Homework for Day 4
Transcript of Homework for Day 4
HelpHomework for Day 4
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it. You are welcome to try it as a learning exercise.
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the answers here are my own work.
Question 1
(Difficulty: ) Let
be a DTFT transform-pair. Assume to be differentiable, compute the inverse DTFT of
.
Hint : use integration by parts
You should write your answer in term of and elementary functions and constants, for
example would be written :
pi/2*x[n]
⋆⋆ ⋆⋆ ⋆ ⋆
⋆ ⋆
x[n] ↔ X( )ejω
X
j X( )ddω
ejω
x[n]
x[n]π2
Question 2
(Difficulty: ) Which property of the DTFT allows you to compute easily the inverse DTFT of
?
Remember the result you obtained in the previous question.
Question 3
(Difficulty: ) A discrete sequence has for DTFT .
The real and imaginary parts of are:
By visual inspection of the plots, tick all the true statements about the sequence :
is Hermitian-symmetric .
is real valued.
is 0-mean, i.e. .
⋆
X( )/π − 2ddω
ejω
⋆ x[n] X( )ejω
X( )ejω
x[n]
x[n] x[n] = [−n]x∗
x[n]
x[n] x[n] = 0∑n∈Z
Preview
Question 4
(Difficulty: ) Consider the following signal
and its DFT defined as .
Compute the mathematical expression for
You can find here a short guide for answering quizzes requiring an equation as answer. In
particular we underline that in the Coursera platform the symbol (capital i) is used for the
imaginary unit instead of , that is the symbol used in class. Moreover, you can use both the
exponential function and the exponentiation of the Euler's number (instead of the
classic ). Pi number is defined as . Apart from the usual mathematical constants and
functions you should only use the (case-sensitive) variable names
k M L
(do not forget to validate your syntax by clicking "Preview")
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Question 5
(Difficulty: ) Consider a finite-length discrete signal of length ( )
and its DFT .
Consider its periodization and with its DFS .
Which of the following statements are true?
,
for all
(assume )
⋆ x[l] = { ,10
0 ≤ l ≤ M − 1M ≤ l ≤ L − 1
X[k] = x[l]∑L−1l=0 e−j lk
2π
L
X[k]
I
j
exp(⋅) E
e pi
⋆ x[n] N n = 0, . . . , N − 1
X[k]
[n] = x[n mod N]x~ [k]X~
[l] = X[l mod N]X~
l ∈ Z[−2] = X[2]X
~
N > 2
[k + lN] = X[k]~