Question

我正在尝试对加速计数据进行fft。但是，当我绘制幅度（以db为单位）与频率对比时，db scale不会从零开始。它从20开始直到-80。我想我按照他们的方式遵循所有步骤。

将信号零填充到下一个更高的2的幂（如果需要）
应用窗口功能
采用FFT
通过缩放将双面FFT输出转换为单面输出
应用窗口校正因子（相干增益）
查找信号幅度 - 绝对值和db

知道我可能做错了什么吗？我的整个代码如下所示。

library(signal)
library(pracma)

signal_in <- pz$ACC[1:1000]
fs <- 5000
zero_padding <- FALSE
windowfn <- "rectangle"
demean <- FALSE
detrend <- FALSE

# Length of the signal

L <- length(signal_in)

# Removing the trend or mean as required

if (detrend == TRUE) {
    signal_in <- detrend(signal_in, "linear")
} else if (demean == TRUE) {
    signal_in <- detrend(signal_in, "constant")
}

# Various window functions

if (windowfn == "hanning") {
    window_coeff <- hanning(L)
    amp_correction_factor <- mean(window_coeff)
} else if (windowfn == "hamming") {
    window_coeff <- hamming(L)
    amp_correction_factor <- mean(window_coeff)
} else if (windowfn == "flattop") {
    window_coeff <- flattopwin(L, "symmetric")
    amp_correction_factor <- mean(window_coeff)
} else if (windowfn == "rectangle") {
    window_coeff <- rep(1, L)
    amp_correction_factor <- mean(window_coeff)
}

# Apply the window function and if required, zero padding
# L is the length of the original signal
# N is the length of the transformed signal or the number of points in the FFT (to the next higher power of 2)

if (zero_padding == TRUE) {

    N <- 2^nextpow2(L)

    if (L < N) {
        zero_pad <- rep(0, (N - L))
        signal_in_w <- signal_in * window_coeff
        signal_in_w <- c(signal_in_w, zero_pad)
    } else if (L == N) {
        signal_in_w <- signal_in * window_coeff
        N <- L
    }

} else {

    if (L %% 2 != 0) {
        signal_in_w <- signal_in * window_coeff
        signal_in_w <- c(signal_in_w, 0)
        N <- L + 1
    } else {
        signal_in_w <- signal_in * window_coeff
        N <- L
    }

}

# Double-sided FFT

signal.fft <- fft(signal_in_w)

# Single-sided FFT with normalisation - signal amplitude (absolute)

signal.amp.abs <- (2/L) * Mod(signal.fft[1:(length(signal.fft)/2)])

# the DC component should not be doubled so halving it

signal.amp.abs[1] <- signal.amp.abs[1] / 2

# window scaling (coherent gain)

signal.amp.abs <- signal.amp.abs / amp_correction_factor

# signal amplitude (dB)

signal.amp.db <- 20*log10(signal.amp.abs)

# Frequency vector

signal.freq <- seq(from = 0, to = fs/2, length.out = length(signal.fft)/2)

# Plot signal amplitude (absolute) vs frequency (Hz)

plot(signal.amp.abs ~ signal.freq, t = "l")

# Plot signal amplitude (dB) vs frequency (Hz)

plot(signal.amp.db ~ signal.freq, t = "l")

我的计算在db中找到幅度是否正确？

对于FFT幅度，dB标度不从零开始

0 个答案: