6.02 spring 2010: quiz 1 name: ii./hgas -...

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6.02 Spring 2010: Quiz 1 NAME: Sa II./h gAS 6.02 Digital Communication Systems-Spring 2010 Quiz 1 - 7:30-9:30pm (Two Hours) Wednesday, March 4th, 2010 Check your section Section Time Room Rec. Instr. 0 1 10-11 13-5101 Lizhong Zheng 0 2 11-12 13-5101 Lizhong Zheng 0 3 1-2 38-166 Tania Khanna D 4 2-3 38-166 Tania Khanna Directions: The exam consists of 5 problems on 14 pages. Please make sure you have all the pages. Enter all your work and your answers directly in the spaces provided on the printed pages of this exam. Please make sure your name is on all sheets. DO IT NOW! All sketches must be adequately labeled. Unless indicated otherwise, answers must be derived or explained in the space provided, not just simply written down. This examination is closed book, but students may use one 8 1/2 x 11 sheet of paper for reference. Calculators may not be used. The cumulative distribution function for a zero-mean unit standard deviation Gaussian (Normal) random variable is: .i>: x 1 ~d' <P(x) == 100 ~e 2 . X. A brief table of values for <D is x <D(x) -0.5 0.31 -l.0 0.16 -l.5 0.067 -2.0 0.023 -3.0 0.0013 -4.0 0.000032 Please leave the rest of this page blank for use by the graders: I Problem I No. of points I Score I Grader I 1 25 2 31 3 16 4 14 5 14 Total 100 1

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  • 6.02 Spring 2010: Quiz 1 NAME: Sa II./h gAS6.02 Digital Communication Systems-Spring 2010

    Quiz 1 - 7:30-9:30pm (Two Hours)Wednesday, March 4th, 2010

    Check your section Section Time Room Rec. Instr.0 1 10-11 13-5101 Lizhong Zheng0 2 11-12 13-5101 Lizhong Zheng0 3 1-2 38-166 Tania KhannaD 4 2-3 38-166 Tania Khanna

    Directions: The exam consists of 5 problems on 14 pages. Please make sure you have all the pages.Enter all your work and your answers directly in the spaces provided on the printed pagesof this exam. Please make sure your name is on all sheets. DO IT NOW! All sketches mustbe adequately labeled. Unless indicated otherwise, answers must be derived or explained in thespace provided, not just simply written down. This examination is closed book, but students mayuse one 8 1/2 x 11 sheet of paper for reference. Calculators may not be used.

    The cumulative distribution function for a zero-mean unit standard deviation Gaussian (Normal)random variable is:

    .i>:

    x 1 ~d'

  • 6.02 Spring 2010: Quiz 1 NAME: ?5 Q (It .sProblem 1 (25 points)

    The following parts of this question refer to a channel whose input is a transmitted sequence ofvoltage samples denoted X, and whose output is a received sequence of voltage samples, Y. Thenth sample values of the X and Y sequences are denoted x[n] and y[n] respectively.lA. (8 points) Suppose a linear time-invariant channel has a unit sample response given by

    h[n] =~ n = 0,1,2h[n] =0 otherwise.

    If the input to the channel is

    x[n] -~ n = 2,3,42x[n] = 0, otherwise

    (1)

    (2)

    please determine the maximum value of the output of the channel, and please determine the sampleindex where that maximum occurs.

    hCjJLX('ll

    ~z..

    2...> 4- n

    ~" Y[J\] M~

    max., y[nJ ~ -----~--

    Sample index, n, of maximum = -----~-----

    y C

  • 6.02 Spring 2010: Quiz 1 NAME: 5 0 { ft5IB. (8 points) Suppose the receiver of a noisy channel receives a voltage sample equal to 0.9 +noise volts when the transmitter is sending a ' l ' bit, and a sample equal to 0.1 + noise volts whenthe transmitter is sending a '0' bit (so no intersymbol effects). If the threshold voltage is 0.5 volts,the noise is Gaussian with standard deviation a, and the probability of a bit error is 0.023, pleasedetermine a numerical value for a (note there is a cumulative distribution function table on thefront page of this exam, and you may assume' l' bits and '0' bits are equally likely).

    T f(e-rr-or-/ hl~~\1') 4- i 'P{etror-' ~\t~\01) z:1.. !P (- ~ )+1 (,~'_

  • 6.02 Spring 2010: Quiz 1 NAME: 5AsProblem 2 ( 31 points)

    For this problem, please consider three linear and time-invariant channels, channel one, channeltwo, and channel three. The unit sample response for each ofthese three channels are plotted below.Please use these plots to answer all the parts of this question.

    USR of Channell

    0.51" ..w 0.4·····~ 0.3":g 0.2::> 0.1

    QO[ 1-0.1L-~~-7o~~~~~~~~2~~~~~~~4~~~~~~~-6L-~~~~~~~8~~~~~~~10Sample Number

    USf

  • 6.02 SprinR 7010: Quiz I NAME: 59W------2A. (7 points) Which channel (1, 2, or 3) has the following step response, and what is the value ofmaximum value of the step response?

    Step Response

    ""

    \ Qtge~f~ \)(V\f

    ~_hO:I '-r

  • 6.02 Spring 2010: Quiz 1 NAME: 2B. (8 points) Which channel (1, 2, or 3) produced the pair of transmitted and received samplesin the graph below, and what is the value of voltage sample number 24 (assuming the transmittedsamples have the value of either one volt or zero volts)?

    Tra nsm itted Sa n: pies

    I-..............•-.-.•---•.-.-....•-_.-.--0_.-'._-1 J__.

    I •o 15

    Sample Number

    III

    5 10 20-- .. -- •........•... - !.. -L-----' --.--O.... .•__o-

    25

    r~eceivecJ Sanlples 1_-~~~~~~---,----------- I

    ~ a:'/ J c....te; est- ~f.lPA,~r~ .

    Channel = ~ _

    15 20Sample Number

    ( Ft. r-st r..t.~ \ ~S\ J ~tdf s\ o....t- , i!I:! ~~. J I,) ~~ if (;1

    e t va.J J#\ttIJ if\t ff.t ,,~. i~)

    ::L h [tV)) k [ltr_~J z: ~ [~] tot hC'1- \t hey "*~t:)) ~ItL,Ctt]·v

    ~

    6 - O. L -t Ol¥t~~2..

    o 5

    y[24] = 011..7;y [~.••.J

    25

  • 6.02 Spring 2010: Quiz 1 NAME: 59 [,,~2C.(8 points) Which channel (1, 2, or 3) produced the eye diagram below (based on 4 samples perbit), and how wide open is the eye at its widest (lowest voltage associated with a transmitted' l 'bit - highest voltage associated with a transmitted '0' bit)?

    ,) r~

    //

    \ t t ~VIi

    1Ii,

    () t \ f//: "*/it .-~,~_-.._ ....~..

  • 6.02 Spring 2010: Quiz 1 NAME: :; Q l122D.(8 points) Suppose an approximate deconvolver was generated for each of channels, using thedifference equation given by

    win] ~ •.'0' : • '" (Yniln + I] + noiseln + 1]- f him +1]wln - m]) (5)where Ynf[n] = I:~~:~h[m]x[n - m] is the noise-free output of the particular channel beingdeconvolved. When the deconvolver was used for two of the channels, limn->oo Iw[n,ll approachedinfinity, but when the deconvolver was used with the remaining channel, it worked reasonably well.For which channel did the approximate deconvolver work well, (1, 2, or 3)7 You must justify youranswer to receive full credit. /'" I. l

    Channel for which the approximate deconvolver worked well = --L._/d..gj\i\e \ (,

    Justification (Needed for full credit):

    ~ ( LJ+..~t3] +.)~~[If:], __Cl.to~~ i ~ I.L~) + I-1C9Jf t- 1, ~.1 Y ~11 .." •.....M lGj •• _~~--~""_"" •••••••••• ~;.=x"'''''•.•''k-•.•~',.,.'-.~-'''''.,-.' __ $

  • 6.02 Spring 2010: Quiz I NAME: 5 Q (ASProblem 3 (16 points)

    Suppose a channel has both noise and intersymbol interference, and further suppose the voltageat the receiver is:

    8.0 + noise volts when the transmitter sends a '1' bit preceded by a ' l' bit,6.0 + noise volts when the transmitter sends all' bit preceded by a '0' bit,2.0 + noise volts when the transmitter sends a a '0' bit preceded by a '1' bit,0.0 + noise volts when the transmitter sends a '0' bit preceded by a '0' bit,In answering the following parts, please assume the receiver uses 4.0 volts as the threshold

    for deciding the bit value.3A.(8 points) Suppose noise is Gaussian with standard deviation a = 1, and all bit patterns areequally likely. Using the the tables at the beginning of the exam, please determine the probability

    Ofablle~!(2_4-J) + i;: lt-l{'--L» t i; ~&-)1t!~:: -t ~ (~4-) 1'" {: i (-2.) =- ± t!- ('IO·~

    Probability of a bit error = Q~.J?JLSlb ~'+ i@. (n.'3)3B.(8 points) Again suppose noise is Gaussian with standard deviation a = 1, and suppose thatfor a particular set of transmitted data, which we will refer to as checkerboard data, there is anincreased probability of unequal contiguous bits. That is, for checkerboard data, the probability oftwo '0' bits in a row is ~, the probability of two' l' bits in a row is ~, the probability of a '1' bitpreceded by a '0' bit is 1and the probability of a '0' bit preceded by a '1' bit is ~. Approximatelywhat is the ratio of the probability of bit error for the checkerboard case to the probability of biterror for the uniform (all bit patterns equally likely) case. Be sure to justify your answer.

    PI"'!,;:;::::'~':::::'.dN ~j'£Justification: ,

    .-'":>

    -s- 0 (z c. O. 0 ~>)- 1- ~ ~ ;./'iP c7t.l + 3 :i"y')

    -' . "'1'-:"'''''''''' -""'-" "' '0 -- •?'" . 1\ \. ';~'!i""1'_ - 14•••_ . .-1 . ..., .', ,.,', ~ '''. ,.""",Y1.. !~lt -t ~ ~k-?)~ ~~

    9

  • 6.02 Spring 2010: Quiz 1 NAME: 2st (1t5Problem 4 ( 14 points)

    In this problem you will be answering questions about a linear time-invariant channel charac-terized by its response to a five-sample pulse, denoted P5[nJ.

    Five Sample Pulse

    1.0 I· r0.81

    I~ 0.61-2I:g 0.4

    ::~J0

    1

    I~-.--~ I • • • • .• •• 0 ---Ii " • __ a •

    10Sample Number

    15 20

    1.0····

    Five Sample Pulse Reponse p,),;,.

    I

    t...J.......•.•U.__J.·

    0.8

    'J.> 0.6 ,..t»2g 0.4'

    0.2 ..

    0.0J_" 1. • .000. _ _____ ..••... ___.----$--- ....••.... - ..•

    o 10Sample Number

    15 20

    10

  • 6.02 Spring 2010: Quiz 1 NAME: 5QL~ .24A.(5 points) Suppose the input to the channel is as plotted below. Plot the output of the channelon the axes provided beneath the input.

    Channel Input, rOll:

    •• 0.6C>.5§ 0.4'

    1.0 .- • • • II 'I IIII • ., fl •

    0.8

    0.2 .

    0.0

    o 10Sample Number

    Channel Output

    ~ 0.6···Bg 0.4'

    1.0

    0.8·"

    0.2

    10Sample Number

    xthJy [~J

    -_.

    IS

    ,t

    15

    --p. (i l~e[tt)TPs ("J

    P IJ Ise en -~[P-t2]5

    ~>9--l~-

    :;:

    11

    20

    20

    \~J

  • 6.02 Sp,ing 2010 .. Quiz 1 NAME:;50 [_I\~!.>=. _4B.(9 points) The unit sample response,h[n], can be related to the step response, s[n] by the for-mula h[n] = s[n]-s[n-l], Please derive a similar formula for the relation between the five-samplepulse response, P5[n], and the unit sample response (an infinite series is an acceptable form for the ,answer). il It) II' \$ 0 [A-'I ] -' , . i

    l (11 - r '-""' , ~1"'-'~~""'", u,«Y'v 5 e ! ',I lJ \ " =

  • 6.02 Sp,ing 20JO.· Quiz 1 NAME: 501---+-n---#-,cs~----Problem 5 ( 14 points)

    Suppose the perfect deconvolving difference equation for a linear time-invariant channel is

    1 1w[n] = -y[n] - - (2w[n - 1] + w[n - 2]) .2 2

    That is, if x[n] is the input to the channel and y[nJ is the channel output, in the noise-free case,w[nJ will be exactly x[nJ.SA.(S points) On the axes below, plot the unit sample response of the linear time-invariant channel.

    6

    Step Response

    .5

    4'"

    .,0>s 3··~

    2

    1

    o ~..~...I.~-~----#----~-.-~.-~-.. ~ ...Sample Number

    l \J [t\J'-v\1(9)

    o 4 106 8

    -t ~ \N o-\]+~W [A -1.J =-y[~~[O hc?-]

    13

  • 6.02 Spring 2010: Quiz 1 NAME: 5~tl\1S5B.(9 points) Suppose a noise spike is added to the output of the channel before deconvolving, asIn

    w[n] = ~(y[n] + o[n]) - ~ (2w[n - 1] + w[n - 2]) ,where o[n] is the unit sample (6['17,] = 1 when n = 0 and zero otherwise). Suppose the input tothe channel, x[n], is as plotted below. Please determine the first three values of the deconvolveroutput, w[O], w[l], and w[2]. (think before you compute, or you will spend 20 minutes doingalgebra. Recall that if not for the noise spike, you would have a perfect deconvolver).

    2

    1 1· .(\)

    ..... ID;'3 0g I .1 "-1

    -2-

    1 2Sample Number

    3 4 5-1 o

    '·f \w [0] = ..y