# why ifft is complex coefficient?

조회 수: 2 (최근 30일)
sangwoo ha 2021년 12월 2일
답변: Nadia Shaik 2022년 2월 3일
i have to multiply complex number to fft of signal
but the symmetric property is eliminated when complex number is multiply
So the ifft is also complex number
and i do multiply on half of fft
and make symmetric property using filplr and conj
but eventhough i match the symmetric property, the ifft of that frequency spectrum is complex number
code is below
why that happened?
help me
clear all;
close all;
fs = 1000;
ts = 1/ fs;
f = 200;
gain = exp(i*10);
t = 0 : ts : 0.1;
x = sin(2*pi*f*t); % original signal
X = fft(x);
X1 = X(1:51)*gain; %multiply complex number
% X1(1) = 1; %when i do this the coeffcient is real
X2 = fliplr(X1); %make symetric
X3 = [X1 conj(X2)];
X4 = X3(1:end-1);
x4 = ifft(X4);
figure(1)
plot(x4);
X1 = X(1:51)*gain;
X2 = fliplr(X1);
X3 = [X1 conj(X2)];
X4 = X3(1:end-1);
x4 = ifft(X4) % coefficient is complex number
figure(2)
plot(x4);

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### 답변 (1개)

Hi,
From my understanding you are getting the expected ifft result when the first element of the input is a real number and a different result when the first element is a complex number.
In ifft , the conjugate symmetry holds good from 2nd element to last element of the input. For a real signal, the first element of fft is always real.
So, while computing ifft, the same is expected. The required real signal can be obtained when the first element is a real value.
Hope it helps!

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