A classical example of the use of feedback is the phase-locked loop used to demodulate frequency-modulated signals (Figure E.27).
The excitation, x(t) , is a frequency-modulated sinusoid. The phase detector detects the phase difference between the excitation and the signal produced by the voltage-controlled oscillator. The response of the phase detector is a voltage proportional to phase difference. The loop filter filters that voltage. Then the loop filter response controls the frequency of the voltage-controlled oscillator. When the excitation is at a constant frequency and the loop is “locked” the phase difference between the two phase-detector excitation signals is zero. (In an actual phase detector the phase difference is near 90° at lock. But that is not significant in this analysis since that only causes a near 90° phase shift and has no impact on system performance or stability.) As the frequency of the excitation, x (t) , varies, the loop detects the accompanying phase variation and tracks it. The overall response signal, y (t) , is a signal proportional to the frequency of the excitation.
The actual excitation, in a system sense, of this system is not x(t) , but rather the phase of x(t) , ?x(t) , because the phase detector detects differences in phase, not voltage. Let the frequency of x(t) be fx(t) . The relation between phase and frequency can be seen by examining a sinusoid.
Let x(t) = A cos (2? f0t) . The phase of this cosine is 2? f0t and, for a simple sinusoid ( f0 constant), it increases linearly with time. The frequency is f0, the derivative of the phase.
Therefore the relation between phase and frequency for a frequency-modulated signal is
Let the frequency of the excitation be 100 MHz. Let the transfer function of the voltage-controlled oscillator be 
Let the transfer function of the loop filter be 
Let the transfer function of the phase detector be 1 V/radian. If the frequency of the excitation signal suddenly changes to 100.001MHz, graph the change in the output signal ? y (t) .
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