0
$\begingroup$

The block diagram below represents a linear modulation system operating at the frequency of $1000 Hz$, $f_C = 1000 Hz$, transmitting the message $m(t) = 2\cos(400πt)$

Figure 1

At point B, i got the signal: $$cos(1600 \pi t) + cos(2400 \pi t)$$

At point C, I got the following signal: $$\frac{1}{2} (\cos(380 π t) + \cos(420 π t) + \cos(3620 π t) + \cos(4420 π t)) $$

However, I am having difficulty acquiring $y(t)$. Would the filter $H(f)$ eliminate the entire signal from point C through?

$\endgroup$
10
  • $\begingroup$ In my head, I'm note even getting closely the same thing as you at (C); maybe I'm wrong though. How did you arrive at your term? Please edit your question and include your derivation! $\endgroup$ Jul 17 '21 at 17:31
  • $\begingroup$ In point B, i multiplied $m(t)cos2000\pi t$. In point C, i multiplied $x_{1}(t)cos(2020 \pi t)$. $\endgroup$
    – July H.
    Jul 17 '21 at 17:34
  • $\begingroup$ Hint: it's easier if you don't take the 2 out of $2\pi t$; anyway, I don't see how you arrive at $1600\pi t$, could you also explain that in your question? $\endgroup$ Jul 17 '21 at 17:41
  • $\begingroup$ aaaaah I now see that $m(t) =\cos(2\pi 200t)$, OK, that can make more sense. $\endgroup$ Jul 17 '21 at 17:42
  • $\begingroup$ So, I recommend making your life easier: 1. set $a=2\pi t$.Lug that around and only insert it at the very end. Especially, do not multiply the 2 from $2\pi$ with the other numbers! Remember what the frequency of $\cos(2\pi f t)$ is! $\endgroup$ Jul 17 '21 at 17:44
0
$\begingroup$

Your diagnosis seems right. When I just precompute the product of the two oscillators in my head, that gives me frequency components at $\pm 10$ and at 2010, so mixing a tone with frequency 200 with that yields frequency components at

  • 190,
  • 210,
  • -190,
  • -210,
  • 1810,
  • 2210,
  • -1810 and
  • -2210,

But nothing in $[-90;90]$.

PS: it's better to keep the 2 with the $\pi$, as it is part of the factor between frequency and angular frequency.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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