# 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 :) Nov 11 '20 at 17:41
• @MarcusMüller I think you are right but I couldn't prove that. Nov 11 '20 at 17:42

## 1 Answer

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 $\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}(x^2(t) + \hat{x}^2(t))dt$ ? 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 $x(t)$. 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? 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. 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. Nov 11 '20 at 21:31