For the control system below with Kr = 150 & a = 500: Write the open loop transf
ID: 1857615 • Letter: F
Question
For the control system below with Kr = 150 & a = 500: Write the open loop transfer function for this system in terms of the system gains & parameters(Kp, Ki, Kr & a) Select Ki to achieve an open loop phase margin of 45 degree @ Wx = 50 rad/sec where is the open loop BW or crossover frequency, (hint: use the Bode method) Write an equation from which Kp could be determined in terms of Ki, Kr & a and calculate Kp to achieve an open loop crossover, Wx = 50 rad/sec. Sketch the open loop bode plot in standard form showing approximate slopes, breakpoints, etc. Indicate the phase margin (PM) and gain margin (GM) on the plot. Write the closed loop transfer function for Yo / Yi in terms of the system gains & parameters (Kp, Ki, Kr & a) and sketch the frequency response (very approximate -magnitude only).Explanation / Answer
1. a) let G1(s) = Kr*/s
G(s) = open loop transfer function
= Kp*(s+Ki)*[ G1(s)/(1+G1(s))]*1/s
=Kp*(s+Ki)[kr/(s +Kr)]
=Kp*(s+Ki)*kr/(s +Kr)
=
b) put s = Jw rad/s in G(s)
then G(jw) = Kp*(jw+Ki)*kr/(jw +Kr)
phase of G(jw) = tan^-1(w/ki)-tan^1(w/kr) = tan^-1(w/ki)-tan^1(w/50)
pahse margin = phase of G(jw) +180
phase of G(jw) = -180 +45
at w= 50
tan^-(50/ki)-tan^1(50/50) = -180+45
tan^-(50/ki) -tan^1(50/50) = -135
tan^1(50/ki) = -90
ki = 0
c) loop gain = G(s)*a/(s+a)
=Kp*(s+Ki)*kr*s/[(s+a)(s +Kr)]
=Kp*(s+Ki)*kr*s/[(s+a)(s +Kr)]
=Kp*(s+ki)*50*s/[(s+a)(s +50)]
=Kp*(s)*50*s/[(s+a)(s +50)]
gain = magnitude of G(jw)
=Kp*(w)*50*w/[(w^2+a^2)(w^2+50^2)]^0.5
at w= 50 , gain =1
Kp*(50)*50*50/[(50^2+a^2)(50^2+50^2)]^0.5 = 1
kp* 50^2 /[(50^2+a^2)*2]^0.5 =1
kp =[(50^2+a^2)*2]^0.5/50^2 , put a =500
=[(50^2+500^2)*2]^0.5/50^2
= 0.284