# E12 Assignment 08

Description:

You are duly notified: this assignment is due on Wednesday due to the exam you will do on Monday.

Problems:
1. System Identification, First Order

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))?

Solution:

a) A=3, τ=2, α=0.5;

b)

c)

d)

2. s-land

The plot below represents the Laplace plane (i.e., the variable s).  The horizontal axis is the real part of s, the vertical axis is the imaginary part such that s=σ+jω.  If the poles of a transfer function existed at the center of a square, sketch the corresponding response in the square (for complex values of s, assume that there is a corresponding complex conjugate pole).  For the square that is already filled in σ=-2 and ω=2.   This square shows a decreasing function due to a response of the form e(σ+jω)t (plus a corresponding conjugate term).  Clearly indicate similarities and differences in the expected responses corresponding to differences in σ and ω. You don't need to try to figure things out in terms of ζ and ω0.

Solution:

3. Time and Magnitude of Peak of Underdamped System

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:

Solution:

a) Differentiate and set to zero to find peaks.

The right side is equal to zero when

The first peak occurs when n=1

The solution above is quite long - during problem session a student developed a much shorter way to get the solution by differentiating y(t) in the Laplace Domain (i.e., multiply Y(s) by s, and then take the inverse transform). This obviates the need for the trig identities.

b)

4. Underdamped Pole Movements

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, which arrow describes the pole motion as ζ varies? ...and is ζ increasing or decreasing?  What if you vary α?

Note: you can't vary only one of the parameter without effecting any of the others, but you can vary ζ while holding ω0 constant, but both α and ωd will vary. Likewise, you can vary ωd while holding α constant but ζ and ω0 will change. ...

ii) For each of your answers above describe, in one or two sentences, how this would effect the shape of the step response.

Solution:

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).

5. Dominant Pole and Zeta

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 is the corresponding value 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?

Solution:

a)

b)

c)

```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

6. System Identification, 2nd Order System

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) From the graph, 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...)

Solution:

a)

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.

7. Op Amp DIfferentiator - The easy way

Consider the circuit shown (which uses an ideal op-amp)

Since the positive terminal of the op-amp is at 0V, so is the negative terminal. So, as far as ein(t) is concerned it is seeing C in series with Ri then connected to ground, so the circuit has a time constant τ=RiC.

a) It the input is a unit step, find the find the output voltage at t=0+, i.e., what is eout(0+)? Use the fact that the voltage across the capacitor is zero at t=0+.

b) It the input is a unit step, what is eout(∞)? Do this by inspection - no complex circuit analysis is necessary.

c) For the transfer function below, if ein(t) is a unit step, find τ, eout(0+) and eout(∞) in terms of K, α and β. Don't perform an inverse Laplace Transform.

d) Use these results to write the transfer function of the op-amp circuit.

e) Verify the result using the fact that the transfer function is H(s) = -Zf/Zi where Zf is the feedback impedance (between output and inverting input) and Zi is the input impedance (between ein(t) and the inverting input).

Solution:

a) At t=0+, the voltage to the left of Ri is 1 volt. Ri and Rf form an inverting amplifier with a gain of -Rf/Ri. So eout(0+) = -Rf/Ri.

b) As t→∞ the capacitor becomes an open circuit and the voltage to the left of Ri is 0 volt, so eout(∞) = 0.

c)

d)

e)