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INFR3720U Assignment#1 Solution Winter 2013
Question 1)
A communication channel has a bandwidth of 10 MHz and an SNR(db) of 30 db. What is the theoretical maximum data rate (R) on this channel? How many signal levels do we need to achieve this data rate? What will be the symbol rate (S) and the number of bits per symbol in this case?
Answer-
Theoretical maximum data rate is determined by Shannon’s limit:
R = B × log
2
(1+SNR) = 10 MHz × log
2
(1+10^(30/10)) = 99.7 ~ 100 MbpsAlternatively, you could also use the approximate Shannon formula:R = B × (SNR_db) /3= 10 MHz × 30/3 = 100 MbpsApplying the Nyquist theorem, we could achieve a symbol rate of S = 2 × B = 20 M Symbols/s. Inorder to have a data rate of 100 Mbps, we need to have:R = 2 × B × log2 (L)100 = 2× 10 × log2 (L)log2 (L) = 5L =2^5 = 32
Question 2)
Consider the following network with given line speeds and lengths: Assume the speed of light is 2×10
8
m/s, and that the processing and queuing delays are negligible. Node A sends a 1500-byte frame to node C through node B. Suppose Node A starts sending the first bit of the frame at t=0. Answer the following questions:
a.
At what time (t=?) does node B receives the frame in full (i.e. the last bit)?
Answer
–
t
B
= (transmission time over link A-B) +(Propagation time over link A-B)
t
B
= ( 1500 × 8 / 4 (Mbps) ) + ( 1000 × 1000 / 2×10
8
m/s )
t
B
= 3 milliseconds + 5 milliseconds = 8 milliseconds
b.
After node B receives the frame in full, then it will start sending the frame to node C. Determinethe time (t=?) when node C receives the frame in full.
Answer
–
t
C
=
t
B
+ (transmission time over link B-C) +(Propagation time over link B-C)
t
C
= 8 milliseconds + ( 1500 × 8 / 1 (Mbps) ) + (200 × 1000 / 2×10
8
m/s )
t
C
= 8 + 12 + 1 = 21 millisecondsA
B
C
1000 KM4 Mbps
200 KM
1 Mbps
INFR3720U Assignment#1 Solution Winter 2013
c.
Suppose at this time node C sends a small acknowledgment back to node A. The size of theacknowledgement frame is small enough that its transmission time is negligible. Calculate at what times (t=?) the acknowledgment arrives at nodes B and A.
Answer-
t
Ack-B
=
t
C
+ (Propagation time on C-B) = 21 + 1 = 22 milliseconds
t
Ack-A
=
t
Ack-B
+ (Propagation time on B-A) = 22 + 5 = 27 milliseconds
d.
Assume after sending its frame, node A wait for the acknowledgment back from node C beforesending the next frame. What is the average throughput (number of transmitted bits per second)of the data flow from node A to node C?
Answer- Based on the results from (d), Node A sends 1500 Bytes to Node C every 27 milliseconds.Therefore the average throughput = (1500 × 8) / (0.027 seconds) = 444.44 Kbps
Question 3)
A low-pass signal with a bandwidth of 250 KHz was sampled at the Nyquist rate using 512 levels of quantization to generate a PCM signal.a. Calculate the bit rate of the digitized signal b. Calculate the quantization error as SNR(db)c. What is the bandwidth of the PCM signal?
Answer-a.
Sampling rate = 2 × W =2 × 250 = 500 Kilo Samples/sn
b
= log
2
(L) = log2 (512) = 9R (Bit rate) = 9 × 500 K = 4.5 Mbpsb.
SNR(db) = 6.03 × n
b
+ 1.76 = 56.03 dbc.
B
PCM
= n
b
× B
Analog
= 250 × 9 = 2.25 MHz

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