Question

Homework problem! Need help fast

Example 2: Figure below shows the principle of a vibration damper. The main platform (mass mi) is suspended on a spring/damper system. The suspension has the spring conn Di and the friction coefficient RF. On the platform is a device that produces oscillatioss such as a vibrating sieve (indicated by the dashed box in Figure). The oscillabory force im- posed on the platform by the device can be described as Fit)-Fn(t) D. m. D2 The actual vibration absorber is a free-hanging mass rma attached to the platform through the second spring with the spring constant D2. The design goal is to select m2 and D2 in such a way that n(t) -o in the steady state when the vibeating force is F(t)F.sin(ttt). (a) Find the Laplace-domain transfer function between the input variable F(s) and the out- put variable Yi(s). (b) Determine the Laplace transform of the platform vibration Y(o) when F(o) (c) Assume f/2 10Hz, m 30kg, D1 -100kN/m, and Rp -3kNs/m. Furthermore, the initial prototype design uses D-20kN/m and m2 = 50kg. Draw the pole-zero diagram of the system. Add to this diagram those poles contributed by the oscillatory force in such a way that the system poles and the input signal poles can be easily distinguished.

Answer 1

(a):

The equations of motion of the system are:

$m_1\ddot{y}_1 = F(t)-D_1y_1-R_F\dot{y}_1+D_2(y_2-y_1)$

$m_2\ddot{y}_2 = D_2(y_1-y_2)$

Taking the laplace transform of above equations, we get:

From the second equation, we get:

$Y_2 (s)= \frac{D_2}{m_2s^2+D_2}Y_1(s)$

Substitutin this in the first equation gives:

$(m_1s^2+R_Fs+D_1+D_2)Y_1(s) = F(s)+\frac{D_2^2}{m_2s^2+D_2}Y_1(s)$

$(m_1s^2+R_Fs+D_1+D_2-\frac{D_2^2}{m_2s^2+D_2})Y_1(s) = F(s)$

$Y_1(s) = \frac{m_2s^2+D_2}{m_1m_2s^4+R_Fm_2s^3+D_1m_2s^2+D_2m_2s^2+D_2m_1s^2+R_FsD_2+D_1D_2}F(s)$

(b):

With

$F(s)=\frac{\hat{F}_w\omega}{s^2+\omega^2}$

$Y_1(s) = \frac{m_2s^2+D_2}{m_1m_2s^4+R_Fm_2s^3+D_1m_2s^2+D_2m_2s^2+D_2m_1s^2+R_FsD_2+D_1D_2}\frac{\hat{F}_w\omega}{s^2+\omega^2}$

(c):

$f=\omega/2\pi = 10 => \omega = 20\pi$

Substituting the given values, we get:

$Y_1(s) = \frac{0.050s^2+20}{1.5s^4+150s^3+5000s^2+1000s^2+600s^2+60000s+2000000}\frac{20\pi\hat{F}_w}{s^2+400\pi^2}$ $Y_1(s) = \frac{0.050s^2+20}{1.5s^4+150s^3+6600s^2+60000s+2000000}\frac{20\pi\hat{F}_w}{s^2+400\pi^2}$

The transfer function of the system is:

$\frac{0.050s^2+20}{1.5s^4+150s^3+6600s^2+60000s+2000000}$

The characteristic equation of this transfer function is:

$1.5s^4+150s^3+6600s^2+60000s+2000000=0$

whose roots, and hence the poles of the system, are:

$s = -49.1856 - 38.4508i; s = -49.1856 + 38.4508i \\s = -0.814439-18.4776i ; s = -0.814439+18.4776i$

The poles of the forcing function are obtained by the characteristic equation:

$s^2+400\pi^2 = 0$

The poles of the forcing function are:

$s = \pm 20\pi i$

The following matlab code can be used to plot the poles.

sys = tf([1 ],[1.5 150 6600 60000 2000000]); % System transfer function
Ff = tf([1],[1 0 400*pi^2]); % Forcing function transfer function
pzplot(sys)
hold on
pzplot(Ff,'r')
xlim([-60 10])
ylim([-70 70])

Since, we only need to plots, the laplace functions of the system and the forcing function are defined with the numerator as 1. This prevents plotting of zeros by the pzplot funciton.

The figure is show below:

The red markers show the poles of the forcing function where as the blue markers show the poles of the system transfer function