about the amplitude of spectrum by fft

As for cos(2*pi*t), the peaks locates at -1 and 1 Hz with a value of 0.5. It's correct.

    t = 0:0.1:10; 
    x = cos(2*pi*t);
    y = fft(x)/length(x);
    real(fftshift(y))

As for sin(2*pi*t), the peak locations are correct but the polarity is reversed from the theoretical predition.

    t = 0:0.1:10; 
    x = sin(2*pi*t);
    y = fft(x)/length(x);
    imag(fftshift(y))

We know that the fourier transform of exp(-pi*t^2) is exp(-pi*f^2). Thus, the peak amplitude is expected to be 1.

    t = -10:0.2:10; 
    t = ifftshift(t);
    x = exp(-pi*t.^2);
    y = fft(x)/length(x);
    real(fftshift(y))

but the peak of the amplitude spectrum by fft is 0.05, not 1 as predicted by the analytic solution.

Answer 1

[Honglei Chen]

why do you say for sin(2*pi*t) the polarity is reversed? The theoretical equation has 1/(2*1i) as the scaling factor, so the result is correct.

As to the Gaussian signal, the relation you quoted is for continuous Fourier transform, not for FFT. Since you don't sample every point on the curve, your frequency domain sample does not end up at 1. The value at f=0 is simply sum(x)/length(x), which is approximately 0.05, matching what you get.

Another way to look at it is that the point in your frequency domain result represents the integration of a small frequency bin, and the width of each bin is fs/N where fs is the sampling rate and N is the number points. In your example, fs is 5 and N is length(x). Therefore, the value is 1*5/length(x).

[Ray Lee]

why the spectral amplitude of sin and cos is not scaled?

[Honglei Chen]

Because they are single frequency so you only get energy at that very point, there is no energy spread in the bin