# Handout 8 - SEAS - University of Pennsylvania ```Tcom 370: Principles of Data Communications
University of Pennsylvania
Handout 8
Spring Semester 2000
Midterm Examination Solutions
In class, closed book, 80 minutes.
All problems carry equal credit. Do any four of the five problems.
Begin each problem on a separate page.
1. Consider a communication network for a city of one million inhabitants.
a. Suppose that each inhabitant, during the busiest hour of the day, is involved in an
average of 4 data transactions per hour that use this network. Each transaction,
on average, causes 4 packets of 1000 bits each to be sent. Each packet travels over
an average of 3 links. What is the aggregate average number of bits per second
carried by this network? How many 64 kbit/sec voice telephone links are required
to carry this tra c?
b. Suppose that the inhabitants use their telephones an average of 10% of the time
during the busy hour. How many voice telephone links are required for this?
Assume that all calls are within the city and travel over an average of 3 links.
Solution.
a. Data are generated at an average rate of
106 &times; 4
transactions
packets
bits
1
hour
&times;4
&times; 1000
&times;
hour
transaction
packet
3600 seconds
≈ 4.44 &times; 106
bits
.
second
As each bit has to travel over three links, the tra c supported by the network on
average is approximately 1.332 &times;107 bits per second. The number of 64 kbps voice
telephone links needed to carry this tra c is hence
&raquo;
1.332 &times; 107
64 &times; 103
&frac14;
= 209.
b. On average, at any given instant there are 106 &times;0.1 = 105 individuals on the phones.
This requires 105 /2 = 5 &times; 104 end-to-end voice links as all calls are between two
people (assuming no conference calls). In circuit switching all voice links are
dedicated for the duration of the call. As each such call requires three links, it
follows that 1.5 &times; 105 voice links are needed. Compare with the answer for data
transfer.
2. In the Fibre Distributed Data Interface (FDDI) every 4-bit block of data is encoded
using a 5-bit code, sometimes called a 4B/5B code, which is constructed so that there
is never more than one leading zero and no more than two trailing zeros in each 5-bit
codeword. The resulting 5-bit codewords are then transmitted using NRZ-I encoding.
1
Tcom 370: Principles of Data Communications
University of Pennsylvania
Handout 8
Spring Semester 2000
The 4B/5B code table is shown below.
4-bit data
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
5-bit codeword
11110
01001
10100
10101
01010
01011
01110
01111
10010
10011
10110
10111
11010
11011
11100
11101
a. The data bit sequence 0010011110000100 is to be transmitted over an FDDI network. Sketch the transmitted waveform.
b. What are the advantages of the FDDI encoding technique? Comment on its bandwidth e ciency vis à vis Manchester encoding and on whether long strings of 0s
in the data stream will cause synchronisation problems. What if there is a long
string of 1s in the data stream?
Solution.
a. Breaking up the data sequence into four bit blocks, the corresponding coded sequence can be written down directly by inspection from the 4B/5B code table.
Spaces have been introduced for reading clarity.
Data sequence:
0010
0111
1000
0100
Coded sequence:
10100 01111 10010 01010
The corresponding NRZ-I waveform is shown below.
b. The FDDI scheme almost doubles the bandwidth e ciency compared to Manchester encoding. (Recall that NRZ requires only one-half the bandwidth of Manchester.) The improvement in bandwith e ciency over Manchester encoding is not
2
Tcom 370: Principles of Data Communications
University of Pennsylvania
Handout 8
Spring Semester 2000
quite 100% because five bits are transmitted for every four data bits. Thus, the
improvement in bandwidth e ciency over Manchester encoding is 80%.
Long strings of 1s do not pose a synchronisation or DC problem in FDDI which inherits its immunity from the di erential encoding NRZ-I. In addition, unlike plain
vanilla NRZ-I, FDDI also enjoys immunity from synchronisation or DC problems
caused by long strings of 0s by virtue of the 4B/5B encoding—observe that the
encoding guarantees that there will be no more than 3 contiguous 0s in the coded
sequence. Thus, neither long strings of 1s nor long strings of 0s cause synchronisation or DC di culties in FDDI. This improvement over NRZ-I is at a cost of just
20% in bandwidth.
3. A linear, time-invariant channel has the following response to a pulse input.
a. Sketch the output waveform y(t) of the channel when the input waveform x(t) is
as shown below. Hint: What would be the output waveform if p(t − τ1 ) + p(t − τ2 )
is input for any fixed τ1 and τ2 ?
b. Suppose NRZ-L signalling is used to encode 3-bit sequences of data using the pulse
p(t). Are the resulting waveforms received distortion-free by the receiver? Can the
receiver decode the original bit stream without errors? Assume the receiver has
perfect synchronisation and that the channel is noise-free.
Solution.
a. The input waveform x(t) is just the superposition of shifted versions of the pulse
p(t):
x(t) = p(t) − p(t − T ) + p(t − 2T ).
As the system is LTI, it follows that the output waveform is given by
y(t) = q(t) − q(t − T ) + q(t − 2T )
and sketched below.
3
Tcom 370: Principles of Data Communications
University of Pennsylvania
Handout 8
Spring Semester 2000
b. In addition to a T -unit delay, the channel causes a spreading of each pulse waveform so that successive bits encoded via NRZ-L are subject to intersymbol interference. Thus, all output waveforms are distorted. Nonetheless, in the absence of
noise and assuming perfect synchrony, the receiver can ambiguously decode the
original bit stream without errors. To see this observe that each input bit string
results in a unique output waveform so that the receiver could actually maintain
a look-up table which maps each possible output waveform to the unique input
bit string that causes it. Much less e ort is needed, however, and the receiver can
actually decode the input bit string sequentially from the output waveform. How
would he go about doing so?
4. Consider the periodic rectangular pulse train x(t) shown below. Suppose T = 1 ms.
The pulse train x(t) is input to a linear, time-invariant transmission medium which
introduces a delay of 1/6 ms. The medium has a flat passband between 2000 Hz and
4000 Hz and blocks all other frequencies. Determine the channel output signal y(t)
and sketch it. Hint: Use Euler’s formul&aelig; for the trigonometric functions to put the
complex exponential Fourier series in a familiar trigonometric form first.
Solution. As usual, set f0 = 1/T . The DC (or zero frequency) coe
series is then given by
a0 =
1
T
while, for n 6= 0, the Fourier coe
1
an =
T
Z T/2
−T/2
x(t)e
−j2πnf 0 t
Z T/2
x(t) dt =
−T/2
1
T
Z T/2
0
dt =
cient of the Fourier
1
,
2
cients are given by
1
dt =
T
Z T/2
e
−j2πnf 0 t
0
=
4
&macr;T/2
e−j2πnf 0 t &macr;&macr;
dt =
−j2πnf0 T &macr;0
1
1 − cos(nπ)
(1 − e−jπnf 0 T ) =
.
j2πnf0 T
j2πn
Tcom 370: Principles of Data Communications
University of Pennsylvania
Thus, when n 6= 0,
an =
&plusmn;
Handout 8
Spring Semester 2000
1/jπn if n is odd,
0
if n is even.
Thus, the pulse train has the Fourier series representation
x(t) =
∞
∞
X
X
1
ej2πnf 0 t
1
2
+
= +
sin(2πnf0 t)
2 n =−∞ jπn
2
πn
n =1
n odd
n odd
by combining the corresponding negative and positive terms in the complex exponential Fourier series and using Euler’s formula. This looks very familiar. And indeed it
is. Look up the odd square wave we’d analysed in class. How is it related to the pulse
train shown here?
If we pass x(t) through an LTI system with transfer function H(f) we obtain the output
waveform
y(t) =
∞
X
1
H(nf0 )ej2πnf 0 t
H(0) +
.
2
jπn
n =− ∞
n odd
Write H(f) = A(f)e
jθ (f)
. We’re given
A(f) =
&plusmn;
1 if 2000 &lt; |f| &lt; 4000,
0 otherwise,
and, as the medium introduces a constant delay τ = T/6 = 1/6 ms,
θ(f) = −2πfτ.
Observe, in particular, as f0 = 1/T = 1000 Hz,
A(nf0 ) =
&plusmn;
1 if n = −3 or n = 3,
0 otherwise.
so that it follows that all harmonics nf0 for |n| 6= 3 are blocked and the only terms in
the Fourier series that survive are the terms n = −3 and the term n = 3. Thus,
y(t) =
&cent;
&cent;
1 &iexcl;
1 &iexcl; j6πf 0 t−j6πf 0 τ
H(3f0 )ej6πf 0 t − H(−3f0 )e−j6πf 0 t =
e
− e−j6πf 0 t+j6πf 0 τ
j3π
j3π
2
2
=
sin(6πf0 t − 6πf0 τ) =
sin(6πf0 t − π)
3π
3π
as f0 τ = 1/6. It follows that
y(t) = −
2
sin(6πf0 t).
3π
5. A particular noisy communication link has a length of 10 km. In a digital signalling
scheme, each d km length of the link may be modelled as a binary symmetric channel
(n)
with raw bit error probability p = p(d) for any 0 ≤ d ≤ 10. Let Pbit = Pbit denote the
end-to-end bit error probability when n uniformly spaced repeaters are deployed in
the link. For our purposes the receiver himself counts as a repeater.
5
Tcom 370: Principles of Data Communications
University of Pennsylvania
(1)
(2)
Handout 8
Spring Semester 2000
(3)
a. Determine Pbit , Pbit , and Pbit when p(d) = d2 /1000 for 0 ≤ d ≤ 10.
b. Repeat when p(d) = d/100 for 0 ≤ d ≤ 10. Comment on the two cases.
Solution. For any n, the direct approach shows us that
(n )
Pbit
n &micro; &para;
X
&iexcl;
&cent;
n k
=
p (1 − pn−k ) = 12 1 − (1 − 2p)n ,
k
k=0
k odd
where p = p(d) = p(D/n) where D = 10 km is the end-to-end distance along the link,
and d = D/n is the link length between successive repeaters.
a. When p(d) = d2 /1000, direct substitution shows that
(1)
&iexcl;
&cent;
1 − (1 − 2 &times; 102 /1000)1 = 0.1,
&iexcl;
&cent;
= 12 1 − (1 − 2 &times; 52 /1000)2 = 0.04875,
&iexcl;
&cent;
= 12 1 − (1 − 2 &times; 102 /32 &times; 1000)3 = 0.0325981.
Pbit =
(2)
Pbit
(3)
Pbit
1
2
b. When p(d) = d/100, direct substitution again yields
(1)
&iexcl;
&cent;
1 − (1 − 2 &times; 10/100)1 = 0.1,
&iexcl;
&cent;
= 12 1 − (1 − 2 &times; 5/100)2 = 0.095,
&iexcl;
&cent;
= 12 1 − (1 − 2 &times; 10/3 &times; 100)3 = 0.0934815.
Pbit =
(2)
Pbit
(3)
Pbit
1
2
It is clear that increasing the number of repeaters brings more benefit when the raw
bit error probability increases faster than linearly with link length.
6
```