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I have an exercise where I have to calculate the FFT of a complex signal $x=l+j\cdot r$ using only one single call to a complex FFT algorithm. $l$ is the left, $r$ the right real valued vector of a stereo audio signal.

Now after the transform, how do I separate the right and left channel into its parts in the frequency domain to get the same result as if I would calculate the FFT of $l$ and $r$ separately with two tranforms?

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This trick uses the conjugate symmetric property of DFT. We know that a real-valued signal, $x(n)$, has a conjugate symmetric DFT $X(k)$: $$ X(k) = X^*(N-k), \ \ \ k=1, \ldots, N-1 $$ The real part of $X(k)$ has even symmetry about $N/2$, and the imaginary part of $X(k)$ has odd symmetry about $N/2$.

If $x(n)$ is a real-valued signal, now we perform the FFT of a signal $y(n)=jx(n)$. The FFT of $y(n)$, denoted $Y(k)$, ends up being conjugate anti-symmetric: $$ Y(k) = -Y^*(N-k), \ \ \ k=1, \ldots, N-1 $$ Observe that the real part of $Y(k)=jX(k)$ has even symmetry about $N/2$, and the imaginary part of $Y(k)=jX(k)$ has odd symmetry about $N/2$.

Let $y(n)=x_1(n)+jx_2(n)$ where $x_1(n)$ and $x_2(n)$ are both real signals. When we take the FFT of the composite signal, we get $$ Y(k) = X_1(k)+jX_2(k) $$

Now we split $Y(k)$ into real part and imaginary part as $$ Y(k) = R(k)+jI(k) $$ where $R(k)=\mathcal{Re}[Y(k)]$ and $I(k) = \mathcal{Im}[Y(k)]$.

Because of the symmetry properties presented above, we can obtain $X_1(k)$ and $X_2(k)$ from $Y(k)$ without loss of data. For $k=1, \ldots, N-1$, we have $$ \begin{aligned} & \mathcal{Re}[X_1(k)] = \frac{1}{2} [R(k) + R(N-k)]\\ & \mathcal{Im}[X_1(k)] = \frac{1}{2} [I(k) - I(N-k)]\\ & \mathcal{Re}[X_2(k)] = \mathcal{Im}[jX_2(k)] = \frac{1}{2} [I(k) + I(N-k)]\\ & \mathcal{Im}[X_2(k)] = -\mathcal{Re}[jX_2(k)] = \frac{1}{2} [R(N-k) - R(k)] \end{aligned} $$

We know the periodicity property of DFT that $X(k+N) = X(k)$, thus $X(N)=X(0)$. Therefore, for $k=0$ $$ \begin{aligned} & \mathcal{Re}[X_1(0)] = \frac{1}{2} [R(0) + R(N)] = R(0)\\ & \mathcal{Im}[X_1(0)] = \frac{1}{2} [I(0) - I(N)] = 0\\ & \mathcal{Re}[X_2(0)] = \frac{1}{2} [I(0) + I(N)] = I(0)\\ & \mathcal{Im}[X_2(0)] = \frac{1}{2} [R(N) - R(0)] = 0 \end{aligned} $$

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  • $\begingroup$ Thank you very much for your detailed explanation. How do you get to the expressions after "Because of the symmetry properties presented above..."? $\endgroup$
    – user230754
    Aug 31 at 12:49
  • $\begingroup$ @user230754 The real part of $X(k)$ has even symmetry about $N/2$ and the imaginary part has odd symmetry about $N/2$. And $Y(k)$ is the opposite. Write down these two properties and you’ll see. $\endgroup$
    – ZR Han
    Aug 31 at 13:07
  • $\begingroup$ Ok, thank you very much for your help! $\endgroup$
    – user230754
    Aug 31 at 13:22
  • $\begingroup$ @user230754 glad to help you out. The case of IFFT leaves for you as a further exercise. $\endgroup$
    – ZR Han
    Aug 31 at 13:31

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