How can I get the FFT of a flat signal with a peak?

Hello everyone,

I am an aerospace engineering student. I am working on a project in which I have a certain aircraft that must be solicited with an impulse that makes the elevator move with respect to its equilibrium position. I have assigned the command law in this way

global dE_eq %this value comes from a previous simulation that finds the trimmed position of the elevator

t_f = 120; %lenght of the command law

vBreakPointsT = [0,     t_f/240, t_f/120,   (3/200)*t_f, t_f/60,...
                        t_f/48,  t_f/40,    t_f/24,      t_f/12,...
                        t_f/6,   t_f*(5/6), t_f];

vBreakPointsD = [dE_eq, dE_eq,     dE_eq,       dE_eq,          2.5*dE_eq,...  
                        dE_eq,     dE_eq,       dE_eq,          dE_eq,...
                        dE_eq,     dE_eq,       dE_eq];  


dE = @(t) interp1(vBreakPointsT, vBreakPointsD, t, "pchip");

and I would like to make a spectral analysis of dE(t), looking for the dominant frequency. Physically, does it make sense to implement an analysis with the default FFT tool? Thank you

Answer 1

[William Rose]

@Leonardo Molino,

You are correct to have doubts about the interpretation of the FFT of dE(t), given that dE(t) is computed the way you have done it. You use the pchip method for doing the interpolation. The FFT of dE(t) will provide information about the frequency content of pchip rather than learning about the aircraft response. You can demonstrate that this is true by trying other methods for the interpolation, such as spline, linear, nearest.... You will see that the FFT is different when you use a different interpolation method.

[William Rose]

@Leonardo Molino, Here is a demonstration of what I meant in my answer. Notice that the plots of the FFTs of the differently interpolated signals are different at high frequencies (top right plot, below), but the FFTs are essentially the same at low frequencies - just a simple spike at frequency=0 (bottom right plot, below). The FFTs are not telling us about the aircraft response. They are just telling us about the interpolation method we used and, to some extent, they are telling us about where we chose the breakpoints.

dE_eq=1;        %trimmed elevator position
t_f = 120;      %length of command law (s)
vBreakPointsT = [0,1/240,1/120,3/200,1/60,1/48,...
                 1/40,1/24,1/12,1/6,5/6,1]*t_f;
vBreakPointsD = [1,1,1,1,2.5,1,1,1,1,1,1,1]*dE_eq;  
N=2400;             %number of points in interpolated signal
t=(0:N-1)*t_f/N;    %time values (s)
dE1 = interp1(vBreakPointsT, vBreakPointsD, t, "pchip");
dE2 = interp1(vBreakPointsT, vBreakPointsD, t, "linear");
dE3 = interp1(vBreakPointsT, vBreakPointsD, t, "nearest");
% Compute the FFTs
DE1 = fft(dE1);
DE2 = fft(dE2);
DE3 = fft(dE3);
df  = 1/t_f;        %frequency spacing of FFT (Hz)
frq =(0:N/2)*df;    %frequencies of lower half of FFT (Hz)

%% Plot results
%  Lower plots are zoomed-in versions of the upper plots.
figure;
subplot(221)
plot(vBreakPointsT,vBreakPointsD,'ks',t,dE1,'-r.',t,dE2,'-g.',t,dE3,'-b.')
xlabel('Time (s)'); ylabel('dE'); grid on
legend('Break points','pchip','linear','nearest')
subplot(222)
semilogy(frq,abs(DE1(1:N/2+1)),'-r.',frq,abs(DE2(1:N/2+1)),'-g.',frq,abs(DE3(1:N/2+1)),'-b.')
xlabel('Frequency (Hz)'); ylabel('|DE|'); grid on; yticks(10.^(-2:3))
legend('pchip','linear','nearest')
subplot(223)
plot(vBreakPointsT,vBreakPointsD,'ks',t,dE1,'-r.',t,dE2,'-g.',t,dE3,'-b.')
xlabel('Time (s)'); ylabel('dE'); grid on; xlim([1,3])
%legend('Break points','pchip','linear','nearest')
subplot(224)
semilogy(frq,abs(DE1(1:N/2+1)),'-r.',frq,abs(DE2(1:N/2+1)),'-g.',frq,abs(DE3(1:N/2+1)),'-b.')
xlabel('Frequency (Hz)'); ylabel('|DE|'); grid on; xlim([0,.1]); yticks(10.^(1:3))
legend('pchip','linear','nearest')

Try the above. Good luck.

[Leonardo Molino]

Thank your very much for you answer, it is more than what I've asked!

[William Rose]

You're wlcome @Leonardo Molino.