Two-Sided Frequency Spectrum

Question

I am trying to make FFT simulation in Matlab by generating noise added two sinus waves in 60Hz and 100Hz.

After adding the noise into these signals then I have applied the FFT as I put my Matlab code below.

But I am in difficulty interpreting the FFT plot which shows two spectral peaks in the plot diagram below.

Could you please explain how to interpret this spectrum? Why we see two peaks in the spectrum? How can we reduce the spectral to show only the frequencies of 60Hz and 100Hz?

%%%% Noise_Added_Two_Sinus%%%
>> f1=60;
>> f2=100;
>> fs=512;
>> t=0:1/fs:2-1/fs;
>> x1=2.4*sin(2*pi*f1*t);
>> x2=0.96*sin(2*pi*f2*t);
>> y=x1+x2+randn(size(t));
>> F=fft(y);
>> plot(abs(F))

Answer 1

First of all, you should totally fix the $x$-axis :)

Now, as you only have two sine waves, you should expect to have peaks at their exact frequencies (that is, at $\pm 60\textrm{ Hz}$ and $\pm 100\textrm{ Hz}$). The FFT function in MATLAB gives you the Discrete Fourier Transform of your signal within $0$ and $F_s$, where $F_s$ is your sampling frequency -and given that you represent your signal with samples, even if you are not aware of it, you have one. Anyway, to make visualization easier, we focus on the interval $[-\frac{F_s}{2},\frac{F_s}{2}]$; that way, your spectrum will be centered around the $0$ frequency -use the function fftshift for that. So, now, once you fix the $x$-axis according to your sampling frequency, you will find that the peaks are at the frequencies you have defined for your sine waves.

When it comes to the little ripples in the lower part of your spectrum, that's just disturbance caused by the noise. Since you are adding noise to the signal by the randn function, your noise follows a Gaussian distribution, translating into white noise for your signal. That is why your entire spectrum is showing sings of being slightly disturbed.

By the way, after fftshift, just get rid of the first half of your vector and you'll end up with positive frequencies.

Answer 2

The Fourier transform of a real-valued signal has Hermitian Symmetry to $f=0$. Always.

This is a direct result of applying the Fourier transform. This really should not surprise you; if it does, I can only encourage you to manually calculate the Fourier transform of e.g. $\cos(2\pi f_{0} t)$ by considering the trigonometric functions as sum of complex oscillations, i.e. considering Euler's formula:

$$\begin{align*} \cos(t) &= \frac 12 \left(e^{it}+e^{-it}\right)\\ \mathcal{F}\left\{\cos(t)\right\}&= \int\limits_{-\infty}^{\infty}\cos(t)e^{i2\pi f t}\,dt\\ &=\frac12\int\limits_{-\infty}^{\infty}\left(e^{it}+e^{-it}\right)e^{i2\pi f t}\,dt\tag{*} \end{align*}$$

The last line shows that the result of the integral must be conjugate-symmetrical to $f=0$.

Answer 3

Using a simple code for displaying the FFT of real signals (FFTR.m), in a one-sided fashion, trying to directly have amplitudes on the $y$-axis, let us just change the final part of your code:

%%%% Noise_Added_Two_Sinus%%%
f1=60;
f2=100;
fs=512;
t=0:1/fs:2-1/fs;
x1=2.4*sin(2*pi*f1*t);
x2=0.96*sin(2*pi*f2*t);
y=x1+x2+randn(size(t));

time_Sampling = 1/fs;
[fft_R,fft_Axe] = FFTR(y',time_Sampling);
[fft_R_Max,fft_R_Idx] = max(fft_R);
figure; hold on
h1 = plot(fft_Axe,fft_R,'x');axis tight;grid on;xlabel('Frequency (Hz)');ylabel('Amplitude (a. u.)');
h2 = plot(fft_Axe(fft_R_Idx),fft_R(fft_R_Idx),'ro');
set([h1,h2],'LineWidth',6)

you get:

I try to keep single points, to avoid interpret linear segments between frequency bins.