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The file with size 6 MB is transmitted in 100 seconds using a constellation coder QAM with a constellation of 6 bit/symbol

$ { M = 10^6, B = 8} $

Tasks:

1. What is the signal bandwidth if the band exceedance factor is 0.5?

$W_m = (1+α)R_s$

$W_m = (1+0.5)*80 000 = 120k$

2. What is the signal sampling frequency at the modulator output if the oversampling ratio equals 3.

$ T_S = 6*0.0000020833 = 0.0000125 $

$ n = T_s / T_p $

$ 3 = 0.0000125 / T_p = 0.00000416666 = 4.17 µs$

$ F_p = 1/T_s => F_p = 239808.15 Hz = 240 kHz $

3. How numerous is the set of symbols of the constellation used?

$ M = 2^k $

$ k = 6 $

$ M = 2^6 = 64 $

4. What is the modulation performance?

$ R_b / W_m = 480k / 120k = 4 $

5. How long will the signal transmission last if the band exceedance factor is 1.

And this is the only task I have no idea how to do. Any suggestions?

I would also be very grateful I you check the other tasks, if they were done correctly, or if I made any mistakes. Thank you in advance!

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  • $\begingroup$ How did you solve question 2? The complex baseband signal has bandwidth 120 kHz, so it needs to be sampled at 240 kHz, and oversampling by a factor of 3 should result in 720 kHz, doesn't it? $\endgroup$ – MBaz Feb 20 at 2:59
  • $\begingroup$ If question 2 refers to samples per symbol, then you're correct. You could have solved it easily like this: if the oversampling factor is $k$, then $f_s = kR_s=3\cdot80\times 10^3 = 240 \times 10^3$. $\endgroup$ – MBaz Feb 20 at 3:17
  • $\begingroup$ @MBaz oh yeah, thank you very much! :) I made a slightly different approach, but now I see, your method is much easier! I just calculated my filze size to bits: 6MB = 6*8 = 48*10^6 bits. Now I calculated my Rb = 48*10^6 / 100 = 480 k. And because I have Rb I can easily get Tb = 1/480k = 0,00000208333. And Ts is just: K*Tp so: 6*0,00000208333 = 0,0000125 $\endgroup$ – JimPanse Feb 20 at 4:12
  • $\begingroup$ Unfortunately, it isn't mentioned if it's about samples per symbol, but I guess it is. $\endgroup$ – JimPanse Feb 20 at 4:18
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Since the bandwidth stays constant at 120 kHz, the symbol rate needs to be reduced: $$R_s = W_m/(1+\alpha) = 120000/2 = 60000 \text{ symbols per second.}$$

Then the bit rate is $R_b = 6R_s=360,000 \text{ bits per second.}$

The transmission duration is then $$T = \frac{48 \times 10^6 \text{ b}}{360 \times 10^3 \text{ b/s}} = 133.3 \text {s}.$$

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  • $\begingroup$ Thank you for this answer! I have only 2 questions: 1. How do you know that the bandwidth stays constant? 2. Why the symbol rate needs to be reduced because of this? $\endgroup$ – JimPanse Feb 20 at 15:41
  • $\begingroup$ 1. It's the only interpretation that makes sense, but the question could definitely have been stated more clearly. 2. As your own formula shows, the symbol rate and the bandwidth are related. If $\alpha$ increases while keeping $W_m$ constant, then $R_s$ has to decrease. $\endgroup$ – MBaz Feb 20 at 16:47

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