The graph below
shows experimental data for a system with unit step applied at t=0. The response of a system and can be described by the equation (for t≥0).
a) What are approximate values for A and τ (rounded to the nearest integer).
b) What is the Laplace Transform of the step response?
c) Express the transfer function as a ratio of polynomials?
d) What is the differential equation that describes the system (input=f(t), output=y(t))?
a) A=3, τ=2, α=0.5;
For the circuit shown (which uses an ideal op-amp)
a) Find the transfer function H(s)=Eout(s)/Ein(s) in terms of Rf, Ri, and C.
b) Find an expression for the unit step response in the time domain.
c) Sketch the step response (there are no component values, so you can only show the shape of the response). Give expressions for the initial value (t=0+), final value (t=∞) and time constant in terms of circuit parameters.
c) Initial value=-Rf/Ri, final value=0, τ=CRi.
An underdamped system has poles as shown by "x" on the diagram.
i) If the poles move as indicated by the arrows labeled a, b, c, and d, describe which one of the quantities (α, ωd, ζ, ω0) is being changed, and whether it is increasing or decreasing. For example, if you vary ζ which arrow describes the pole motion? What if you vary α?
ii) For each of your answers above describe, in one or two sentences, how this would effect the shape of the step response.
a) ω0 is increasing. The shape of the response will remain unchanged, but the speed will increase in proportion to ω0. If ω0 doubles, so will the speed of the response.
b) α increases. This will cause the exponential envelope of the step response to decay more rapidly, but the frequency of the oscillation (determined by ωd) will be unchanged.
c) ωd decreases. This will cause the frequency of oscillations to decrease (oscillations are slower), but the exponential envelope (determined by α) will be unchanged.
d) ζ decreases resulting in a response with less damping. The frequency of oscillation will change (but not by much) and the oscillations will continue for a longer time (the envelope will decay less rapidly).
The step response of an underdamped second order system is given by:
a) Find the time, tmax, at which the first (biggest) peak occurs. Keep in mind that since the sine function is periodic, you should find that there are many extrema (maxima and minima); you want to pick the first maximum value for t>0.
b) Show that the value of the maximum overshoot is given by:
a) Differentiate and set to zero to find peaks.
The right side is equal to zero when
The first peak occurs when n=1
For an overdamped (ζ>1) system defined by the transfer function :
a) Find an expression for the two time constants of the step response in terms of ζ and ω0. Identify which is the faster and which is the slower time constant, i.e., τfast and τslow.
b) If we decide to apply the dominant pole approximation if τslow≥5·τfast, what are the corresponding values for ζ?
c) For ω0=1, ζ=2 plot the exact unit step response of the system and the response of the system represented by the dominant pole approximation. They should be similar in shape and size.
d) For ζ=2, what is the ratio of time constants?
Hexact=tf(1,[1 4 1]); Hdompole=tf(1,[1 0.268])/3.73; step(Hexact,Hdompole); legend('exact','dominant pole');
d) 3.73/0.268 = 13.9179
The system shown
has a unit step response (fa(t)=γ(t)) given by the following graph
a) Write a transfer function for H(s)=X(s)/Fa(s).
b) Find the values of the decay coefficient, α, and the damped frequency, ωd.
c) What are ω0 and ζ? You can assume that ζ is small (so √(1-ζ²)≈1). Find the exact value for ζ by examining the decay between the given peak values. Hint: it is easier to find α and ωd directly from the data than it is to find ω0 and ζ.
d) What are m, k, and b? Hint, use all the information about the response that you know (ωd, α, ω0, ζ, initial and final values...)
b) Transfer function is same form as standard second order system, , if we multiply by k. In other words we can write
The step response is given by
We find ωd from graph which shows that two periods take 7.92-2.59=6.33 seconds, so one period=T=3.16 seconds and ωd=2π/T=1.99. We also know the size of the "envelope" at 1.59 seconds=0.865, and at 7.92 seconds is 0.603. We know (from graph) that the final value is 0.5, but we don't know the amplitude of the envelope (call it A). So:
c) We can solve for ζ and then ω0.
As expected ωd and ω0 are very close (because ζ is small). Much of this work is avoided if you make the assumption that √(1-ζ²)≈1
d) From expression for step response we know final value=1/k=0.5, so k=2. We also know ω0=√(k/m)=2, so m=½. And finally 2·ζ·ω0=0.4=b/m so b=0.2.