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43 views8 pagesAE December 2019 16 Elec B1
Uploaded by
revelation991Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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/ 8
National Exams December 2019
16-Elec-B1,
igital Signal Processing
3 hours duration
1. If doubt exists as to the interpretation of any question, the
candidate is urged to submit with the answer paper, a clear statement
of any assumptions made.
2. This is a Closed Book exam.
Candidates may use one of two calculators, the Casio or Sharp
approved models. They are also entitled to one aid sheet with tables &
formulas written both sides. No textbook excerpts or examples solved.
3. FIVE (5) questions constitute a complete exam:
Clearly indicate your choice of any five of the six questions given
otherwise the first five answers found will be considered your pick.
4. All questions are worth 12 points.
See below for a detailed breakdown of the marking.
Marking Scheme
1. (a) 3, (b) 3, (c) 3, (d) 3, total = 12
(a) 3, (b) 3, (c) 3, (a) 3, total = 12.
(a) 3, (b) 6, (c) 3, total = 12
(a) 3, (6)4, (€) 5, total = 12
(2) 2, (6) 2, (€) 2, @) 2, (€) 2, (9 2, total = 12
(a) 4, (b) 4, (c) 4, total = 12
‘The number beside each part above indicates the points that partis worth
Page 1 of 8 16-Elec-B1, December 2019 Page 2 of 8
1. Consider the representation of the process of sampling followed by reconstruction shown in
the figure below.
s(t)= > 8(t-nT)
|}—>-
x(t) xO
Assume that the input signal is
xe(t) = 2cos(L00nt — n/4) + cos(300nt + 1/3), —o 50,
B = 40.5842(A — 21) + 0.07886(A— 21), 21<.A<50, where A= —20logiod,
0.0, A<21
m= =" here Aw is the transition band width in the design specifications.
2.28540
(a) What is the highest value of the tolerance 6 that can be used to meet the
specifications? Explain.
(b) What is the corresponding value of B?
(©) What is the highest value of Aw to be used to meet the specifications? Explain
(@) Find the length of the shortest impulse response hfn], ie. least number of filter
coefficients, that will satisfy the specifications above. Explain.
(e) What is the delay introduced by the resulting filter in number of samples?
(f) Provide the ideal impulse response hg[n] to which the Kaiser window should be
applied to obtain h[n]. 16-Elec-B1, December 2019 Page 5 of 8
6 (a)
b)
©
During the design of an IR digital filter using the bilinear transformation, the s-plane
poles obtained in the design of the Butterworth analogue filter prototype based on
\nGayP
py = 0.710032 ps = 0.71e/9/32 py = 0.7 10/70/82
Pz = 0.71 6/90/12 Po = 0710/18/12 Pro = 0.71¢/39"/12
Py = O.71e/58/2 Py = O.71e/ 138/12 Pus = O.71e/22"/12
py = 0.710792 pa = 0.71¢/18H/12 Daz = 0.710)29H/32
(Which poles would you choose to form the analogue filter response H(3)?
Explain why.
(ii) How would you then find the digital filter response H(z) from H(3)?
You are to design a lowpass IR digital filter using the impulse invariance
transformation. During the design of the Butterworth analogue filter prototype you
solve a system of two equations determined by the desired frequency response for the
pass-band comer frequency and the stop-band corner frequency.
‘The values resulting for the Butterworth analogue filter parameters are:
0, =0.815, and N =5.305
(@_ What should your next step be?
ii) If you were to exactly meet the desired frequency response specifications for the
passband comer frequency or the stopband comer frequency, which one of the two
would you choose to meet? Explain why.
‘There are four types of linear-phase FIR digital filters. Not all of them are able to
accommodate all types of frequency selective filters; Jowpass, highpass, bandpass and
bandstop.
In the table below mark with an X those implementations that are not possible to occur
and justify why not based on the z-plane forced zero locations of H(2) for each filter
type.
FIR Filter Lowpass Highpass Bandpass Bandstop
Type
Type II
‘Type Il
| Type IV 16-Elec-B1, December 2019
Page 6 of 8
Additional Information
(Not all of this information is necessarily required today!)
DTFT Synthesis Equation
1%
l= 3 Jxce)ei"a0
Parseval’s Theorem
E= Eps? = alae | deo
DTFT Analysis Equation
x(e®) = Yatme
N-point DFT
a 20
XU) = xin] wi, Wy = eT
ms
Ztransform of a sequence xn]
X@) = Yale
‘Sinusoidal response of LT! systems, real h[n]
yln] = |H(e!)| cos (wun + xH(el**))
‘SOME 2-TRANSFORM PROPERTIES
Property Section
Number Reference Sequence ‘Transform ROC
x(a x@ Ry
alm] Xi(z) Rx
alm) Xa) Rey
1 341 amla] +dmli] aXy(z) + bXe(z) Contains Re, Ry
2 342 xln—nol 2MX(e) Rx, except for the possible
‘addition or deletion of
the origin or oo
3 343 hate) X(/en) Ieol Rx
344 asta] =. Ry
5 345 xt] x Ry
6 Retxla}) FEE) +X*(2)] Contains Ry
7 intx(n}) phe = x+(z*)] Contains Ry
8 3. xn] Xta/st) 1/Re
9 3470 mba) XX Contains Rey Rey 16-Elec-B1, December 2019
Page 7 of 8
Additional Information (cont'd)
Geometric Sum
No
Geometric Series
Properties of the Discrete Fourier Transform
Finite-Length Sequence (Length N)
N-point DFT (Length N)
1. xn]
2. xi[n], 22[n]
3. axfn] + bln]
4. X[nj
5. x{(2—m))w]
6 )
We
wa
1. Soa oprle mw]
m0
8. xu fn}xaln]
9. x*[n)
10. x*[((-n))w]
1. Retx{nl}
12, jFmix{n})
13. xepln) = jn] +27)
14. Xophr] = Fb] - 1-2) w)}
Properties 15-17 apply only when x[7] is real
15, Symmetry properties
16. xeplr] = 3(xln] +a1(—n)) WI}
17. Xopln] = 3(xla] — a{((—n))w})
XI
AUK, Xalk)
aX(i) + bX2{K]
Nx{(-%)n]
wh XU
X((k-O)]
Al Xo)
Ne
HR O*AK- Ov]
t
(4)
x1)
Xeolk] = }1X(CO)w] + X1(-H) wd}
XoplA] = FLAT] ~ X71) wd
Ret X[K)
JIm{X [KD
AIK = X(-#)n}
Re X[K) = Re{ X(—K)w}}
Fn X{k) = —ImiX[(—H)w)
[XU = |AU(-)w Il
<(XTK)) = ~11(-W)) I
Ret X (KI)
jIm{ X[K) 16-Elec-B1, December 2019 Page 8 of 8
SOME COMMON z-TRANSFORM PAIRS
Sequence ‘Transform ROC
1. dln] Az
2. uln] I2dl>1
3. -u[-n —1] ld<1
4, 8{n— mi All z except 0 (if m > 0)
or co (ifm < 0)
5. auf] |2l > lal
6. -a"u[-n—1] lel < lal
7. na" u(r] lel > lal
8. —na"ul-n — 1] lel < lal
9. [cos wor]uln] ld>
. [singlet 1
10. [sin won} ctemsre =
» 1 = [reoso}e~ .
11. [r" cos won]ufn] To Presoolet re? Idl>r
h [rsin wolz! :
12. fr" sinwonluln) TPosodeteree RP?
lel > 0
a", O