What is the DFT of DFT of 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
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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
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Duplicate of How do you interpret FFT of an FFT of a discrete signal?
– Dilip Sarwate
2 days ago
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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
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
discrete-signals fourier-transform dft
asked Dec 6 at 19:49
Mert Ege
133
133
3
Duplicate of How do you interpret FFT of an FFT of a discrete signal?
– Dilip Sarwate
2 days ago
add a comment |
3
Duplicate of How do you interpret FFT of an FFT of a discrete signal?
– Dilip Sarwate
2 days ago
3
3
Duplicate of How do you interpret FFT of an FFT of a discrete signal?
– Dilip Sarwate
2 days ago
Duplicate of How do you interpret FFT of an FFT of a discrete signal?
– Dilip Sarwate
2 days ago
add a comment |
2 Answers
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5
down vote
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
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 $n$.
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
yesterday
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).
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
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
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 $n$.
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
yesterday
add a comment |
up vote
5
down vote
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
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 $n$.
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
yesterday
add a comment |
up vote
5
down vote
up vote
5
down vote
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
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 $n$.
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
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
10.4k31448
10.4k31448
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
yesterday
add a comment |
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
yesterday
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
yesterday
Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
yesterday
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).
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).
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).
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).
edited Dec 6 at 20:13
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 |
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3
Duplicate of How do you interpret FFT of an FFT of a discrete signal?
– Dilip Sarwate
2 days ago