If you do an FFT plot of a simple signal, like:

t = 0:0.01:1 ;
N = max(size(t));
x = 1 + sin( 2*pi*t ) ;
y = abs( fft( x ) ) ;
stem( N*t, y )

1Hz sinusoid + DC


FFT of above


I understand that the number in the first bin is "how much DC" there is in the signal.

y(1)  %DC
  > 101.0000

The number in the second bin should be "how much 1-cycle over the whole signal" there is:

y(2)  %1 cycle in the N samples
  > 50.6665

But it's not 101! It's about 50.5.

There's another entry at the end of the fft signal, equal in magnitude:

  > 50.2971

So 50.5 again.

My question is, why is the FFT mirrored like this? Why isn't it just a 101 in y(2) (which would of course mean, all 101 bins of your signal have a 1 Hz sinusoid in it?)

Would it be accurate to do:

mid = round( N/2 ) ;

% Prepend y(1), then add y(2:middle) with the mirror FLIPPED vector
% from y(middle+1:end)
z = [ y(1), y( 2:mid ) + fliplr( y(mid+1:end) ) ];

stem( z )

Flip and add-in the second half of the FFT vector

enter image description here

I thought now, the mirrored part on the right hand side is added in correctly, giving me the desired "all 101 bins of the FFT contain a 1Hz sinusoid"

>> z(2)

ans =

  • 1
    $\begingroup$ A similar question has been answered here: dsp.stackexchange.com/questions/3466/… $\endgroup$ Commented Oct 28, 2012 at 17:35
  • 2
    $\begingroup$ But this is specifically about the symmetry (I believe it's called Hermetian symmetry?) of the signal. $\endgroup$
    – bobobobo
    Commented Oct 28, 2012 at 17:44
  • 2
    $\begingroup$ For a pure real signals F(k)=conj(F(N-k)), this is why the Fourier transform of a pure real signal is symmetric. $\endgroup$
    – WebMonster
    Commented Oct 28, 2012 at 18:02
  • 4
    $\begingroup$ Because a Fourier transform breaks up a signal into complex exponentials, and a sine wave is the sum of 2 complex exponentials. dsp.stackexchange.com/a/449/29 $\endgroup$
    – endolith
    Commented Oct 28, 2012 at 19:55
  • 2
    $\begingroup$ What concerns me a bit is that the two bins have magnitudes $50.6665$ and $50.2971$. Theoretically, they should have equal magnitude $50.5$ as you say, but the difference is a little larger than I feel is attributable to round-off error. Could you re-run your program and tell us what the actual complex values are in those two bins? (Theoretically, the two bin values should be complex conjugates with imaginary parts close to $\pm 50.5$ and small real parts). $\endgroup$ Commented Oct 31, 2012 at 18:31

3 Answers 3


Real signals are "mirrored" in the real and negative halves of the Fourier transform because of the nature of the Fourier transform. The Fourier transform is defined as the following-

$$H(f) = \int h(t)e^{-j2\pi ft}dt$$

Basically it correlates the signal with a bunch of complex sinusoids, each with its own frequency. So what do those complex sinusoids look like? The picture below illustrates one complex sinusoid.

enter image description here enter image description here enter image description here

The "corkscrew" is the rotating complex sinusoid in time, while the two sinusoids that follow it are the extracted real and imaginary components of the complex sinusoid. The astute reader will note that the real and imaginary components are the exact same, only they are out of phase with each other by 90 degrees ($\frac{\pi}{2}$). Because they are 90 degrees out of phase they are orthogonal and can "catch" any component of the signal at that frequency.

The relationship between the exponential and the cosine/sine is given by Euler's formula-

$$e^{jx} = \cos(x) + j\cdot\sin(x)$$

This allows us to modify the Fourier transform as follows- $$ H(f) = \int h(t)e^{-j2\pi ft}dt \\ = \int h(t)(\cos(2\pi ft) - j\cdot\sin(2\pi ft))dt $$

At the negative frequencies the Fourier transform becomes the following- $$ H(-f) = \int h(t)(\cos(2\pi (-f)t) - j\sin(2\pi (-f)t))dt \\ = \int h(t)(\cos(2\pi ft) + j\cdot\sin(2\pi ft))dt $$

Comparing the negative frequency version with the positive frequency version shows that the cosine is the same while the sine is inverted. They are still 90 degrees out of phase with each other, though, allowing them to catch any signal component at that (negative) frequency.

Because both the positive and negative frequency sinusoids are 90 degrees out of phase and have the same magnitude, they will both respond to real signals in the same way. Or rather, the magnitude of their response will be the same, but the correlation phase will be different.

EDIT: Specifically, the negative frequency correlation is the conjugate of the positive frequency correlation (due to the inverted imaginary sine component) for real signals. In mathematical terms, this is, as Dilip pointed out, the following

$H(-f) = [H(f)]^*$

Another way to think about it:

Imaginary components are just that..Imaginary! They are a tool, which allows the employ of an extra plane to view things on and makes much of digital (and analog) signal processing possible, if not much easier than using differential equations!

But we can't break the logical laws of nature, we can't do anything 'real' with the imaginary content$^\dagger$ and so it must effectively cancel itself out before returning to reality. How does this look in the Fourier Transform of a time based signal(complex frequency domain)? If we add/sum the positive and negative frequency components of the signal the imaginary parts cancel, this is what we mean by saying the positive and negative elements are conjugate to each-other. Notice that when an FT is taken of a time-signal there exists these conjugate signals, with the 'real' part of each sharing the magnitude, half in the positive domain, half in the negative, so in effect adding the conjugates together removes the imaginary content and provides the real content only.

$^\dagger$ Meaning we can't create a voltage that is $5i$ volts. Obviously, we can use imaginary numbers to represent real-world signals that are two-vector-valued, such as circularly polarized EM waves.

  • 1
    $\begingroup$ Good answer - one slight nitpick though, I am not on-board with "Because they are the same, anything that one correlates with, the other will too with the exact same magnitude and a 90 degree phase shift.". I know what you are trying to say, however (as you know), a sine correlates with a sine (score 1), but wont correlate at all with a cosine at all, (score 0). They are the same signal, but with different phases afterall. $\endgroup$
    – Spacey
    Commented Oct 29, 2012 at 12:54
  • $\begingroup$ You're right. There's another more serious problem too. I will fix it later. $\endgroup$
    – Jim Clay
    Commented Oct 29, 2012 at 13:23
  • $\begingroup$ It would be nice if you could edit your answer to be more responsive to the question which is about DFTs (though it says FFT in the title) rather than giving the general theory of Fourier transforms. $\endgroup$ Commented Oct 29, 2012 at 22:35
  • $\begingroup$ @DilipSarwate My goal is to help the questioner understand, and I think my approach is best for that. I have upvoted your answer, though, for doing the discrete math. $\endgroup$
    – Jim Clay
    Commented Oct 29, 2012 at 23:01
  • $\begingroup$ @JimClay Your approach is greatly appreciated by the entire readership of dsp.SE, and I hope that you will find the time to make your answer a truly great answer by explicitly including in your answer what is currently left for the reader to deduce: viz. that the equations show that $H(-f) = [H(f)]^*$ (and hence $|H(-f)| = |H(f)|$) when $x(t)$ is a real-valued signal and that this is the "mirroring" that the OP was asking about. In other words, I request that you edit your answer to be more responsive to the question actually asked (as I requested in my previous comment). $\endgroup$ Commented Oct 30, 2012 at 12:42

The FFT (or Fast Fourier Transform) is actually an algorithm for the computation of the Discrete Fourier Transform or DFT. The typical implementation achieves speed-up over the conventional computation of the DFT by exploiting the fact that $N$, the number of data points, is a composite integer which is not the case here since $101$ is a prime number. (While FFTs exist for the case when $N$ is a prime, they use a different formulation that might or might not be implemented in MATLAB). Indeed, many people deliberately choose $N$ to be of the form $2^k$ or $4^k$ so as to speed up the DFT computation via the FFT.

Turning to the question as to why the mirroring occurs, hotpaw2 has essentially stated the reason, and so the following is just a filling in of the details. The DFT of a sequence $\mathbf x = \bigr(x[0], x[1], x[2], \ldots, x[N-1]\bigr)$ of $N$ data points is defined to be a sequence $\mathbf X =\bigr(X[0], X[1], X[2], \ldots, X[N-1]\bigr)$ where $$X[m] = \sum_{n=0}^{N-1} x[n]\left(\exp\left(-j2\pi \frac{m}{N}\right)\right)^n, m = 0, 1, \ldots, N-1$$ where $j = \sqrt{-1}$. It will be obvious that $\mathbf X$ is, in general a complex-valued sequence even when $\mathbf x$ is a real-valued sequence. But note that when $\mathbf x$ is a real-valued sequence, $\displaystyle X[0]=\sum_{n=0}^{N-1} x[n]$ is a real number. Furthermore, if $N$ is an even number, then, since $\exp(-j\pi) = -1$, we also have that $$X\left[\frac{N}{2}\right] = \sum_{n=0}^{N-1} x[n]\left(\exp\left(-j2\pi \frac{N/2}{N}\right)\right)^n = \sum_{n=0}^{N-1} x[n](-1)^n$$ is a real number. But, regardless of whether $N$ is odd or even, the DFT $\mathbf X$ of a real-valued sequence $\mathbf x$ has Hermitian symmetry property that you have mentioned in a comment. We have for any fixed $m$, $1 \leq m \leq N-1$, $$\begin{align*} X[m] &= \sum_{n=0}^{N-1} x[n]\left(\exp\left(-j2\pi \frac{m}{N}\right)\right)^n\\ X[N-m] &= \sum_{n=0}^{N-1} x[n]\left(\exp\left(-j2\pi \frac{N-m}{N}\right)\right)^n\\ &= \sum_{n=0}^{N-1} x[n]\left(\exp\left(-j2\pi + j2\pi\frac{m}{N}\right)\right)^n\\ &= \sum_{n=0}^{N-1} x[n]\left(\exp\left(j2\pi\frac{m}{N}\right)\right)^n\\ &= \left(X[m]\right)^* \end{align*}$$ Thus, for $1 \leq m \leq N-1$, $X[N-m] = \left(X[m]\right)^*$. As a special case of this, note that if we choose $m = N/2$ when $N$ is even, we get that $X[N/2] = \left(X[N/2]\right)^*$, thus confirming our earlier conclusion that $X[N/2]$ is a real number. Note that an effect of the Hermitian symmetry property is that

the $m$-th bin in the DFT of a real-valued sequence has the same magnitude as the $(N-m)$-th bin.

MATLABi people will need to translate this to account for the fact that MATLAB arrays are numbered from $1$ upwards.

Turning to your actual data, your $\mathbf x$ is a DC value of $1$ plus slightly more than one period of a sinusoid of frequency $1$ Hz. Indeed, what you are getting is $$x[n] = 1 + \sin(2\pi (0.01n)), ~ 0 \leq n \leq 100$$ where $x[0] = x[100] = 1$. Thus, the first and the last of $101$ samples has the same value. The DFT that you are computing is thus given by $$X[m] = \sum_{n=0}^{100} \left(1+\sin\left(2\pi \left(\frac{n}{100}\right)\right)\right)\left(\exp\left(-j2\pi \frac{m}{101}\right)\right)^n$$ The mismatch between $100$ and $101$ causes clutter in the DFT: the values of $X[m]$ for $2 \leq m \leq 99$ are nonzero, albeit small. On the other hand, suppose you were to adjust the array t in your MATLAB program to have $100$ samples taken at $t=0, 0.01, 0.02, \ldots, 0.99$ so that what you have is $$x[n] = 1 + \sin(2\pi (0.01n)), ~ 0 \leq n \leq 99.$$ Then the DFT is $$X[m] = \sum_{n=0}^{99} \left(1+\sin\left(2\pi \left(\frac{n}{100}\right)\right)\right)\left(\exp\left(-j2\pi \frac{m}{100}\right)\right)^n,$$ you will see that your DFT will be exactly $\mathbf X = (100, -50j, 0, 0, \ldots, 0, 50j)$ (or at least within round-off error), and the inverse DFT will give that for $0 \leq n \leq 99$, $$\begin{align*} x[n] &= \frac{1}{100}\sum_{m=0}^{99}X[m]\left(\exp\left(j2\pi \frac{n}{100}\right)\right)^m\\ &= \frac{1}{100}\left[100 - 50j\exp\left(j2\pi \frac{n}{100}\right)^1 + 50j \left(\exp\left(j2\pi \frac{n}{100}\right)\right)^{99}\right]\\ &= 1 + \frac{1}{2j}\left[\exp\left(j2\pi \frac{n}{100}\right) - \exp\left(j2\pi \frac{-n}{100}\right)\right]\\ &= 1 + \sin(2\pi (0.01n)) \end{align*}$$ which is precisely what you started from.

  • $\begingroup$ So, is it possible to tell from the FFT if a signal is periodic or not? $\endgroup$ Commented Sep 18, 2016 at 14:01
  • $\begingroup$ @displayname That is a separate question that should be asked in its own right (and perhaps has been asked and answered already). $\endgroup$ Commented Sep 20, 2016 at 15:58
  • $\begingroup$ When I carefully pry out the conjugate symmetrical bins [By writing a 0 + 0i into them] and reconstruct the time domain signal using ifft, the magnitude of the reconstructed time domain signal has halved. Is this natural or is it a tooling problem? I do take care of FFT output normalization and its reverse after iFFT. $\endgroup$
    – Raj
    Commented Aug 15, 2017 at 4:51
  • $\begingroup$ So since |X[1]| = |X[N-1]|, |X[2]| = |X[N-2]|, ..., X[ceil(N/2-1/2)] as a consequence of Hermitian Symmetry for real signals x, there are only ceil(N/2-1/2) unique frequency bins. Is that somehow related to fs = 1/2, the value most plots of ffts of a signal go up to? $\endgroup$
    – Minh Tran
    Commented Oct 5, 2020 at 17:53
  • $\begingroup$ Frequency resolution of an fft is fs/N so the highest frequency represented by k = ceil((N-1)/2) is k*fs/N. To simplify, I'm just going to say N is selected such that ceil(N/2-1/2) = N/2. Then k*fs/N = fs/2. Plots usually are given in terms of normalized frequency so that fs/2 corresponds to 1/2. So that's how they're related. Does that sound right? $\endgroup$
    – Minh Tran
    Commented Oct 5, 2020 at 18:04

Note that an FFT result is mirrored (as in conjugate symmetric) only if the input data is real.

For strictly real input data, the two conjugate mirror images in the FFT result cancel out the imaginary parts of any complex sinusoids, and thus sum to a strictly real sinusoid (except for tiny numerical rounding noise), thus leaving you with a representation of strictly real sine waves.

If the FFT result wasn't conjugate mirrored, it would represent a waveform that had complex values (non-zero imaginary components), not something strictly real valued.


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