreads noise power, giving FM its high-fidelity, low-noise reception.

Correct answer: the stronger signal dominates after the limiter..

The limiter output level is fixed by the stronger of two co-channel FM signals; the discriminator then responds essentially only to that signal's frequency excursions, suppressing the weaker one. In AM, two signals simply add in amplitude and both are heard simultaneously (or produce heterodyne whistles).

Correct answer: integral + PM = FM..

If a phase modulator is driven by ∫m(t)dt, then φ(t) = k_p∫m(t)dt and the instantaneous frequency becomes (1/2π)dφ/dt = (k_p/2π)m(t) — exactly the FM law. This lets a crystal-stabilized NBFM signal be generated and then frequency-multiplied up to the desired wideband deviation.

Correct answer: small β, AM-like bandwidth..

For β ≪ 1, J_0(β) ≈ 1 and J_1(β) ≈ β/2 with higher orders negligible, so the spectrum collapses to a carrier plus one small sideband pair — bandwidth ≈ 2f_m, much like AM. Only with β ≫ 1 does FM gain its noise-immunity advantage, at the cost of wide bandwidth.

Correct answer: Δf = 24 kHz, β = 6..

Δf = k_f A_m = 8 × 3 = 24 kHz; β = Δf/f_m = 24/4 = 6. The instantaneous frequency swings between 90 MHz − 24 kHz = 89.976 MHz and 90 MHz + 24 kHz = 90.024 MHz. Option (a) miscomputes β as f_m/Δf; (b) misuses k_f = A_m × f_m; (d) doubles β.

Correct answer: 56 kHz..

BW ≈ 2(Δf + f_m) = 2(24 + 4) = 56 kHz. The number of significant sideband pairs is about β + 1 = 7, i.e. spectral lines out to f_c ± 7f_m = f_c ± 28 kHz on each side. Option (d) wrongly uses 2(Δf + 2f_m).

Correct answer: BW = 180 kHz, D = 5..

D = Δf_max/W = 75/15 = 5; BW ≈ 2(75 + 15) = 180 kHz. This is why FM broadcast channels are allocated 200 kHz spacing, leaving a small guard band between adjacent stations. Option (b) inverts the deviation ratio.

Correct answer: constant at 64 W..

P_T = A_c²/(2R) = 80²/(2×50) = 64 W, whatever the value of β. Raising β only redistributes the same fixed power among the carrier and more sidebands (Σ J_n²(β) = 1). This constant-power property contrasts with AM, where P_T grows with m_a.

Correct answer: Δf = 40 kHz; sidebands at 100 MHz ± 10 kHz..

Comparing with s(t) = A_c cos[2πf_c t + β sin(2πf_m t)]: f_c = 100 MHz, β = 4, f_m = 10 kHz. Hence Δf = βf_m = 40 kHz, and the first sideband pair sits at f_c ± f_m = 100 MHz ± 10 kHz (the ±40 kHz positions would be the fourth pair, at f_c ± βf_m).

Correct answer: 10.003 MHz..

f_i(t) = (1/2π)dθ/dt = 10⁷ + 3×10³ cos(2π×10³ t) Hz. At t = 0, cos(0) = 1, so f_i(0) = 10 MHz + 3 kHz = 10.003 MHz. The deviation here is Δf = βf_m = 3 × 1 kHz = 3 kHz.