QAM-spectrum with uneven real & non-odd imaginary part

Question

Despite the signal being real, as is seen in the last line of the first picture:

, the exemplary spectrum of the signal shows neither an even real nor an odd imaginary part:

But from my theorems I know that for every real x(t) the real part of the Fourier-Transform is even and the imaginary part of it is odd?

Answer 1

You are right in that the Fourier Transform (FT) of every real $x(t)$ has even real part and odd imaginary part. However, your figures only show the absolute value of the bandpass spectrum, so you cant see real- and imaginary part separately. Note that the above figure (showing $|S(f)|$ is the baseband signal, which is indeed non-even and hence complex valued. But, after upconversion, the signal becomes real and if you write a small script that does the calculation of the upconversion, you will see, the FT $S_{BP}(f)$ of the bandpass signal $s_{BP}(t)$ will have mentioned properties. I have once written a walkthrough article of up-down-conversion here which might also help in understanding this process.

To clarify a bit more in math. terms:

1. $\underline{s}(t)=s_I(t)+js_Q(t)$ is a complex baseband signal, which consists of the real part $s_I(t)$ and the imaginary part $s_Q(t)$.
2. $S(f)=\mathcal{F}\{\underline{s}(t)\}$ is the Fourier Transform of the baseband signal. This $S(f)$ is arbitrary (i.e. not even or odd) since $\underline{s}(t)$ is a complex-valued baseband signal.
3. $s_{BP}(t)$ is the upconverted version of $\underline{s}(t)$, and is a real-valued signal.
4. Hence, $S_{BP}(f)=\mathcal{F}\{s_{BP}(t)\}$ has even real part and odd imaginary part.