TI中文支持网我用的28335，调用controlsuilt的定点FFT库例程计算512点FFT，代码如下，输出的是幅值的平方，而且是Q30格式，
fft.mag(&fft);/* Q31 format (abs(ipcbsrc)/2^16).^2 */ ，请问怎么将输出的幅值还原成真实值？
//######################################################################################
// $TI Release: C28x Fixed Point Library v1.01 $
// $Release Date: January 11,2011 $
//######################################################################################
#include "DSP28x_Project.h"
#include <fft.h>
#include "math.h"
#include "float.h"
/* Create an Instance of FFT module*/
#defineN512 // FFT size
#pragma DATA_SECTION(ipcb, "FFTipcb"); // Input/output memory allocation
#pragma DATA_SECTION(ipcbsrc, "FFTipcbsrc");
long ipcbsrc[2*N];
long ipcb[2*N];
RFFT32 fft=RFFT32_512P_DEFAULTS; // Header structure definition
/* Define window Co-efficient Array and place the
.constant section in ROM memory */
const long win[N/2]=HAMMING32; // Select window, not used in this example
int xn,yn;
float RadStep = 0.1963495408494f; // Simulated signal generation parameter
float Rad = 0.0f; // Initial value of Radstep
void main()
{unsigned long i;InitSysCtrl();
DINT;
InitPieCtrl();
IER = 0x0000;
IFR = 0x0000;
InitPieVectTable();
EINT;// Enable Global interrupt INTM
ERTM;// Enable Global realtime interrupt DBGM// Generate sample waveforms:
Rad = 0.0f;
//Clean up input/output buffer
for(i=0; i < (N*2); i=i+2)
{
ipcb[i] =0;
ipcb[i+1] = 0;
}
//Simulated input signal
for(i=0; i < N; i++)
{
ipcbsrc[i] =(long)(2147483648*(sin(Rad) + cos(Rad*2.3567))/2); //Q31
Rad = Rad + RadStep;
}
/*--------------------------------------------------------------------------- FFT Calculation
----------------------------------------------------------------------------*/
RFFT32_brev(ipcbsrc, ipcb, N); /* real FFT bit reversing*/
fft.ipcbptr=ipcb; /* FFT computation buffer */
fft.magptr=ipcbsrc; /* Magnitude output buffer */
fft.winptr=(long *)win; /* Window coefficient array */
fft.init(&fft); /* Twiddle factor pointer initialization */
fft.calc(&fft); /* Compute the FFT */
fft.mag(&fft); /* Q31 format (abs(ipcbsrc)/2^16).^2 */
//asm("ESTOP0");
for(;;);
} /* End: main() */
fw yang：同问
我用的28335，调用controlsuilt的定点FFT库例程计算512点FFT，代码如下，输出的是幅值的平方，而且是Q30格式，
fft.mag(&fft);/* Q31 format (abs(ipcbsrc)/2^16).^2 */ ，请问怎么将输出的幅值还原成真实值？
//######################################################################################
// $TI Release: C28x Fixed Point Library v1.01 $
// $Release Date: January 11,2011 $
//######################################################################################
#include "DSP28x_Project.h"
#include <fft.h>
#include "math.h"
#include "float.h"
/* Create an Instance of FFT module*/
#defineN512 // FFT size
#pragma DATA_SECTION(ipcb, "FFTipcb"); // Input/output memory allocation
#pragma DATA_SECTION(ipcbsrc, "FFTipcbsrc");
long ipcbsrc[2*N];
long ipcb[2*N];
RFFT32 fft=RFFT32_512P_DEFAULTS; // Header structure definition
/* Define window Co-efficient Array and place the
.constant section in ROM memory */
const long win[N/2]=HAMMING32; // Select window, not used in this example
int xn,yn;
float RadStep = 0.1963495408494f; // Simulated signal generation parameter
float Rad = 0.0f; // Initial value of Radstep
void main()
{unsigned long i;InitSysCtrl();
DINT;
InitPieCtrl();
IER = 0x0000;
IFR = 0x0000;
InitPieVectTable();
EINT;// Enable Global interrupt INTM
ERTM;// Enable Global realtime interrupt DBGM// Generate sample waveforms:
Rad = 0.0f;
//Clean up input/output buffer
for(i=0; i < (N*2); i=i+2)
{
ipcb[i] =0;
ipcb[i+1] = 0;
}
//Simulated input signal
for(i=0; i < N; i++)
{
ipcbsrc[i] =(long)(2147483648*(sin(Rad) + cos(Rad*2.3567))/2); //Q31
Rad = Rad + RadStep;
}
/*--------------------------------------------------------------------------- FFT Calculation
----------------------------------------------------------------------------*/
RFFT32_brev(ipcbsrc, ipcb, N); /* real FFT bit reversing*/
fft.ipcbptr=ipcb; /* FFT computation buffer */
fft.magptr=ipcbsrc; /* Magnitude output buffer */
fft.winptr=(long *)win; /* Window coefficient array */
fft.init(&fft); /* Twiddle factor pointer initialization */
fft.calc(&fft); /* Compute the FFT */
fft.mag(&fft); /* Q31 format (abs(ipcbsrc)/2^16).^2 */
//asm("ESTOP0");
for(;;);
} /* End: main() */
fw yang：
用的是定点库的2833x_FixedPoint_RFFT
原始信号幅值分别为1和6，除以7，总幅值为1；
但调用函数后计算结果再开根号分别为0.0714和0.4284，加起来幅值为0.5001，为原总幅值的一半，是否库有问题？请问如何复原幅值？
从目前结果看是乘以2可解决，但我也不确定到底是不是这样
