Homework 5 - CPET 355 - Carson Hunter
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Purdue University, Fort Wayne *
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Apr 3, 2024
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Homework 5 – Carson Hunter
6-4:
We need to use synchronous TDM and combine 20 digital sources, each of
100Kbps. Each output slot carries 1 bit from each digital source, but one extra bit is added to each frame from synchronization. Answer the following
questions:
a.)
What is the size of an output frame in bits?
There is 1 bit from each source with an extra bit for synchronization so:
Frame size = 20*1+1 = 21 bits
b.) What is the output frame rate?
Frame rate = 100,000 frames/s
c.) What is the duration of the output frame?
Frame duration = 1/Frame Rate
1/100,000 = 10 microseconds
Frame duration = 10 microseconds
d.) What is the output data rate?
100,000 frames/s * 21 bits/frame = 2.1Mbps
Data Rate = 2.1 Mbps
e.) What is the efficiency of the system (ratio of useful bits to the total bits)?
Efficiency = 20/21 = 95%
6-6:
We have 14 sources, each creating 500 8-bit characters per second. Since only some of these sources are active at any moment, we use
statistical TDM to combine these sources using character interleaving. Each frame carries 6 slots at a time, but we need to add 4-bits addresses to each slot. Answer the following questions:
a.)What is the size of an output frame in bits?
6*(8+4) = 72
Frame size = 72 bits
b.)What is the output frame rate?
500 frames/s
c.) What is the duration of the output frame?
Frame duration = 1/Frame Rate
1/500 = 2ms
Frame duration = 2ms
d.)What is the output data rate?
500 frames/s * 72 bits/frame = 36 kbps
Data rate = 36 kbps
6-12:
Figure 6.34 shows a multiplexer in a synchronous TDM system. Each output slot is only 10 bits long (3 bits taken from each input plus 1 framing bit). What is the output stream? The bits arrive at the multiplexer as shown by the arrows.
6-13:
Figure 6.35 shows a demultiplexer in a synchronous TDM. If the input
slot is 16 bits long (no framing bits). What is the bit stream in each output? The bits arrive at the demultiplexer as shown by the arrows.
6-15:
What is the minimum number of bits in a PN sequence is we use FHSS with a channel bandwidth of B=4KHz and Bss=100KHz?
100 KHz / 4 KHz = 25
log2(25) = 4.65 bits
(Rounding up to nearest bit) = 5 bits
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