17. What is the channel capacity for a telephone channel with a 4000-Hz bandwidt
ID: 3676280 • Letter: 1
Question
17. What is the channel capacity for a telephone channel with a 4000-Hz bandwidth and a signal-to-noise ratio of 127? 18. An amplifier has an output of 0.1 W. Express its output in dBm. 19. Can we transmit digital signal in free space? Why or why not? 20. What key factors affect channel capacity? 21. Determine the height of an antenna for a TV station that must be able to reach customers up to 100 km away 22. Calculate the power at the receiver given the following information of a 2.4 GHz microwave system: Transmitter Power 30 dBm Transmitter antenna Gain 30 d8 Receiver antenna Gain 30 dB Frees pace Loss Transmitter Cable and connector Loss Receiver Cable and connector Loss 130 dB 6 dB 6 dB 23. A digital signaling system is required to operate at 19200 bps. If a signal element encodes a 4-bit word, what is the minimum required bandwidth of the channel as per Nyquist?Explanation / Answer
17) Here is the formula to calculate bandwidth ..
C = B log2(1+SNR) ==> 4000 log2(1+127) ==> 4000 x 7 log2 2 ==> 4000 x 7 = 28000 bps ==> 28 kbps
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18)
Here is the formula
P(dBW) = 10 · log10( P(W) / 0.0011W) ====> 10 x log10(0.1/0.001) = 10 x 2 = 20 dbm
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19) Since in space we have vaccum space, so we cannot transmit signal.
we can transmit only radio waves.
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20) factors are .. datarate, bandwidth, error rate.
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21) formula ..
d=3.57 square root(Kh) , with K = 4/3,
So squring both sides results
100 ^2 = (3.57)^2 * 1.33 X h
===> 10000/ (3.57)^2 * 1.33 == h
h = 10000/ 16.95 ===> 589.97 ~~ 590m
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22)
receiver power = transmitted power + gain - losses
===> 30 x 1000 + 30 + 30 - (130 + 6 + 6)
==> 30060 - 142 ==> 29.91 dbm
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23)
It required bandwidth is 19200 x 4 ==> 76800 ==> 76.8 kbs
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