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?










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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
    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?










    share|improve this 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.
















      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?










      share|improve this question














      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






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      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.






















          2 Answers
          2






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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 $n$.






          share|improve this answer




















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

















          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).






          share|improve this answer


















          • 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

















          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 $n$.






          share|improve this answer




















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














          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 $n$.






          share|improve this answer




















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












          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 $n$.






          share|improve this answer












          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 $n$.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          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
















          • 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










          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).






          share|improve this answer


















          • 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














          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).






          share|improve this answer


















          • 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












          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).






          share|improve this answer














          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).







          share|improve this answer














          share|improve this answer



          share|improve this answer








          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












          • 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


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