Problem 1. Consider the following closed-loop system:
(a) Can you design a compensator C(s) such that the closed-loop system track a unit step reference
input (i.e., r(t) is unit step) with zero steady-state error?
(b) If your answer to (a) is yes, suggest a form of a compensator C(s) to achieve your objective, i.e.,
to drive the steady-state error to a step input to zero.
(c) For your suggested form of compensator in Part (b), find the four closed-loop transfer functions
Hry(s) : r → y, Hwy(s) : w → y, Hru(s) : r → u, Hwu(s) : w → u. Compare these transfer
functions and discuss.
Problem 2. Consider the following closed-loop system where w1(t) and w2(t) denote disturbance
signals respectively at the input and the output of the plant.
(a) Assume our plant model is represented by a pure integrator (i.e., P(s) = 1
). Ignore the disturbance
at the input (i.e., w1(t) = 0). Our goal is to track a step reference input (i.e., r(t) = u(t)) with
zero steady-state error in the presence of a step disturbance at the output of the plant (i.e.,
w2(t) = u(t)). Can you suggest the simplest compensator to achieve that?
(b) With the same plant model, what if we want to also reject a step disturbance at the plant input
as well?, i.e., we want to have ess = e(∞) = 0.0 when r(t) = w1(t) = w2(t) = u(t)? Can your
compensator in Part (a) achieve that? If not, why? What would be the simplest form of the
compensator you can propose to achieve that? Compare with Part (a) and discuss. (hint: Don’t
forget that your closed-loop system should remain stable. Think of PI compensators).
These homework problems are compiled using the different textbooks listed on the course syllabus
ECE141 – Principles of Feedback Control Homework 4 2
(c) With the same plant model, and the same compensator you designed for Part (b), what would
be the seady-state tracking error at the output when the reference input is a ramp, while both
disturbance signals are still modeled as step signals?, i.e., find ess, given r(t) = tu(t) and w1(t) =
w2(t) = u(t).
(d) With the same plant model, and the same compensator you designed for Part (b), what would be
the System Type for this system? What would be the value of the corresponding Error Constant?
(e) Use the same assumptions as Part (c), except this time, the disturbance at the plant output also
happens to behave like a ramp. What would be the steady-state tracking error in this case? i.e.,
find ess, given r(t) = w2(t) = tu(t) and w1(t) = u(t).
(f) Use the same assumptions as Part (c), except this time, we want to reject ramp-like disturbances
at both the input and the output of the plant, i.e., our design objective is to have ess = 0.0
when r(t) = w1(t) = w2(t) = tu(t). Can we achieve zero steady-state tracking error with the
same compensator as the one in Part (b)? If not, why? How would you propose to modify your
compensator in Part (b) to achieve that?
(g) Now, assume the plant model is P(s) = 1
s+1 and we use a pure integrator as our compensator, i.e.,
C(s) = 1
. Assume our reference input is: r(t) = (1 + sin(t))u(t), the disturbance at the output
is: w2(t) = cos(0.01t)u(t), and the disturbance at the input to the plant is ignored (w1(t) = 0.0).
Find the steady-state output response yss.
Problem 3. Sensitivity to variations in Feedback vs Feedforward Elements: We have discussed how the sensitivity of any parameter M to variations in some parameter K may be defined
Relative/fractional change in M
Relative/fractional change in K
We also considered a unity-feedback system and showed how the sensitivity function of the closed-loop
transfer function T(s) to the changes in the the loop transfer function L(s) can be obtained as:
1 + L
And we therefore concluded that for the purpose of reducing the closed-loop sensitivity to loop gain
variations, we would desire |L(jω)| 1 at those frequencies where we would expect the loop gain
variations to occur. Now, what if we have elements in the feedback path as well? How sensitivie would
our closed-loop transfer function be to the changes in the parameters of the feedback elements vs the
feedforward elements? Specifically, let’s consider the following closed-loop feedback system:
Let: Hcl(s) = Hry(s) = Y (s)
be the closed-loop transfer function, and L(s) , H(s)F(s) be our
loop transfer function.
ECE141 – Principles of Feedback Control Homework 4 3
(a) Obtain Hcl(s) in terms of H(s) and F(s).
(b) Obtain the sensitivity of Hcl(s) to changes in the feedforward path transfer function H(s), i.e.,
H , and write it in terms of the loop transfer function L(s).
(c) Obtain the sensitivity of Hcl(s) to changes in the feedback path transfer function F(s), i.e., S
and write it in terms of the loop transfer function L(s).
(d) Compare the two results, and discuss how the magnitude of our loop transfer function L(jω) would
impact our closed-loop sensitivity to variations in the feedforward elements versus the feedback
elements. Based on this result, say, if you are designing a feedback amplifier, where would you put
your more accurate components?
Problem 4. Building an inverse system: Consider the closed loop system in the previous problem
(a) Let H(S) = H be a constant gain. For what range of values of H, can our closed-loop transfer function be approximated as 1
? This, in fact, can be considered as a basic technique to
implement an inverse of a system.
(b) Interestingly, the result in (a) would still hold even if we have a nonlinear element in the feedback
path. Specifically, consider the following closed loop system:
where H(s) = H is a constant gain in the feedforward path, and f(.) is a nonlinear function in
the feedback path. Let’s assume that f
(.) exists, i.e., if x = f(y), one could obtain y = f
Also assume that our closed-loop system remains stable in the presence of nonlinearity. Write the
input-output relationship for this closed-loop system. The loop gain in this case may be defined
. For what values of the this loop gain, will our closed-loop system act like the inverse of
the feedback element, i.e., y(t) = f
(c) A nice exampe application of the result in Part (b) is in designing logarithmic amplifiers where
the output voltage is equal to K times the natural logarithm of the input voltage. From the result
in Part (b), we know that one can build a feedback system with proper feedforward gain values
in order to implement the inverse of the characteristics in the feedback path. So to implement
a logarithmic amplifier, we need to get an exponential characteristics in our feedback path. And
that is exactly what the following basic logarithmic amplifier circuit does:
ECE141 – Principles of Feedback Control Homework 4 4
The voltage-current relationship of the diode may be approximated as:
Id ≈ Ise
where Id is the diode current in the forward direction, Vd is the voltage across the diode, Is is the
so-called reverse saturation current of the diode, and VT is a constant called thermal voltage. This
is a decent approximation when the diode is operating in the so-called forward bias region and the
voltage across it is much bigger than the thermal voltage VT . Assume an ideal OpAmp with very
large open-loop gain A, so the current flowing into the OpAmp terminals as well as the differential
voltage at the input to the OpAmp may be assumed negligible. Show that the input-output voltage
relationship for the above amplifier can be obtained as follows:
Vout = −VT ln Vin
Problem 5. Robust Stability (The solution is Extra Credit to submit but mandatory to
learn!) Consider a flexible beam, connected to a mass and controlled by a motor as shown below:
The nominal rigid body model from the motor control input to the beam tip deflection is given by:
P(s) = 6.28
In order to stabilize the beam and keep the mass at the beam tip at a desired position, the following
compensator in a unity feedback loop (shown below) had been designed:
C(s) = 500(s + 10)
s + 100
ECE141 – Principles of Feedback Control Homework 4 5
(a) What is the Loop Transfer Function L(s)?
(b) What is the Sensitivity Function S(s)?
(c) Plot the magnitude of |S(jω)| over frequency. What is the crossover frequency for S(jω) in rad/sec?
Remember that the crossover frequency is the frequency ωsc at which |S(jω)| = 1 = 0dB. (Use
bodemag function in MATLAB to plot |S(jω)|. Then use margin function to find the crossover
frequency. Call it in this format [gm,pm,wcg,wsc] = margin(S) and the last value in the returned
vector would be the magnitude crossover frequency.)
(d) Based on your answer to Part (c), if we had any disturbance modeled at the input to the plant,
which disturbance frequencies would be attenuated by feedback?
(e) What is the closed-loop transfer function T(s) from the reference input to the system output?
Remember T(s) is also the Complementary Sensitivity Function. Where are the closed-loop poles
located? Is the feedback design based on the nominal rigid body model of the beam stable? (You
can use the feedback function, followed by pole and zero functions in MATLAB).
(f) Plot the step response of the closed-loop system. (Use step function in MATLAB).
(g) Notice the time scale and how fast the step response is. It turns out that the bandwidth specification was not part of the design specification for the above compensator. As a result, this
compensator would target too fast of a response such that it would excite the additional flexible
modes of the beam which were not accounted for in the rigid body model of the beam. Assume
the ”true” model for the beam was:
P1(s) = 6.28
2 + 0.707s + 28
where the second term shows the unmodeled dynamics of the beam when designing the above
compensator C(s). Find out the multiplicative plant perturbation δp(s).
(h) Find the closed-loop transfer function with P1(s) and the same compensator C(s) above. Where
are the closed-loop poles located? Is the closed-loop system still stable? (Similar to Part (e), you
can use MATLAB here).
(i) Plot the magnitudes of 1/|T(jω)| as well as |δp(jω)| over frequency on the same figure. Was the
Robust Stability condition satisfied for all frequencies? Please discuss. (You can again use bodemag
function in MATLAB here, and don’t forget to use hold on to keep both plots on the same figure).
(j) Now, let’s try a new compensator:
C1(s) = (5 × 10−4
)(s + 0.01)
s + 0.1
Find the new closed-loop transfer function T1(s) assuming compensator C1(s) and the original
nominal rigid body model for the beam P(s) = 6.28
2 . Where are the closed-loop poles located?
Confirm that the nominal closed-loop system is still stable with this new compensator.
ECE141 – Principles of Feedback Control Homework 4 6
(k) On the same figure as Part (i), add the magnitude plot for 1/|T1(jω)|. (Again, use bodemag
function in MATLAB). Does the perturbation due to the unmodeled dynamics, as given in Part
(g), satisfy the robust stability condition now with this new compensator?
(l) Plot the step response of T1(s). Compare the time scale with the step response in Part (f), and
discuss. (Use step command in MATLAB).
(m) Finally, obtain the closed-loop transfer function T2(s) with the new compensator C1(s) and the
”true” beam model P1(s). Where are the closed-loop poles located? Notice how this new compensator has led to a closed-loop system that is robust enough to ensure stability despite the
unmodeled flexible modes of the beam.