# simple interview question (1) on PLL and answer, but...

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#### ethan

##### Member level 4 Q:
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)
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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!

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