关于CE018_FFT_DSPlib问题

你知道在SuffRealEdgCpxx（）函数之后，你观察到这个阶段的复傅立叶变换结果的平方模吗？为了使图表变得更像Apple谱密度图，需要从复制的值中提取平方根。此外，你应该只考虑图的一半，因为平方和SuffDuffCxx（）将复数的值替换为两倍短的相同的实数值。您应该检查傅立叶变换数据的对齐方式和内存类型，包括信号和旋转系数。当我把一个例子转移到MPLABX和XC16时，这是我的第一个问题。检查旋转因子和信号被放置在相应的ROM/RAM段（X存储器、Y存储器或程序存储器）中，并保证它们是最接近两个功率的。它的起始地址的所有数据都是零位。例如，如果向量具有64个复数值的长度，则它占据128个分数值，这是256个字节。由于字节寻址，地址长度将是Log2（256）＝8比特。这就是为什么正确对齐向量的第一个元素应该具有最后一个地址字节零。如果矢量较大，对准也应该更大。



      以下为原文

    Do you understanding that after the SquareMagnitudeCplx() function you're observing squared modulus of the complex Fourier transform result at this stage? To make the diagram become more usual like a power spectral density diagram you need take a square root from the values you're copied. Also you should take in account only the half of a diagram because the SquareMagnitudeCplx() replaces complex values by same number real values which are twice shorter.
Indeed the diagram is not representing clear square wave spectrum. You should check the alignment and the memory type of the Fourier transforms data, both signal and twiddle coefficients. It was my first problems when I've transferred an example into MPLABX and XC16. Check the twiddle factors and signal are placed in corresponding ROM/RAM segments (X-memory, Y-memory or program memory) and ensure they are aligned for closest power of two. It is mandatory for the start address to have all zero bits. For example, if the vector have length of 64 complex values, it occupies 128 fractional values which are 256 bytes. Due to byte addressing the address length will be log2(256)=8 bits. That is why the first element of correctly aligned vector should have last address byte zero. If the vector is bigger the alignment should be bigger too.