complex multiplier in divide and combine FFT












1














I am studying radix 2 algorith from Proakis' book.



But I'm a bit confusied why 1st DFT $G_1$ is not multiplied by complex entity while 2nd DFT $G_2$ is being multiplied by complex entity $W$ as shown highlighted in attached figure.



enter image description here






fft complex







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  • 1




    Hi! Have you tried wiriting the DFT of x[n] by dividing it into even and odd sampled sequences g1[n] and g2[n] ? The weight will come from that DFT.
    – Fat32
    Dec 2 '18 at 17:48
















1














I am studying radix 2 algorith from Proakis' book.



But I'm a bit confusied why 1st DFT $G_1$ is not multiplied by complex entity while 2nd DFT $G_2$ is being multiplied by complex entity $W$ as shown highlighted in attached figure.



enter image description here






fft complex







share|improve this question




















  • 1




    Hi! Have you tried wiriting the DFT of x[n] by dividing it into even and odd sampled sequences g1[n] and g2[n] ? The weight will come from that DFT.
    – Fat32
    Dec 2 '18 at 17:48














1












1








1







I am studying radix 2 algorith from Proakis' book.



But I'm a bit confusied why 1st DFT $G_1$ is not multiplied by complex entity while 2nd DFT $G_2$ is being multiplied by complex entity $W$ as shown highlighted in attached figure.



enter image description here






fft complex







share|improve this question















I am studying radix 2 algorith from Proakis' book.



But I'm a bit confusied why 1st DFT $G_1$ is not multiplied by complex entity while 2nd DFT $G_2$ is being multiplied by complex entity $W$ as shown highlighted in attached figure.



enter image description here






fft complex




fft complex






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 2 '18 at 19:05









Marcus Müller

11.6k41431




11.6k41431










asked Dec 2 '18 at 17:32









student

314




314








  • 1




    Hi! Have you tried wiriting the DFT of x[n] by dividing it into even and odd sampled sequences g1[n] and g2[n] ? The weight will come from that DFT.
    – Fat32
    Dec 2 '18 at 17:48














  • 1




    Hi! Have you tried wiriting the DFT of x[n] by dividing it into even and odd sampled sequences g1[n] and g2[n] ? The weight will come from that DFT.
    – Fat32
    Dec 2 '18 at 17:48








1




1




Hi! Have you tried wiriting the DFT of x[n] by dividing it into even and odd sampled sequences g1[n] and g2[n] ? The weight will come from that DFT.
– Fat32
Dec 2 '18 at 17:48




Hi! Have you tried wiriting the DFT of x[n] by dividing it into even and odd sampled sequences g1[n] and g2[n] ? The weight will come from that DFT.
– Fat32
Dec 2 '18 at 17:48










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Now as the document says, let $x[n]$ be a length $N$ (even) sequence whose even and odd indexed samples are denoted as $g_1[n]$ and $g_2[n]$ of length $N/2$ each.



Then the $N$-point DFT $X[k]$ of $x[n]$ can be written as folows:



$$begin{align}
X[k] &= sum_{n=0}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} ~~~,~~ k = 0,1,...,N-1 \\
&= sum_{n=0,2,4}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} + sum_{n=1,3,5}^{N-1} x[n] e^{ -j frac{2pi}{N} n k}\\
&= sum_{m=0}^{N/2-1} x[2m] e^{ -j frac{2pi}{N} 2m k} + sum_{m=0}^{N/2-1} x[2m+1] e^{ -j frac{2pi}{N} (2m+1) k}\\
&= sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} + e^{ -j frac{2pi}{N} k} sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} \\
&= left( sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} right) + e^{ -j frac{2pi}{N} k} left( sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} right) \\
&= G_1[k] + W_N^k G_2[k] ~~~,~~ k = 0,1,2...,N-1 \
end{align}$$



Where we have recognized the summations inside the parenthesis as the $N/2$ point DFTs of the sequences $g_1[n]$ and $g_2[n]$ respectively. Note that when $k$ spans the range $k=0,1,...,N-1$, the $N/2$ point DFTs will repeat twice.






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    3














    Now as the document says, let $x[n]$ be a length $N$ (even) sequence whose even and odd indexed samples are denoted as $g_1[n]$ and $g_2[n]$ of length $N/2$ each.



    Then the $N$-point DFT $X[k]$ of $x[n]$ can be written as folows:



    $$begin{align}
    X[k] &= sum_{n=0}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} ~~~,~~ k = 0,1,...,N-1 \\
    &= sum_{n=0,2,4}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} + sum_{n=1,3,5}^{N-1} x[n] e^{ -j frac{2pi}{N} n k}\\
    &= sum_{m=0}^{N/2-1} x[2m] e^{ -j frac{2pi}{N} 2m k} + sum_{m=0}^{N/2-1} x[2m+1] e^{ -j frac{2pi}{N} (2m+1) k}\\
    &= sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} + e^{ -j frac{2pi}{N} k} sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} \\
    &= left( sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} right) + e^{ -j frac{2pi}{N} k} left( sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} right) \\
    &= G_1[k] + W_N^k G_2[k] ~~~,~~ k = 0,1,2...,N-1 \
    end{align}$$



    Where we have recognized the summations inside the parenthesis as the $N/2$ point DFTs of the sequences $g_1[n]$ and $g_2[n]$ respectively. Note that when $k$ spans the range $k=0,1,...,N-1$, the $N/2$ point DFTs will repeat twice.






    share|improve this answer




























      3














      Now as the document says, let $x[n]$ be a length $N$ (even) sequence whose even and odd indexed samples are denoted as $g_1[n]$ and $g_2[n]$ of length $N/2$ each.



      Then the $N$-point DFT $X[k]$ of $x[n]$ can be written as folows:



      $$begin{align}
      X[k] &= sum_{n=0}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} ~~~,~~ k = 0,1,...,N-1 \\
      &= sum_{n=0,2,4}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} + sum_{n=1,3,5}^{N-1} x[n] e^{ -j frac{2pi}{N} n k}\\
      &= sum_{m=0}^{N/2-1} x[2m] e^{ -j frac{2pi}{N} 2m k} + sum_{m=0}^{N/2-1} x[2m+1] e^{ -j frac{2pi}{N} (2m+1) k}\\
      &= sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} + e^{ -j frac{2pi}{N} k} sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} \\
      &= left( sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} right) + e^{ -j frac{2pi}{N} k} left( sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} right) \\
      &= G_1[k] + W_N^k G_2[k] ~~~,~~ k = 0,1,2...,N-1 \
      end{align}$$



      Where we have recognized the summations inside the parenthesis as the $N/2$ point DFTs of the sequences $g_1[n]$ and $g_2[n]$ respectively. Note that when $k$ spans the range $k=0,1,...,N-1$, the $N/2$ point DFTs will repeat twice.






      share|improve this answer


























        3












        3








        3






        Now as the document says, let $x[n]$ be a length $N$ (even) sequence whose even and odd indexed samples are denoted as $g_1[n]$ and $g_2[n]$ of length $N/2$ each.



        Then the $N$-point DFT $X[k]$ of $x[n]$ can be written as folows:



        $$begin{align}
        X[k] &= sum_{n=0}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} ~~~,~~ k = 0,1,...,N-1 \\
        &= sum_{n=0,2,4}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} + sum_{n=1,3,5}^{N-1} x[n] e^{ -j frac{2pi}{N} n k}\\
        &= sum_{m=0}^{N/2-1} x[2m] e^{ -j frac{2pi}{N} 2m k} + sum_{m=0}^{N/2-1} x[2m+1] e^{ -j frac{2pi}{N} (2m+1) k}\\
        &= sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} + e^{ -j frac{2pi}{N} k} sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} \\
        &= left( sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} right) + e^{ -j frac{2pi}{N} k} left( sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} right) \\
        &= G_1[k] + W_N^k G_2[k] ~~~,~~ k = 0,1,2...,N-1 \
        end{align}$$



        Where we have recognized the summations inside the parenthesis as the $N/2$ point DFTs of the sequences $g_1[n]$ and $g_2[n]$ respectively. Note that when $k$ spans the range $k=0,1,...,N-1$, the $N/2$ point DFTs will repeat twice.






        share|improve this answer














        Now as the document says, let $x[n]$ be a length $N$ (even) sequence whose even and odd indexed samples are denoted as $g_1[n]$ and $g_2[n]$ of length $N/2$ each.



        Then the $N$-point DFT $X[k]$ of $x[n]$ can be written as folows:



        $$begin{align}
        X[k] &= sum_{n=0}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} ~~~,~~ k = 0,1,...,N-1 \\
        &= sum_{n=0,2,4}^{N-1} x[n] e^{ -j frac{2pi}{N} n k} + sum_{n=1,3,5}^{N-1} x[n] e^{ -j frac{2pi}{N} n k}\\
        &= sum_{m=0}^{N/2-1} x[2m] e^{ -j frac{2pi}{N} 2m k} + sum_{m=0}^{N/2-1} x[2m+1] e^{ -j frac{2pi}{N} (2m+1) k}\\
        &= sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} + e^{ -j frac{2pi}{N} k} sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} \\
        &= left( sum_{m=0}^{N/2-1} g_1[m] e^{ -j frac{2pi}{N/2} m k} right) + e^{ -j frac{2pi}{N} k} left( sum_{m=1}^{N/2-1} g_2[m] e^{ -j frac{2pi}{N/2} m k} right) \\
        &= G_1[k] + W_N^k G_2[k] ~~~,~~ k = 0,1,2...,N-1 \
        end{align}$$



        Where we have recognized the summations inside the parenthesis as the $N/2$ point DFTs of the sequences $g_1[n]$ and $g_2[n]$ respectively. Note that when $k$ spans the range $k=0,1,...,N-1$, the $N/2$ point DFTs will repeat twice.







        share|improve this answer














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        share|improve this answer








        edited Dec 2 '18 at 18:14

























        answered Dec 2 '18 at 18:07









        Fat32

        14.3k31129




        14.3k31129






























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