Application of time reversal property of DFT

Question

I came across the time reversal property of $DFT$ that states:

\begin{equation*} x(<-n>_{N})\rightarrow X^{*}(k) \end{equation*}

I can't seem to understand though how can this property be used practically, lets say in a case where you know a sequence $x(n)=\{e_1,e_2,e_3\}$ and its DFT $X(k)=\{e_{11},e_{22},e_{33}\}$ and you want to find the DFT of the sequence $x(-n)$.

Yes I know...I didn't realised that I had mentioned shifting in the title...I will correct it. — Adam, Dec 10 '15 at 19:09

score 0 · Answer 1 · answered Dec 10 '15 at 18:24

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Time reversal property states that if signal is reversed in time, the fourier coefficient is conjugate of $F(x[n])$ i.e $X(e^{(-j\omega)})$ or simply $X(-j\omega)$. (assuming real x[n])

So, if you mark fourier coefficients on a circle based on phase angle, $\omega$. To find coefficients of fourier transform of $x[n]$ you read anti-clockwise on circle and for $x[-n]$ clockwise

For the example you stated above DFT of $x[-n] = \bar{e11}, \bar{e22}, \bar{e33}$

answered Dec 10 '15 at 18:24

Umang Gupta

143

So $X^{*}(k)$ will have the same elements as $X(k)$ but shifted? – Adam Dec 10 '15 at 18:39
I made a mistake in my notation please check again my previous comment. I obviously know about $x(n)$ and $x(-n)$. – Adam Dec 10 '15 at 18:43
$X^*(k)$ is conjugate of $X(k)$. $X(k)$ can be imaginary. – Umang Gupta Dec 10 '15 at 18:56
I know. In your answer you are talking about the $x(n)$ and $x(-n)$ and you don't even mention the $X(k)$ and $X^{*}(k)$. – Adam Dec 10 '15 at 18:58
Then what do $X(e^{-j\omega})$ mean ? or $X(e^{j\omega})$ mean ? Replace k by $j\omega$. This is another notation that is it. – Umang Gupta Dec 11 '15 at 05:54
So $X^*(k)$ will have the same elements as $X(k)$ but shifted?
$X^*(k)$ is conjugate of $X(k)$.
– Umang Gupta Dec 11 '15 at 05:55

score 0 · Answer 2 · answered Dec 10 '15 at 19:22

To try explain it as simple as possible. You can think of it as mirroring each sine and cosine in the Fourier Transform in the middle point.

The cosines (real part of complex exponential) are even ($\cos(wx) = \cos(-wx)$), so they don't change.
The sines (imaginary part of complex exponential) are odd ($\sin(-wx) = -\sin(wx)$) so they will switch sign.

Therefore the imaginary part of each Fourier coefficient will switch sign but the real part will remain the same.

Application of time reversal property of DFT

2 Answers2