In an FM radio receiver, a PLL is used to demodulate the signal. What physical quantity directly appears as the demodulated audio output?
AThe phase detector output voltage, which represents the instantaneous phase difference
BThe VCO control voltage (loop filter output) that keeps the PLL locked, which tracks instantaneous frequency deviations
CThe frequency of the VCO itself, which is converted to audio by a frequency-to-voltage converter
DThe phase of the VCO output, which is proportional to the audio signal amplitude
In FM demodulation using a PLL, the VCO is forced to track the instantaneous frequency of the FM-modulated carrier. The control voltage that drives the VCO — the output of the loop filter — must exactly represent the instantaneous frequency deviation of the input signal. Since FM encodes the audio message as instantaneous frequency deviation, the control voltage that keeps the VCO locked IS the demodulated audio signal. The feedback action that maintains lock also recovers the message, with no additional demodulation step required.
Question 2 Multiple Choice
A PLL designer is choosing loop filter bandwidth. Which statement correctly characterizes the narrow-bandwidth (slow loop) choice?
ANarrow bandwidth tracks rapid frequency changes well but passes more phase noise to the output
BNarrow bandwidth rejects high-frequency phase noise but responds slowly to frequency changes in the input signal
CNarrow bandwidth eliminates steady-state phase error but creates loop instability at low frequencies
DNarrow bandwidth forces the VCO to run at its free-running frequency, making frequency synthesis impossible
Loop bandwidth determines what the PLL tracks vs. what it filters. Within the bandwidth, the PLL follows input phase variations — so narrow bandwidth means slow response to legitimate frequency changes in the input. Outside the bandwidth, the VCO runs near its free-running frequency and the loop averages out (rejects) phase noise. Narrow bandwidth is chosen when the reference is stable but noisy, and you want a cleaned-up output. Wide bandwidth is needed when you must track rapid frequency changes. The bandwidth choice is fundamentally a tracking-vs-noise tradeoff.
Question 3 True / False
A PLL is a Type 1 feedback system (one integrator in the forward path — the VCO) and will therefore track a constant phase offset between input and VCO with zero steady-state phase error.
TTrue
FFalse
Answer: True
The VCO is an integrator in the phase domain: its output phase is the time-integral of its control voltage. This integrator in the forward path makes the PLL a Type 1 system. In control theory, a Type 1 system tracks a step input (constant reference) with zero steady-state error — which here means zero steady-state phase error between input and VCO output. A Type 2 PLL (with a second integrator in the loop filter) tracks a frequency ramp (linearly increasing phase) with zero steady-state phase error.
Question 4 True / False
A PLL synchronizes by matching the frequency of the local VCO to the input signal; once frequency is matched, phase is irrelevant to the loop's operation.
TTrue
FFalse
Answer: False
This gets the mechanism backwards. A PLL is fundamentally a phase-feedback system — it controls phase, not frequency directly. The phase detector measures phase difference and drives the VCO to null this phase error. When the loop is locked, the phase difference is held constant (ideally zero), which also implies frequency match. 'Frequency match' is a consequence of 'phase lock,' not the target itself. The PLL ensures coherent phase alignment, which is what makes it useful for demodulation and clock recovery where phase information carries the signal.
Question 5 Short Answer
How does placing a frequency divider by N in the PLL feedback path enable frequency synthesis at N times the reference frequency?
Think about your answer, then reveal below.
Model answer: When a divider by N is placed in the feedback loop, the phase detector compares the input reference (at frequency fref) against the VCO output divided by N (at frequency fVCO/N). The loop drives this phase difference to zero, which forces fVCO/N = fref, so fVCO = N·fref. By changing N, you can program the VCO to any integer multiple of the stable reference frequency. Since the reference is typically a precise crystal oscillator, the VCO output inherits its long-term frequency accuracy while reaching much higher frequencies than the crystal alone could generate.
The key insight is that the PLL 'sees' the divided frequency and locks it to the reference — it doesn't know the VCO is running N times faster. The feedback loop does the arithmetic implicitly: nulling the phase error between fref and fVCO/N forces fVCO = N·fref. This is the foundation of all modern frequency synthesizers in radios, cellular phones, and digital clocks — a single stable crystal at (say) 10 MHz can generate precise frequencies at hundreds of MHz or GHz by programming N.