Final Exam @ 2021-01-12 12:51:19-08:00
EECS 16B Designing Information Devices and Systems II
Fall 2019 UC Berkeley Final Exam
Exam location: Proctoring Office
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Section 0: Pre-exam questions (4 points)
1. Tell us about something you are proud of this semester. (2 pt)
2. What are you looking forward to in winter break? Describe how you will feel. (2 pts)
Do not turn this page until the proctor tells you to do so. You can work on Section 0 above before time starts.
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3. Transistor Behavior (12 pts)
For all NMOS devices in this problem, Vtn = 0.5V. For all PMOS devices in this problem, |Vt p|= 0.6V.
(a) (4 pts) Which is the equivalent circuit for the right-hand side of the circuit? Fill in the correct bubble.
−+1V
−+ 2V
Ron, N
−+ 2V
Circuit A
Ron, N
−+ 2V
Circuit B
A B
Equivalent Circuit © ©
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(b) (4 pts) Which is the equivalent circuit for the right-hand side of the circuit? Fill in the correct bubble.
−+0.4V RL
−+ 2V
RL
−+ 2V
Circuit A
RL
Ron,P
−+ 2V
Circuit B
RL
Ron,P
−+ 2V
Circuit C
A B C
Equivalent Circuit © © ©
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(c) (4 pts) Which is the equivalent circuit for the right-hand side of the circuit? Fill in the correct bubble.
−+0.7V
−+ 2V
Ron,N
Ron,P
−+ 2V
Circuit A
Ron,N
Ron,P
−+ 2V
Circuit B
Ron,N
Ron,P
−+ 2V
Circuit C
Ron,N
Ron,P
−+ 2V
Circuit D
A B C D
Equivalent Circuit © © © ©
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4. Filter Design and Bode Plots (28 pts)
On the Bode plots below, we have plotted the magnitude responses of first-order low pass filters and high
pass filters using the example of cutoff frequency ω0 = 106.
103 104 105 106 107 108 109
10−4
10−3
10−2
10−1
100
101
ω
|H
(
jω
)|
Lowpass Bode Plot (ω0 = 106)
103 104 105 106 107 108 109
10−4
10−3
10−2
10−1
100
101
ω
|H
(
jω
)|
Highpass Bode Plot (ω0 = 106)
Recall that the transfer functions for such simple low pass filters and high pass filters are:
Hlowpass( jω) =
1
1+ jωω0
; Hhighpass( jω) =
jω
ω0
1+ jωω0
.
(a) (6 pts) We want to design a bandpass filter that can pass through a 2.4GHz WiFi signal while blocking
other interfering signals — FM radio at 100MHz and WiGig at 60GHz. (Recall: Mega = 106 and Giga
= 109.) We will achieve this by cascading lowpass and highpass filters, using ideal op-amp buffers in
between to prevent any loading effects.
Unfortunately, when we look in the lab, we only see inductors, 1kΩ resistors, and op-amps.
We will start by cascading a single highpass filter followed by a single lowpass filter, with an op-amp
buffer in between. Using only op-amps, two inductors, and resistors (as many as needed), draw
the full band-pass filter. Label Vin and Vout and label the two inductors with L1 and L2. Do not worry
about picking the values for L1 and L2 in this part.
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(b) (8 pts) One interfering signal that we want to block is the WiGig signal at 60GHz. If we want to atten-
uate/reduce the magnitude of the WiGig signal by a factor of about
√
101≈ 10, What is a candidate
‘cutoff frequency’ (in Hz) desired for this lowpass filter?
What inductance value should we use for the lowpass filter? Recall that we only have resistors with
1kΩ resistance. It is fine to give your inductance as a formula — you don’t have to simplify it.
For your convenience, here are some calculations that may or may not be relevant:
60×109
2pi = 9.549×109 2.4×10
9
2pi = 382×106 100×10
6
2pi = 15.9×106
60×109×2pi = 377×109 2.4×109×2pi = 15.08×109 100×106×2pi = 628×106
(HINT: Look at the relevant Bode plot and read off how far away in frequency from ω0 you need to be
to reduce the magnitude by the desired factor of around 10.)
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(c) (14 pts) Another interfering signal that we want to block is FM radio at 100MHz and we want to reduce
its magnitude by a factor of around 100. We decide to use multiple highpass filters in a row (separated
by ideal op-amp buffers) to attenuate the FM radio signal more strongly. We design the system with
the highpass filter cutoff frequencies all at 1GHz. In this case, what inductor value should each of
the highpass filters use? Recall that we only have resistors with 1kΩ resistance. It is fine to give your
inductance as a formula — you don’t have to simplify it.
For your convenience, here are some calculations that may or may not be relevant:
60×109
2pi = 9.549×109 2.4×10
9
2pi = 382×106 100×10
6
2pi = 15.9×106
60×109×2pi = 377×109 2.4×109×2pi = 15.08×109 100×106×2pi = 628×106
How many highpass filters must we cascade in order to attenuate the FM signal at 100MHz by a
factor of around 100?
Draw the full circuit for your complete filter including op-amp buffers, the lowpass filter, and
the highpass filters.
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5. Controllable Canonical Form (26 pts)
Suppose that we have a two-dimensional continuous time system governed by:
d
dt
~x(t) =
[
1 −1
0 −4
]
~x(t)+
[
1
1
]
u(t)
We would like to put this system into Controllable Canonical Form (CCF) to use state feedback to place the
eigenvalues at desired locations. For your convenience, the characteristic polynomial det
λ I−[1 −1
0 −4
]=
(λ −1)(λ +4) = λ 2+3λ −4.
(a) (4 pts) Is the system stable? Why or why not?
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(b) (16 pts) Recall that our original system is:
d
dt
~x(t) =
[
1 −1
0 −4
]
~x(t)+
[
1
1
]
u(t) with det
λ I−[1 −1
0 −4
]= λ 2+3λ −4.
We would like to change coordinates to bring the system into CCF:
d
dt
~z(t) =
[
0 1
a0 a1
]
~z(t)+
[
0
1
]
u(t)
Compute the T basis such that~z(t) = T−1~x(t) or equivalently (if you want), give us T−1 directly.
To help you along, here are some calculations already done for you:[
1 0
1 −4
]−1
=
1
4
[
4 0
1 −1
]
[
0 1
1 a
]−1
=
[
−a 1
1 0
]
What is a0? What is a1?
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(c) (6 pts) Using state feedback
u(t) =
[
k˜0 k˜1
]
~z(t)
place the closed-loop eigenvalues at λ1 =−1, λ2 =−2. What is k˜0? What is k˜1?
(Notice that we are asking for the feedback gains in terms of~z(t) not the original ~x(t). If you have
time, feel free to check your work by the original~x(t).)
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6. You knew this was coming (16 pts)
In the last midterm, you began solving a minimum norm problem in which we had a wide matrix A and
wanted to solve A~x=~y while minimizing the norm ‖C~x‖ whereC was also a wide matrix. You then focused
on some special cases where certain matrices were assumed to have orthonormal columns. The nonorthonor-
mal case wasn’t resolved. This question is about the heart of what remains to be resolved.
Suppose that we have a matrix A f , but we don’t know whether or not the columns of A f are linearly in-
dependent or orthonormal. How can you use the results from the SVD A f = U fΣ fV>f = ∑
r
i=1σ f ,i~u f ,i~v>f ,i
to help project a vector ~s onto the subspace spanned by the columns of A f ? Here, the σ f ,1, . . . ,σ f ,r are all
strictly positive. Give an expression for a matrix Pf such that Pf~s is the projection of~s in the subspace
spanned by the columns of A f .
Also give an expression for an~x f so that A f~x f = Pf~s.
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7. Movie Ratings and PCA (36 pts)
Recall from the lecture on PCA that we can think of movie ratings as a structured set of data. For every
person i and movie j, we have that person’s rating Ri, j (thought of as a real number).
Suppose that there are m movies and n people. Let’s think about arranging this data into a big n×m matrix
R with people corresponding to rows and movies corresponding to columns. So the entry in the i-th row and
j-th column should be Ri, j above.
(a) (6 pts) Suppose we believe that there is actually an underlying pattern to this rating data and that a
separate study has revealed that every movie is characterized by a set of characteristics: say action and
comedy. This means that every movie j has a pair of numbers a[ j],c[ j] that define it. At the same time,
every person i has a pair of sensitivities sa[i],sc[i] that essentially define that person’s preferences. A
person i will rate the movie j as Ri, j = sa[i]a[ j]+ sc[i]c[ j].
If we arrange the sensitivities into a pair of n-dimensional vectors ~sa,~sc, and the movie characteristics
into a pair of m-dimensional vectors~a,~c, use outer products to express the rating matrix R in terms
of these vectors ~sa,~sc,~a,~c.
Recall that the outer product of a real vector ~u with a real vector ~v is ~u~v> — the reverse order as
compared to the usual inner product.
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(b) (6 pts) Now suppose that we want to discover the underlying nature of movies from the data R itself.
Suppose for this part, that you have four observed rating data vectors (corresponding to four different
movies being rated by six individuals).
All of the movie data vectors just happened to be multiples of the following 6-dimensional vector
~v=

2
−2
3
−4
4
0

. (For your convenience, note that ‖~v‖= 7.)
You arrange the movie data vectors as the columns of a matrix R given by:
R=
−~v −2~v 2~v 4~v
 (1)
You want to perform PCA (for movies) using the SVD of the matrix R to better understand the pattern
in your data.
The first “principal component vector” is a unit vector that tells which direction we would want to
project the columns of R onto to get the best rank-1 approximation for R.
Find this first principal component vector of R to explain the nature of your movie data points.
(HINT: You don’t need to actually compute any SVDs in this simple case. Also, be sure to think about
what size vector you want as the answer. Don’t forget that you want a unit vector!)
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(c) (12 pts) Suppose that now, we have two more data points (corresponding to two more movies being
rated by the same set of six people) that are multiples of a different vector ~p where:
~p=

6
3
−2
0
0
0

. (For your convenience, note that ‖~p‖= 7 and that ~p>~v= 0.)
We augment our ratings data matrix with these two new data points to get:
R=
−~v −2~v 2~v 4~v −3~p 3~p
 (2)
Find the first two principal components corresponding to the nonzero singular values of R. This is
what we would use to best project the movie data points onto a two-dimensional subspace.
What is the first principal component vector? What is the second principal component vector?
Justify your answer.
(Hint: Think about the inner product of~v and ~p and what that implies for being able to appropriately
decompose R. Again, very little computation is required here.)
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(d) (12 pts) In the previous part, you had
R=
−~v −2~v 2~v 4~v −3~p 3~p

with ‖~v‖= 7 and ‖~p‖= 7, satisfying ~p>~v= 0.
If we use~ri to denote the i-th column of R, plot the movie data points~ri projected onto the first and
second principal component vectors. The coordinate along the first principal component should be
represented by horizontal axis and the coordinate along the second principal component should be the
vertical axis. Label each point.
−32 −28 −24 −20 −16 −12 −8 −4 4 8 12 16 20 24 28 32
−24
−20
−16
−12
−8
−4
4
8
12
16
20
24
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8. Latch (46 pts)
The circuit below is a type of latch, which is one of the fundamental components of memory in many
digital systems. The latch is a bistable circuit, which means that there are two possible stable states: one
representing a stored ‘1’ bit and the other a stored ‘0’ bit.
R
C
R
VDD
C
Vout1 Vout2
Figure 7: Simplified latch: the gate capacitances have been pulled out explicitly.
(a) (6 pts) To get a basic understanding of the stable operating points for the latch, consider the following
simplified circuit using the pure switch model for MOSFETs (and a threshold voltage of VDD2 ).
Vout2 ≥VDD/2
Vout1
R
Vout1 ≥VDD/2
Vout2
R
VDD
Figure 8: Pure switch model for a latch
First assume that Vout1 = 0. What is Vout2? Are the left and right switches open or closed?
Open or VDD Closed or 0
Left Switch © ©
Right Switch © ©
Vout2 © ©
Suppose that Vout1 =VDD. What is Vout2? Are the left and right switches open or closed?
Open or VDD Closed or 0
Left Switch © ©
Right Switch © ©
Vout2 © ©
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(b) (10 pts) To get an understanding of latch dynamics, we will now break it down into smaller pieces.
Below is one half of the latch circuit.
Vin
Ids
R
VDD
Vout
C
Figure 9: Latch half-circuit
Write a differential equation for the voltage Vout in terms of the drain to source current Ids. Treat
Ids as some specified input signal and treat the transistor as a current source connected to ground.
(i.e. There is no dependence on Vin in this part. In this part, treat the Ids as a constant that you are
given.)
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(c) (10 pts)
R
C
R
VDD
C
Vout1 Vout2
For this circuit, we care about more detailed analog characteristics of the MOSFETs, so we will model
their behavior more accurately as current sources that are controlled by their gate voltages Vin with the
following equation:
Ids = g(Vin)
Where g(Vin) is a some nonlinear function. Using this Ids expression together with the result from
the previous part, write down a system of differential equations forVout1 andVout2 in vector form:
d
dt
[
Vout1(t)
Vout2(t)
]
= ~f
[Vout1(t)
Vout2(t)
] .
(Hint: Notice that the latch above can be constructed by taking two of the circuit in Figure ??, and
connecting the Vout of one to the gate Vin of the other and vice-versa.)
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(d) (20 pts) For the rest of this problem, assume that your analysis yields the following system of nonlinear
differential equations: [
dVout1
dt
dVout2
dt
]
=
[
1−Vout1−g(Vout2)
1−Vout2−g(Vout1)
]
Suppose that you put this latch into an ideal circuit simulator, and measure g(Vin) and
dg
dt (Vin). The
results from these measurements are shown in the graphs below. From your simulations, you also can
see that for the following initial conditions, the latch voltages do not change over time:[
V ∗out1
V ∗out2
]
=
[
1
0
]
,
[
0
1
]
,
[
0.5
0.5
]
Use the graphs below to linearize the differential equations around the three operating points.
Write a linearized system of differential equations around each of those operating points
[
V ∗out1
V ∗out2
]
.
For which of the provided
[
V ∗out1
V ∗out2
]
points is the latch locally stable? For which of the provided
points is the latch locally unstable? Why?
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Vin
g(
V i
n)
g(Vin)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Vin
dg d
t(
V i
n)
dg
dt (Vin)
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9. Choosing cost functions for learning classification (12 pts)
In this problem, we want to design classifiers for data that comes from two classes “+” and “-”. We have
labeled data {(~xi, `i)} from which we want to learn a vector ~w so that we can use the sign of ~w>~x to classify
~x. To do this, we are going to be trying to minimize a sum ctotal(~w) = ∑i c`i(~x>i ~w).
The cost functions we are considering include:
• Squared loss: c+(p) = (p−1)2 and c−(p) = (p− (−1))2
• Exponential loss: c+(p) = e−p and c−(p) = e+p
• Logistic loss: c+(p) = ln
(
1+ e−p
)
and c−(p) = ln
(
1+ e+p
)
For the plotted data, which cost functions will give reasonable answers? Select all that apply.
Squared Exponential Logistic
Reasonable Choice © © ©
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Squared Exponential Logistic
Reasonable Choice © © ©
Squared Exponential Logistic
Reasonable Choice © © ©
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Squared Exponential Logistic
Reasonable Choice © © ©
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10. DFT and Polynomials (34 pts)
Consider the polynomial-style DFT basis that comes from evaluating the monomials x0,x1,x2,x3 on the 4-th
roots of unity x0 = e j
2pi
4 0 = 1, x1 = e j
2pi
4 1 = j, x2 = e j
2pi
4 2 =−1, x3 = e j 2pi4 3 =− j.
B=

1 1 1 1
1 e j
2pi
4 e j2
2pi
4 e j3
2pi
4
1 e j2
2pi
4 e j4
2pi
4 e j6
2pi
4
1 e j3
2pi
4 e j6
2pi
4 e j9
2pi
4
=

1 1 1 1
1 j −1 − j
1 −1 1 −1
1 − j −1 j
 .
(a) (4 pts) The DFT coefficients ~F are related to a vector of samples ~f by the relationship ~f = B~F . In
other words, ~F represents ~f in the basis given by the columns of B. Similarly for the DFT coefficients
~G and a vector of samples~g — they too satisfy the relationship~g= B~G.
What are the DFT coefficients ~H for~h=α~f +β~g in terms of ~F and ~G. Here, α and β are constant
real numbers.
(b) (8 pts) Explicitly find the DFT coefficients ~F of the vector ~f =

0
1
0
−1
 i.e. we want ~F so that ~f = B~F .
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(c) (10 pts) Show that if ~f is a real vector, then:
• F [0] is always real and so is F [2].
• F [1] = F [3].
(HINT: What do you know about B−1?)
(d) (12 pts) We notice that when we use the standard polynomial interpolation approach fˆ (e jθ ) = F [0]+
F [1]e jθ +F [2]e j2θ +F [3]e j3θ , we often get nonzero imaginary parts even though our samples ~f were
purely real. Here ~F are the DFT coefficients corresponding to ~f so that ~f = B~F .
Give new constants k1,k3 so that
fˆ (e jθ ) = F [0]+F [1]e jk1θ +F [2]
e j2θ + e− j2θ
2
+F [3]e jk3θ
will always return a real number for each θ ∈ [0,2pi], and actually interpolates the data points
— i.e. f [i] = fˆ (e j
2pii
4 ) for i= 0,1,2,3.
Prove that this fˆ (e jθ ) is always real for θ ∈ [0,2pi] no matter what real values the vector ~f has.
(HINT: For proving realness, use what is already established by the previous part (even if you didn’t
get the previous part!) and write out fˆ (e jθ ) in terms of constants, sines, and/or cosines. Using a polar
form for F [1] might be helpful.)
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Final Exam @ 2021-01-12 12:51:19-08:00
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Final Exam @ 2021-01-12 12:51:19-08:00
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