What is the DFT of DFT of discrete signal [duplicate]
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How do you interpret FFT of an FFT of a discrete signal?
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What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
discrete-signals fourier-transform dft
asked Dec 6 at 19:49
Mert Ege
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marked as duplicate by Dilip Sarwate, MBaz, Matt L., lennon310, A_A Dec 11 at 13:02
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up vote
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down vote
favorite
This question already has an answer here:
How do you interpret FFT of an FFT of a discrete signal?
2 answers
What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
discrete-signals fourier-transform dft
asked Dec 6 at 19:49
Mert Ege
153
marked as duplicate by Dilip Sarwate, MBaz, Matt L., lennon310, A_A Dec 11 at 13:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
How do you interpret FFT of an FFT of a discrete signal?
2 answers
What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
discrete-signals fourier-transform dft
asked Dec 6 at 19:49
Mert Ege
153
This question already has an answer here:
How do you interpret FFT of an FFT of a discrete signal?
2 answers
What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
This question already has an answer here:
How do you interpret FFT of an FFT of a discrete signal?
2 answers
discrete-signals fourier-transform dft
discrete-signals fourier-transform dft
asked Dec 6 at 19:49
Mert Ege
153
asked Dec 6 at 19:49
Mert Ege
153
asked Dec 6 at 19:49
Mert Ege
153
asked Dec 6 at 19:49
Mert Ege
153
asked Dec 6 at 19:49
Mert Ege
153
153
marked as duplicate by Dilip Sarwate, MBaz, Matt L., lennon310, A_A Dec 11 at 13:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dilip Sarwate, MBaz, Matt L., lennon310, A_A Dec 11 at 13:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
2 Answers
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let
$$beginalign
X[k] &= mathcalDFT Big x[n] Big \
&triangleq sumlimits_n=0^N-1 x[n] , e^-j2pi nk/N
endalign $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcalDFT Big y[n] Big $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13
add a comment |
up vote
2
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
answered Dec 6 at 20:04
hotpaw2
25.5k53472
1
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
Dec 6 at 20:05
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
let
$$beginalign
X[k] &= mathcalDFT Big x[n] Big \
&triangleq sumlimits_n=0^N-1 x[n] , e^-j2pi nk/N
endalign $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcalDFT Big y[n] Big $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13
add a comment |
up vote
5
down vote
let
$$beginalign
X[k] &= mathcalDFT Big x[n] Big \
&triangleq sumlimits_n=0^N-1 x[n] , e^-j2pi nk/N
endalign $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcalDFT Big y[n] Big $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13
add a comment |
up vote
5
down vote
up vote
5
down vote
let
$$beginalign
X[k] &= mathcalDFT Big x[n] Big \
&triangleq sumlimits_n=0^N-1 x[n] , e^-j2pi nk/N
endalign $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcalDFT Big y[n] Big $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
let
$$beginalign
X[k] &= mathcalDFT Big x[n] Big \
&triangleq sumlimits_n=0^N-1 x[n] , e^-j2pi nk/N
endalign $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcalDFT Big y[n] Big $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
answered Dec 6 at 20:04
robert bristow-johnson
10.5k31448
10.5k31448
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13
add a comment |
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13
add a comment |
up vote
2
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
answered Dec 6 at 20:04
hotpaw2
25.5k53472
1
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
Dec 6 at 20:05
add a comment |
up vote
2
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
answered Dec 6 at 20:04
hotpaw2
25.5k53472
1
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
Dec 6 at 20:05
add a comment |
up vote
2
down vote
up vote
2
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
answered Dec 6 at 20:04
hotpaw2
25.5k53472
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
answered Dec 6 at 20:04
hotpaw2
25.5k53472
edited Dec 6 at 20:13
answered Dec 6 at 20:04
hotpaw2
25.5k53472
answered Dec 6 at 20:04
hotpaw2
25.5k53472
answered Dec 6 at 20:04
hotpaw2
25.5k53472
25.5k53472
1
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
Dec 6 at 20:05
add a comment |
1
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
Dec 6 at 20:05
1
1
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
Dec 6 at 20:05
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
Dec 6 at 20:05
add a comment |
