Homework Answers
Bandwidth = c* N * 1/r = 1/2* 17 * 1/2 = 4.25Mbps
Now the bandwidth is scaled by 1/4
therefore new bandwidth = 4.25/4 = c* N * 1/r = 1/2* N * 1/3
=> N = 6.375 Mbps
Therefore new data rate is 6.375 Mbps
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***This problem must be done using R so please provide
the R code used to find the solution. I have provided the data in
fitbit.txt below the question. I will also give "thumbs-up for
correct R code" Thanks in advance.***
Here is the rest of the data needed to solve the
problem. Please solve using R. Thank you!
charge3= c(5.66,4.67,5.99,4.54,4.51,5.26,4.59,
5.1,5.46,4.57,4.41,5.58,5.39,5,5.22,
4.5,4.69,5.38,4.6,4.93)
charge4= c(6.05,5.67,7.71,6.46,5.4,5.12,5.76,
4.96,6.22,5.45,4.76,4.45,7.03,5.88,
6.71,6.04,6.32,6.43,5.66,5.94)
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If a 6 Mbps LTE data rate on the downlink is desired, what would the minimum bandwidth be and why, if BPSK modulation is being used and the entire bandwidth is able to be allocated to a single user if needed? Recall that LTE has 1.4, 3, 5, 10, 15, and 20 MHz BW capabilities 1.
Problem 2: An analog signal is to be converted into a PCM signal that is a binary polar NRZ line code. The signal is transmitted over a channel that is absolutely bandlimited to B-8 kHz. Assume that the PCM quantizer has 32 steps and that the overall equivalent system transfer function He is of the raised cosine-rolloff type with r 0.5. Find the maximum symbol rate D that can be supported over this channel without causing ISI a) c) Find...
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***This problem must be done using R so please provide
the R code used to find the solution. I have provided the data in
fitbit.txt below the question. I will also give "thumbs-up for
correct R code" Thanks in advance.***
Here is the rest of the data needed to solve the
problem. Please solve using R. Thank you!
charge3= c(5.66,4.67,5.99,4.54,4.51,5.26,4.59,
5.1,5.46,4.57,4.41,5.58,5.39,5,5.22,
4.5,4.69,5.38,4.6,4.93)
charge4= c(6.05,5.67,7.71,6.46,5.4,5.12,5.76,
4.96,6.22,5.45,4.76,4.45,7.03,5.88,
6.71,6.04,6.32,6.43,5.66,5.94)
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