|
Имеется оптимизированная либа C28x Floating Point Unit Library, содержащая следующие функции
RFFT_f32 void RFFT_f32(RFFT_F32_STRUCT *);
This module computes a 32-bit floating-point real FFT including input bit
reversing. This version of the function requires input buffer memory alignment. If
you do not wish to align the input buffer, then use the RFFT_f32u function.
RFFT_f32u void RFFT_f32u(RFFT_F32_STRUCT *);
This module computes a 32-bit floating-point real FFT including input bit
reversing. This version of the function does not have any buffer alignment
requirements. If you can align the input buffer, then use the RFFT_f32 function
for improved performance.
RFFT_f32_mag void RFFT_f32_mag(RFFT_F32_STRUCT *);
This module computes the real FFT magnitude.
RFFT_f32s_mag void RFFT_f32s_mag(RFFT_F32_STRUCT *);
This module computes the scaled real FFT magnitude.
RFFT_f32_phase void RFFT_f32_phase(RFFT_F32_STRUCT *);
This module computes FFT Phase
RFFT_f32_sincostable void RFFT_f32_sincostable(RFFT_F32_STRUCT *);
This module generates the twiddle factors used by the real FFT.
typedef struct {
float32 *InBuf;//Input buffer
float32 *OutBuf; //Output buffer
float32 *CosSinBuf;
float32 *MagBuf;
float32 *PhaseBuf;
Uint16 FFTSize;
Uint16 FFTStages;
} RFFT_F32_STRUCT;
По поводу output buffer сказано:
Result of RFFT_f32. This buffer is used as
the input to the magnitude and phase
calculations. The output order for FFTSize
= N is:
OutBuf[0] = real[0]
OutBuf[1] = real[1]
OutBuf[2] = real[2]
………
OutBuf[N/2] = real[N/2]
OutBuf[N/2+1] = imag[N/2-1]
………
OutBuf[N-3] = imag[3]
OutBuf[N-2] = imag[2]
OutBuf[N-1] = imag[1]
Почему на выходе _REAL_ FFT видим imag, т.е. мнимую часть?
Каков смысл FFT magnitude и phase?
Как используя эти функции реализовать обратное преобразование?
--------------------
The truth is out there...
|
c28х FPU library beta1, помогите разобраться с real FFT
Sep 16 2009, 19:33
