In the simulations, we will demonstrate the duality between power-minimizing problem and rate-maximizing problem. In the following examples, the number of subchannels M is 4. The noise vector q is assumed to be complex white Gaussian with E[qq†] = I4. The symbol error rate constraint ǫ is assumed to be 10−4. In examples 1-2, we use a fixed 4 × 4 MIMO channel as shown in example 1. In examples 3-4, the results are averaged over random channels. For the problems Apow and Arate, we use the solutions in Section 5.3.1. For Apow,int and Arate,int, we use the solutions in Section 5.3.2.
Example 1. Duality between Apow and Arate. In this example, we will demonstrate the results in Theorem 5.1 and Theorem 5.2. Consider a 4 × 4 channel H that is given by
(a)
Given target bit rate B0, we use (5.11) to find the optimal transceiver, and (2.1) to compute the corresponding transmit power P∗(B0) for the problem Apow. Table 5.1(a) shows the minimal transmit power P∗(B0) when the target bit rates are B0 = 2, 4, 6, 8, 10 bits. Using the minimized power in Table 5.1(a) as power constraint, Table 5.1(b) shows the maximal bit rate for the rate maximizing problem Arate. The rates are computed using (2.20) for the optimal transceiver in (5.12). We can see that B∗(P∗(B0)) = B0 and the solution of the power-minimizing problem is also optimal for the rate-maximizing problem as we have shown in Theorem 5.1.
Table 5.2(a) shows the maximal bit rate B∗(P0) for Arate when the power constraints are P0 = 2, 4, 8, 16, 32 dB. Table 5.2(b) shows the minimal power P∗(B0) for the problem Apowwhen the target bit rates are equal to the maximized rate in Table 5.2(a). We can see that P∗(B∗(P0)) = P0 and the solution of rate-maximizing problem is also a solution of the power-minimizing problem as we have shown in Theorem 5.2.
Example 2. ZF and MMSE receivers for Apow,int. When there is no integer bit constraint, the optimal solution with ZF receiver and the optimal solution with MMSE receiver are the same. When there is integer constraint, the solutions are in general different as we demonstrated in this example. We use the same channel as in example 1. Let us compute the optimal integer bit
(a)
allocation and transceiver of Apow,int when MMSE reception is considered. We use the method mentioned in section 5.3.2. The target bit rate B0 is set to be 8 bits. In this case the optimal integer bit allocation is
b =
and the minimal transmit power is 13.891 dB. The optimal transmitter is given by
and the overall transfer function T is
T = GHF =
We can see that the overall transfer function for the MMSE receiver is not di-agonal. For the ZF case. For the ZF case, the minimized power is 13.8915 dB, which is very close to the MMSE case.
Example 3. Duality between Apow and Arate. In this example, we use random channels to demonstrate the connections between power minimization and rate maximization problems. The channel is of size 4 × 4 and the elements
are complex Gaussian random variables whose real and imaginary parts are in-dependent with zero mean and variance 1/2. Monte Carlo simulation using 106 channel realizations is used to generate the following results. For each channel, we compute the optimal solutions of Apow and Arate using (5.11) and (5.12) in Section 5.3.1. Fig. 5.2 shows the maximal transmission rates B∗(P0) of Arate as a function of power constraint. Fig. 5.3 shows the minimal transmit power P∗(B0) of Apow as a function of target bit rate. We can observe the duality between that the power-minimizing and rate-maximizing problems from Fig. 5.2 and Fig. 5.3.
For example, the minimal power of Apow is 9 dB when the target bit rate is 5 bits. When we set the power constraint in Arate to be 9 dB, the maximal bit rate is 5 bits. On the other hand, the maximal bit rate of Arate is 9 bits when the power constraint is 15 dB. When we set the target bit rate in Apow to be 9 bits, the minimal power is 15 dB.
0 5 10 15 20
0 5 10 15
Maximal bit rate
Transmit power constraint P
0 (dB) B*(P
0)
Figure 5.2: Maximal bit rate B∗(P0) for Arate as a function of power constraint P0 without integer constraint.
Example 4. Minimal power for Apow,int and maximal bit rate for Arate,int. We use the same random channel as in example 3. In Table 5.3, we compute the minimal transmit power of Apow,int with MMSE and ZF receivers.
0 5 10 15 0
5 10 15 20
Target bit rate B
0 (bits)
Minimal transmit power (dB)
P*(B
0)
Figure 5.3: Minimal transmit power P∗(B0) for Apow as a function of target bit rate B0 without integer constraint.
When the target bit rate is B0, the minimal transmit powers of the MMSE case and the ZF case are denoted by Pint,mmse∗ (B0) and Pint,zf∗ (B0), respectively. For comparison, we also show the transmit power P∗(B0) of Apow (without integer constraint). We can see that the gap between Pint,mmse∗ (B0) and Pint,zf∗ (B0) is small. Also, the difference between Pint,mmse∗ (B0) and P∗(B0) is smaller than 0.21 dB. In Table 5.4, we compute the maximal bit rate of Arate,int for the MMSE and ZF receivers. The maximal bit rate for the MMSE and ZF cases are denoted respectively by Bint,mmse∗ (P0) and Bint,zf∗ (P0). Also shown in Table 5.4 is the maximal bit rate B∗(P0) of Arate (without integer constraint). We can see that Bint,zf∗ (P0) is close to Bint,mmse∗ (P0). The difference between B∗int,mmse(P0) and B∗(P0) is smaller than 0.6 bits. This gap is less than 0.15 bits per symbols.
5.5 Summary
In this chapter, we consider two commonly used transceiver design criteria: power minimization criterion and rate maximization criterion. The duality are derived
B0 (bits) P∗(B0) (dB) Pint,zf∗ (B0) (dB) Pint,mmse∗ (B0) (dB)
2 2.1167 2.3322 2.3254
4 7.1634 7.3107 7.3078
6 10.6614 10.7832 10.7826
8 13.4708 13.5796 13.5788
10 15.9114 16.0175 16.0170
12 18.1311 18.2314 18.2312
14 20.1976 20.2972 20.2970
Table 5.3: Transmit power of Apow(without integer bit allocation), Apow,int (ZF), and Apow,int (MMSE) when the target bit rate is B0 = 2, 4, 6, 8, 10, 12 bits.
P0 (dB) B∗(P0) (dB) Bint,zf∗ (P0) (bits) Bint,mmse∗ (P0) (bits)
2 2.0305 1.4549 1.4572
4 2.7096 2.1549 2.1557
6 3.5429 2.9858 2.9865
8 4.5391 3.9689 3.9700
10 5.7103 5.1326 5.1333
12 7.0629 6.4788 6.4794
14 8.5888 8.0033 8.0037
16 10.2714 9.6811 9.6815
Table 5.4: Bit rate of Arate (without integer bit allocation), Arate,int (ZF), and Arate,int (MMSE) when the target bit rate is P0 = 2, 4, 6, 8, 10, 12, 14, 16 dB.
for these two problems. If there is no integer bit constraint, the optimal solution in either one solution is also optimal in the other problem. When there is an integer bit constraint, we have shown the rate-maximizing problem is equivalent to the power-minimizing problem with power modifications. Using the duality, the optimal solution of the rate maximization problem with integer bit constraint can be found using the solution of the power minimization problem. Simulation results have been shown to demonstrate the duality of these two problems.
Chapter 6
Overview of Multicarrier Systems
Multicarrier systems have found many applications in DMT systems and OFDM systems. For the multicarrier systems, the frequency band of the channel is divided into a number of subchannels and information is transmitted on each of the subchannel. In this chapter, we will introduce the multicarrier systems and find the filterbank representation of the multicarrier systems.
6.1 DFT Based Multicarrier System
Figure 6.1: Block diagram of the DFT based multicarrier system.
The block diagram of the DFT based multicarrier system is as shown in Fig. 6.1. The input of the transmitter s is an M × 1 vector of modulation symbols. The symbol vector s are assumed to be zero-mean and uncorrelated.
The autocorrelation matrix of the input vector s is assumed to be
Rs = εsIM. (6.1)
The channel is modeled as an FIR filter of order L, i.e., C(z) =
L
X
n=0
c(n)z−n. (6.2)
The channel noise q(n) is assumed to be a circularly symmetric complex Gaussian random process with zero mean and variance N0. The channel noise q(n) is assumed to be uncorrelated with the symbols sk(n). At the transmitter, IDFT is applied to the input symbol vector s and the output vector x is
x = W†s, (6.3)
where W denotes the M × M normalized DFT matrix given by [W]mn= 1
√Me−j2πmnM , for 0 ≤ m, n ≤ M − 1.
The outputs are converted to a block of M serial samples by the parallel to serial operation (P/S). Then a cyclic prefix of length ν is inserted by copying the last ν samples of the block to the beginning. The length of the cyclic prefix ν is chosen so that ν ≥ L, which ensures that inter-block-interference (IBI) can be removed easily by discarding the prefix at the receiver.
At the receiver, after prefix removal the samples are blocked into M by 1 vectors r by the serial to parallel operation (S/P). When there is no channel noise, it can be shown that the transfer matrix from x to r is the M ×M circulant matrix given by
In the presence of channel noise, the received vector r is
r = Ccircx + q, (6.5)
where q is the blocked channel noise vector of size M. Then the DFT matrix is applied, i.e.,
y = Wr (6.6)
= WCcircW†s + Wq. (6.7)
From [41], we know that circulant matrices Ccirc can be diagonalized using DFT and IDFT matrices,
Ccirc= W†ΛW, (6.8)
where Λ is a diagonal matrix. The diagonal element λk of Λ corresponding to the M-point DFT of the channel impulse response, i.e.,
λk= [Λ]kk = C(z)|z=e−j2πk/M. (6.9) The DFT output vector becomes
y = Λs + Wq. (6.10)
Then the scalar multipliers 1/λk, which are called frequency domain equalizers (FEQ), are applied to y. The transceiver is ISI free and the receiver is a zero-forcing receiver. The receiver outputs are identical to the inputs of the transmitter in the absence of channel noise. From (6.10), the signal to noise ratio (SNR) of the k-th subchannel is given by
βk = |λk|2εs N0
. (6.11)
Transmission rate: For the QAM modulation, suppose the target symbol error rate ǫk are given. Then the number of bits loaded on the k-th subchannel can be computed by (2.19), i.e.,
βk
Suppose the sampling time of the system is Ts. Then the transmission rate is given by
R =
PM −1 k=0 bk
NTs
. (6.13)
Complexity:
The main computations of the transceiver are those of the IDFT and DFT matri-ces, for which fast algorithms can be applied. The complexity of the transmitter is simply that of an IDFT matrix and the complexity of the receiver is that of a DFT matrix plus M multiplications for FEQs. Moreover, except for the FEQs, the computations are channel independent.