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

Bandpass-signal Power-spectrum of QAM-signal

, the exemplary spectrum of the signal shows neither an even real nor an odd imaginary part: Exemplary plot of Power-spectrum of QAM-signal

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?


1 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.
  • $\begingroup$ But how can the baseband-signal S(f) be non-even, suggesting that there's a complex signal, when the bandpass-signal clearly is stated as $\ s(t) = s_I(t)cos(2\pif_ct) - s_Q(t)sin(2\pif_ct) \$ and then Fourier-transformed as S(f)? $\endgroup$
    – Starhowl
    Feb 27, 2017 at 9:47
  • $\begingroup$ I added some text. Is it still unclear? $\endgroup$ Feb 27, 2017 at 10:03
  • $\begingroup$ Since the set of {s_I(t), s_Q(t), cos(2\pif_ct), sin(2\pif_ct)} doesn't contain any complex elements, the function s_{BP}(t) is real itself. But every real signal is complex-conjugate symmetric, which isn't the case in the depicted spectrum? $\endgroup$
    – Starhowl
    Feb 27, 2017 at 10:06
  • $\begingroup$ no, $s(t)$ is complex, according to $s(t)=s_I(t)+\mathbf{j}s_Q(t)$. This is the baseband signal, which is complex. Only after upconversion to $s_{BP}(t)$ it becomes real-only. $\endgroup$ Feb 27, 2017 at 10:08

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