ECE 3050
Signals and Systems
Autumn 2015
MINI-PROJECT #1 SOLUTIONS
Problems:
1. (a) The method to solve this problem is outlined in the lecture notes. We use the command Xmat
= convmtx(x, Lh) to set u
ECE 3050
Signals and Systems
Autumn 2015
MINI-PROJECT #1
Due Mon. Oct. 5, 2015 at 5:00pm
You have just purchased a used speaker. However, once it is set up you realize that the speaker
distorts the so
ECE 3050 Signals and Systems Spring 2015
Ebb.18,2015
MIDTERM EXAMINATION #1
Name:
50L 0 T/oA/S
Instructions:
0 Do not turn over this cover page until instructed to do so.
a You will have 55 minutes to
ECE 3050
Signals and Systems
Spring 2015
MINI-PROJECT #2
Due Wed., Mar. 25, 2015 at 5:00pm
Problem description:
Download the audio le audio.wav from Carmen. We can load this le into MATLAB using the
f
ECE 3050
Signals and Systems
Spring 2015
MINI-PROJECT #1 SOLUTIONS
Problems:
1. The method to solve this problem is outlines in the lecture notes.
>
>
>
>
>
Lh = 6;
Xmat_pilot = convmtx(x_pilot, Lh);
ECE 3050
Signals and Systems
Spring 2015
MINI-PROJECT #2
Due Wed., Mar. 25, 2015 at 5:00pm
Problem description:
Download the audio le audio.wav from Carmen. We can load this le into MATLAB using the
f
ECE 3050
Signals and Systems
Spring 2015
HOMEWORK ASSIGNMENT #7 SOLUTIONS
Problems:
1. (a) From tables, X(j) = 1 if | < W and X(j) = 0 otherwise.
(b) Note that
x[n] = x(t) =
sin(W T n)
.
T n
(1)
From
ECE 3050
Signals and Systems
Spring 2015
MINI-PROJECT #2 SOLUTIONS
Problems:
1. (a) From transform tables, the impulse response of an ideal low-pass lter is
h[n] =
where fc =
c
2
sin(c n)
= 2fc sinc(2
ECE 3050
Signals and Systems
Spring 2015
MIDTERM EXAM #2, REVIEW
Midterm exam #2 will be held on Monday, April 6.
The exam is closed book. Calculators are allowed but internet enabled devices are no
ECE 3050
Signals and Systems
Spring 2015
FINAL PRACTICE QUESTIONS
Problems:
1. A discrete-time LTI system is known to have impulse response h[n] = [1, 1, 1, 1, 1, 1, 1].
(a) Find the output of this sy
ECE 3050 Signals and Systems Spring 2015
Apr. 6, 2015
NIIDTERM EXAMINATION #2
Name:
SOLU’THNUS
Instructions:
0 Do not turn over this cover page until instructed to do so.
0 You will have 55 minutes to
ECE 3050
Issued: August 30, 2013
Assignment # A2
Due: September 6, 2013
Readings:
All materials are posted to the Carmen web page
Oppenheim & Willsky: sections 2.02.6.
1 Computing the convolution su
ECE 3050
Signals and Systems
Spring 2015
MIDTERM EXAM #2, REVIEW
Midterm exam #2 will be held on Monday, April 6.
The exam is closed book. Calculators are allowed but internet enabled devices are no
ECE3050
Assignment # 1
Solutions & Comments
1 Discrete-time system model example: amortization table
(a) The balance at month n is the output and is denoted y[n]. This balance is the previous
months b
ECE3050
Homework Set 1
1. For V = 18 V, R1 = 39 k, R2 = 43 k, and R3 = 11 k, use Ohms Law, voltage division,
and current division to solve for V1 , V2 , I1 , I2 , and I3 .
V1 = 18
I1 =
39 k
= 14.7 V
3
ECE 3050
Signals and Systems
Spring 2015
HOMEWORK ASSIGNMENT #4
Due Fri. Feb. 13, 2015 (in class)
Suggested Reading: Oppenheim, Chapter 3, Sections 3.33.4, pp. 177202.
Problems:
1. Determine the Fouri
Properties of the Fourier Transform
Properties of the Fourier Transform
Reference:
Sections 2.2 - 2.3 of
Professor Deepa Kundur
S. Haykin and M. Moher, Introduction to Analog & Digital
Communications,
ECE 3050 Lecture Notes, Lesson #6: Time & Frequency
We have learned that LTI systems are characterized in the time-domain by
convolution with an impulse response.
We have also learned several Fourie
FINAL PILAC Tlce Gugswords
Io . - l L
PT (muoluhm 7“} I I I I if}
_w. _ n.4,. —. -f—. up“
IL) A n. ?‘ T l, t c
gm]: 2’ Mn] = 2 \ :‘nH ogmﬁé r
K30 L10 a i
e ‘ I
at = ? 1176 i
VI 40 f0 ‘l “J 4 5’ 8 3
ECE 3050
Signals and Systems
Spring 2015
FINAL PRACTICE QUESTIONS
Problems:
1. A discrete-time LTI system is known to have impulse response h[n] = [1, 1, 1, 1, 1, 1, 1].
(a) Find the output of this sy
EEL205 : Signals and Systems
Kushal K. Shah
Asst. Prof. @ EE, IIT Delhi
Email : [email protected]
Web : http:/web.iitd.ac.in/~kkshah
Fourier Series
Response of LTI systems to complex exponentials
c
ECE 3050
Signals and Systems
Spring 2015
MINI-PROJECT #1
Due Mon. Feb. 16, 2015 at 5:00pm
Consider a wireless communications system in which a transmitter sends a message signal, x[n], this
message is
ECE 3050
Signals and Systems
Spring 2015
MINI-PROJECT #3 SOLUTIONS
Problems:
1. Since this is a rst order system, we know that
H(j) =
b0 + b1 j
.
1 + a1 j
From the Bode plot, we can see that we have a
ECE 3050
Signals and Systems
Spring 2015
MINI-PROJECT #3
Due Fri., May 1, 2015 at 5:00pm
Problem description:
In this project, we study the sampling system illustrated in Fig. 1. This system is compri
ECE 3050 Lecture Notes, Lesson #2: Convolution
The concepts of linearity and time-invariance lead us to convolution, the central
idea in the time-domain study of systems.
Topics:
1. Impulse response,
ECE 3050
Signals and Systems
Spring 2015
MIDTERM EXAM #1, REVIEW
Midterm exam #1 will be held on Wednesday, February 18.
The exam is closed book. Calculators are allowed but internet enabled devices
The Ohio State University
Department of Electrical and Computer Engineering
ECE 3050 SIGNALS AND SYSTEMS SP 2015
Instructor:
Prof. Lori Dalton
664 Dreese Laboratories
[email protected]
Webpage:
http:
ECE 3050 Lecture Notes, Lesson #5: Discrete-time FT
For LTI systems:
Time domain: convolution (Chapter 2)
We may write the input as a weighted sum of delayed impulses.
The output is then a weighted s
Quiz 6
j
Problem 1. X1 (e ) is the DTFT of x1 [n].
Write an equation for x2 [n], with the DTFT shown below, in terms of x1 [n].
Problem 2. Let x3 [n] = x2 [n] h[n], where h[n] =
X3 (ej ) from 2 to 2.
Quiz 5
Problem 1. h[n] has the DTFT shown below. Find the DTFT of (1)n h[n].
1
.
.
2 2+wc
wc
0 wc
2wc 2
w
Hint: ej = 1.
sin( 5 ) 2
2
Problem 2. Compute
d.
1
sin( 2 )
Z
ECE 3050 Quizes
08
Solution for HW2
1. Computing convolution
These can be done from the convolution sum, graphically, or with Matlab. By Matlab:
(a) h=[3 2 -1 0 1]; x=[2 -1]; y=conv(h,x)
%observe y=[6 1 -4 1 2 -1]
(b) h