English: The simulation uses 3000 discrete time steps. Each time increment is 2,7 ps (that is the line pulse period of 100 ps divided by 37).

After 200 time steps (i.e. after .2,7 * 37 = 543 ps) the loop - that was until then open (because LOS was = 1) starts its acquisition phase ( LOS = 0). The input phase takes in that moment a positive step of 1.3 rad. The input transitions appear with an average density of 50 %, but otherwise in a fully random sequence.. The acquisition lasts for a little over 3 ns in this example. It ends when the phase error settles around its final value (settles around 0 rad in this example). The pattern of pulses in the signal that drives the VCO (black trace) is, although random, constantly made only of positive pulses during the acquisition of lock.

It may be noted that the VCO has a ffr lower than fp (in actuality lower by 5750 ppm), as the VCO phase lags behind the input phase (lower negative slope in the red trace) every time the VCO drive signal it at its intermediate level, both before detection of the input signal and when there is a lack of transition (mid level of the VCO drive signal). The steeper slopes of the red trace, positive and negative, tell when the VCO is driven to its top and bottom frequencies (1,00575 1010 ±2,39 108 Hz ) by the positive and negative pulses of its drive signal. The drive signal is exponentially smoothed when it goes down to zero, because there is a parasitic low-pass (at 6,28 1011 rad/sec = ten time ti line pulse frequency) just after the ternary phase detector. It is sharp at its top and bottom corners because the it is clamped by an instantaneous circuit.

The acquisition time would vary every time with the randomness of the transitions at the signal appearance, and be significantly longer than in the example with a non-zero probability.

If the transitions density had been equal to 100%, the acquisition would have been faster and predictable.