Polytechnic University, Dept. Electrical and Computer Engineering
EE3414 Multimedia Communication System I
Spring 2006, Yao Wang
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Homework 11 (Digital Communication)
(Solution)
Written Assignment:
1.
Consider the two digital modulation schemes shown in the figure below. For each scheme, determine the minimum
distance between two symbols dmin , the average energy per symbol Eav, and the average energy per transmitted bit Eb.
Which scheme is more efficient in terms of transmission energy usage?
10 x
11
x
-3A
10
00
01
x
x
-A
A
x
(a) 4-ary PAM
3A
11
x
A x 00
A
x
01
(b) 4-QAM
Solution:
For 4-ary PAM, the distance between any two adjacent symbols (e.g. “A” and “3A”) is 2A, therefore dmin=2A. The
energy for symbols “A” and “-A” are both A2 ; and the energy for symbols “3A” and “-3A” are both 9A2 . Therefore the
average energy per symbol is Eav =(A2+9A2)/2=5A2. Because each symbol carries 2 bits, the average energy per bit is
Eb = Eav /2=5A2/2.
For 4-QAM, the distance between two closest symbols (say “00” and “01”) is 2A, therefore dmin=2A. The length from
the origin to each symbol is 2 A . So the energy for each symbol is 2A2 . Therefore the average energy per symbol is
Eav =2A2. Because each symbol carries 2 bits, the average energy per bit is Eb = Eav /2=A2.
Because the two schemes have the same dmin the probability of one symbol be decoded to its nearest neighboring
symbol because of channel noise is roughly equal. Because in both schemes, each symbol contains 2 bits, the
probability of bit error will be roughly the same. But 4-QAM requires less energy per bit, and therefore is more efficient
in making use of transmission energy.
2.
What is the function of parity check? Consider the scheme where we append a parity bit to each group of 7 bits to
generate 1 byte so that each byte has an “even” parity. What is the parity bit you will add to “1010100”? With this
added parity bit, can you detect that some bits in a received byte get inverted? What cases of errors can be detected and
what cases cannot be detected?
Solution: Parity check is a simple method of detecting bit errors. For the particular example, because there are 3 “1”s in the
given 7 bits, to make the number of “1”s to be even, we should add a “1”. The resulting sequence is “10101001”. Upon
receiving 8 bits, if the number of “1”s is odd, then we know some bits must be wrong. In fact, it could be 1 bit (any one of
the 8 bits) or 3 bit or 7 bits got flipped, which will all cause the total number of “1”s being odd, rather than even. Therefore,
this method can detect all error patterns containing odd number of inverted bits. But it cannot detect any error pattern
containing even number of erroneous bits.
3. Consider repetition coding with a redundancy factor of 3, i.e., to send an information bit “0”, we send “000”; to send an
information bit “1”, we send “111”. Suppose the probability that any one bit gets inverted during transmission is p=0.1,
what is the probability that an information bit is decoded wrong?
Solution: If the information bit is “0”, we will send “000”. If the received 3 bits has 2 or 3 bits that are “0”, then we will
decode the information bits to “0”, otherwise, to “1”. Therefore, “0” will be decoded to “1” only if 2 or 3 bits are inverted.
That is “000” is changed to either of the following: “111” (3 bit error, with probability p^3), “011”,”101”,”110” (2 bit
errors, 3 possibilities, each with probability p^2 (1-p) ). So the total probability is
pe  p 3  3 p 2 (1  p ). For p=0.1, pe  0.028.
4.
Consider a channel with bandwidth 1MHz.
(a) Suppose the channel is noiseless, what is the maximum number of bits that can be transmitted reliably if you use 2ASK for modulation? What if you use 8-ASK?
(b) Suppose the channel suffers from noise with a signal to noise ratio of 10 dB. What is the maximum number of bits
that can be transmitted reliably? What if the channel SNR is 30 dB?
Solution: (a) when the channel is noiseless, the channel capacity C=2Bm, where B is the channel bandwidth, m is the
number of bits per modulation symbol. In our case, B=1MHz. With 2-ASK, m=1, C=2 M bit/s. With 8-ASK, m=3, C= 6
Mbit/s.
(b) When the channel is noisy, the channel capacity = C=B log2 (1+SNR) bits/second.
With SNR=10 dB=10, C=10^6 log2(11)=
With SNR=30 dB = 1000, C=10^6 log2(1001)=