Two signals have arrived with different phases. The sum signal is given by

$$x(t) = A\cos(2\pi f_ct) + B\cos(2\pi f_ct + \theta)$$

What is the complex envelope of $x(t)$? Need advice on how to get started.

What is the analytic signal $x_+(t)$? For this, can I not apply the $x(t) + \mathscr{H}\big\{x(t)\big\}$ to both $A\cos(2\pi f_ct)$ and $B\cos(2\pi f_ct + \theta)$, and just add the result?

  • 2
    $\begingroup$ Hi and welcome. A rule here is that any question with homework problems needs to contain the attempts at solving them. Please edit your question with what you've tried so far and where you're stuck, and I'm sure you'll get help and guidance! $\endgroup$
    – Jdip
    Commented Feb 1, 2023 at 2:53
  • 2
    $\begingroup$ Homework? Please edit your question with the work you've done so far, or if you're utterly stuck, tell what you do know. $\endgroup$
    – TimWescott
    Commented Feb 1, 2023 at 2:54
  • 1
    $\begingroup$ Are $A$ and $B$ constant? Doesn't look like your envelope has much shape. $\endgroup$ Commented Feb 1, 2023 at 3:24
  • $\begingroup$ @robertbristow-johnson yes, A & B are constants... same signals, just the 2nd one comes in with a phase $\endgroup$
    – Gotz2bril
    Commented Feb 1, 2023 at 3:30
  • 1
    $\begingroup$ There are trig identities that you should make use of. This is hardly a complex envelope problem. Not a complex-valued signal (i suppose $A$ and $B$ could be complex?) and the envelope is a constant function. It's just a sinusoid with an amplitude and a phase. $\endgroup$ Commented Feb 1, 2023 at 3:44

2 Answers 2


Using the fact that


it is straightforward to write down the analytic signal of $x(t)$.

The analytic signal has the form


and you have


where $x_c(t)$ is the complex envelope. Judging from the comments under the question, you know everything you need in order to obtain the analytic signal as well as the complex envelope.

  • $\begingroup$ what if I have two different fc values ie f1 and f2? $\endgroup$
    – Gotz2bril
    Commented Feb 24, 2023 at 21:30
  • $\begingroup$ @Gotz2bril: You could better ask that as a new question. This question has been answered as far as I can tell. $\endgroup$
    – Matt L.
    Commented Feb 26, 2023 at 17:41

Perhaps the third row of Table 1 at:


would be helpful to you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.