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;

b)

c)

d)

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

a) Find the transfer function H(s)=E_{out}(s)/E_{in}(s)
in terms of R_{f}, R_{i}, 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.

a)

b)

c) Initial
value=-R_{f}/R_{i}, final value=0, τ=CR_{i}.

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, ω_{d}, ζ, ω_{0}) 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) ω_{0} is increasing. The shape of
the response will remain unchanged, but the speed will increase in
proportion to ω_{0}. If ω_{0} 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 ω_{d}) will be unchanged.

c) ω_{d} 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, t_{max}, 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

b)

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 ω_{0}.
Identify which is the faster and which is the slower time constant, i.e.,
τ_{fast}
and τ_{slow}.

b) If we decide to apply the dominant pole approximation if
τ_{slow}≥5·τ_{fast},
what are the corresponding values for ζ?

c) For ω_{0}=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?

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

The system shown

has a unit step response (f_{a}(t)=γ(t)) given by the
following graph

a) Write a transfer function for H(s)=X(s)/F_{a}(s).

b) Find the values of the decay coefficient, α, and the damped
frequency, ω_{d}.

c) What are ω_{0} 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 ω_{d}
directly from the data than it is to find ω_{0} and ζ.

d) What are m, k, and b? Hint, use all the information about the
response that you know (ω_{d}, α,
ω_{0}, ζ, initial and final values...)

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 ω_{d} from graph which shows that two periods take
7.92-2.59=6.33 seconds, so one period=T=3.16 seconds and ω_{d}=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 ω_{0}.

As expected ω_{d} and ω_{0} 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 ω_{0}=√(k/m)=2,
so m=½. And finally 2·ζ·ω_{0}=0.4=b/m so b=0.2.