COSC 3213: Computer Networks
Assignment # 2  SOLUTIONS
Winter 2005
Problem 1:
(8x10)inch
2
* (800x800)pixels/inch
2
* 8 bits/pixel = 409.6 Mbits per picture
(8x10)inch
2
* (800x800)pixels/inch
2
* 24 bits/pixel = 1.228 Gbits per picture
Problem 2:
Note:
SNR in dB = 10 log
10
(SNR on a linear scale)
a.
BW = 2.4kHz, noiseless channel,
M
= 16
⇒
C
= 2 × 2400 × log
2
(M) = 19,200 bps.
b.
BW = 2.4kHz, SNR = 20dB or 100 on linear scale
⇒
C
= 2400 × log
2
(1+100) = 15,979 bps.
c.
BW = 3.0kHz, SNR = 40dB or 10
4
on linear scale
⇒
C
= 3000 × log
2
(1+10
4
) = 39,863 bps.
Problem 3:
a.
The time taken to fully transmit and receive a packet of length
L
is composed of a
propagation delay
T
p
and a transmission delay
T
t
:
T
= 100ms =
T
p
+
T
t
The propagation delay is obtained by dividing the length of the channel 1 by the signal speed:
T
p
= l/
c
= 4800 km / 200,000 km/s = 24ms
Accordingly,
T
t
= 100ms – 24ms = 76ms =
L
/
R
Hence, the data rate on the given channel turns out to be:
R
=
L/T
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 Fall '10
 A
 Computer Networks, Bit rate, Orders of magnitude, English Channel, Kilometre, The Signal, Phaseshift keying

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