# Relation between power of the baseband signal and the low pass equivalent

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 $$(1)$$.

• 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 :) – Marcus Müller Nov 11 '20 at 17:41
• @MarcusMüller I think you are right but I couldn't prove that. – S.H.W Nov 11 '20 at 17:42

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.

• How can we prove that? I mean what's the next step after $\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}(x^2(t) + \hat{x}^2(t))dt$ ? – S.H.W Nov 11 '20 at 20:55
• @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 $x(t)$. – Matt L. Nov 11 '20 at 20:57
• 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? – S.H.W Nov 11 '20 at 21:13
• @S.H.W: The magnitude of $\textrm{sgn}(f)$ equals $1$, so the power spectrum is not changed by the Hilbert transform. – Matt L. Nov 11 '20 at 21:22
• 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. – S.H.W Nov 11 '20 at 21:31