3
這裏是場景：使用頻譜分析儀我有輸入值和輸出值。採樣的數量爲32000，採樣率爲2000採樣/秒，輸入爲50 hz的正弦波，輸入爲電流，輸出爲壓力，單位爲psi。在MATLAB中使用FFT的頻率響應
如何使用MATLAB中的FFT函數使用MATLAB， 來計算該數據的頻率響應。
我能夠產生一個正弦波，其給出了在幅度和相角，在這裏的是，我使用的代碼：我從相位曲線的頻率響應在90度
%FFT Analysis to calculate the frequency response for the raw data
%The FFT allows you to efficiently estimate component frequencies in data from a discrete set of values sampled at a fixed rate
% Sampling frequency(Hz)
Fs = 2000;
% Time vector of 16 second
t = 0:1/Fs:16-1;
% Create a sine wave of 50 Hz.
x = sin(2*pi*t*50);
% Use next highest power of 2 greater than or equal to length(x) to calculate FFT.
nfft = pow2(nextpow2(length(x)))
% Take fft, padding with zeros so that length(fftx) is equal to nfft
fftx = fft(x,nfft);
% Calculate the number of unique points
NumUniquePts = ceil((nfft+1)/2);
% FFT is symmetric, throw away second half
fftx = fftx(1:NumUniquePts);
% Take the magnitude of fft of x and scale the fft so that it is not a function of the length of x
mx = abs(fftx)/length(x);
% Take the square of the magnitude of fft of x.
mx = mx.^2;
% Since we dropped half the FFT, we multiply mx by 2 to keep the same energy.
% The DC component and Nyquist component, if it exists, are unique and should not be multiplied by 2.
if rem(nfft, 2) % odd nfft excludes Nyquist point
mx(2:end) = mx(2:end)*2;
else
mx(2:end -1) = mx(2:end -1)*2;
end
% This is an evenly spaced frequency vector with NumUniquePts points.
f = (0:NumUniquePts-1)*Fs/nfft;
% Generate the plot, title and labels.
subplot(211),plot(f,mx);
title('Power Spectrum of a 50Hz Sine Wave');
xlabel('Frequency (Hz)');
ylabel('Power');
% returns the phase angles, in radians, for each element of complex array fftx
phase = unwrap(angle(fftx));
PHA = phase*180/pi;
subplot(212),plot(f,PHA),title('frequency response');
xlabel('Frequency (Hz)')
ylabel('Phase (Degrees)')
grid on
相角，這是計算頻率響應的正確方法嗎？
如何將此響應與從分析儀獲取的值進行比較？這是一個交叉檢查，看看分析器邏輯是否有意義。
謝謝保羅，感謝您的幫助.... – Jerry 2010-10-22 14:43:50