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The source code has a few subtle differences that are not revealed in the pseudocode. The pseudocode in [link] requires an array of complex numbers x n for input, but the source code requires a reference to an array of complex numbers with a stride A stride of n would indicate that only every n t h term is being referred to. – this avoids copying x n into three separate arrays, viz. x 2 n 2 , x 4 n 4 + 1 and x 4 n 4 - 1 , with every invocation of [link] . The subtle complication arises due to the cyclic shifting of the x 4 n 4 - 1 term; the negative shifting results in pointers that reference data before the start of the array. Rather than immediately wrapping the references around to end of the array such that they always point to validdata, the recursion proceeds until the base cases are reached before any adjustment is performed. Once at the leaves of the recursion, any pointers that reference data lying before the start of the input array are incremented by N elements, In this case, N refers to the size of the outer most transform rather than the size of the sub-transform. so as to point to the correct data.

  IF  N = 1     RETURN  x 0   ELSIF  N = 2      X 0 x 0 + x 1      X 1 x 0 - x 1   ELSE      U k 2 = 0 , , N / 2 - 1 TANGENTFFT 4 N / 2 ( x 2 n 2 )      Z k 4 = 0 , , N / 4 - 1 TANGENTFFT 8 N / 4 ( x 4 n 4 + 1 )      Z k 4 = 0 , , N / 4 - 1 ' TANGENTFFT 8 N / 4 ( x 4 n 4 - 1 )     FOR  k = 0  to  N / 4 - 1        X k U k + ( ω N k s N / 4 , k Z k + ω N - k s N / 4 , k Z k ' )        X k + N / 2 U k - ( ω N k s N / 4 , k Z k + ω N - k s N / 4 , k Z k ' )        X k + N / 4 U k + N / 4 - i ( ω N k s N / 4 , k Z k - ω N - k s N / 4 , k Z k ' )        X k + 3 N / 4 U k + N / 4 + i ( ω N k s N / 4 , k Z k - ω N - k s N / 4 , k Z k ' )     END FOR   ENDIF  RETURN  X k
TANGENTFFT 4 N ( x n )
  IF  N = 1     RETURN  x 0   ELSIF  N = 2      X 0 x 0 + x 1      X 1 x 0 - x 1   ELSIF  N = 4      T k 2 = 0 , 1 TANGENTFFT 8 2 ( x 2 n 2 )      T k 2 = 0 , 1 ' TANGENTFFT 8 2 ( x 2 n 2 + 1 )      X 0 T 0 + T 0 '      X 2 T 0 - T 0 '      X 1 T 1 + T 1 '      X 3 T 1 - T 1 '   ELSE      U k 4 = 0 , , N / 4 - 1 TANGENTFFT 8 N / 4 ( x 4 n 4 )      Y k 8 = 0 , , N / 8 - 1 TANGENTFFT 8 N / 8 ( x 8 n 8 + 2 )      Y k 8 = 0 , , N / 8 - 1 ' TANGENTFFT 8 N / 8 ( x 8 n 8 - 2 )      Z k 4 = 0 , , N / 4 - 1 TANGENTFFT 8 N / 4 ( x 4 n 4 + 1 )      Z k 4 = 0 , , N / 4 - 1 ' TANGENTFFT 8 N / 4 ( x 4 n 4 - 1 )     FOR  k = 0  to  N / 8 - 1        α N , k s N / 4 , k / s N , k        β N , k s N / 8 , k / s N / 2 , k        γ N , k s N / 4 , k + N / 8 / s N , k + N / 8        δ N , k s N / 2 , k / s N , k        ϵ N , k s N / 2 , k + N / 8 / s N , k + N / 8        Ω 0 ω N k * α N , k        Ω 1 ω N k + N / 8 * γ N , k        Ω 2 ω N 2 k * β N , k        T 0 ( Ω 2 Y k + Ω ¯ 2 Y k ) * δ N , k        T 1 i ( Ω 2 Y k - Ω ¯ 2 Y k ) * ϵ N , k        X k U k * α N , k + T 0 + ( Ω 0 Z k + Ω ¯ 0 Z k ' )        X k + N / 2 U k * α N , k + T 0 - ( Ω 0 Z k + Ω ¯ 0 Z k ' )        X k + N / 4 U k * α N , k - T 0 - i ( Ω 0 Z k - Ω ¯ 0 Z k ' )        X k + 3 N / 4 U k * α N , k - T 0 + i ( Ω 0 Z k - Ω ¯ 0 Z k ' )        X k + N / 8 U k + N / 8 * γ N , k - T 1 + ( Ω 1 Z k + N / 8 + Ω ¯ 0 Z k + N / 8 ' )        X k + 3 N / 8 U k + N / 8 * γ N , k + T 1 - i ( Ω 1 Z k + N / 8 - Ω ¯ 0 Z k + N / 8 ' )        X k + 5 N / 8 U k + N / 8 * γ N , k - T 1 - ( Ω 1 Z k + N / 8 + Ω ¯ 0 Z k + N / 8 ' )        X k + 7 N / 8 U k + N / 8 * γ N , k + T 1 + i ( Ω 1 Z k + N / 8 - Ω ¯ 0 Z k + N / 8 ' )     END FOR   ENDIF  RETURN  X k
TANGENTFFT 8 N ( x n )

Tangent

The tangent FFT is divided into two functions, described with pseudocode in [link] and [link] . If the tangent FFT is computed prior to convolution in the frequency domain, theconvolution kernel can absorb the final scaling and only [link] is required. Otherwise [link] is used as a wrapper around [link] to perform the rescaling, and the result X k is in the correct basis.

[link] is similar to [link] , except that Z k and Z k ' are computed with [link] , and thus scaled by 1 / s N / 4 , k . Because Z k and Z k ' are respectively multiplied by the coefficients ω N k and ω N - k , the results are scaled into the correct basis by absorbing s N / 4 , k into the coefficients.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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what is inorganic
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Computing the fast fourier transform on simd microprocessors. OpenStax CNX. Jul 15, 2012 Download for free at http://cnx.org/content/col11438/1.2
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