Signal power after frequency shift with different frequency than the carrier frequency

I have this received signal in passband

$$\begin{split} r(t)=&\Re\left\{\tilde{x}_1(t)e^{j2\pi f_1t}+\tilde{x}_2(t)e^{j2\pi f_2t}\right\}\\ =&\Re\left\{e^{j2\pi f_1t}\underbrace{\left[\tilde{x}_1(t)+\tilde{x}_2(t)e^{j2\pi (f_2-f_1)t}\right]}_{\tilde{r}(t)}\right\} \end{split}$$

Suppose the power spectral density (PSD) of $$\tilde{x}_i(t)$$ is $$S_i(f)$$. From above we can say that the power spectrum density of $$r(t)$$ and $$\tilde{r}(t)$$ are

\begin{align} S_{r}(f)=&\frac{1}{2}\left[S_1(f-f_1)+S_1(f+f_1)\right]+\frac{1}{2}\left[S_2(f-f_2)+S_2(f+f_2)\right]\\ S_{\tilde{r}}(f)=&S_1(f)+S_2(f-f_2+f_1) \end{align}

Does this mean that after down-converting the received signal $$r(t)$$ by $$f_1$$, the power of $$\tilde{x}_2(t)e^{j2\pi(f_2-f_1)t}$$ is the same as the power of $$\tilde{x}_2(t)$$, or it's not since $$S_2(f-f_2)$$ is already scaled by $$0.5$$, and thus $$S_2(f-f_2+f_1)$$ is scaled by $$0.5$$ as well?

• I think the second term in your second equation is wrong. Shouldn't that be $\frac{1}{2}\left[S_2(f-(f_2-f_1))+S_2(f+(f_2-f_1))\right]$ ? Oct 24, 2022 at 14:51
• @Hilmar Why? I don't take the real part of $\tilde{r}(t)$. $S_{\tilde{r}}(f)$ is the PSD of $\tilde{r}(t)$, and multiplying the signal with exponential in the time-domain is translated into frequency shift in the frequency domain, right? Oct 25, 2022 at 8:04

The power $$P_x$$ of a signal $$x(t)$$ can be determined from its power spectral density (PSD) $$S_x(f)$$ as $$P_x = \int_{-\infty}^{-\infty} S_x(f) \mathrm{d}f,$$ so it is the area under the PSD curve.