Python 如何用FFT确定频变幅值

我试图测量和计算音频设备的频率响应，只需生成输入信号，然后用扫频测量输出.

以下是我试图在伪代码中实现的:

for each frequency f in (10-20k):
    generate reference signal s with frequency f
    async play s and record result r
    determine amplitude a of r using FFT
    add tuple (f,a) to result

在Python中:

import numpy as np
from matplotlib import pyplot as plt
import sounddevice as sd
import wave
from math import log10, ceil
from scipy.fft import fft, rfft, rfftfreq, fftfreq

SAMPLE_RATE = 44100     # Hertz
DURATION = 3            # Seconds

def generate_sine_wave(freq, sample_rate, duration):
    x = np.linspace(0, duration, int(sample_rate * duration), endpoint=False)
    frequencies = x * freq
    y = np.sin((2 * np.pi) * frequencies)
    return x, y

def main():
    # info about our device
    print(sd.query_devices(device="Studio 24c"))

    # default device settings
    sd.default.device = 'Studio 24c'
    sd.default.samplerate = SAMPLE_RATE

    f_start = 10
    f_end = 20000
    samples_per_decade = 10
    ndecades = ceil(log10(f_end) - log10(f_start))
    npoints = ndecades * samples_per_decade
    freqs = np.logspace(log10(f_start), log10(f_end), num=npoints, endpoint=True, base=10)
    
    measure_duration = 0.25 # seconds
    peaks = []
    for f in freqs:
        _, y = generate_sine_wave(f, SAMPLE_RATE, measure_duration)
        rec = sd.playrec(y, SAMPLE_RATE, input_mapping=[2], output_mapping=[2])
        sd.wait()
        yf = np.fft.rfft(rec, axis=0)
        yf_abs = 1 / rec.size * np.abs(yf)
        xf = np.fft.rfftfreq(rec.size, d=1./SAMPLE_RATE)
        peaks.append(np.max(yf_abs))

    plt.xscale("log")
    plt.scatter(freqs,peaks)
    plt.grid()
    plt.show()

if __name__ == "__main__":
    main()

在测量实际设备之前，我简单地将输出信号循环到音频接口的输入端，以基本上"校准"我正在做的事情.我以为振幅在所有频率上都是相等的.然而，这就是我得到的:

为什么所有的振幅都是到处都是，即使产生的正弦波是相同的？在扫描测量阶段，我没有改变音频接口上的任何东西.