Doubt in FFT and one-third octave analysis

I am trying to do FFT of signal in time domain and would like to check IFFT, so that my FFT is correct. But my IFFT doesnot matches original signal. Here is the code attached. Further I am trying to do one third octave band value of the signal (dB) vs frequency, but no idea about one-third octave analysis.

clc;clear all;close all;

%% read data

data_1=xlsread('Force_pier.xlsx');% Signal data in Time-Domain

t_com=xlsread('time.xlsx');% time data

figure(1);plot(t_com,data_1,'linewidth',2);title('Time response');% response in time domain

N=length(data_1);% Number Of Samples

Ts=mean(diff(t_com));% Sampling Interval

Fs=1/Ts;% Sampling Frequency

Fn=Fs/2;% Nyquist Frequency

%% FFT

FT_signal=fft(data_1)/N;% Normalized Fourier Transform Of Data

Fv=linspace(0,1,fix(N/2)+1)*Fn; % Frequency Vector (For ‘plot’ Call)

Iv=1:length(Fv);% Index Vector (Matches ‘Fv’)

figure(2);plot(Fv,abs(FT_signal(Iv))*2,'linewidth',2);title('frequency response');% FFT of the signal in frequency domain

%% IFFT

yy=ifft((FT_signal)); % inverse fft

N2=length(FT_signal); % determine the length of the signal

dt2=1/Fs; % determine the time increment

tim=0:dt2:(N2-1)*dt2; %create the time axis

figure(3);plot(tim,yy,'linewidth',2);title('back to time domain');

[Mathieu NOE]

hello

file missing : time.xlsx

[Pooja Kumari]

Can you share the xlsx file?

[David Goodmanson]

Hi SP

if the time array is not equally spaced, you should not expect to get agreement. Leaving that aside, you have

FT_signal=fft(data_1)/N;                % Normalized Fourier Transform Of Data
yy=ifft((FT_signal));                   % inverse fft

But fft and ifft are inverses of each other, so if you multiply the fft by 1/N (which is commonly done), then to get back to the original signal you have to multiply the ifft output (yy here) by N. Otherwise you're off by a factor of N.

[Susmita Panda]

Thanks!