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**Unformatted text preview: **L> Problem F.1: jzf
p \/a) (8 pts) Consider the following equation: 7\/§cos (En) — 73in (En) = Acos (u'Jt + 43) Find A, a, and ¢. ,_ ;_ CosCJg: ,\ Sgt)
? :2 are
I 3 leé) 3.771:
(P; a r? e =— ¥ 6 3 Jj 9+3: :N’g—JF'JJ s :4 estate) .1 .L
m ”“9
9% A) (7 pts) Simplify gi {t3u(t + 3)}. 3199 0&th «(— £3 801+?)
$6 or (g 33 + (—95%) 8 63%) ._1 ._2
W (5 pts) Simplify 3,4 :;_5. (Do not touch your calculator when working this part. Show your work to convince your grader that you were able to do this without a calculator.)
\ _+_ {/4 (
~3 * U" ‘5 (”3‘7“) ’1' (Ac? 13> f: 631 41— (j)
’l ”V 3 ~2‘ \
: 3i: —-—J Problem F.2: (a) (4 pts) Fill in the blank with the largest possible number that makes this a true statemet: A large number of biomedical data sets are available at the website bsp.pdx.edu/Data.
The electrocardiogram data was taken using equipment sampling at rate of 500 samples per
second. Assuming that their equipment used ideal continuous-to-discrete converters (they
didn’t, but we’ll pretend they are for this problem), the scientists who took the data could
sample frequencies up to (but not including) _______ Hertz without aliasing. 4g: 900 Mia/QC. % 260%?3 (b) (10 pts) Suppose the output of an ideal continuous-to—discrete converter with a sampling rate
of f3 = 2000 samples/sec is m[n] = 8cos(0.257m — 27r/9). Specify three possible input signals
of the typical cosine form :1:(t) = Acos(wt + (b) that could have given that output. Restrict
your frequencies w to be in the range —70007r rad/sec < w < 70007r rad/sec. k): (%+ 117.).515400 Lalo 1:55:23: 1: @6) Now consider the following piece of MATLAB code: 99 ma
tt = 0:(1/20000):3;
xx = sin(2*pi*800*tt + 1.293); A
soundsc (xx , 5000) ; ((2
(c) (3 pts) Determine the analog frequency (in Hertz) that will be heard. 8 é”,- /% we???) nggs$é°f§w (d) (3 pts) Determine the duration (in seconds) of the ﬁnal played tone. 3 Q/éQ 9W [:2 $0 ‘1
:2- “.5
f . Problem F.3: Consider a zero-DC sawtooth wave with Fourier series coefﬁcients given by a _ j1+k fork¢0
k—
0 fork=0 In all parts of this problem, assume the sawtooth wave has a fundamental period of To =
0.001 seconds. (a) (7 pts) Suppose our sawtooth wave z(t) is input to a cascade of an continuous-time—to-discrete-
time converter (C-to-D) and an ideal discrete-time—to-continuous-time (D-to-C) converter,
each running at a sampling frequency of f3 samples per second: :1:(t) Ideal Ideal y(t)
C-to-D D-to-C
Converter Converter f5 =1/T5 Is there an famin such that y(t) = $(t) if we sample at a rate f3 > fmin? If so, give the
smallest possible fsmgn (in Hertz). If no such fsmin exists, explain why. MWW (b) (13 pts) So you don’t have to go ﬂipping back to the previous page, here’s the Fourier series
coeﬂ'icients of that sawtooth again: a _ 1.71“: forkaéO
k—
0 fork: Recall that our the sawtooth wave :c(t) has a fundamental period of To = 0.001 seconds. In this part, suppose our sawtooth wave :c(t) is input to a continuous-time linear time-invariant
system with an impulse response given by
sin(Bt) h(t)=A 7rt . Specify A and all possible B such that the system output v(t) is v(t) = a:(t) * h(t) = 20 sin(20007rt), (I’m using 1) here instead of 3/ so you won’t get this things mixed up with the 3; used in part (20-) LG) .-—-—— Problem F.4: Choose your tools carefully; use the particular techniques that that are best suited
to each individual part.
Suppose a continuous-time system has the impulse response h(t) = 5e-4°°"‘u(t) (a) (10 pts) Compute the convolution y(t) = [4 + 5cos(4007rt — 27r/3)] * h(t) .£ 00
LLB-tr"? e, Lmbolek) Q Ca‘, [5pc Tré— 1.3—) _.... 5:: JG) Ave/:52, (b) (10 pts) Compute the convolution y(t) = [36(t — 7)] * h(t). 3 mm») Problem F.5: A cascade of two discrete-time systems is depicted by the following block diagram: LTI
System #1 h1[n], H1(z) LTI
System #2 h2[ﬂ], H2(Z) yin] System #1 is deﬁned by the difference equation
y[n] = 0.9y[n — 1] + 0.2y[n —- 2] + 55r[n -— 4]
and System #2 has the system function H2[n] = 1 — 0.42‘1. W5 pts) Find H1(z), the system function of System #1. —- 0- 959+ 0. l$2+ (9% (b) (15 pts) Find the the equivalent impulse response of the full cascade, i.e., find h[n] such that
y[n] = h[n] * a:[n]. (Remember Aaron promised that you wouldn’t have to do partial fraction
expansions on the ﬁnal. So if you ﬁnd yourself doing a partial fraction, guess what! You’re
doing it the wrong way.) It’s winter time, so let’s all sing along! “Factoring through the snow, in a one horse open
sleigh, over the denominator we go, factoring all the way! Ho ho ho!” Lila“: Ciro-hit) (0- 5;: + 0-111???) ’W%- Problem E7: 8( w )+
'tt W’ °
as were (Y Q.) 1“”) cos(2007rt) In the above modulation/ﬁltering system, assume that the input signal a:(t) has a bandlimited Fourier transform X (jw) as depicted in below.
XUW)
A — 7T 50w w (a) First give the general equation that expresses W(jw), the Fourier transform of
w(t) = x(t)[2 cos(507rt) + cos(2001rt)], in terms of X(jw). Wounds («r—<70) riot (clw We) + «— runway 17 x() {-0039 (b) Now carefully plot the Fourier transform W (jw) for the speciﬁc input :1:(t) whose Fourier transform X (jw) is given above in - Note that the negative frequency portion of the
Fourier transform X (jw) is shaded. Mark the corresponding region or regions in your plot of
W(jw), and be sure to carefully label both amplitudes and frequencies. 011’ — Orr
(c) The frequency response of the ﬁlter is 1 |w| 3 50K
0 [col > 50w HUw) ={ Plot the Fourier transform Y(jw) below for the X ( jw) given in above. Be sure to carefully label both amplitudes and frequencies and be sure to s ta e tie region corresponding
to the negative frequencies of the input. Problem F.8: (2 pts each) Some screen capture segments of the continuous-time convolution GUI cconvdemo
are shown below on the next page. Match the inputs on the left with the outputs on the right by
writing the number of the corresponding output by the graph of the corresponding input. Note
that while the horzizontal axes of the outputs all have the same spacing, the spacing of the inputs
changes, so be careful to read the numbering on the axes of the input graphs. You do not need to provide any explanations. This is not an equation crunching sort of
problem; use your intuitive “feel” for convolution. V my" 1mm» @% [LL QM V [hp—Munoaponu ‘ 7'2 32 @LM "LJ #2 """wwlnﬂl ' _._, mMaReuicnu I I
i 2U]: u; 2.
(c)L LL WW._,..WT_ 12.”; 2111: 1’51 MERMunsoM" A i » _-_- _ _ (e) -p.5 o 115 177721':s;r_m-§W.'ui_‘_ § 11115: #5 , Problem F.9: (a) (1 pt each) For each of the following impulse responses, write the name of the corresponding
pole-zero plot from the Page of Pole—Zero Plots that appears just after Question 10 in one of the corners of the impulse response plot. (You need not give any explanations; feel free to
use your intuition!) i051
:0.I . I.[ .I .T .f.7‘.' will“ LA, 57 ,IO ,15 ”20,, 25 Hi urHt...‘
im‘ww I or ..._s "10' -ls; 2.0 "2;: [[[llllllll Illltlm'" o ,_ as V ao'___1:_izos..,,zai ( ) (2 pts) F 1nd the inverse z-transform of H (z) 1 + :24 . »-\ m. 0C?
L '1 :2; / < i I?) \L i? 2
OQCnl 4.0,; => :QPuCA/j. (2 f, This is the Page of Pole-Zero Plots that you will use to answer Problems 9 and 10. ..... ...... i ...... o ..... ”L; Boo U no {-1 116 Problem F.10: (a) (1 pt each) For each of the following frequency responses (only magnitudes are shown), write
the name of the corresponding pole-zero plot from the Page of Pole-Zero Plots that appears
just after Question 10 in one of the corners of the frequency response plot. To start you
off, I’ll give you one of the answers: the upperleftmost panel is “bacon.” (You need not
give any explanations; feel free to use your intuition!) (b) (2 pts) Suppose that for a particular discrete-time LTI system, the phase the frequency
response evaluated at —7r/4 is 0.6, i.e. AH(—j1r/4) = 0.6. If Mn] is real-valued, what is
[H (177/4), i.e. what is the phase of the frequency respone evaluated at 1r/4? (c) (2 pts) [University of Georgia Question] A continuous-time LTI highpass ﬁlter passes what
kind of frequencies? (Circle only one) (a) High frequencies (b) Low frequencies (c) None of the above ...

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