Description
Problem 1. Think about the different systems around you . . .
1. Identify three feedback systems.
2. For each feedback system, draw a simple high-level block diagram, and identify the plant/process,
the inputs/outputs, the sensing mechanism, the actuating mechanism, the control mechanism,
as well as any possible disturbances.
3. Also for each system, determine whether it uses manual or automatic control, whether it is singleinput-single-output (SISO) or multi-input-multi-output (MIMO), and whether it is a tracking or
a regulator system.
Problem 2. A control systems that has always fascinated me is the system used to stabilize camera/lens motion which enables photographers to shoot at lower shutter speeds (well, as long as the
subject itself is steady), and thus be able to shoot at lower light while still keeping the ISO sensitivity
level low and thus maintain low levels of noise in the image.
Here is a New York Times article about this technology: NY Times Article on Image Stabilization
Different camera manufacturers call it with different names. For example, Nikon calls it ”Vibration
Reduction (VR)” whereas Canon calls it ”Image Stabilization (IS)”. The basic control system is shown
in the following picture (taken from the Nikon website linked below):
And here is a description of how the control system works on the Nikon website: Nikon Vibration
Reduction Technology
Draw a block diagram for the feedback system, and identify inputs/outputs (including any disturbance and noise inputs), as well as sensing and actuating mechanisms.
These homework problems are compiled using the different textbooks listed on the course syllabus
1
ECE141 – Principles of Feedback Control Homework 1 2
Problem 3.1. Recall the Bilateral (two-sided) Laplace Transform:
L(x(t)) = X(s) = Z ∞
−∞
x(t)e
−stdt, s ∈ C, X(s) ∈ C
Find the bilateral Laplace transform for x1(t) = e
−atu(t) and x2(t) = −e
−atu(−t), where u(t) is the
unit step function. For each case, determine the associated Region of Convergence (ROC) for the
Laplace transform. That is the region in the complex s-plane where the Laplace integral converges
and thus the Laplace transform exists.
Problem 3.2. Now, let:
H(s) = 5
s
2 + s − 6
Using Partial Fraction Expansion, find the different possible Inverse Laplace Transforms (i.e., h(t)
functions) and, for each h(t), find the associated ROC, and discuss both causality and stability of the
system whose impulse response is given by h(t). Once you are done, be happy that we mostly have to
deal with causal systems, and can use single-sided Laplace transforms.
Problem 4. Consider a causal signal x(t) with the following Laplace transform:
X(s) = 2s
2 + ω
2
s(s
2 + ω2)
1. Using the Final Value Theorem, find limt→∞ x(t)
2. Find the inverse Laplace transform x(t) = L
−1
(X(s)) and see what its value will be as t → ∞.
Compare with the previous result and discuss.
Problem 5. Consider a causal LTI system with a rational Transfer Function H(s) = b(s)
a(s)
. Find a
necessary and sufficient condition for the associated Impulse Response of this system (h(t)) to be
continuous at t = 0.0. (hint: Initial Value Theorem)
Problem 6. Dynamical Signficance of Poles and Zeros: In the second lecture, we discussed
how the poles of the transfer function determine the modes of the system and the form of its natural
response. We also saw how the zeros impact the amplitudes of both the forced response and the
natural response of the system. We also discussed the significance of the exponential inputs e
st as the
eigenfunctions of LTI systems. The objective of this problem is to summarize the key points on the
dynamical significance of poles and zeros. Consider the following proper rational transfer function:
H(s) = b(s)
a(s)
where b(s) and a(s) are polynomials in s and the order of b(s) is equal or smaller than the order of a(s)
(hence a proper transfer function) and b(s) and a(s) have no common factors (i.e., no shared roots).
Recall that the output of the system to an input x(t) ←→L X(s) can be written as:
y(t) ←→L
Y (s) = b(s)
a(s)
X(s) + I(s)
a(s)
where I(s) is another polynmial in s of order n − 1 where n is the order of a(s), and whose coefficients
are all determined by the initial conditions of input and output, i.e.,
y(0−), y˙(0−), y¨(0−), . . . . , x(0−), x˙(0−), x¨(0−), . . .
ECE141 – Principles of Feedback Control Homework 1 3
(a) Assume an exponential input is applied to the sytem at time t = 0, i.e., x(t) = e
s0tu(t), where
u(t) is the unit step function, and s0 is not a pole of the tranfer function. Show that the Laplace
transform of the output can be written as follows:
Y (s) = K0
s − s0
+
β(s)
a(s)
+
I(s)
a(s)
where β(s) is a polynomial in s of order n − 1. What is the value for K0? Discuss how you can
conclude that the output initial conditions y(0−), y˙(0−), . . . can indeed be set such that system
response is obtained as:
y(t) = H(s0)e
s0tu(t)
(b) What will now happen if s0 happens to be a zero of the transfer function, i.e., b(s0) = 0? This is
the main reason why transfer function zeros are sometimes called transmission zeros.
(c) Now, assume the input to the system is identically 0, i.e., x(t) = 0, ∀t. Therefore, the Laplace
transform of the zero-input system response can be written as:
Y (s) = I(s)
a(s)
.
Assume s = p is a pole of the transfer function, i.e, a(s)|s=p = 0 (for simplicity, lets assume it is
not a repeated pole). Show that the output initial conditions can be set such that:
y(t) = Keptu(t)
where K is a constant, and, as before, u(t) is the unit step function.
Problem 7. Use the (single-sided) Laplace transform to solve the following Ordinary Differential
Equation and find y(t) given the initial conditions:
y¨(t) + 6 ˙y(t) + 5y(t) = 6u(t), y˙(0−) = y(0−) = 1.0, u(t) : unit step function
Problem 8. Find the Laplace Transforms for the following functions (u(t) is the unit step function):
(a) x(t) = (e
−2t + t cos(2t))u(t)
(b) x(t) = 4te−4tu(t)
Problem 9. Find the (causal) Inverse Laplace Transforms for each of the following:
(a) X(s) = 10
(s+1)2(s+3)
(b) X(s) = 100(s+2)
(s
2+0.4s+0.4)(s+1)(s+5)e
−4s
Problem 10. Time Delay: Consider the transfer function in Part (b) of the previous problem again.
As you know by now, we end up with an exponential term in the transfer function due to presence
of time delay somewhere in our loop. This is one main exception where we would have to deal with
non-rational transfer functions.
(a) Ignore the time delay for a moment, and use zp2tf command in MATLAB followed by step
command to plot the step response of the system with no time delay.
(b) Now, use the InputDelay option in the tf command to form the transfer function with the 4 sec
delay. Use step command again to plot the delayed step response on the same figure. (use hold
command to keep the current figure handle active)
ECE141 – Principles of Feedback Control Homework 1 4
(c) To deal with the delay-induced exponential term in transfer functions, one option could just be to
approximate it using the power series expansion of the exponential term:
e
−sT ≈ 1 − sT +
1
2
(sT)
2 −
1
3 !(sT)
3 + . . .
But it turns out that a better approximation would be obtained using the so-called Pad´e approximation which approximates the exponential term (or almost any function for that matter) with a
rational transfer function. The 1st-order and 2nd-order Pad´e approximations for the exponential
term are given below:
e
−sT ≈
1− sT
2
1+ sT
2
e
−sT ≈
1− sT
2 +
(sT )
2
12
1+ sT
2 +
(sT )2
12
(1)
So by substituting the Pad´e approximation for the exponential term, one can turn the transfer
function into a rational transfer function again. Thankfully MATLAB makes our life a bit easier
here too. Use pade(sys,n) command where sys is the same system (transfer function) you just
created in Part (b) above, and n is the desired order for Pad´e approximation. Try it with both
1st-order (n = 1) and 3rd-order (n = 3), and then use step command again to plot step responses
of the approximated systems for both cases on the same figure as Part (b). As one would expect,
note how increasing the order of Pad´e approximation improves the accuracy of the approximated
step response. Also note how using Pad´e approximation can lead to a zero on the right-half-plane
(RHP) and thus create a non-minimum phase system with a negative slope for the step response
at t = 0.0.
