I know that we can write a real bandpass signal x(t) as:

$x(t)=x_c(t)\cos\omega_ct + x_s(t)\sin\omega_ct$

$x_c(t) = 2x(t)\cos\omega_ct$ for $|f|<W$

$x_s(t) = 2x(t)\sin\omega_ct$ for $|f|<W$

where $x_c(t)$ and $x_s(t)$ are real and imaginary parts of $x_l(t)$ (complex envelope of $x(t)$) respectively.

Suppose we know PSD of $x(t)$ is $S_x(f)$. I want to find PSD of both $x_c(t)$ and $x_s(t)$ according to $S_x(f)$. I saw in a video that PSD of both $x_c(t)$ and $x_s(t)$ is

$S_x(f+f_c) + S_x(f-f_c)$ for $|f|<W$

(where $W$ is the bandwidth of $x(t)$.) But I can't figure out how it is obtained. As the spectrum of actually what I want is a exact mathematical proof and not an intuitive answer. Thank you.

  • $\begingroup$ Please review the definition of the first three functions -- as you wrote them, $x(t)$ contains sin and cos squared terms. $\endgroup$
    – MBaz
    Commented Jan 9 at 18:18
  • $\begingroup$ @MBaz sorry, I forgot to mention that after x(t) is multiplied by sin/cos, passes through a low-pass filter to generate $x_c(t)$ and $x_s(t)$. $\endgroup$
    – Ang
    Commented Jan 9 at 20:19
  • $\begingroup$ "I want to find PSD of both" -- do you mean, the PSD of each? I don't think it's possible in general. On the other hand, it's easy to obtain the PSD of $x_c(t)+jx_s(t)$. $\endgroup$
    – MBaz
    Commented Jan 9 at 23:47

1 Answer 1


I don't think the conclusion stated by the OP is correct unless the bandpass spectrum itself is symmetric about the carrier frequency $\omega_c$ (which should not generally be assumed), and here is why (in case someone else spots my own error!).

power spectral densities

What I show above is the spectrum components that are translated to baseband, given by $S_x(f-f_c)$ and $S_x(f+f_c)$, are reversed, so the sum of the two cannot equal the equivalent baseband spectrum (complex envelope) unless the spectrum was symmetric about the carrier.

The analytic signal is given as

$$x_a(t) = x(t) + j \hat{x}(t)$$

Where $\hat{x}(t)$ is the Hilbert Transform of $x(t)$. It would be a one sided spectrum as just the positive frequency only portions of $x(t)$ with twice the gain, as depicted in the second line above with the PSD as $S_{x_a}(f)$.

I believe the OP is representing the baseband equivalent signal which is done by frequency translating the analytic signal to baseband as:

$$x_I(t) = LPF\{x_a(t) e^{-j\omega_c t}\}$$

Where $LPF\{\}$ represents a low pass filter operation only passing frequencies below $W$, which would result in a signal that can be described by it's complex envelope independent of any carrier frequency. We could achieve this by low pass filtering the third spectrum shown in the plot above for $S_x(f-f_c)$, and multiply by a gain of 2 to exactly match the operation I gave mathematically.

The issue I show is that the other spectrum given by $S_x(f+f_c)$ would be reversed and so would not constructively add to the desired baseband signal to be $x_I(t)$.

If we did want to derive a relationship with both the sum and difference as the OP proceeds to do, the spectrum reversal can be resolved by instead shifting the power spectral density of the complex conjugate of $x(t)$ (as $S_x^*(f+f_c)$), since the complex conjugate would have a reverse spectrum.


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