(b) Given a signal (so, S₁,... SN-1) of length N, we may define the cyclical shift of the signal as: cycShift (so,..., SN-1) = (S1, S2,..., SN-1, 50). This can also be written as: cycShift (So,...,SN-1)j = $mod (j+1,N). Based on your results in (a), give an equation relating the IDFT of a signal (so,...,$3) and the IDFT of its cyclical shift cycShift (so,...,83). (c) Using the definition of IDFT in Equation (3), prove the equation that you wrote for (b). (d) Generalize your answer to (b) and give an equation that relates the IDFT of a signal (so,... , SN-1) and the IDFT of its cyclical shift cycShift(So,...,SN-1). (e) Using the definition of IDFT, prove the equation that you wrote for (d).
(b) Given a signal (so, S₁,... SN-1) of length N, we may define the cyclical shift of the signal as: cycShift (so,..., SN-1) = (S1, S2,..., SN-1, 50). This can also be written as: cycShift (So,...,SN-1)j = $mod (j+1,N). Based on your results in (a), give an equation relating the IDFT of a signal (so,...,$3) and the IDFT of its cyclical shift cycShift (so,...,83). (c) Using the definition of IDFT in Equation (3), prove the equation that you wrote for (b). (d) Generalize your answer to (b) and give an equation that relates the IDFT of a signal (so,... , SN-1) and the IDFT of its cyclical shift cycShift(So,...,SN-1). (e) Using the definition of IDFT, prove the equation that you wrote for (d).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please do Exercise 2 part B-E and I have attached the answer to part A. Please show step by step and explain
![Use the 2-stage process expressed in Equation
(10) and (11) to find the following IDFT'S
X(0), X(1), x(2), x(3)
a) (i) IDFT (1,0,0,0)
inh
x(n) = 4 [x(0) + X(1) e i/an + X(2)e mh +x(3) en ]
x(0) = 4 [x(0) + X(1) + X(2) + X(3)]
= 4√ [ 1+0+0+0]
4
X (1) = = [X (0) + X (1) + x(2) ² + x(3) e 15%]
iπT
= = [₁ +0+0+0]
=&
|x (2) = + [1] = 1/4
|x(3) = 1/4 [1] = 1/4
x(n) = 4 [0 + en +0+0
_x(n) = 4₁ [en]
(ii) IDFT(0,1,0,0)
[x(n) = = [X (0) + x (1) eilan+x(2)e + x(3) e
(iii) IDET (0,0,1,0)
Xx (n) = ²₁ [x (0) + X(1)e
x (n) = = [ein
3
1
x(n) = 4 2₁ x (k) e 15 kn
in
K=O
x(0) = 4/1
x ( 1 ) = = [e₁ = ] = 4 [cos (# )+ i sin (5)] = [1]]
|x (2) = + [e²"] = +4 [cos (T) +isin (π)] = + (-1]
x(3) = [ei] = [cos (3/2) +isin (3)] = [-1]
X(n) = ( 4₁ 1/4, 1₁, 1/4)
63
X ₁ = ( 17₁ 172 177₁ 177)
4,4)
(iv) IDFT (0,0,0,1)
x(n) == [ej]
+x(2)e+x(3) e
n
x (0) =
|x (1) = ₁ [e¹¹ ] = ₁ [cos(IT) + i Sin(π)] = [1] =
x ( 2 ) = + [ e ² 1 ] ²4 [cos(an) +i Sin(2n)] = √ [1] = 4
ап
आ
+
|x ( 3 ) = = [e³¹¹ ] = [cos (3r) + i sin (3r)] = 4 [- 1] = -1/4
x(n) = ( + + +
13 n
IDFT (1,0,0,0) = + (1,1,1,1)
IDFT(0,1,0,0) = + (1,₁ 1,-1,-1)
IDFT (0,0,1,0) = + (1,-1, 1,-1)
IDF T(0,0,0,1)= 4 (1₁-1₁ -1, 1)
x(0) =
x (1) =
x(2) =
[e²²₁ | = √ [Cos (3) + sin(35)] = [[ - ₁] = -√2/4
= + 4
[en] =4 [cos (31) risin(3r)] = [- 1] = -1/4
[ei] [cos (2) +isin (4)] = + [i] = √3/4
√
|x (n) = ( 4₁ =1/4, 1/4 ₁/4)
x(3) =
=
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2Fd28a4689-d58f-42b2-9826-30daa59dd669%2Flh8drss_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Use the 2-stage process expressed in Equation
(10) and (11) to find the following IDFT'S
X(0), X(1), x(2), x(3)
a) (i) IDFT (1,0,0,0)
inh
x(n) = 4 [x(0) + X(1) e i/an + X(2)e mh +x(3) en ]
x(0) = 4 [x(0) + X(1) + X(2) + X(3)]
= 4√ [ 1+0+0+0]
4
X (1) = = [X (0) + X (1) + x(2) ² + x(3) e 15%]
iπT
= = [₁ +0+0+0]
=&
|x (2) = + [1] = 1/4
|x(3) = 1/4 [1] = 1/4
x(n) = 4 [0 + en +0+0
_x(n) = 4₁ [en]
(ii) IDFT(0,1,0,0)
[x(n) = = [X (0) + x (1) eilan+x(2)e + x(3) e
(iii) IDET (0,0,1,0)
Xx (n) = ²₁ [x (0) + X(1)e
x (n) = = [ein
3
1
x(n) = 4 2₁ x (k) e 15 kn
in
K=O
x(0) = 4/1
x ( 1 ) = = [e₁ = ] = 4 [cos (# )+ i sin (5)] = [1]]
|x (2) = + [e²"] = +4 [cos (T) +isin (π)] = + (-1]
x(3) = [ei] = [cos (3/2) +isin (3)] = [-1]
X(n) = ( 4₁ 1/4, 1₁, 1/4)
63
X ₁ = ( 17₁ 172 177₁ 177)
4,4)
(iv) IDFT (0,0,0,1)
x(n) == [ej]
+x(2)e+x(3) e
n
x (0) =
|x (1) = ₁ [e¹¹ ] = ₁ [cos(IT) + i Sin(π)] = [1] =
x ( 2 ) = + [ e ² 1 ] ²4 [cos(an) +i Sin(2n)] = √ [1] = 4
ап
आ
+
|x ( 3 ) = = [e³¹¹ ] = [cos (3r) + i sin (3r)] = 4 [- 1] = -1/4
x(n) = ( + + +
13 n
IDFT (1,0,0,0) = + (1,1,1,1)
IDFT(0,1,0,0) = + (1,₁ 1,-1,-1)
IDFT (0,0,1,0) = + (1,-1, 1,-1)
IDF T(0,0,0,1)= 4 (1₁-1₁ -1, 1)
x(0) =
x (1) =
x(2) =
[e²²₁ | = √ [Cos (3) + sin(35)] = [[ - ₁] = -√2/4
= + 4
[en] =4 [cos (31) risin(3r)] = [- 1] = -1/4
[ei] [cos (2) +isin (4)] = + [i] = √3/4
√
|x (n) = ( 4₁ =1/4, 1/4 ₁/4)
x(3) =
=
-
Transcribed Image Text:In the previous assignment, we derived the equations:
N-1
sj = Σ axelmikj/N
k=0
an =
1
N
IDFT (So,... SN-1)n ==
N-1
Σ se
j=0
This can also be written as:
N-1
DFT (αo, N-1); := ²kj/N
e-2mijn/N
where these equations hold for any sequence of N complex numbers a0,₁,... N-1. In fact, instead of
starting with the a's we could start with an arbitrary sequence of complex numbers So,... SN-1 and define
the sequence {a} in terms of {s;} by using Equation (2). We can then recover the si's from these an's from
Equation (1). (Note we have written s; instead of s(j) to emphasize the remarkably symmetric situation
between the two sequences (an) in terms of {s;}.)'
In fact, what we have just derived is the complex form of the famous discrete Fourier transform, abbrevi-
ated as DFT. Equation (1) expresses that the sequence {s;} is the DFT of {a}, while Equation (1) expresses
that the sequence {an} is the inverse discrete Fourier transform (abbreviated IDFT) of {s;}. For later con-
venience, we will introduce the notation:
k=0
1 N-1
Σ se-
-2mijn/N
j=0
j=0,..., N-1
(1)
n = 0,..., N-1
(2)
(3)
(4)
The subscripts j and n on the left-hand sides of these equations reflects the fact that the DFT (or IDFT) of a
sequence of N complex numbers is itself a sequence of N complex numbers. The sums on the right-hand
sides are the formulas for the different components of the DFT and IDFT.
With this notation, we may write Equations (1) and (2) very succinctly as:
{sj} = DFT({a;}) ⇒ {aj} = IDFT({s;}),
(5)
where {s} and {a} denote the sequences so,... SN-1 and ao,... N-1, respectively. The notation DFT and
IDFT recalls the notion of function and inverse function: in fact, the N-point DFT may be considered as a
function from CN to CN and the N-point IDFT is its inverse.
It is impossible to underestimate the importance of the DFT in modern science, engineering, and math-
ematics. It pops up everywhere, even in number theory!
Exercise 2. Use the 2-stage process expressed in Equations (10) and (11) to find the following IDFT's:
(a) (i) IDFT(1,0,0,0) (ii) IDFT (0,1,0,0) (iii) IDFT (0,0,1,0) (iv) IDFT(0,0,0,1).
(b) Given a signal (S0, S₁,... SN-1) of length N, we may define the cyclical shift of the signal as:
cycShift (So,...,SN-1) = ($₁,$2,..., SN-1,50).
cycShift (So,...,SN-1); = $mod (j+1,N).
Based on your results in (a), give an equation relating the IDFT of a signal (so,...,s3) and the IDFT of
its cyclical shift cycShift (So,...,$3).
(c) Using the definition of IDFT in Equation (3), prove the equation that you wrote for (b).
(d) Generalize your answer to (b) and give an equation that relates the IDFT of a signal (So,..., SN-1) and
the IDFT of its cyclical shift cycShift (so,...,SN-1).
(e) Using the definition of IDFT, prove the equation that you wrote for (d).
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