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?

  • $\begingroup$ 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

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.


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