ECE 3250
HOMEWORK ASSIGNMENT I
Fall 2009
1. Find, read, and understand a proof of the Schrder-Bernstein Theorem, which states: o Let A and B be sets. If there exists an injective mapping f : A B and an injective mapping g : B A, then there exists a biject

ECE 3250
HOMEWORK ASSIGNMENT IX
Fall 2009
1. Let x1 be a 1024-point signal and x2 be a 1025-point signal. Note that x1 x2 is a 2048-point signal. (a) Show that it takes a worst-case 512.5 211 multiplications to compute x1 x2 directly. (Suggestion: it take

ECE 3250
HOMEWORK VI SOLUTIONS
Fall 2009
1. By denition, a LTI system has a frequency response if and only if for all R the signal t ej t is an admissible input for the system. A signal x is an admissible input for a system such as the one in the problem

ECE 3250
HOMEWORK ASSIGNMENT VI
Fall 2009
1. Show that every continuous-time system with input space X consisting of all decent signals and and system mapping of the form S (x) = Convh0 (x) +
N X k=1
dk Shifttk (x) ,
where h0 is decent and has nite durati

ECE 3250
HOMEWORK I SOLUTIONS
Fall 2009
1. Most of the proofs Ive seen involve drawing pictures and dening bijective mappings between various subsets of A and various subsets of B (where A and B are the two sets youre trying to dene a bijection between).

ECE 3250
HOMEWORK ASSIGNMENT VIII
Fall 2009
1. Note that N = ej N is already in polar form.
k (a) Show that cfw_N : 0 k < N is the set of so-called N th roots of unity. I.e., this is the complete set of solutions to the polynomial equation z N 1 = 0. (b)

ECE 3250
HOMEWORK I SOLUTIONS
Fall 2011
1. Most of the proofs Ive seen involve drawing pictures and dening bijective mappings
between various subsets of A and various subsets of B (where A and B are the two sets
youre trying to dene a bijection between).

ECE 3250
HOMEWORK VI SOLUTIONS
Fall 2011
1. In each case, S (x) = Convh (x).
(a) S (x) = Convh (x), so
Z
S (x)(t)
h(t )x( )d
=
Z
=
e3(t ) u(t )e5 u( )d
Z
=
e3(t ) u(t )e5 d
3t R t 2
e 0 e d if t 0
0
if t < 0
(1/2)(e5t e3t ) if t 0
0
if t < 0 .
0
=
=
The

ECE 3250
HOMEWORK VII SOLUTIONS
Fall 2011
1.
(a) |a + b|2 = (a + b)(a + b), so
|a + b|2
=
aa + bb + ab + ba
=
|a|2 + |b|2 + 2Recfw_ab .
To get the last line, I used the fact that z + z = 2Recfw_z for any complex number
z . You can use the same argument t

ECE 3250
HOMEWORK X SOLUTIONS
Fall 2011
1. Ill solve all these using the prototype examples and the time-shift rule.
(a) Note that x(n) = 2 (n + 1) + 2n u(n) for all n Z. By prototype example 0 and
the time-shift rule,
z
n 2 (n + 1)
2z 0 < |z | <
and by

ECE 3250
HOMEWORK VIII SOLUTIONS
Fall 2011
1.
b
(a) Let H be the frequency response of the composite system. You can see in at least
b
bb
two ways that H = H1 H2 .
b
If you put t ej t into the composite system, then t H ()ej t comes
b
out. When you put t

CHAPTER 1
Numbers
Engineers, scientists, and applied mathematicians think about numbers all the
time, but usually in a utilitarian way. We manipulate them, calculate with them,
make plans based on them, drive cars and y in airplanes whose design depended

ECE 3250
HOMEWORK V SOLUTIONS
Fall 2011
1.
(a) By linearity, S (x) = S (x + d) = y + S (d). Now, h is an l1 signal and d is an l
signal, so, by Criterion 3,
S (x) y
Since d
=
= S (d)
= hd
h
1
d
.
and
h
1
=
X
|h(n)| =
n=0
1/7
= 5/42 ,
1 + 1/5
it follows th

ECE 3250
HOMEWORK IX SOLUTIONS
Fall 2011
`
1. By the Modulation Rule for Fourier transforms, since cos(c t) = ej c t + ej c t /2,
1b
1b
b
Z () = X ( c ) + X ( + c ) for all R .
2
2
b
b
That is, Z looks like shifted replicas of X of half the height centere

ECE 3250
HOMEWORK IV SOLUTIONS
Fall 2011
1. Note that h has nite duration.
(a) By Criterion 1, Dh = FZ because h has nite duration.
(b) S (x) = Convh (x), so for every n Z,
`
10
X
X
3 1 711
14 `
k
S (x)(n) =
h(k)x(n k) =
3(7 ) =
1 711 ; .
=
1 1/7
3
k=
k=0

ECE 3250
HOMEWORK ASSIGNMENT IV
Fall 2011
1. A certain LTI system has impulse response h whose specication is
n
7
if 0 n < 11
h(n) =
0
otherwise.
(a) What is Dh , the set of all x FZ for which Convh (x) exists?
(b) Find the output S (x) of the system whe

ECE 3250
HOMEWORK ASSIGNMENT V
Fall 2011
1. A certain discrete-time LTI system with system mapping S has impulse response h
with specication
(5)n
h(n) =
u(n) for all n Z .
7
(a) Suppose that x is a bounded signal, and that S (x) = y . Suppose d is another

ECE 3250
HOMEWORK ASSIGNMENT VII
Fall 2011
1. In Chapters 5 and 9 of the monograph I asserted that l2 is a vector space and, in fact,
an inner-product space. Heres how to prove it.
(a) Show that for any complex numbers a and b, we have
|a + b|2 = |a|2 + |

ECE 3250
HOMEWORK ASSIGNMENT VI
Fall 2011
1. In each case, h is the impulse response of a LTI system and x is an input signal. Your
task is to nd S (x).
(a)
(b)
(c)
(d)
h(t) = e3t u(t) for all t R and x(t) = e5t u(t) for all t R.
h(t) = e3t u(t) for all t

ECE 3250
HOMEWORK ASSIGNMENT VIII
Fall 2011
b
b
1. Let H1 and H2 be the frequency responses of ideal low-pass and band-pass lters.
Specically, let
1 if | 3700
b
H1 () =
0 otherwise
and
1 if 1737 | 7140
b 2 () =
H
0 otherwise.
(a) Show that the LTI system

ECE 3250
HOMEWORK ASSIGNMENT X
Fall 2011
1. Find the z -transform (formula for X (z ) along with (ROC)X ), if the transform exists,
for each signal x specied below. If the transform fails to exist, explain why.
x(n) = 2n u(n + 1) for all n Z.
x(n) = n3(n+

ECE 3250
HOMEWORK II SOLUTIONS
Fall 2011
1. Heres a relatively easy way to see that the answer is Yes. Given that the sequence
converges, we know that it has a limit c. By denition of the limit, there exists some
integer N > 0 so that |cn c| < 1 for all n

ECE 3250
HOMEWORK III SOLUTIONS
Fall 2011
Note: Ive expressed the solutions to the problems involving calculating convolutions in
particular ways that match up well with some stu well be doing later in the course. Your
answers might not look exactly like

ECE 3250
HOMEWORK ASSIGNMENT III
Fall 2011
For many of the problems, you might nd the following geometric-series identities useful.
In the equations, is a real or complex number.
(
M
Xm
M + 1 if = 1
=
1 M +1
if = 1 .
1
m=0
Furthermore,
X
m =
m=0
1
1
when

CHAPTER 1
Numbers
Engineers, scientists, and applied mathematicians think about numbers all the
time, but usually in a utilitarian way. We manipulate them, calculate with them,
make plans based on them, drive cars and y in airplanes whose design depended

CHAPTER 1
Numbers
Engineers, scientists, and applied mathematicians think about numbers all the
time, but usually in a utilitarian way. We manipulate them, calculate with them,
make plans based on them, drive cars and y in airplanes whose design depended

ECE 3250
HOMEWORK ASSIGNMENT II
Fall 2016
1. If cfw_cn is a convergent sequence of real numbers, does there necessarily exist R > 0
such that |cn | R for every n 2 N? Equivalently, is cfw_cn : n 2 N necessarily a bounded
set of real numbers? Explain why

ECE 3250
HOMEWORK III SOLUTIONS
Fall 2016
Note: Ive expressed the solutions to the problems involving calculating convolutions in
particular ways that match up well with some stuff well be doing later in the course. Your
answers might not look exactly lik

ECE 3250
HOMEWORK IV SOLUTIONS
Fall 2016
1. Note that h has finite duration.
(a) By Criterion 1, Dh = FZ because h has finite duration.
(b) S(x) = h x, so for every n Z,
`
10
X
X
7 1 311
21 `
k
S(x)(n) =
=
h(k)x(n k) =
1 311 ; .
7(3 ) =
1 1/3
2
k=
k=0
Not

ECE 3250
PRELIM I
Fall 2015
Problems 1 through 11 are worth 8 points each. Problem 12 is worth 12 points. Throughout, denotes convolution.
1. A is a set, B is a proper subset of A (i.e. B A and B 6= A), and f : A B is a
mapping. You can be certain that
(i

ECE 3250
HOMEWORK II SOLUTIONS
Fall 2016
1. Heres a relatively easy way to see that the answer is Yes. Given that the sequence
converges, we know that it has a limit c. By definition of the limit, there exists some
integer N > 0 so that |cn c| < 1 for all

ECE 3250
HOMEWORK ASSIGNMENT I
Fall 2016
1. Find, read, and try to understand a proof of the Schr
oder-Bernstein Theorem, which
states: Let A and B be sets. If there exists an injective mapping f : A ! B and an
injective mapping g : B ! A, then there exis