Question

3. A mass on frictionless rollers is attached to a wall using a spring-damper mechanism: 99m where input u is force and output y is position. Consider a unity-feedback control scheme as shown in question 2, but where the controller and plant are the following: C(s)1 P(s) Note that the parameters M1 and K 2 have been substituted, but that damping coefficient B is unknown. Suppose the damper slowly wears out over time. Draw a root-locus plot illustrating how the closed-loop poles vary as B changes (assume that the damper is passive and, thus, B 0). How low can B get before the system becomes unstable? [5 marks]

Answer 1

This is a simple problem, where we have to find the relation between the poles and the coefficient B.

Now we know that, the transfer function of this system would be given by,

T(s) = \frac{P(s)}{C(s)} = \frac{s}{s^2 + Bs + 2}

Now the poles of the this transfer function are those values of s for which the denominator of the above function is zero.

Hence there would be two poles since the denominator is a quadratic equation and these two poles are given by,

s^2 + Bs + 2 = 0

This gives us two values for s as,

s_1 = \frac{-B + \sqrt{B^2 - 4*2}}{2} = \frac{-B + \sqrt{B^2 - 8}}{2}

and similarly,

s_2 = \frac{-B - \sqrt{B^2 - 8}}{2}

Now we can create a plot between the two variables as follows,

Note : The plot has been constructed using matlab.

The code to create the plot is as follows,

2- 3 4- B= 0:0.01:10; 무for í = 1:length (B) 31(1) 32(1) (-B(1) (-3(1) sqrt ( (B (1)^2 ) -sqrt ( (B (1)^2) -8) ) /2; -8))/2; = end 9subplot (1,2,1) 10 plot (B,sl) xlabel ('coefficient B') ylabel ('pole sl') 12- 13 title'Pole si vs B 14 15subplot (1,2,2) 16lot (B, s2) 17xlabel ('coefficient B') 18ylabel ('poles s2) 19t title ('Pole s2 vs B')