complex multiplier in divide and combine FFT
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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.
fft complex
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up vote
1
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favorite
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.
fft complex
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 at 17:48
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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.
fft complex
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.
fft complex
fft complex
edited Dec 2 at 19:05
Marcus Müller
11.4k41431
11.4k41431
asked Dec 2 at 17:32
Abu Bakar Talha Jalil
273
273
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 at 17:48
add a comment |
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 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 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 at 17:48
add a comment |
1 Answer
1
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oldest
votes
up vote
3
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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.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
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.
add a comment |
up vote
3
down vote
accepted
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.
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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.
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.
edited Dec 2 at 18:14
answered Dec 2 at 18:07
Fat32
14.1k31128
14.1k31128
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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 at 17:48