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?

  • $\begingroup$ 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] $ ? $\endgroup$
    – Hilmar
    Oct 24, 2022 at 14:51
  • $\begingroup$ @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? $\endgroup$ Oct 25, 2022 at 8:04

1 Answer 1


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.

When multiplying a signal by a complex exponential, you are shifting and possibly rotating the spectrum, i.e. you are shifting the PSD. A horizontal shift of the curve will not change the area under the curve, therefore the power does not change.


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