This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Autumn 2007, ECE 161A, Midterm  Tuesday, November 20, 2005 ECE 161A Midterm II
Tuesday, 20 November, 2007 Name: (Last) (First) _____ (MI)— Problem 1: Problem 2: Problem 3: Problem 4: Problem 5: TOTAL: [100
____—___—____—_—u.—._
Rules: 1. By writing your name above, you certify that the solution is your own work. Cheating ChmPLUM \J in any form (copying another student's work, using unauthorized materials, etc.)
results in an immediate score of zero for the exam, and further disciplinary actions by
the School of Engineering. . The exam is open book and notes. . The exam time is exactly 75 minutes. . No calculators or any other electronic devices. . Write clearly. That is necessary to ensure that you receive partial credit. . Cross out any work you do NOT want to be graded. Wrong answers cause you to loose points. . Partial credit is given ONLY to correct solution techniques, not to correct answers from wrong procedures. . The exam is 12 pages and 5 Problems. Autumn 2007, ECE 161A, Midterm ll Tuesday. November 20, 2005 Problem 1 When an input
x[n] = (5[n] + 0.85[n — 1]— 0.26m — 2] is appiied to a causal linear timeinvariant system, the output is y[n] = 65M]
a) Find the transfer function H(z) of this system. Answer: b) Use the inverse ztransform to find the impulse response h[n] of this system.
Answer: c) Find the difference equation corresponding to this system. Answer:
_ —'2_
CL) Y(2):é ) X(Z): [+0.32 holE 4,,
. 6,1
HQ): 6 —2_: ‘2'
i+52"_.2_2 22+.82.a¢
is H Z” + 5%
i (2) (Z “2) (2+0 Autumn 2007, ECE 161A, Midterm ll Tuesday, November 20, 2005 0 M2): iii); 5% way—Mama) W Y(2:)(M+o,<gi'2f) A m) 133%] +‘ 9 35413 .., 9150123ch m]. Autumn 2007, ECE 161A, Midterm Ii Tuesday, November 20, 2005
Problem 2 Consider a causal filter with real and positive M impulse response h[n]
and transfer function H(z). Let y[n] be the output of this system when the input signal is x[n]= 1:322]. Show that y[n] is increasing with n and limy[n]= H(1). Autumn 2007. ECE 161A, Midterm  Tuesday, November 20, 2005 Problem 3 Consider x[n]= aLmJﬂn], for a such that a<1, and where H is the ﬂoor functionl.
Find the DTFI' of this signal. (Hint: first draw the signal.) Answer: lafl <i (217W Xz(2):X(Z)£—l =___Z—__ midliadﬁa 12317194 23¢
PM?
I 2
E
X43(%)" 22,0L + Zia .
23a ‘33“
1.. ___e__ﬂ_ﬂ.__ 1. if.
X60.) :: X(%) 2:331 “' (aidﬂ ._ 0L ezjﬁru 11f you do not recall the deﬁnition of this function, see the last page of the exam. 5 Autumn 2007, ECE 161A, Midterm [1 Tuesday. November 20, 2005 Autumn 2007, ECE 161A, Midterm  Tuesday, November 20, 2005 Problem 4 a) Draw the impulse response of the system with the following impulse response: b) What is the DTFI'? Answer: Tm om MDHMAE‘ \OMJL ““1300 SDI"RK]  FF“ lad; Fig/i
kzm we ‘W/‘i
H(JL):%£: S<~Q_%E(_L)
by
: nLgrﬂ“Q
K: Autumn 2007, ECE 161A, Midterm  Tuesday, November 20, 2005 Autumn 2007, ECE 161A, Midterm H Tuesday, November 20, 2005 Problem 5 Consider a ﬁlter whose frequency response is of the form
11(9) = 1 4005(29) + 2a sin(SQ). a) What is the length of impulse response of this ﬁlter (notice that this is a non
causal FIR)? ‘5: HM): Answer:
b) Find a such that the energy of impulse response is 11. (Hint: there are more than one such a). Answer: c) Find a such that 1) the energy of impulse response is 11, and 2) the filter has real
coefficients (impulse response). (Hint: there are more than one such a). Answer: d) If the input is x[n]= cos(%n), what is the output of the filter in part ((2)? Answer: Autumn 2007, ECE 161A, Midterm ll Tuesday, November 20, 2005 0L) HUL): I— 4C3) 1J1 +élq 350511
*5le “'9'“ 9331 _ 33‘“
HQ.) :29... a. a e +l w a e a +8.37%“;
J
1‘ T r T 1?.
M135] h [—2] h [.9] M1] 1“ L9] I l ”Em. \mgttl 049% WW ~44 i ‘ I] r Cum Arqu' 5‘“ T.
Hag.» “ng ) A) ~Fbr gym? PicL (1:3 HM): lﬂq_c'3&_§L+2J'§l,p‘3.ﬂ_ H(.Q='_E_): I’Ll‘chzt—jt Ajﬂ‘nig a —9\+J. 3 “4+3 Autumn 2007, ECE 161A, Midterm  Tuesday. November 20, 2005 11 Autumn 2007, ECE 161A, Midterm ll Tuesday, November 20, 2005 f[n]=a"u[n]#~F(z)= , lz>al
za f[n]=—a“u[—n—1]e—£—>F(z)= Z , lzl<lal z—a Some useful Ztransforms: f[n]= a"u[n—1]‘#>F(z)= a. , lz > al z—a fln]=—a"u[n]+—~Z——»F(z)= a , Izl<lal z—a
Z f[n]= all #Fm = “z + , al<lzl<llfa
l—az z—a f[n] a “[n](%F(Q) —1 1 ”n ._. lZ I>l a
— (16
we".:2 1
',a 1 F 9 = + .
Some useful DTFTtransforms: f[n]= a < ( ) l—aem l—ae'lQ fln]= 36m —kN]<—”»F(9)=— 26(Q——) Ion—co k._oo f[n*no]%e‘j""‘aF(9) Energy of a signal Ex = Elﬁnf
Floor Function: [x] = maximum integer smaller than or equal to x 12 ...
View
Full Document
 Spring '10
 Javaidi

Click to edit the document details