Transient duration of PLL

Hi,

Can anybody explain for me why 1st order PLL has smaller transient duration than 2nd order loop for the same value of loop noise equivalent bandwidth??

Thanx

Hi!

This is from my knowledge of transient theory - please correct me if I'm wrong!

A first-order PLL would only have one time constant so its transient response to a sudden change in the measured/controlled function would only be a simple decaying exponential dying away to the desired final value.

A second-order PLL with two time-constants can exhibit a response that can be either a surge made up of the difference between two decaying exponentials, a critically-damped surge or a decaying oscillatory surge depending on the amount
of first-order damping present in the circuit, and in many cases the time taken for these elements to decay to the final controlled value is a longer time period than would be achieved by a first order PLL which can be calculated to have a short time-constant & therefore rapid stabilisation.

Sorry if this is a bit long winded - I've tried to explain it as best I can without resorting to lots of revolting maths!

Chris Williams

Don't know. Are you talking real-world or simulated by someone's software?

2nd order loop settles until there is a zero phase error. A 1st order loop will settle to a fixed error phase. Maybe that is the difference--a different end point?

Thanx for explanation. But what I am comparing (by simulating in Matlab)is a 2nd order loop with damping ratio of 0.707 and an input of step function for both 1st order and 2nd order loop.

Tell us more. I would expect it to be the other way around, as a 2nd order loop with an integrator in the loop filter will have much higher open loop gain below the open loop bandwidth frequency.

What exactly do you mean by 1st order and 2nd order? Does your 2nd order have an integrator, and the 1st order just a passive loop filter? Do both phase detectors have exactly the same gain constant, comparison frequency?

This is the model I am trying to simulate. Here F(Z) is is a digital filter.
What I mean by first order loop is F(Z) = A i.e. any constant(obtained from loop bandwidth)
and
for 2nd order loop F(Z) = (aZ-b)/(Z-1). a and b also obtained from loop bandwidth and damping ratio.

Added after 2 minutes:

Sorry I forgot to attach the file.
Here it is.

I recommend you ask the same question on the DSP edaboard. We're not too good with Z's here. We do understand S's.

Thanx