# How to separate real and imaginary part of a complex FFT?

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?

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}

• Thank you very much for your detailed explanation. How do you get to the expressions after "Because of the symmetry properties presented above..."? Aug 31 at 12:49
• @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. Aug 31 at 13:07
• Ok, thank you very much for your help! Aug 31 at 13:22
• @user230754 glad to help you out. The case of IFFT leaves for you as a further exercise. Aug 31 at 13:31