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If we do FT of cosine wave, then sine wave will be orthogonal. So, the imag parts of FT will be 0. This is my think. But, the result isn't show that. Like this. enter image description here

Figure 1. Real and Imag parts (Y label is Frequency)

Below figure is about FFT result and phase(radian). enter image description here

Figure 2.

Code: fs=128; t=-4*pi:1/fs:4*pi; x=1*cos(0.5*t); X=fft(x); N=length(x); n=0:N-1; f=n*fs/N;

enter image description here

Figure 3. code: a=cos(2*pi*t); plot(t,a) A=fft(a); N=length(a); n=0:N-1; f=fs * n/N; phs=angle(fftshift(a));

enter image description here

Figure 4. 1st: Real and Imag parts 2nd: Real with Frequency 3rd: imag with Frequency

Please let me know why Imag part occur in cosine FT.

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