Question

In the phase-locked-loop (PLL) circuit shown below, the output of the VCO (voltage-controlled oscillator) is a digital square wave.

If the input is a square wave at \(10\) kHz, the steady state frequency of the output is kHz (rounded off to the nearest integer).

Hint:

In lock, the two signals reaching the phase detector must have the same frequency. Divide the input by \(20\) and the output by \(1024\), then set those two equal and solve for the output frequency.

Answer 1

Correct Answer: 512

Step 1: Understand what a PLL locks onto.
A phase locked loop drives its voltage-controlled oscillator (VCO) until the two signals reaching the phase detector have the same frequency, with a fixed phase relationship between them. Once the loop is in lock, we only need to match the frequency at the two phase detector inputs.

Step 2: Identify the two signals entering the phase detector.
The input square wave first passes through a \(\div 20\) counter before it reaches the phase detector, so that input to the phase detector runs at \(f_{in}/20\).
The VCO output is fed back through a \(\div 1024\) counter before it reaches the phase detector, so that input runs at \(f_{out}/1024\).

Step 3: Write the lock condition and solve for the output frequency.
In lock, these two phase detector inputs are equal:
\[ \frac{f_{in}}{20} = \frac{f_{out}}{1024} \]
Solving for \(f_{out}\):
\[ f_{out} = f_{in} \times \frac{1024}{20} \]
With \(f_{in} = 10\) kHz:
\[ f_{out} = 10 \times \frac{1024}{20} = 10 \times 51.2 = 512 \text{ kHz} \]

Final Answer:
The steady state output frequency of the VCO is \(512\) kHz. \[ \boxed{512 \text{ kHz}} \]