Convert Time Domain Signal Data into Frequency Domain, How to handle the imaginary terms?

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Tan
Tan 2012-6-2
回答： jagadeesh jagadeesh 2019-10-28
Hi Everyone,
I'm a beginner of MATLAB. I'm having some problems of converting time domain signal into frequency domain.
In time domain, is this the way to plot the graph?
value = [42007935 111212895 184546560 238219725 238219725 184546560 111212895 42007935]; time = [0, 1, 2, 3, 4, 5, 6, 7, 8 ]; plot (time, value);
Then, when i want to plot it in frequency domain, i use the following codes:
fft_Conv = fft (value); plot (time, fft_Conv);
I get a warning: Warning: Imaginary parts of complex X and/or Y arguments ignored.
Could anyone guide me how to plot signal in time domain and frequency domain?
Regards, HJ.C
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shiva prasad kattula
shiva prasad kattula 2016-3-5
编辑：shiva prasad kattula 2016-3-5
The frequency-domain representation of a signal carries information about the signal's magnitude and phase at each frequency. This is why the output of the FFT computation is complex. A complex number, x, has a real, x_r, and an imaginary part, x_i, such that x = x_r + ix_i. The magnitude of is computed as sqrt{(x_r^2+x_i^2)}, and the phase of x is computed as arctan{(x_i/x_r)}. You can use MATLAB functions abs and angle to respectively get the magnitude and phase of any complex number.

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Wayne King
Wayne King 2012-6-2
You want to plot the magnitude and phase separately for the complex-valued data. Think of the polar form of a complex number. A few other things: you want to create a frequency vector to use in plotting the frequency domain data (DFT). You may or may not want to center 0 frequency in your Fourier transform, I do this below. Because the mean of your time data is so large, you are going to get a large 0 frequency magnitude in your Fourier transform.
value = [42007935 111212895 184546560 238219725 238219725 184546560 111212895 42007935];
time = 0:7;
stem(time,value,'markerfacecolor',[0 0 1])
title('Time Data'); xlabel('Time'); ylabel('Amplitude');
dftvalue = fft(value);
freq = -pi:(2*pi)/length(value):pi-(2*pi)/length(value);
figure;
stem(freq,abs(fftshift(dftvalue)),'markerfacecolor',[0 0 1])
title('Magnitude of DFT'); xlabel('Frequency'); ylabel('Magnitude');
figure;
stem(freq,angle(fftshift(dftvalue)),'markerfacecolor',[0 0 1])
title('Phase of DFT');
xlabel('Frequency'); ylabel('Magnitude');
% or stem(freq,unwrap(angle(dftvalue)),'markerfacecolor',[0 0 1])
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Tan
Tan 2012-6-2
The values in the value array are actually the result of an impulse with 1V passing through a low pass filtered with frequency coefficients shown below.
Numerator: Denominator:
[3.91719822748777E-02, [1.000]
0.103629842929331,
0.171922134825388,
0.221881476438683,
0.221881476438683,
0.171922134825388,
0.103629842929331,
3.91719822748777E-02]
I intend to see if the frequency response of the impulse after pass through the filter really response like what i expect.
Tan
Tan 2012-6-2
Numerator:
[3.91719822748777E-02,
0.103629842929331,
0.171922134825388,
0.221881476438683,
0.221881476438683,
0.171922134825388,
0.103629842929331,
3.91719822748777E-02]
Denominator:
[1.000]

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更多回答（3 个）

Wayne King
Wayne King 2012-6-2
Then you want to use freqz()
b = [3.91719822748777E-02,
0.103629842929331,
0.171922134825388,
0.221881476438683,
0.221881476438683,
0.171922134825388,
0.103629842929331,
3.91719822748777E-02];
[H,W] = freqz(b,1);
plot(W,20*log10(abs(H)))
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Wayne King
Wayne King 2012-6-2
If you feed it the impulse AND the system is LTI, then yes, the Fourier transform of the impulse response is the frequency response. Just keep in mind that by giving me that b vector above you are asserting that is the impulse response and that system has a finite impulse response
Tan
Tan 2012-6-2
Yes. My system a linear shift-invariant system and the input signal is an impulse.
So if i want to plot the response out when we are just given:
value = [42007935 111212895 184546560 238219725 238219725 184546560 111212895 42007935];
How should i do?
I'm sorry if i ask a repeating question, because i have vague idea on this

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Wayne King
Wayne King 2012-6-2
Then you are telling me that
b = [42007935 111212895 184546560 238219725 238219725 184546560 111212895 42007935];
% so
[h,f] = freqz(b,1);
plot(f,20*log10(abs(h)));
Of course if you know the sampling frequency, then you can use that in freqz() to get the response in Hz.
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jagadeesh jagadeesh
jagadeesh jagadeesh 2019-10-28
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sample time
L = 1000; % Length of signal
t = (0:L-1)*T; % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
y = x + 2*randn(size(t)); % Sinusoids plus noise
figure(1)
plot(Fs*t(1:50),y(1:50))
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('time (milliseconds)'
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
% Plot single-sided amplitude spectrum.
figure(2)
plot(f,2*abs(Y(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')

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