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Tutorial 5 – Frequency Diversity

Question 1
For typical wireless channels, why is the delay spread in general much less than the coherence time? Give
an example. What are the implications of this observation on: a) an OFDM system and b) a direct-sequence
spread-spectrum (DSSS) system with Rake combining?
Solution
In outdoor propagation environments, the difference in the distances travelled by different paths is of
the order of a few hundred meters. For example, for a path difference of 300m, the delay spread is
Td = 300m/c = 1µs. As such, Td is generally in the order of a microsecond. On the other hand, for a
mobile speed of 60km/h and a carrier frequency of 1GHz, the Doppler shift is Ds = fc v/c ≈ 55.55Hz
and correspondingly the coherence time is Tc ≈ 1/4Ds ≈ 4.5ms. As such, Tc is generally in the order of
a millisecond. This indicates that Td ≤ Tc (underspread) in typical scenarios.
a) OFDM. OFDM assumes that the channel remains constant during the transmission of an OFDM
block, which requires N << Tc W . The overhead (both in time and power) incurred by the use of the
cyclic prefix is L/(L + N ), where L = Td W is the number of taps. To obtain a small overhead, we need
L << N , which implies Td ≤ Tc . As such, the underspread condition is required for a small overhead.
b) DSSS. The Rake receiver requires the channel to remain constant during the transmission of the
spreading sequence of length n, so n << Tc W . Also we require n >> L for the ISI to be negligible.
These two conditions together require that Td ≤ Tc .

Question 2
Consider a linear equalizer with N = L = 2.
a) Express the received signals.
b) Over the two symbol times (time 1 and time 2), the ISI channel can be viewed as a 2 × 2 MIMO
channel between the input and output symbols. Identify the channel matrix H.
Solution
a) The received signals for N = L = 2 can be expressed as
   
y[1] [ ] w[1]
  x[1] x[2] x[3]  
 y[2]  = [h0 h1 ] +  w[2]  . (1)
0 x[1] x[2]
y[3] w[3]
As such, the received signal for time 1 is given by y[1] = h0 x[1] + w[1], and the received signal for time
2 is given by y[2] = h0 x[2] + h1 x[1] + w[2].
b) Over the two symbol times (time 1 and time 2), the channel matrix H can be equivalently expressed
as
[ ]
h0 0
H= . (2)
h1 h0
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Question 3
Let us describe a simple OFDM system with the following parameters.
a) If the total number of samples for DFT and IDFT is N , what is the DFT and IDFT matrix?
b) Suppose that the sample vector at the output of IDFT at the transmitter is [a1 , · · · , aN , b1 , · · · , bN ],
what is the relationship between the cyclic prefix signals and the value of N ?
c) Suppose that two CP signals are added to each OFDM symbol. Assume that the received signal is
[yN −1 , yN , y1 , · · · , yN , zN −1 , zN , z1 , · · · , zN ] and one OFDM symbol is added before data transmission to
facilitate channel estimation. Detail the channel estimation process and the data decoding process.
Solution
a) The DFT matrix ix expressed as a Vandermonde matrix given by
 
1 1 1 1 ··· 1
 1 w2 w3 ··· wN −1 
 w 
 
1  1 w2 w4 w6 ··· w2(N −1) 
W=√  
,
 (3)
N 1 w3 w6 w9 ··· w3(N −1) 
 .. .. .. .. .. .. 
 . . . . . . 
1 wN −1 w2(N −1) w3(N −1) · · · w(N −1)(N −1)
√
where w = exp(−2πi/N ) with i = −1. The IDFT matrix is W−1 .
As presented in the lecture notes, the 2-point DFT matrix is
[ ]
1 1 1
W=√ , (4)
2 1 −1
and the 4-point DFT matrix is
 
1 1 1 1
1
 1 −i −1 i 

W=  . (5)
2 1 −1 1 −1 
1 i −1 −i

b) If the number of the cyclic prefix signals is L, then aN to aN −L+1 should be added before a1 . For
example, if L = 1, the transmit vector is [aN , a1 , · · · , aN , bN , b1 , · · · , bN ], and if L = 2, the transmit
vector is [aN −1 , aN , a1 , · · · , aN , bN −1 , bN , b1 , · · · , bN ].
c) At the receiver, the CP is first removed from the received signal, resulting in [y1 , · · · , yN , z1 , · · · , zN ].
As described in the question, y = [y1 , · · · , yN ]T are transmitted for channel estimation and z = [z1 , · · · , zN ]T
are the transmitted data. As such, the estimated channel is obtained via f = Wy. The transmitted data is
obtained via ż = Wz.
After obtaining these parameters, we note that the data żn is transmitted via the channel fn . As such,
the data is decoded as z̈n = fn∗ żn .
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Question 4
Consider an OFDM system with total passband bandwidth B = 1 MHz. A single carrier system would
have symbol time Ts = 1/B = 1µs. The channel has a maximum delay spread of Td = 5µs, so with Ts
and Td there would clearly be severe ISI. Assume an OFDM system with 16-QAM modulation applied
to each subchannel. To keep a small overhead, the OFDM system uses N = 128 subcarriers to mitigate
ISI. So TN = N Ts = 128µs. The length of the cyclic prefix is set to η = 8 > Td /Ts to insure no ISI
between OFDM symbols. For these parameters, find the subchannel bandwidth, the total transmission time
associated with each OFDM symbol, the overhead of the cyclic prefix, and the data rate of the system.
Solution
a) The subchannel bandwidth is BN = 1/TN = 7.8125 KHz. As such, we see BN << Wc = 1/2Td =
100 KHz, which insures negligible ISI.
b) The total transmission time for each OFDM symbol is T = TN + ηTs = 128 + 8 = 136µs.
c) The overhead associated with the cyclic prefix is 8/136 which is roughly 5.88%.
d) The system transmits log2 16 = 4 bits/subcarrier every T seconds, so the data rate is 128 × 4/136 ×
−6
10 = 3.76 Mbps. We see that this data rate is slightly less than 4B = 4 MHz due to the cyclic prefix
overhead.
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Home Work
Requirements
1. Write down your full name, student number, and signature.
2. Detail the steps of your analysis/calculations/solutions. A single answer without derivations or expla-
nations is not acceptable.
3. Hand in a hard copy in the following class.

Question 1
Consider a linear equalizer with N = 4 and L = 3.
a) Express the received signals for all symbol times.
b) Identify the channel matrix H over all symbol times.
c) If MLSD is adopted to recover the received signals, detail the receiving process and explain why the
full diversity gain is achieved.
d) If zero-forcing (decorrelating) equalizer is adopted to recover the received signals, which receive filter
is used at the receiver? How does the achievable diversity gain change?

Question 2
Consider a system with total passband bandwidth B = 1.25 MHz and the signal travels along a path of
1.2 km. For this system, OFDM with 256 subcarriers is applied to mitigate ISI and 64-QAM modulation
is applied to each subchannel.
a) If the length of the cyclic prefix is set to η = 10, find the subchannel bandwidth, the total transmission
time associated with each OFDM symbol, the overhead of the cyclic prefix, and the data rate of the
system.
b) Using the same length of the cyclic prefix as a) but applying 128-QAM to each subchannel, what
changes will occur to the system configuration parameters obtained in a)?

Question 3
Consider an OFDM system. Assume that the total number of samples for DFT and IDFT is 2. The total
number of subcarriers is 2. One sample of cyclic prefix is added to each OFDM symbol. Each packet
consists of one OFDM symbol for channel estimation followed by one or more OFDM symbols for data
transmission. In data symbols, QPSK with {1 + j, −1 + j, −1 − j, 1 − j} is as the modulation scheme.
Suppose that the received signals are given by

1.6−1.1j, −1.2+0.5j, 0.9−1.7j, 0.2+2.3j, −0.4−1.7j, 1.2+0.3j, −0.5−0.5j, −1.1+1.2j, 0.2−0.1j. (6)

What is estimated channel? Assuming that the estimated channel is correct, what are the transmitted
signals?