fft

How to perform a *Fast* DCT (Discrete Cosine Transform) in R?

ぐ巨炮叔叔 提交于 2020-01-31 03:50:51
问题 Using Rprof revealed the dct in the dtt package to be the main offender in a piece of R code that was running quite slowly. Swapping it out for fft in the stats package (which is not the same transformation, but should take the same time to compute) my run time improved dramatically. In fact 2/3 of my Rprof lines were dct calls previously and a mere 3 lines of about 600 were fft calls after making the switch. Is the dct implementation in the dtt package not done using the fast discrete

MATLAB界面编程

寵の児 提交于 2020-01-27 00:01:49
MATLAB 界面编程 Reference to non-exiting field! 控件没有添加Tag属性: 直接双击fig文件启动(会缺少初始化操作)和在m文件执行开始启动结果不一样: 编译fig程序为EXE文件可同时执行多个实例: 选择APP里面的Compiler: 选择自己的工程并build,最后会在对应的文件目录下会生出EXE文件: 设置背景颜色为白,方便截图(另存为emf再粘贴到word比较清晰) 附件: 实现FFT的MATLAB图形界面: function varargout = easy_fft_1(varargin) % EASY_FFT_1 MATLAB code for easy_fft_1.fig % EASY_FFT_1, by itself, creates a new EASY_FFT_1 or raises the existing % singleton*. % % H = EASY_FFT_1 returns the handle to a new EASY_FFT_1 or the handle to % the existing singleton*. % % EASY_FFT_1('CALLBACK',hObject,eventData,handles,...) calls the local % function named

另一道不知道哪里来的FFT题

て烟熏妆下的殇ゞ 提交于 2020-01-26 04:33:44
给定一个序列,求出这个序列的k阶前缀和,模998244353,n<=1e5。 k阶前缀和可以看成一个一个n*k的平面上的二维行走问题。 第i项对第j项的贡献是从(i,0)走到(j,k)的NE Latice Path的方案数。 大力FFT一下即可。 来源: https://www.cnblogs.com/Creed-qwq/p/10765359.html

FFT based image registration in python

谁说胖子不能爱 提交于 2020-01-24 13:47:41
问题 I found simple code in python for image registration here in simple case of translation we have: def translation(im0, im1): """Return translation vector to register images.""" shape = im0.shape f0 = fft2(im0) f1 = fft2(im1) ir = abs(ifft2((f0 * f1.conjugate()) / (abs(f0) * abs(f1)))) t0, t1 = numpy.unravel_index(numpy.argmax(ir), shape) if t0 > shape[0] // 2: t0 -= shape[0] if t1 > shape[1] // 2: t1 -= shape[1] return [t0, t1] but I don't understand this part: if t0 > shape[0] // 2: t0 -=

Get music tempo or BPM?

雨燕双飞 提交于 2020-01-24 07:50:26
问题 I'm currently building a game in windows phone 7 using xna i'm trying to get beat per minute from song that played in background song, i also not quite sure if what i want is BPM, what i want is something like pace or tempo in music, faster the pace ,faster the sprites is moving. What i'm thinking right now, BPM is how much a frequency from music hits a range of defined constant, e.g 20 Mhz - 30 MHz, Feel free to correct me if i'm wrong, i'm not really familiar with audio thing, i have tried

python实现歌曲相似度比较(超详细,总有收获)

时光总嘲笑我的痴心妄想 提交于 2020-01-23 18:21:07
python实现歌曲相似度比较 2019/9/20 最近学信号与系统,想着弄个小项目来提高学习兴趣。特此记录一下。 这是大概想到的准备工作,一边推进,一边学吧!!! 2019/9/21 频域信号处理 FFT变换所得的复数的含义: 下标为0的实数表示时域信号的直流部分 下标为i的复数为a+bj表示时域信号中周期为N/i个取样值的正弦波和余弦波的成分,其中a表示余弦波形的成分,b表示正弦波形的成分 复数的模的两倍为对应频率的余弦波的振幅 复数的辐角表示对应频率的余弦波的相位 import numpy as np from scipy.fftpack import fft, ifft import matplotlib.pyplot as plt from matplotlib.pylab import mpl x = np.arange(0, 2*np.pi, 2*np.pi/128) y = 0.3*np.cos(x) + 0.5*np.cos(2*x+np.pi/4) + 0.8*np.cos(3*x-np.pi/3) + np.sin(4*x) + np.cos(x) yf = fft(y)/len(y) print(np.array_str(yf[:5], suppress_small=True)) for ii in range(0, 5): print(np.abs(yf

Understanding Fourier for Seasonality

跟風遠走 提交于 2020-01-23 15:28:51
问题 I am using the auto.arima from the forecast package in R to determine the optimal K-terms for fourier series. After I do that, I want to then calculate the seasonality and plug that one seasonality variable into a multiple regression model. Using the dataset from the forecast package, I was able to extract the optimal amount of fourier terms: library(forecast) ##Public dataset from the forecast package head(gas) ##Choose Optimal Amount of K-Terms bestfit <- list(aicc=Inf) for(i in 1:6) { fit

Understanding Fourier for Seasonality

泄露秘密 提交于 2020-01-23 15:28:10
问题 I am using the auto.arima from the forecast package in R to determine the optimal K-terms for fourier series. After I do that, I want to then calculate the seasonality and plug that one seasonality variable into a multiple regression model. Using the dataset from the forecast package, I was able to extract the optimal amount of fourier terms: library(forecast) ##Public dataset from the forecast package head(gas) ##Choose Optimal Amount of K-Terms bestfit <- list(aicc=Inf) for(i in 1:6) { fit

Choosing Real vs Complex 2D FFTs using apple Accelerate framework

帅比萌擦擦* 提交于 2020-01-23 11:41:40
问题 Can anyone advise on the correct FFT to be using (Real or Complex)? I have looked here but still have questions. I want to do image correlation to identify the location of a sub image within a master image. I understand the basics of FFTs and iFFTs. The plan: Perform an FFT on a master image. 512x512 Take complex conjugate of sub image. Perform an FFT on the sub image. 30x30 but padded with zeros to 512x512 Complex Multiply the two resulting matrixes Perform iFFT on result Even though the

频谱泄漏和拖尾现象

a 夏天 提交于 2020-01-23 05:18:18
吉布斯现象 (英语:Gibbs phenomenon),由 Henry Wilbraham 于1848年最先提出 [1] ,并由 约西亚·吉布斯 于1899年证明 [2] 。在工程应用时常用有限正弦项 正弦波 叠加逼近原周期信号。所用的谐波次数N的大小决定逼近原波形的程度,N增加,逼近的精度不断改善。但是由于对于具有不连续点的周期信号会发生一种现象:当选取的 傅里叶级数 的项数N增加时,合成的波形虽然更逼近原函数,但在不连续点附近会出现一个固定高度的过冲,N越大,过冲的最大值越靠近不连续点,但其峰值并不下降,而是大约等于原函数在不连续点处跳变值的9%,且在不连续点两侧呈现衰减振荡的形式 图像 维基百科 在做信号处理时,经常涉及到“泄漏”。那泄漏是什么,是什么原因造成了泄漏呢?在这将告诉您答案。 1. 信号截断 一次FFT分析截取1帧长度的时域信号,这1帧的长度总是有限的,因为FFT分析一次只能分析有限长度的时域信号。而实际采集的时域信号总时间很长,因此,需要将采样时间很长的时域信号截断成一帧一帧长度的数据块。这个截取过程叫做 信号截断 。 假设有一段10s的时域信号,取1帧的长度 T =1s,无重叠,则该信号将被截断为10帧,如下图所示。按此规律进行FFT计算,将得到10个瞬时频谱,如果将这些瞬时频谱进行平均,那么平均次数为10次,最终的FFT分析结果为这10个瞬时频谱的平均结果。