Understanding this FFT Graph

I have plotted a graph using FFT function; using a sweep sine wave.

The code is generic NFFT = 2^nextpow2(L) and using absolute values to plot FFT values on the Y axis.

My question is; what would the label be for the Y-axis and secondly, is the X axis just frequency in the time domain?

Any help greatly appreciated.

T=5; %size of window
fs=44100; %sampling frequency
df=1/T; %frequency res
dt=1/fs; %time resolution
t=(0:+dt:T-dt); %time vector
df_t=500; %swept rate (Hz/seconds)

% pre-allocate size of t:

sweptsin = zeros(size(t));
for i=1:+1:length(t)
    %i=i+1;
    
    if(i==1)      %initialise f and t.
        f=20; ti=0;  
    else    
        ti=ti+dt; %time increment
        f=f+df_t*dt; %freq increment 
    end
    
    w=2*pi*f; %omega
    sweptsin(i)=sin(w*ti); %swept sine wave

end

sound(sweptsin, fs)


figure
y = cos(2*pi*f*t)+randn(size(t));
L = 1000;
NFFT = 2^nextpow2(L);
Y = abs(fft(y,NFFT))/L;
f = fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y(1:NFFT/2+1)))
xlim([0 22050])
ylim([0 0.8])

Answer 1

[Star Strider]

‘My question is; what would the label be for the Y-axis and secondly, is the X axis just frequency in the time domain?’

The Y-axis here would be labeled ‘Amplitude’. It represents the amplitude of the signal components at each frequency. (If it were in decibels, it would be either 'dB' or 'Power (dB)' since converting it to decibels implicitly squares the amplitude creating units of power, and then takes the base-10 logarithm of it.)

The X-axis is the frequency axis, and the units are in cycles/(time unit). The frequency vector is calculated to extend from D-C (zero cycles/(time unit)) to the Nyquist frequency (one-half the sampling frequency), the highest frequency that can be uniquely determined in a sampled signal. The sampling frequency (samples/(time unit)) is the inverse of the sampling interval ((time units)/sample), relating the time vector to the frequency vector.

In a ‘normalised’ signal, the frequency extends from zero to π randians/(time unit). The frequency vector is calculated with respect to the length of the fft result, so that each element of the fft result is assigned a specific frequency between zero and the Nyquist frequency (or π). If the time unit is in units of seconds, the frequency vector is units of Hz.

T=5; %size of window
fs=44100; %sampling frequency
df=1/T; %frequency res
dt=1/fs; %time resolution
t=(0:+dt:T-dt); %time vector
df_t=500; %swept rate (Hz/seconds)

% pre-allocate size of t:

sweptsin = zeros(size(t));
for i=1:+1:length(t)
    %i=i+1;
    
    if(i==1)      %initialise f and t.
        f=20; ti=0;  
    else    
        ti=ti+dt; %time increment
        f=f+df_t*dt; %freq increment 
    end
    
    w=2*pi*f; %omega
    sweptsin(i)=sin(w*ti); %swept sine wave

end

sound(sweptsin, fs)


figure
y = cos(2*pi*f*t)+randn(size(t));
L = 1000;
NFFT = 2^nextpow2(L);
Y = abs(fft(y,NFFT))/L;
f = fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y(1:NFFT/2+1)))
xlim([0 22050])
ylim([0 0.8])
xlabel('Frequency')
ylabel('Amplitude')

.

Answer 2

[David Kendal]

I did assume it was amplitude earlier on Y. I’m getting very tired it’s late here and I don’t code well when I’m falling asleep 😆…on note to your reply - that’s very informative and given me a few things to think about in terms of the Nyquist frequency. Thanks again.

[Star Strider]

As always, my pleasure!