Trying to find the integral under the curve at a certain bandwidth

clear all
close all
clc
L=10;
n=1.45;
c=2.9979e8;
dt = 6e-12;
T=10*2*L*n/c;

t = (-T/2/dt:1:T/2/dt)*dt;
Nt=round(T/dt);
fsine = 1e9;
vsine = 1;
phi = vsine*sin(2*pi*fsine*t);
EL1t=1.274e7*exp(1i*phi);
FP=fft(phi);
fs=1/dt/Nt;
Fs=(-1/dt/2:fs:1/dt/2-1);
Z=plot(Fs,fftshift(abs(fft(EL1t))));

%xlim([-0.5e10 0.5e10]);
pow = fftshift(abs(fft(EL1t)));
freq_f = 16e9;
% nnz(Fs==freq_f) % returns 0
[val,ind] = min(abs(Fs-freq_f)); % Find freq closest to 16 GHz
disp(Fs(ind));

   1.6001e+10

disp(pow(ind));

   3.2278e+07

I wanted to find the area under the curve at 16GHz with 50MHz on the right and left side range.

I am trying to use trapz command but still a bit confused...

[Mathieu NOE]

hello

not sure to understand what your goal is

if your signal is a sine wave at f = fsine = 1e9, there should be zero power around 16 GHz +/- 50 MHz

[Yogesh]

Hii , I am trying to find the area under the fft curve mentioned above at 16GHz +/-50MHz..

Answer 1

[Star Strider]

Try this —

clear all
close all
clc
L=10;
n=1.45;
c=2.9979e8;
dt = 6e-12;
T=10*2*L*n/c;

t = (-T/2/dt:1:T/2/dt)*dt;
Nt=round(T/dt);
fsine = 1e9;
vsine = 1;
phi = vsine*sin(2*pi*fsine*t);
EL1t=1.274e7*exp(1i*phi);
FP=fft(phi);
fs=1/dt/Nt;
Fs=(-1/dt/2:fs:1/dt/2-1);

figure
Z=plot(Fs,fftshift(abs(fft(EL1t))));
xline([-1 1]*50E+6 + 16E+9, '--r', 'Requested Frequency Range')

figure
Z=plot(Fs,fftshift(abs(fft(EL1t))));
xlim([-1 1]*50E+6 + 16E+9)                                         % Set 'xlim' Values To Show The Desired Frequency Range
title('Signal In Requested Frequency Range')

freqlims = 16E+9 + [-1 1]*50E+6;
% idxrng = (Fs >= 16E+9-50E+6) & (Fs <= 16E+9-50E+6)
% q = nnz(idxrng)

magvals = interp1(Fs, fftshift(abs(fft(EL1t))), freqlims);      % Interpolate To Get The Required Magnitude Values In The Desired Frequency Range

AUC = trapz(freqlims, magvals);
fprintf('\nThe area under the curve between %.5E Hz and %.5E Hz is %.5E units.\n', freqlims, AUC)

The area under the curve between 1.59500E+10 Hz and 1.60500E+10 Hz is 3.22793E+15 units.

%xlim([-0.5e10 0.5e10]);
pow = fftshift(abs(fft(EL1t)));
freq_f = 16e9;
% nnz(Fs==freq_f) % returns 0
[val,ind] = min(abs(Fs-freq_f)); % Find freq closest to 16 GHz
disp(Fs(ind));

   1.6001e+10

disp(pow(ind));

   3.2278e+07

There are no index values in the requested frequency range, so the only option is to interpolate the values that are, and use them as arguments to trapz. This approach would work for any frequency range within ‘Fs’.

.

[Mathieu NOE]

yeap

there is simply no energy in that frequency band

[Yogesh]

Thank you @Star Strider...

You were really helpful!!!...

[Star Strider]

@Yogesh — As always, my pleasure!

Thank you!