NAME: SOLUTIONS EEE 203 HW 1

Similar documents
Mathematical Preliminaries for Transforms, Subbands, and Wavelets

EEE 304 Test 1 NAME:

Problem 2. Describe the following signals in terms of elementary functions (δ, u,r, ) and compute. x(t+2) x(2-t) RT_1[x] -3-2 = 1 2 = 1

( A) ( B) ( C) ( D) ( E)

Response of LTI Systems to Complex Exponentials

Fourier Series: main points

, R we have. x x. ) 1 x. R and is a positive bounded. det. International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:10 No:06 11

EEE 304 Test 1 NAME: solutions

Part B: Transform Methods. Professor E. Ambikairajah UNSW, Australia

c. What is the average rate of change of f on the interval [, ]? Answer: d. What is a local minimum value of f? Answer: 5 e. On what interval(s) is f

From Fourier Series towards Fourier Transform

A L A BA M A L A W R E V IE W

1a.- Solution: 1a.- (5 points) Plot ONLY three full periods of the square wave MUST include the principal region.

ELECTROMAGNETIC COMPATIBILITY HANDBOOK 1. Chapter 12: Spectra of Periodic and Aperiodic Signals

Review Topics from Chapter 3&4. Fourier Series Fourier Transform Linear Time Invariant (LTI) Systems Energy-Type Signals Power-Type Signals

AE57/AC51/AT57 SIGNALS AND SYSTEMS DECEMBER 2012

1973 AP Calculus BC: Section I

EE 350 Signals and Systems Spring 2005 Sample Exam #2 - Solutions

82A Engineering Mathematics

Fooling Newton s Method a) Find a formula for the Newton sequence, and verify that it converges to a nonzero of f. A Stirling-like Inequality

Digital Image Processing

Gavilan JCCD Trustee Areas Plan Adopted October 13, 2015

EEE 303: Signals and Linear Systems

Face Detection and Recognition. Linear Algebra and Face Recognition. Face Recognition. Face Recognition. Dimension reduction

Chapter 11 INTEGRAL EQUATIONS

Numerical Simulation for the 2-D Heat Equation with Derivative Boundary Conditions

Chapter 2: Time-Domain Representations of Linear Time-Invariant Systems. Chih-Wei Liu

AR(1) Process. The first-order autoregressive process, AR(1) is. where e t is WN(0, σ 2 )

Exercises for lectures 23 Discrete systems

Chapter 7 INTEGRAL EQUATIONS

Physics 160 Lecture 3. R. Johnson April 6, 2015

The Exile Began. Family Journal Page. God Called Jeremiah Jeremiah 1. Preschool. below. Tell. them too. Kids. Ke Passage: Ezekiel 37:27

Approximation of Functions Belonging to. Lipschitz Class by Triangular Matrix Method. of Fourier Series

Use precise language and domain-specific vocabulary to inform about or explain the topic. CCSS.ELA-LITERACY.WHST D

ECE351: Signals and Systems I. Thinh Nguyen

The z-transform. Dept. of Electronics Eng. -1- DH26029 Signals and Systems

Chapter Taylor Theorem Revisited

Chapter 3 Linear Equations of Higher Order (Page # 144)

T h e C S E T I P r o j e c t

MAT3700. Tutorial Letter 201/2/2016. Mathematics III (Engineering) Semester 2. Department of Mathematical sciences MAT3700/201/2/2016

) and furthermore all X. The definition of the term stationary requires that the distribution fulfills the condition:

Instructors Solution for Assignment 3 Chapter 3: Time Domain Analysis of LTIC Systems

Fourier series. (sine and cosine)( ) ... : w h ere 2 (1 1)

r/lt.i Ml s." ifcr ' W ATI II. The fnncrnl.icniccs of Mr*. John We mil uppn our tcpiiblicnn rcprc Died.

W i n t e r r e m e m b e r t h e W O O L L E N S. W rite to the M anageress RIDGE LAUNDRY, ST. H E LE N S. A uction Sale.

H A M M IG K S L IM IT E D, ' i. - I f

.-I;-;;, '.-irc'afr?*. P ublic Notices. TiffiATRE, H. aiety

OH BOY! Story. N a r r a t iv e a n d o bj e c t s th ea t e r Fo r a l l a g e s, fr o m th e a ge of 9

Wireless & Hybrid Fire Solutions

ECEN620: Network Theory Broadband Circuit Design Fall 2014

Fourier Techniques Chapters 2 & 3, Part I

Nikesh Bajaj. Fourier Analysis and Synthesis Tool. Guess.? Question??? History. Fourier Series. Fourier. Nikesh Bajaj

Option 3. b) xe dx = and therefore the series is convergent. 12 a) Divergent b) Convergent Proof 15 For. p = 1 1so the series diverges.

Taylor and Maclaurin Series

Agenda Rationale for ETG S eek ing I d eas ETG fram ew ork and res u lts 2

Worksheet: Taylor Series, Lagrange Error Bound ilearnmath.net

2 T. or T. DSP First, 2/e. This Lecture: Lecture 7C Fourier Series Examples: Appendix C, Section C-2 Various Fourier Series

AP Calculus BC AP Exam Problems Chapters 1 3

[ m] x = 0.25cos 20 t sin 20 t m

SECTION 7: STEADY-STATE ERROR. ESE 499 Feedback Control Systems

EEE 184: Introduction to feedback systems

This is a repository copy of Analysis of nonlinear oscillators using volterra series in the frequency domain Part I : convergence limits.

Continous system: differential equations

Frequency Measurement in Noise

Department of Electronics & Telecommunication Engineering C.V.Raman College of Engineering

Math 3301 Homework Set 6 Solutions 10 Points. = +. The guess for the particular P ( ) ( ) ( ) ( ) ( ) ( ) ( ) cos 2 t : 4D= 2

Practice papers A and B, produced by Edexcel in 2009, with mark schemes. Practice Paper A. 5 cosh x 2 sinh x = 11,

EE Control Systems LECTURE 11

P a g e 5 1 of R e p o r t P B 4 / 0 9

Chapter 5. Long Waves

Mixing time with Coupling

Gavilan JCCD Trustee Areas Plan Adopted November 10, 2015

Approximate solutions for the time-space fractional nonlinear of partial differential equations using reduced differential transform method

TMA4329 Intro til vitensk. beregn. V2017

C o r p o r a t e l i f e i n A n c i e n t I n d i a e x p r e s s e d i t s e l f

Linear Systems Analysis in the Time Domain

- Prefixes 'mono', 'uni', 'bi' and 'du' - Why are there no asprins in the jungle? Because the parrots ate them all.

THIS PAGE DECLASSIFIED IAW EO IRIS u blic Record. Key I fo mation. Ma n: AIR MATERIEL COMM ND. Adm ni trative Mar ings.

Chapter 2 Limits and Continuity

Chapter 5 Transient Analysis

EKOLOGIE EN SYSTEMATIEK. T h is p a p e r n o t to be c i t e d w ith o u t p r i o r r e f e r e n c e to th e a u th o r. PRIMARY PRODUCTIVITY.

PURE MATHEMATICS A-LEVEL PAPER 1

Signals & Systems - Chapter 3

( ) ( ) + = ( ) + ( )

Geometric Predicates P r og r a m s need t o t es t r ela t ive p os it ions of p oint s b a s ed on t heir coor d ina t es. S im p le exa m p les ( i

ARC 202L. Not e s : I n s t r u c t o r s : D e J a r n e t t, L i n, O r t e n b e r g, P a n g, P r i t c h a r d - S c h m i t z b e r g e r

Form and content. Iowa Research Online. University of Iowa. Ann A Rahim Khan University of Iowa. Theses and Dissertations

Poisson Arrival Process

Polynomial expansions in the Borel region

Mathematical Statistics. Chapter VIII Sampling Distributions and the Central Limit Theorem

EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions

REFUGEE AND FORCED MIGRATION STUDIES

Use precise language and domain-specific vocabulary to inform about or explain the topic. CCSS.ELA-LITERACY.WHST D

Lecture 21 : Graphene Bandstructure

ECE-314 Fall 2012 Review Questions

On the Existence and uniqueness for solution of system Fractional Differential Equations

by Martin Mendez, UASLP Copyright 2006 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Control Systems. Transient and Steady State Response.

Windowing in FIR Filter Design. Design Summary and Examples

P a g e 3 6 of R e p o r t P B 4 / 0 9

Transcription:

NAME: SOLUIONS EEE W Problm. Cosir sigal os grap is so blo. Sc folloig sigals: -, -, R, r R os rflcio opraio a os sif la opraio b. - - R - Problm. Dscrib folloig sigals i rms of lmar fcios,,r, a comp a. r r r

NAME: EEE ES Problm. 4ps Cosir sigal os grap is so blo. Sc folloig sigals: -, -, R, r R os rflcio opraio a os sif la opraio b. - - R R Problm. 6ps Dscrib folloig sigal i rms of lmar fcios,,r, a comp a. r r /

NAME: EEE ES Problm. 6ps Cosir sigal -rr-. Sc folloig sigals:, -, -, -, - R, r R os rflcio opraio a os sif la opraio b. Problm. 4ps Dscrib folloig sigal i rms of lmar fcios,,r, a comp a.

NAME: SOLUIONS EEE W Problm. Comp covolio * r-r---, -. is i sp Uforal, so is a la vrsio of prssio: r--r---. * -* - r--r--- Problm. Cosir filrs: A. B.. i a grap ir impls rsposs. A. B.. i a grap ir sp rsposs. A. sp B. sp L. Wic filrs ar casal? Jsif A. No casal, - is o zro. B. No casal, is ozro i -,. 4. Wic filrs ar sabl? Jsif A. Sabl sic i is IR. is absoll smmabl. Σ 4 < if. B. Usabl sic ivrgig implis a ivrgs. sp

NAME: SOLUIONS EEE ES W. 9/7/6, clos boos&os Problm. Comp covolio * - -,. r r Problm. Cosir filrs: A. B.. i a grap ir impls rsposs. A. B.. Wic filrs ar casal? Jsif A. Casal bcas, for all <. B. Casal bcas, for all <.. Wic filrs ar sabl? Jsif A. Sabl bcas < B. Usabl bcas lim, ic ivrgs.

EEE W NAME: Problm : L b prioic sigal so i figr blo sqar av i offs. Comp cofficis a of orir sris pasio of. -.5 L b saar sqar av i abls, i 4,.,.5.5 S{ }.5.5 S{ 4 si }.5S{} for for No: S{} for, a oris. Also, prssio for ca b frr simplifi o si.5 Problm : L X b orir rasform of. i X a X. Usig fiiio of orir rasform: X X X No: I is cas, i is also possibl o comp orir rasform of : { } { } { } R{ } R{ 4 } R 4

NAME: SOLUIONS EEE ES //6,, clos boos a os, rasform abls allo Problm : L b prioic saoo av sigal so i figr blo. Comp cofficis a of orir sris pasio of. {} { } {} > < Dirc compaio :, S S S S S S S a No: Parsval s orm sas > < a. Appli o or cas: > < 4 4. Compl compaio o riv ll-o formla for sris /.

NAME SOLUIONS EEE W 4 /8/6 Problm : Cosir filr i impls rspos.. i rasfr fcio. Sc Bo Plo. i orir rasform of op 4. i orir rasform of op si 5. i op a sig covolio, a b aig ivrs orir rasform of or asr o par. 6. i op si a sig covolio, a b aig ivrs orir rasform of or asr o par 4. } { * 5. * } { }* {si } { 4., } {. s aac plos a,.. X X

si cos par R } { si cos Imagiar par si * 6. 4 4 al m m R R R I I

NAME SOLUIONS EEE s 4 CLOSED BOOK & NOES. RANSORM ABLES ALLOWED. Problm : Cosir filr i impls rspos..p i rasfr fcio.p i orir rasform of op si.p i op... 6 6 X X. { } R R R 45 si a si si AL.: 45 si 8 a cos 6 6 6 o o o Problm : Cosir filr i impls rspos si..p i rasfr fcio.p i orir rasform of op si.p i op... X X. { } { }

EEE W 5 NAME: SOLUIONS Problm : i largs samplig irval s o allo prfc rcosrcio of sigals: NOE: * os covolio of a si si. * si si si si * { } { }{ } ma. frq. ra/s, Nqis frq. ra/s, mi.samplig irval si. si si si * { r r r } s ma { }*{ } ma. frq. ra/s, Nqis frq. 4 ra/s, mi. samplig irval si. *si si si * si { si } { } ma. frq. ra/s, Nqis frq. ra/s, mi.samplig irval s ma Nq. i.., zro or a cosa fcio ca b sampl i arbiraril larg samplig irvals Nq s ma. Problm : or a samplig procss i ra ms, a is coff frqc of ial lo-pass filr for rcosrcio? filr coff frqc sol b sam as maimm allo sigal frqc: s ms ma ma ra/s 5 z. s Nq.

EEE ES 5 NAME: SOLUIONS CLOSED BOOK & NOES. RANSORM ABLES ALLOWED. Problm : i largs samplig irval s o allo prfc rcosrcio of sigals: NOE: * os covolio of a si. *cos. si si *cos { } { cos} { }{ } ma. frq. ra/s, Nqis frq. ra/s, mi.samplig irval s ma Nq si. cos. si si cos * cos { } { } { }*{ } ma. frq. ra/s, Nqis frq. 6 ra/s, mi.samplig irval s ma Nq Problm : frqc spcrm of a sigal is so i figr blo. Drmi samplig ra a coff frqc of ial lo-pass filr for rcosrcio. X 4 z maimm frqc i sigal is approimal 4 z, or, 4 ra/s. rfor maimm samplig im, o allo rcosrcio, is /8 /8 s. /f ma ial lo-pass filr coff is sam as maimm frqc i sigal, c 8 ra/s 4 z.

EEE W 6 NAME: Problm : Cosir casal filr scrib b iffrc qaio 4. Drmi rasfr fcio. Comp rspos of filr o. Comp sa sa rspos o 4. Comp sa sa rspos o si / 6 - Problm : Cosir coios im casal filr i rasfr fcio s s s Comp rspos of filr o Problm : Cosir iscr im sabl filr i rasfr fcio z z z.9 z. Comp rspos of filr o

EEE W 6 NAME: SOLUIONS_ Problm : Cosir casal filr scrib b iffrc qaio 4. Drmi rasfr fcio. Comp rspos of filr o. Comp sa sa rspos o 4. Comp sa sa rspos o si / 6 - No: or a sabl ssm if covrgs o a prioic sigal _s, covrgs o a prioic sigal a is ssm rspos o _s. is ca b comp sig orir or. A sfl simplificaio is: o o o cos cos or Coios im: o o o o Ωo Ωo Ωo Ωo Ωo or Discr im: cos Ωo cos Ωo r, s, z ar coios a iscr ssm rasfr fcios, rspcivl... zz zz zz zz zz 4 zz zz 9 zz 4 4 zz zz 4 zz 4 zz zz 4 9 zz zz ZZ 4 9 zz 4 zz 9 zz 4 zz 4 9 4 4 9..47.48. filr is sabl sic pol /4 as magi lss a o. c, sa-sa rspos is ll-fi. or cosa sa-sa, zz cos 4 9 4. ssss Ω si Ω Ω, Ω 6, ssss si aa si 6.4 si 6 6 9oo cos 6 4 si 6 cos 6 4 Problm : Cosir coios im casal filr i rasfr fcio s s s Comp rspos of filr o ss ss ss ssss ss LL ss / ss / ss / /

Problm : Cosir iscr im casal filr i rasfr fcio z z z.9 z. Comp rspos of filr o zz zz zz zz ZZ zz. zz.9. zz. zz zz.9zz. 5.9 5. 5zz ZZ zz.9 5zz zz. 4 5. 4.4 5.

EEE ES 6 NAME: SOLUIONS CLOSED BOOK & NOES. RANSORM ABLES ALLOWED. 6 Problm : or coios im ssm i rasfr fcio s s s s. i ROC, assmig a ssm is sabl.. i sa-sa rspos o.. i ampli of sa-sa op for a sisoi ip of frqc ra/s si.. pols ar a,. Sic ssm is sabl ROC icls -ais a s o ars pol lf a rig. So, ROC R s > - s. ss ss s s s s. ss si Ampl.. 45 s s 4 s Problm : or iscr im ssm i rasfr fcio z z.5 z.5. i ROC, assmig a ssm is sabl.. i ROC, assmig a ssm is casal. or sabl z, comp rspos o a sp.. pols ar a.5,.5. Sic ssm is sabl ROC icls i circl a s o ars pol. So, ROC {z : z >.5}.. ROC of a casal rasfr fcio s from largs i magi pol, o ifii. So, ROC {z : z >.5}. z Z{ } z z. z z X z z.5 z.5 z Z { z}.5.5.5.5 z.5.5.5 z.5 4.5.5 z