ethan
Member level 3
Q:
A PLL as a center frequency of 10^5 rad/s, Ko=10^3 rad/s/v, Kd=1V/rad.
Assume there is no other gain in the loop.
Ask: 1. Determine the loop bandwidth in the first-order loop configuration.
2. Determine the single-pole, loop-filter pole location to give the closed-loop poles located on 45 degree radials from the origin of the complex frequency plane.
Answer: 1. Loop Bandwidth = K =Ko Kp = 10^3 /s. (no problem
2. In order to produce poles at 45 degree to the axis, we add a loop filter pole at Ω1 where
Ω1 = 2 K = 2000 rad/s.
then, the filter transfer function becomes, F(s) = Ω1 / (s+ Ω1)
----------------------------------------
My question here is: why choose Ω1 = 2 K ?
so, the original K = 10^3/s, so the bandwidth (also cross-over bandwidth) will be at 10^3 rad/s (right?)
after adding this filter with one more pole, the new bandwidth will be Ω1 x K,
if Ω1 = 2000 rad/s, then the new cross-over bandwidth K will be 2000 x 1000 = 2 ^6 rad/s (right? )
Thx!
A PLL as a center frequency of 10^5 rad/s, Ko=10^3 rad/s/v, Kd=1V/rad.
Assume there is no other gain in the loop.
Ask: 1. Determine the loop bandwidth in the first-order loop configuration.
2. Determine the single-pole, loop-filter pole location to give the closed-loop poles located on 45 degree radials from the origin of the complex frequency plane.
Answer: 1. Loop Bandwidth = K =Ko Kp = 10^3 /s. (no problem
2. In order to produce poles at 45 degree to the axis, we add a loop filter pole at Ω1 where
Ω1 = 2 K = 2000 rad/s.
then, the filter transfer function becomes, F(s) = Ω1 / (s+ Ω1)
----------------------------------------
My question here is: why choose Ω1 = 2 K ?
so, the original K = 10^3/s, so the bandwidth (also cross-over bandwidth) will be at 10^3 rad/s (right?)
after adding this filter with one more pole, the new bandwidth will be Ω1 x K,
if Ω1 = 2000 rad/s, then the new cross-over bandwidth K will be 2000 x 1000 = 2 ^6 rad/s (right? )
Thx!