The best way is probably to make the clock period of your vpulse source a variable (let's call it clkPer), and the width equal to 0.5*clkPer. Then run a parametric analysis with clkPer as the varied parameter. Plot the output of your DLL and you can see if it has locked or not at each clock frequency.
Have you looked at the datasheet of the 4046 PLL ? The 4046 contains both types of PD.
The type-1 PD implemented as an XOR outputs 0 when both it's input signals are equal and outputs a 1 when they are not. It cannot distinguish between both signals so it cannot detect if Fvco is too high or too low. It can only detect that it is "not the same" as Fin.
At phase = π the signal inverts so at phase = π - delta the PD's output signal is the same as what it is at phase = π + delta. This explains the positive slope changing to a negative slope at phase = π. The input signal inverts but the XOR treats it the same way, it cannot do any better !
"Why would a PLL with this type of PD not lock if the phase difference input to the PD is greater than 2π?"
Your assumption is wrong, it does lock.
Let me explain:
I give you two signals and a 4 channel oscilloscope.
At t = 0 I provide you with 3 signals:
signal A is a 1 kHz sinewave starting at phase = 0
signal B is a 1 kHz sinewave starting at phase = π
signal C is a 1 kHz sinewave starting at phase = 10 π
Now tell me which signal is which !
Think about it before reading any further !
The answer is that you can only tell me which is signal B.
You cannot distinguish signals A and C because a sinewave repeats
itself every 2π of the phase.
Like you, a type-2 PD also cannot distinguish signals which are shifted by 2π
so it will treat a phase of delta the same as a phase of delta + 2π
or delta + 4π. That is why the graph only shows 0 to 2π, the graph repeats itself every 2π just like a sinewave.
It can however distinguish a phase of π - delta from a phase of π + delta !
That is it's advantage over a type-1 PD.
For a type-2 PD it is not the absolute phase that is locked, it is the modulo(2π) of that phase and that is OK as the signal repeats.
Best Answer
If you look at the control voltage into the VCO, its average value (ignoring ripple) is representative of the output frequency produced: -
If this filtered control voltage is stable (not end stopped) then the PLL is in equilibrium or has settled to a constant steady state error (ignoring noise).
So, if you made a step change to the reference frequency, you would see the classic 2nd order response of the control voltage: -
Picture taken from here
But, because there are many, many types of loop filters and amplifiers you could get variations rather like you would with a PID controller: -
Picture taken from here
In other words, with a simple proportional control (kp), there will be a frequency lock error because the loop gain is finite. If the gain were made too high then it could become unstable. So, the integral term becomes useful (ki) and this can reduce the frequency error to zero without necessarily causing instability. The differential term (kd) can act as a "brake" on the control loop and significantly reduce overshoot.
I'm stating all this because it is not 100% clear what your control loop actually is.