 # About Bode plots

#1

How to use bode plots to choose the suitable valve.

In the iyeya.cn/thread-42480-1-1.html peter reference that
"The valve in the picture is only a 3 Hz valve. Look at where the amplitude ratio starts drop from 0 db.

Manufacturers like to rate valves at 90 degrees phase lag but this is not a good rating because anything over 45 degrees isn’t usable.

The valve frequency response should be greater than the natural frequency of the hydraulic cylinder and load.

Sanity check.
At 3 Hz the valve phase delay is about 30 degree.
At 3 Hz a PLC with 10 millisecond scan time adds 10.8 degree of lag.
A hydraulic cylinder and load with a natural frequency of 20 Hz has a phase lag of 8.8 degrees.
Position control systems start with 90 degrees of lag.
90+30+10.8+8.8=139.6 degrees of phase lag.
Phase lag should not be more than 135 degrees for stable closed loop system. At 180 degrees of phase lag the system will oscillate.
Obviously this system cannot accelerate or decelerate quickly.
The derivative gain in a motion controller can add phase lead to reduce the phase lag and make the system more stable."

I have some questions
At 3 Hz a PLC with 10 millisecond scan time adds 10.8 degree of lag.
A hydraulic cylinder and load with a natural frequency of 20 Hz has a phase lag of 8.8 degrees.//////////////////////////////////////////////
Thanks

1:Can you tell me how calculate the red part(10.8 and 8.8)?
2:You think 45 degrees lag is the Lowest Standard in Closed-loop control，Right？
3：In open loop control degrees lag is not important，or Is there a lowest standard
4：In Frequency response characteristic curves，mainly reference Phase lag，Is the Amplitude ratio not important？
Some person said that -3db Amplitude ratio almost equivalent 90 degree lag，Is it right？

#2

The 10.8 degrees phase lag is dead time due to the controller scan cycle.
At 3 Hz a PLC with a scan time of 0.01 seconds has a lag of 30.01360=10.8 degrees.

Controller Lag= 3Hz*0.01sec*360degrees=10.8 degree
The phase lag from the cylinder should be 5.9 degrees not 8.8 degrees. Start with the assumed open loop transfer function.
ω=2π*Hz natural frequency of the cylinder and mass
ζ=1/3 damping factor assumed to be a typical 1/3
T(s)=1/((s/ω)^2+2*ζ*(s/ω)+1) open loop transfer function for the cylinder and mass.
Then substitute:
s= j2π*Hz
The find the angle between the real and imaginary part of the answer. That is the phase lag due to the cylinder and load.