This is my extra credit assignment so don't tell me answers, but please guide me how I should do this. I need to learn.

Question states: Determine the power output of the receiver in watts and in the appropriate measure of dB

let: Pin = 2W

diagram: --> [ G1 = 1/4 ] --> [ G2 = 1/10 ] --> [ G3 = 20 ] -->

How should I start?

After some thinking..

I'm not sure but G might represent the General formula which is G = Pout/Pin


Is this correct?

Pin(dB) = 10log(2W/1W) = 3.01dB --- as input

dB = 10log(1/4) + 10log(1/10) + 10log(20) = -3.01dB --- As output

So the output at the destination is 0?

  • $\begingroup$ What are G1, G2 and G3 and what are the numbers 1/4, 1/10 and 20 associated with them? $\endgroup$ – Chris Taylor Nov 30 '11 at 8:27

You did it correctly! Great job!

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  • 1
    $\begingroup$ Thank you! In my class professor got the same answer. $\endgroup$ – HelpNeeder Nov 30 '11 at 20:55

Hint: Search for transfer function and Bode Plot

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  • $\begingroup$ @HelpNeeder I have a feeling about the question but you provide so little info about the diagram that I don't even know, if we are talking about the same thing. Transfer Function is a function that relates an input to an output. Its bode plot shows how the power is distributed among frequency spectrum but again, this is guessing. $\endgroup$ – user13838 Nov 30 '11 at 17:45
  • $\begingroup$ @HelpNeeder Just tell me what Diagram > means. $\endgroup$ – user13838 Nov 30 '11 at 17:52
  • $\begingroup$ This is basically what looks like on my problem: -->[G1]-->[G2]-->[G3]--> Meaning that I have 3 boxes, first 2 are reducing a signal, and last one acts as amplifier. So I have 2W input and what the output should be after last box. $\endgroup$ – HelpNeeder Nov 30 '11 at 17:55
  • $\begingroup$ @HelpNeeder :D See, this is completely different from what you ask. OK, they are simple gain blocks in series hence you can multiply to obtain just one block and the $G$ blocks are processing the signal I suppose.Do you know how to calculate the power of the signal? $\endgroup$ – user13838 Nov 30 '11 at 18:05
  • $\begingroup$ I have an idea, but I'm not sure if I'm correct. What I know is in my last edit in OP. $\endgroup$ – HelpNeeder Nov 30 '11 at 18:09

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