By Jean-Pierre Elloy
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Extra info for Classical and Modern Control with Worked Examples
The flow q thus represents the set point of a servo system whose mea surement q is the image of the angular position Θ of the cheek of the rudder. We shall assume that the oil circulating in this part of the equipment is non-compressible. Note : the flow q passes through a laminar damping orifice with the coefficient C defined by q = C ΔΡ where ΔΡ is the difference in pressure at the edges of the orifice. The power mechanism of the equipment comprises two motor jacks running on modulus B oil. The oil is distributed by the slide valve whose in put and output flow rates correspond to Q1 - - Q~ = K x,.
The position of the valve is measured using a potentiometer fitted to the motor axle. 5 m. 5 m - \n with X - 20 volts/m : P is graduated from 0 to n Λ output c max - Xn (X - 20 volts/m) c c Amplifiers 1 and 2 : the gains A and A„ are adjustable. 1 s m P : angular position with a voltage proportional K = 1 volt/rad) V Reducer : _v_ _ n m measurement of the drive to Θ (v = K„ Θ m m p m shaft. 1 m /s/rad Θ with a flow rate q - k θ% ^ v v a) Trace the block diagram of the unit. e. 0388 %. determined and supposing Αη to be the values of the coefficients of the open loop function.
S Nyquist test : If the point " - 1 " belongs to the segment OA, then N = - 2 , P = O, therefore Z = 2, the system is unstable in closed loop. If the point " - l " does not belong to the segment OA, then N = O, P = O, therefore Z = O, the system is stable in closed loop. The value K of K and the pulsation ω , for Q which the curve of KG(s) passes through the point " - 1 " corresponds to |KoG(jü)o) | = 1 and /KoG»"o> Now 180°. = /KG(j ) = - 90° - A r c t g ( Ο . J. 02 ω Whence K |K0G(ju>o) /"5Ö /T75" ΓΓ υ 15 = 1 therefore 15 So if K is lower than 15, |KG(jo>o) | < 1 # the system in closed loop is stable.
Classical and Modern Control with Worked Examples by Jean-Pierre Elloy