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Data Communications people? Help with test
Looking for people who know the ins an out on data communication. I am taking a class with that right now and I have a take home final I really need to kick ass on.
Heres the test.
1) The signal to noise ratio, S/N in a communication line is 500.
a)What is the S/N value in decibels?
b)What is the max bit rate of the line if the bandwidth is 4KHZ?
c)What value os S/N would be needed to transmit 56,000 bps?
d) What is the value in part C expressed in decibels?
2)A frequency band extending from 900 KHZ to 1.2 MHZ is to be divided into slots for FM radio stations to be multiplexed over a communication line. The stations will carry an 8KHZ audio signal and guard bands of 5 KHZ will be placed between the stations. Approximately how many stations can exist in this frequency range?
3)In a certain communication line the probability of a single bit error is .05 and data is sent in 8 bit groups.
a)What is the probability that a group of bits arrives correctly?
b) With a parity bit added, what is the probablity that either group of bits arrives with no errors or else, if there are errors, the errors are detected?
There are a few more questions but I know what the answers are. Thanks in advanced
Heres the test.
1) The signal to noise ratio, S/N in a communication line is 500.
a)What is the S/N value in decibels?
b)What is the max bit rate of the line if the bandwidth is 4KHZ?
c)What value os S/N would be needed to transmit 56,000 bps?
d) What is the value in part C expressed in decibels?
2)A frequency band extending from 900 KHZ to 1.2 MHZ is to be divided into slots for FM radio stations to be multiplexed over a communication line. The stations will carry an 8KHZ audio signal and guard bands of 5 KHZ will be placed between the stations. Approximately how many stations can exist in this frequency range?
3)In a certain communication line the probability of a single bit error is .05 and data is sent in 8 bit groups.
a)What is the probability that a group of bits arrives correctly?
b) With a parity bit added, what is the probablity that either group of bits arrives with no errors or else, if there are errors, the errors are detected?
There are a few more questions but I know what the answers are. Thanks in advanced
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Re: Data Communications people? Help with test
Ohcome on...I know there are smart people out there that must remember this stuff from when they took the class. Or I am sure there are people out there that never took the class but know any/all of this for pooing and giggles.
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F1 Driver
Join Date: Jan 2005
Location: DC
Posts: 3,147
Re: Data Communications people? Help with test
[quote author=saluteTOme link=topic=64510.msg1106511#msg1106511 date=1165963701]
Looking for people who know the ins an out on data communication. I am taking a class with that right now and I have a take home final I really need to kick ass on.
Heres the test.
1) The signal to noise ratio, S/N in a communication line is 500.
a)What is the S/N value in decibels?
b)What is the max bit rate of the line if the bandwidth is 4KHZ?
c)What value os S/N would be needed to transmit 56,000 bps?
d) What is the value in part C expressed in decibels?
[/quote]
I do NOT guarantee that this is correct. Use at your own risk. At this time, imagine I'm telling you that you should have paid more attention in class.
1A: you need to know if this is a power ratio or an amplitude ratio. If it's a power ratio, then it's 10*log10 (500). If it's the amplitude ratio, then I believe it's either 20*log10 of either the amplitude ratio or the square root of the amplitude ratio. But it's been a while.
1b: can't help you there
1c: obviously depends on 1b
1d: use equations from 1a
Looking for people who know the ins an out on data communication. I am taking a class with that right now and I have a take home final I really need to kick ass on.
Heres the test.
1) The signal to noise ratio, S/N in a communication line is 500.
a)What is the S/N value in decibels?
b)What is the max bit rate of the line if the bandwidth is 4KHZ?
c)What value os S/N would be needed to transmit 56,000 bps?
d) What is the value in part C expressed in decibels?
[/quote]
I do NOT guarantee that this is correct. Use at your own risk. At this time, imagine I'm telling you that you should have paid more attention in class.
1A: you need to know if this is a power ratio or an amplitude ratio. If it's a power ratio, then it's 10*log10 (500). If it's the amplitude ratio, then I believe it's either 20*log10 of either the amplitude ratio or the square root of the amplitude ratio. But it's been a while.
1b: can't help you there
1c: obviously depends on 1b
1d: use equations from 1a
F1 Driver
Join Date: Jan 2006
Location: Tallahassee, Boynton Beach, and Pensacola, FL
Posts: 12,565
Re: Data Communications people? Help with test
Bump. I wish I could help, but maybe someone will see it and can
F1 Driver
Join Date: Dec 2004
Location: PA, USA
Posts: 5,896
Re: Data Communications people? Help with test
[quote author=saluteTOme link=topic=64510.msg1106511#msg1106511 date=1165963701]
3)In a certain communication line the probability of a single bit error is .05 and data is sent in 8 bit groups.
a)What is the probability that a group of bits arrives correctly?
b) With a parity bit added, what is the probablity that either group of bits arrives with no errors or else, if there are errors, the errors are detected?
[/quote]
Hmm... I'm going to try to wing this off the top of my head because it just looks like a math problem. Can't guarantee at all that this is right.
3a) I can think of the bits as ping pong balls. I have ping pong balls on a table. Every one of them has a .05 (5 in 100) chance of being defective. The chance that I will get a defective one in a set of 8 of them has to be larger than the chance that any individual one will be defective. In this case, the chance is 8 times larger... so the probability that 1 in every 8 balls is defective is 8*.05 = 0.4, or 4 in 10. That means that 4 out of every 10 "groups" of 8 balls will have at least one bad ball.
So the chances of getting a good group of bits is 100% minus the probability of getting a bad group, or 1  0.4 = 0.6
There's a 60% chance that a group of bits arrives correctly.
(by the way, the probability that all 8 bits in a group are bad is very small: 0.5^8 = 0.0039, or 39 in 10,000.)
3b) Well now we have 9 ping pong balls in a group, because we added one more to tell us if the other 8 are good or bad. This additional ball is a ball like every other one, so it too can be bad. The probability that the parity ball is bad is 0.05 as well.
So the probability that a group of balls has no defects means that all 8 normal balls are good AND the additional 9th ball is good. Well the probability that all 8 are good, from part a, is .6
So the probability that all of the balls are good is (1  0.05) * .6 = .57. 57% of the groups will arrive correct with the partiy bit added. This makes sense because 57 is less than 60, as it has to be since we added another bit and increased the probability that we will get a bad bit in a group.
In order to detect errors, the parity bit has to be good (that's the 1  .05 number above, or 95%). We already know that the probability of getting an error in the other 8 bits is .4 or 40%. So 95% of that 40% of the time we can detect errors.... that's .95 * .40 = .38. So 38% of the time we will be detecting errors.
Checking my work:
The probability of a good parity bit is .95, which can be broken down into:
P(good parity, good data) = .57
P(good parity, bad data) = .38
The probability of a bad partity bit is .05, which can be broken down into:
P(bad parity, good data) = .05 * .60 = .03
P(bad parity, bad data) = .05 * .40 = .02
(.57 + .38) + (.03 + .02) = .95 + .05 = 100 % Golden 8)
That's my best guess.
3)In a certain communication line the probability of a single bit error is .05 and data is sent in 8 bit groups.
a)What is the probability that a group of bits arrives correctly?
b) With a parity bit added, what is the probablity that either group of bits arrives with no errors or else, if there are errors, the errors are detected?
[/quote]
Hmm... I'm going to try to wing this off the top of my head because it just looks like a math problem. Can't guarantee at all that this is right.
3a) I can think of the bits as ping pong balls. I have ping pong balls on a table. Every one of them has a .05 (5 in 100) chance of being defective. The chance that I will get a defective one in a set of 8 of them has to be larger than the chance that any individual one will be defective. In this case, the chance is 8 times larger... so the probability that 1 in every 8 balls is defective is 8*.05 = 0.4, or 4 in 10. That means that 4 out of every 10 "groups" of 8 balls will have at least one bad ball.
So the chances of getting a good group of bits is 100% minus the probability of getting a bad group, or 1  0.4 = 0.6
There's a 60% chance that a group of bits arrives correctly.
(by the way, the probability that all 8 bits in a group are bad is very small: 0.5^8 = 0.0039, or 39 in 10,000.)
3b) Well now we have 9 ping pong balls in a group, because we added one more to tell us if the other 8 are good or bad. This additional ball is a ball like every other one, so it too can be bad. The probability that the parity ball is bad is 0.05 as well.
So the probability that a group of balls has no defects means that all 8 normal balls are good AND the additional 9th ball is good. Well the probability that all 8 are good, from part a, is .6
So the probability that all of the balls are good is (1  0.05) * .6 = .57. 57% of the groups will arrive correct with the partiy bit added. This makes sense because 57 is less than 60, as it has to be since we added another bit and increased the probability that we will get a bad bit in a group.
In order to detect errors, the parity bit has to be good (that's the 1  .05 number above, or 95%). We already know that the probability of getting an error in the other 8 bits is .4 or 40%. So 95% of that 40% of the time we can detect errors.... that's .95 * .40 = .38. So 38% of the time we will be detecting errors.
Checking my work:
The probability of a good parity bit is .95, which can be broken down into:
P(good parity, good data) = .57
P(good parity, bad data) = .38
The probability of a bad partity bit is .05, which can be broken down into:
P(bad parity, good data) = .05 * .60 = .03
P(bad parity, bad data) = .05 * .40 = .02
(.57 + .38) + (.03 + .02) = .95 + .05 = 100 % Golden 8)
That's my best guess.
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F1 Driver
Join Date: Dec 2004
Location: PA, USA
Posts: 5,896
Re: Data Communications people? Help with test
You better post the answers to these when you get them! :) I'm curious.
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Re: Data Communications people? Help with test
thanks peeps. haha
I don't think I will get the answers being it's a final. But since it is a take home test...I will try to "check" my answers with other people in the class
I don't think I will get the answers being it's a final. But since it is a take home test...I will try to "check" my answers with other people in the class
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