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Let $x(t)$ be a real bandpass signal and $x_l(t) = (x(t) + j\hat{x}(t))e^{-j2\pi f_0t}$ be the lowpass equivalent. Is there any relation between $P_x$ and $P_{x_l}$ where $P_x$ refers to the average power of $x(t)$ and $P_{x_l}$ refers to the average power of $x_l(t)$? I tried to use the definition $$P_x=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}|x_l(t)|^2dt = \lim_{T\rightarrow\infty}\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}(x^2(t) + \hat{x}^2(t))dt \tag{1}$$ And I got stuck here. Maybe we should add some assumptions on $x(t)$ in order to be able to simplify $(1)$.

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    $\begingroup$ it should be the same (or related with a fixed factor, depending on how you define the mixing), otherwise things wouldn't be very equivalent :) $\endgroup$ Nov 11 '20 at 17:41
  • $\begingroup$ @MarcusMüller I think you are right but I couldn't prove that. $\endgroup$
    – S.H.W
    Nov 11 '20 at 17:42
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Since the Hilbert transform is an allpass transformation, the powers of $x(t)$ and its Hilbert transform $\hat{x}(t)$ are the same. Consequently, the complex baseband signal as defined in your question has twice the power of the real-valued bandpass signal. That's why in some textbooks the bandpass signal is defined as

$$x(t)=\sqrt{2}\textrm{Re}\big\{s(t)e^{j\omega_ct}\big\}\tag{1}$$

where $s(t)$ is the complex baseband signal.

The scaling in $(1)$ ensures that $x(t)$ and $s(t)$ have the same power.

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  • $\begingroup$ How can we prove that? I mean what's the next step after $\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}(x^2(t) + \hat{x}^2(t))dt$ ? $\endgroup$
    – S.H.W
    Nov 11 '20 at 20:55
  • $\begingroup$ @S.H.W: That expression is just the sum of the powers of $x(t)$ and of $\hat{x}(t)$ and since they have the same power it's just twice the power of $x(t)$. $\endgroup$
    – Matt L.
    Nov 11 '20 at 20:57
  • $\begingroup$ I see. We have $\mathcal{F}(\hat{x}(t)) = -jsgn(f)X(f)$. How this implies that powers of $x(t)$ and its Hilbert transform $\hat{x}(t)$ are the same? $\endgroup$
    – S.H.W
    Nov 11 '20 at 21:13
  • $\begingroup$ @S.H.W: The magnitude of $\textrm{sgn}(f)$ equals $1$, so the power spectrum is not changed by the Hilbert transform. $\endgroup$
    – Matt L.
    Nov 11 '20 at 21:22
  • $\begingroup$ It seems I'm missing something. The only definition which I know for the power of signal is $P_x=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}|x(t)|^2dt$. How can this be evaluated in the frequency domain? Parseval's theorem can be used for calculating the energy but it doesn't help us in the case of the power. $\endgroup$
    – S.H.W
    Nov 11 '20 at 21:31

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