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?

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

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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?

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

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?

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

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

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 |

2 Answers
2

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5
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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

10.5k31448

Thanks for the clarification but can we use any property of DFT to find this solution?
– Mert Ege
Dec 8 at 21:13

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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).

edited Dec 6 at 20:13

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

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 $n$.

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

$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$

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

10.5k31448

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

$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$

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

10.5k31448

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

10.5k31448

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

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

add a comment |

up vote
2
down vote

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

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

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

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

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

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

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

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

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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