Question

4. Consider the type I system in Fig. 5.73. We would like to design the compensation D(s) to meet the following requirements: (1) The steady-state value of y due to a constant unit disturbance w should be less than $, and (2) the damping ratio ( = 0.7. Using root-locus techniques: a) Show that proportional control alone is not adequate. b) Show that proportional-derivative control will work. c) Find values of the gains K and K, for D(s) = K +Kis that meet the design specifications --9----- 5(+ 1)

Answer 1

Question 1 and Question 2

consider D(s) = K  

The root locus of the uncompensated system is

the dotted lines are the zeta lines

0.6 Root Locus 0:56 22 6.81 0.89 0 0:2 0.4 -0.95 System: 9 Gainz 0.51 Pole: -0.5 +0.5091. Damping: 0.7 Övershoot (%): 4.58 Frequency (rad/s): 0.714. 0.2 0.988. 0.8 0.6 0.4 0.2 Imaginary Axis (seconds) 0.988 -0.2 -0.4 0:95 0.89 2.81 0.56 0.38 0:2 -0.6 -1.2 -1 -0.8 -0.2 0.2 -0.6 -0.4 Real Axis (seconds)

from the graph the gain required at the zeta value of 0.7 is 0.5

so K = 0.5

now the steady-state value for this K value becomes

steady state value of state value of y for w=113 for W=1/3 for. K = 005 S + y + k D s(s+1) we get [co-y]k4w] fece ] म y o [ + s(s+1) S(St) Đ W yo w s(s+1) kt s(s+1) 1 k + 1 S($+1) take Now w=116 y = Vs Kt 5(8+1) alow state value of ý Ct becomes in Steady stati value ti y(t) = at sy(s) t7x $70 f S90 k +.5.(5+1) k. al D YCH) ㅗ K ㅗ 0.5 2 » 9/5 tx when 1&=0.7 for So steady state value proportional Controller. Hence we cannot we proportional controller Partb 8 - the, kd 'k zaoof when compensator e P(s) = K (S+z) or we can write k s takze Now w=lls LS rzo K(suze) 5(5+1) We geto- [[o-y](&$*+zes) + ] [o(6+1)= 15 y = k (8+2c)+s(s+0) similar to as we derived phe vien previously. - pH yet = L SY(s) twal H S40 So k(s+zc) +s(s+1) kZc k ze So Newly stala police deput en ik and xe steady state Istate = where k and Ze are to be choosen so that Steadystate 2/5 Hence by choosing appropriate kand te we can get the required steady state and & value. So it is possible using bo controllor. O Scanned with CamScanner

Question 3

Design of the controller D(s) = K(s+z_c)

now from the above discussion we can see that the error is inversely proportional to z_c value

so let us choose a large enough value of zero such that error becomes less than 4/5

we see for z_c = 5;  

we get D(s) = K(s+5)

so now we need to find the K value required by plotting the root locus of the compensated system as

open loop function now becomes

$\frac{K(s+5)}{s(s+1)}$

the root locus plot becomes

Root Locus 5 4 System: untitled1 Gain: 7.87 Pole: -4.34 + 4.42i Damping: 0.101 Overshoot (%) 4.58 Frequency (rad/d): 6.19 3 2 Imaginary Axis (seconds) -2 3 -4 1 -5 -18 -16 -14 -12 -4 -2 0 2 -10 -8 -6 Real Axis (seconds)

from the above plot we can see the value of gain for zeta = 0.7 is 7.67

hence we get the K = 7.67

so the controller is D(s) = 7.67(s+5)

now the steady-state value is 1/K*z_c = 1/(7.67*5) << 4/5

hence the steady state value requirement is also satisfied

the code used is

c = tf([1 5],[1])
g = tf([1],[1 1 0])
rlocus(g*c)

in summary

we have found why proportional controller is not useful

we have shown that PD controller can be used to find the required specifications

we have found the controller so that zeta and steady state value is satisfied