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NAME: 2013Digital Signal Processing I Exam 3 Fall 2013Session 41
25 Nov. 2013
Cover Sheet
Test Duration: 60 minutes.Open Book but Closed Notes. One 8.5 x
11 crib sheet allowed
Calculators NOT allowed.This test contains two problems.
All work should be done on the blank pages provided.Your answer
to each part of the exam should be clearly labeled.
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Digital Signal Processing I Exam 3 Fall 2013
Problem 1. [50 pts] For all parts of this problem, let x[n] be a
discrete-time rectangularpulse of length L = 8 and h[n] be a
discrete-time rectangular pulse of length M = 4 asdefined
below:
x[n] = u[n] u[n 8] h[n] = u[n] u[n 4]
Also, for all parts of this problem, the reconstructed spectrum
is computed according toEquation 1 below:
Yr() =N1k=0
YN(k)sin
[N2
( 2k
N
)]
N sin[1
2
( 2k
N
)]ejN12 ( 2pikN ) (1)
(a)-(i) With XN(k) computed as the 16-pt DFT of x[n] and HN(k)
computed as the 16-ptDFT of h[n], the product YN(k) = XN(k)HN(k) is
formed. Determine the N = 16values of the 16-pt Inverse DFT of
YN(k) = XN (k)HN(k).
(a)-(ii) The product sequence, YN(k) = XN(k)HN(k), formed as
directly above with N = 16,is used in Eqn (1). Write a closed-form
expression for the reconstructed spectrumYr().
(b)-(i) With XN(k) computed as the 12-pt DFT of x[n] and HN(k)
computed as the 12-ptDFT of h[n], the product YN(k) = XN(k)HN(k) is
formed. Determine the N = 12values of the 12-pt Inverse DFT of
YN(k) = XN (k)HN(k).
(b)-(ii) The product sequence YN(k) = XN(k)HN(k), formed as
directly above with N = 12, isused in Eqn (1). Write a closed-form
expression for the reconstructed spectrum Yr().
(c)-(i) With XN(k) computed as the 8-pt DFT of x[n] and HN(k)
computed as the 8-pt DFTof h[n], the product YN(k) = XN(k)HN(k) is
formed. Determine the N = 8 values ofthe 8-pt Inverse DFT of YN(k)
= XN(k)HN(k).
(c)-(ii) The product sequence YN(k) = XN (k)HN(k), formed as
directly above with N = 8, isused in Eqn (1). Write a closed-form
expression for the reconstructed spectrum Yr().
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Digital Signal Processing I Exam 3 Fall 2013
Problem 2. [50 points] Consider a causal FIR filter of length M
= 8 with impulseresponse as defined below:
hp[n] =
=
sin[58(n+ 8)
]
(n+ 8){u[n] u[n 8]}
(a) Determine all 8 numerical values of the 8-pt DFT of hp[n],
denoted H8(k), for0 k 7. List the values clearly: H8(0) =?, H8(1)
=?, H8(2) =?, H8(3) =?,H8(4) =?, H8(5) =?, H8(6) =?, H8(7) =?.
(b) Consider the sequence x[n] of length L = 8 below, equal to a
sum of 8 finite-lengthsinewaves.
x[n] =7
k=0
ejk2pi
8n {u[n] u[n 8]}
y8[n] is formed by computing X8(k) as an 8-pt DFT of x[n], H8(k)
as an 8-pt DFT ofhp[n], and then y8[n] as the 8-pt inverse DFT of
Y8(k) = X8(k)H8(k). Express theresult y8[n] as a weighted sum of
finite-length sinewaves similar to how x[n] is writtenabove.
Next, consider a causal FIR filter of length M = 8 with impulse
response as defined below:
hp[n] =
=
8
3
sin[58(n+ 8)
]
(n+ 8)
sin[38(n+ 8)
]
(n + 8){u[n] u[n 8]}
(c) Determine all 8 numerical values of the 8-pt DFT of hp[n],
denoted H8(k), for0 k 7. List the values clearly: H8(0) =?, H8(1)
=?, H8(2) =?, H8(3) =?,H8(4) =?, H8(5) =?, H8(6) =?, H8(7) =?.
(d) Consider the sequence x[n] of length L = 8 below, equal to a
sum of 8 finite-lengthsinewaves.
x[n] =7
k=0
ejk2pi
8n {u[n] u[n 8]}
y8[n] is formed by computing X8(k) as an 8-pt DFT of x[n], H8(k)
as an 8-pt DFT ofhp[n], and then y8[n] as the 8-pt inverse DFT of
Y8(k) = X8(k)H8(k). Express theresult y8[n] as a weighted sum of
finite-length sinewaves similar to how x[n] is writtenabove.
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