6. If x(n) is a discrete, imaginary, and even function, its Fourier transform, X(w), is (A) a real and even function (В) a real and odd function (C) an imaginary and odd function (D) an imaginary and even function 6. The Fourier transform of a discrete signal a(n) is given by Χ(ω)Σε(n) εTψη =x(n)(coswn+ j sinwn) 1 Σ(n) coswn + jΣ-(n) sin wn In the second term, n) is an even function and sin wn is an odd function, so the entire term is equal to zero, and x(n) cos wn+0 -Σ(n)cos un X(w) The function z(n) is imaginary, so X(w) is an imaginary function. The functions x(n) and cos wn are both even, so X(w) is an even function The answer is (D)
6. If x(n) is a discrete, imaginary, and even function, its Fourier transform, X(w), is (A) a real and even function (В) a real and odd function (C) an imaginary and odd function (D) an imaginary and even function 6. The Fourier transform of a discrete signal a(n) is given by Χ(ω)Σε(n) εTψη =x(n)(coswn+ j sinwn) 1 Σ(n) coswn + jΣ-(n) sin wn In the second term, n) is an even function and sin wn is an odd function, so the entire term is equal to zero, and x(n) cos wn+0 -Σ(n)cos un X(w) The function z(n) is imaginary, so X(w) is an imaginary function. The functions x(n) and cos wn are both even, so X(w) is an even function The answer is (D)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Hi,
the attached image shows a question and its answer but still, I couldn't understand it please simply explain it

Transcribed Image Text:6. If x(n) is a discrete, imaginary, and even function, its
Fourier transform, X(w), is
(A)
a real and even function
(В)
a real and odd function
(C)
an imaginary and odd function
(D)
an imaginary and even function

Transcribed Image Text:6. The Fourier transform of a discrete signal a(n) is
given by
Χ(ω)Σε(n) εTψη
=x(n)(coswn+ j sinwn)
1
Σ(n) coswn + jΣ-(n) sin wn
In the second term, n) is an even function and sin wn
is an odd function, so the entire term is equal to zero,
and
x(n) cos wn+0
-Σ(n)cos un
X(w)
The function z(n) is imaginary, so X(w) is an imaginary
function. The functions x(n) and cos wn are both even,
so X(w) is an even function
The answer is (D)
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