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 100 %, which is not realistic but useful to illustrate the condition normally specified for the "block preamble". The acquisition lasts for a little over 1.3 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 made only of positive pulses during the acquisition of lock. 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) when the VCO drive signal it at its intermediate level before detection of the input signal (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. If the transition density was less than 100%, 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 the transitions density equal to 100%, the acquisition is the fastest and its duration is constant every time. Acquisition with a preamble of 101010.. i.e a density of 100%, of adequate duration, are a standard in systems that frequently need to reacquire the signal lock

When DT = 100%, it is even more evident that the tracking of a fp higher than ffr is the cause of a ratio of positive pulses to negative pulses (evident in the VCO drive signal) that is higher than 1.