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Unformatted text preview: EE 351 gin/{42, FINAL EXAM
FALL 2001
Name:
NOTE: 1. Exam is comprised of 5 problems, each with multiple parts. 2. Point values are given in parantheses for each part. 3. You are allowed three 8 1/2 X 11 sheet of paper and a calculator. 4. You MUST show all work and write your answers in the spaces provided. 5. Z—tmnsform tables are provided at the end of the exam sheet. 6. You have 2 hours to complete the exam. DO NOT WRITE BELOW THIS LINE: Problem 1 (20. Points) :
Problem 2 (20 Points) : __
Problem 3 (20 Points) :
Problem 4 (15 Points) : Problem 5 (25 Point TOTAL (100 P mts) ; ________ Problem 1,) All parts below are independent of each other. (20 Points)
’, ‘ a) Useiheﬂéﬂnitionféfthe Z’traﬁ$f§rih tidkéélculatevthe:Z—transforﬁl of m[n] whereéV L
"‘ ’ ; A (3)n+1, n : 0,2,456; 8,... ' ‘ 0 , _ else V Make sure that you include ROC‘in yoﬁr answer. (10".Poihts) ' b) By using the Ztransform table and properties (NOT the deﬁnition), ﬁnd the Ztransform of
’ ’ y[n] where: ““1 ¥ 0%  2)  <2)"  uln — 21 + (4)"  u[—n .. 31 Make sure that you include ROC in your answer. (10 Points) Mn]: ("AMA 1+0an L] + (A) J'Fuﬁhﬂplj
30’] 4 W) (2% w 2347'; (4) ll)mo(((n+2)—U '1' ‘7’ . —Z<
aw, "‘(1)n'UCn3H 2% ,/ Pch _ 7—1—5 . (231 Problem #2 (20 Points): . a) Consider an LTI discrete—time system given below Determine and sketch all possible regions
of convergence (ROG) for H (2) For each ROC, determine Whether the corresponding system 1s: i) causal, anticausal, noncausal and ii) stable, unstable. (12 Points)
111(2) _ 22 — 2.25z + 0.5 __ (z — 2)(z — 0.25)
‘ 23 — 22 + 0.012 + 0.12 _ (z + 0.3)(2 — 0.5)(z — 0.8) [2I708’ 0M 12) <06
Causal Mon CW3 o4 12M 03
A714? Causal Wslat lc 0.5 < I2)<O'Y
Nomadic} Unjlwlala Problem #3 (20 Points): . A discrete—time LTI system is given as: 82 . z
. H(z)_(z_2)(z_4)(z_6) ROC.2<<4
a) Find h[n]. (14 Points)
A
H (20 .. 5' , _ 2 b) What would h[n] be if ROG: 4 _< M < 6. (6 Points) 2%,. Problem #4 (15 Points): All parts below are independent of each other. . a) Assume that you have a DSP chip (the Texas Instruments TMS3ZOC‘31) with an unknown
DT system loaded onto it. You attempt to determine this mysterious DT LTI system by
applying a known signal as input and measuring the corresponding output. Speciﬁcally, when
w[n] = nu[n] you determine that the output of the system is y[n] = u[n] — 5[n] Find the impulse response, h[n], of this system. (8 Points) b) The pole—zero map of an LTI discrete—time system is given below. Note that complex conju
gate poles all have a magnitude of 0.9 and are located at +7r/ 4, —7r/ 4, +37r/4, —37r/4 radians and zeros'are located at z = 1, j, —0.95 and —j. Give an approximate sketch of
the magnitude response for this system. Make sure to mark all key frequencies. (7 Points) lmag. [W539 l l l (a 1 Problem #5 (25 Points): . In Europe, the power distribution frequency is typically 50 Hz. Suppose that it is desired
to ﬁlter out 50 Hz and its harmonic 150 Hz from a received signal without disturbing other frequencies much. Your job is, if you accept, to design a digital ﬁlter which can null the 50 Hz
and 150 Hz without affecting other frequencies much. a) Assume that the sampling frequency is 400 Hz. Find poles and zeros of your digital ﬁlter,
show them on a pole—zero map given below and draw the magnitude response approximately.
Make sure to mark all key frequencies. (9 Points) b) Write down the transfer function H (z) for the ﬁlter you designed. You do NOT need to
simplify the equation and your answer might depend on a gain parameter. If DC gain of 1 is
desired, give an equation for the gain parameter ( do NOT solve it expilicitlyl.) (7 Points) We) H (e 2%) (29%) (223%) (2 23 W) @Dc web’ﬁ “(33’ ":3 c) Let’s assume that for some reason we would like to amplify (NOT inﬁnitely, though !) 50 Hz and 150 Hz frequencies, again without affecting other frequencies much instead of nulling
. them. How would you change your design? Explain clearly and plot poles and zeros of your new digital ﬁlter on a pole—zero map given below. Give an approximate sketch of the magnitude response of your new digital ﬁlter. Make sure to mark all key frequencies. Assume that the sampling frequency is same as before, 400 Hz. Note that you do NOT have to write H (z) or ﬁnd‘the gain factor. (6 Points) d) Now, if you want to ﬁlter out 100 Hz and 300 Hz instead of 50 Hz and 150 Hz, how would
you change your design in a simplest way? You are allowed to change anything you want ! “1;? ﬁt me 1W9 o /e ﬁtJM/h% £7
/ . 1 “ TABLE 3 2 PROPERTIES OF THE ZTFlANSFOFlM Propert‘fi . Tinie Domain I zDomain ‘ T V' ROC. V , .. I I ' .. WW.
Notation . x(n) X (z) 1 ROC: r2 < 12.! < r; ' .
x10?) X1 (z) " ROC;
J€201) _ , X2(Z) ' ’ , "R002
7 Linearity a1x1(n) + a2x2(n) a1X1(z) + a2X2(Z) At least the intersection of ROC;
' _ p, ' and ROC;
' Time shifting x(n —— k) ~z"‘X (z) That of X (2) except z —— 0 if k > 0
. , ' and z— — 00 if k < 0
Scaling in the zdomain a"x(n) X(a‘1z) _ar2 < 12] < lain
Time teversal x(—n) . X(z"‘) 1' < M < ‘1‘
_ L r r
Conjugation x‘(n) X ‘(z’) RIOC ‘ 2
Real part Re{x(n)] HX (z) + X’(Z‘)l includes ROC
Imaginary part Imlﬂnn ﬁx (2)  X‘(z')] ' Includes ROC '
Diffeiientiation in the nx(n) —z dX(Z) r2 < 17,] < f,
z—domain dz .
Convolution x101) an x201) ‘X1(z)X§(z) At least, the intersection of ROC;
and ROC;
v Correlation r;t1 ,1 (l) — x10) 4 x2( —I) RxJlxz (z)= X1(_z)X2(z‘1) At least, the intersection of ROC of Initial value theorem
Multiplication > Parseval’s relation x1(n)x2(n) If x(n) causal ‘ x(0) _ lim X (2) 2—5” 1
2,”: —éX1(U)X2 Zx1(n)x;(n)1= 2—:7ﬁx1 (v')X;(1/v‘)v'1dv Xﬁz) and X2(z“1) (3) rich: At least rum < Izl < nurzu Signal, x(n.)
sol)
u(n)
a".u(n)
mfuoz)
a"u(in — 1)
na"u(n  1)
(cos won)u(n) ‘
(sin won)u (n) (a cos won)u(n) 9. (a" sin won)u(n) TABLE 3.3 SOME COMMON Z'TRANSFORM FAlBS zTransfqrm, X (z) (1  az“)2
1 — 2'1 cos we 1 22" coswo + 2'2 2'1 sin coo 1  2:"1 Cosmo + 22, 1 — az‘1 cos (Do 1  2az'1 cos coo + azz‘2 az‘l sin we .1 1 — 2:22"1 coswo + azz’2 ROC 'AlIz 'lzl >‘1 ' __ lzl > Ial 7 12! > lal
lzl < Ial
Ill < m
z > 1
121 > 1 IZI > lal lzl > lazl 10. ' 11. 12. fin] 5W _ /
sin  no] u[n]
nu[n] anum’
na”u[n]
d'sin(bn)u[n]
a"cos(bn)u[n]
~uL—n — 1],
a”u[n — 1]
W"  1] a“, a < 1 F(‘z) (2  1)2 Z
24 “I. (2  4): zZ—2az<:osb+a2 , zlzacosbl zz2azcosllv+a2 z
z1
Z az sinb I‘d
az (z  a)2 v Z
za z—l/a ROC Allz z¢0, nozl
lzl>1 {2 > 1 IZI > lal
Izl > I'll '
l2 > Ial
lzl > lal
z[ < 1 Izl < Ial
IZI < la'l la < z; < 11/4 ' ...
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 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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