equalization - islamic university of...
TRANSCRIPT
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© Ammar Abu-Hudrouss Islamic University Gaza
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EqualizationEqualization
Slide 2Wireless Communications
ISI can cause an irreducible error floor when the modulation symbol time is on the same order as the channel delay spread.
In a broad sense, equalization defines any signal processing technique used at the receiver to alleviate the ISI problem caused by delay spread.
Signal processing can also be used at the transmitter to make the signal less susceptible to delay spread. These are such as spread spectrum and multicarrier modulation.
Equalizer design must typically balance ISI mitigation with noise enhancement, since both the signal and the noise pass through the equalizer
Introduction
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Slide 3Wireless Communications
goal of equalization is to mitigate the effects of ISI. However, this goal must be balanced so that in the process of removing ISI, the noise power in the received signal is not enhanced.
At the receiver front end white Gaussian noise n(t) is added to the signal, so the signal input to the receiver is
Y (f) = S(f)H(f)+N(f),
Equalizer Noise Enhancement
Slide 4Wireless Communications
Suppose we wish to equalize the received signal so as to completely remove the ISI introduced by the channel. This is easily done by introducing an analog equalizer in the receiver defined by
And the output of the equalizer become
where N(f) is colored Gaussian noise with power spectral density N0 /|H (f )|2.
For small values of H (f) the noise is enhanced.
fHfH eq /1
fNfSfHfNfSfY '/
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Slide 5Wireless Communications
Slide 6Wireless Communications
Equalizer Types
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Slide 7Wireless Communications
Equalization techniques fall into two broad categories: linear and nonlinear. The linear techniques are generally the simplest to implement and to understand conceptually. However, linear equalization techniques typically suffer from more noise enhancement than nonlinear equalizers.
Among nonlinear equalization techniques, decision-feedback equalization (DFE) is the most common, since it is fairly simple to implement and generally performs well.
However, on channels with low SNR, the DFE suffers from error propagation when bits are decoded in error, leading to poor performance.
Slide 8Wireless Communications
The optimal equalization technique is maximum likelihood sequence estimation (MLSE).
Unfortunately, the complexity of this technique grows exponentially with the length of the delay spread, and is therefore impractical on most channels of interest.
Equalizers can also be categorized as symbol-by-symbol (SBS) or sequence estimators (SE).
Linear and nonlinear equalizers are typically implemented using a transversal or lattice structure.
The transversal structure is a filter with N − 1 delay elements and N taps with tunable complex weights.
The lattice filter uses a more complex recursive structure
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Slide 9Wireless Communications
Folded Spectrum and ISI-free Trans
Slide 10Wireless Communications
The symbol dk is passed through pulse shape filter g (t ) and then transmitted over the ISI channel with impulse response c (t).
We define the equivalent channel impulse response h (t) = g (t) ∗c (t), and the transmitted signal is thus given by
d (t) is the train of information symbols and given by
tctgtd **
k sk KTtdtd
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Slide 11Wireless Communications
Let f (t) denote the combined baseband impulse response of the transmitter, channel, and matched filter:
Then the matched filter output is given by
Then sampling y (t) every Ts seconds yields
tgtctgtf m ***
tnkTtfdtgtntftdty gskm ***
nvknfdnyk
k
nvknfdfdnynk
kn
0
Slide 12Wireless Communications
We notice that we get zero ISI if f [n − k] = 0 for k ≠ n, i.e. f [k] = δ [k]f[0].
In this y [n] = dnf [0] + ν [n].
We now show that the condition for ISI-free transmission, f [k]= δ [k]f [0 ]
The function FΣ(f) is often called the folded spectrum, and FΣ (f ) = f [0] implies that the folded spectrum is flat.
01 fTnfF
TfF
n ss
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Slide 13Wireless Communications
To show this equivalence, first note that
We first show that a flat folded spectrum implies that f [k] = δ [k]f [0 ].
If we assume the folded spectrum equal f (0 )
Slide 14Wireless Communications
If we assume f [k] = δ [k]f [0 ]
So f [k] is the inverse Fourier transform of FΣ(f). Therefore, if f [k] = δ [k]f [0 ] and FΣ (f) = f [0].
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Slide 15Wireless Communications
Example: Consider a channel with combined baseband impulse response f (t) = sinc (t/Ts). Find the folded spectrum and determine if this channel exhibits ISI.
The Fourier transform of f (t ) is
Thus
No ISI . This can be seen from f (nTs) = 1 only for n = 0;
TfTfTTfT
fTrectTfF s
s
ss
/5.00/5.05.0/5.0
11
n ss TnfF
TfF
Slide 16Wireless Communications
If FΣ (f ) is not flat, we can use the equalizer Heq(z) to reduce ISI. In this section we assume a linear equalizer implemented via an N = 2L + 1 tap transversal filter:
The length of the equalizer N is typically dictated by implementation considerations, since a large N usually entails higher complexity.
Causal linear equalizers have wi = 0, i < 0.
For a given equalizer size N the equalizer design must specify and update the tap weights {wi} for a given varying channel frequency response.
Linear Equalizer
L
Li
iieq zwzH
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Slide 17Wireless Communications
The input to the equalizer can be represented as
where Ng (z) is the power spectrum of the white noise after passing through the matched filter G∗ (1/z∗) and
This is the discrete-time equivalent to the analog equalizer , and it suffers from the same noise enhancement properties. Specifically, the power spectrum N(z) is given by
Zero Forcing (ZF) Equalizer
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**** /11
/11
)(1
zGzHzGzCzGzFzH ZF
20
2
zHNzZFHzNzN g
Slide 18Wireless Communications
The ZF equalizer defined by HZF (z) = 1/F(z ) may not be implementable as a finite impulse response (FIR) filter. Specifically, it may not be possible to find a finite set of coefficients w−L, . . ., wL such that
IIR approximation can be done or the tap weights can be set to minimize the peak distortion by convex optimization.
zFzw
L
Li
ii
1
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Slide 19Wireless Communications
Example: Consider an channel with impulse response
The channel also has AWGN with power spectral density N0. Find a two-tap ZF equalizer for this channel.
Solution: We have
So
The two tap ZF equalizer therefore has tap weight coefficients w0 = 1 and w1 = -e −(Ts/τ) .
000/
tte
nht
1/11
zezH
sT
1/1 zezH sTeq
Slide 20Wireless Communications
In MMSE equalization the goal of the equalizer design is find the filter coefficients wi that minimize
Since the MMSE is a linear equalizer, its output is a linear combination of the input samples y [k]:
Minimum Mean Square Error (MMSE) Equalizer
2ˆkk ddE
L
Liik ikywd
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Slide 21Wireless Communications
Slide 22Wireless Communications
Consider a linear filter with N = 2L + 1 taps:
Define v = (v [k + L], v [k + L − 1 ] . . . , v [k − L]) = (vk+L, vk+L−1, . . . , vk−L) as a vector of inputs to the filter then
Thus, we want to minimize the mean square error
L
Li
iieq zwzH
wvvwd TTk ˆ
22*2*ˆ
kkHHT
kk ddwvwvvwEddEJ
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Slide 23Wireless Communications
Define Mv = E[vvH] and vd = E[vHdk]. The matrix Mv is an N × N Hermitian matrix and vd is a length N row vector. Assume E|dk|2 = 1. Then the MSE J is
Finding the gradient of J with respect to w and solving for
Which leads to
12 ** wvwMwJ dvT
022,,
dvT
LLw vMw
wJ
wJJ
Td
Tvopt vMw
1
Slide 24Wireless Communications
Substituting in these optimal tap weights we obtain the minimum mean square error as
For an infinite length equalizer, v = (vn+∞, . . . , vn, vn−∞) and w = (w−∞, . . ., w0, . . ., w∞). Then wTMv = vd can be written as [ref]
Taking z transforms
Ref: G.L. St¨uber, Principles of Mobile Communications, 2nd Ed. Kluwer Academic Publishers, 2001. (Ch 7.4)
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jjgijNijfw mi
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eq
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Slide 25Wireless Communications
The full MMSE equalizer when it is not restricted to finite length
1) The ideal infinite length MMSE equalizer cancels out the noise whitening filter.
2) This infinite length equalizer is identical to the ZF filter except for the noise term N0, so in the absence of noise the two equalizers are equivalent.
3) This ideal equalizer design clearly shows a balance between inverting the channel and noise enhancement
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/1
ˆ
NzFzGzH
zHm
eqeq
Slide 26Wireless Communications
It can be shown that the minimum MSE can be expressed in term of the folded spectrum FΣ(f )
Example: Find Jmin when the folded spectrum FΣ(f ) is flat, FΣ(f )= f [0], in the asymptotic limit of high and low SNR.
For high SNR, f0 >> N0 so Jmin ≈ N0/f0 = N0/Es =1/snr
For low SNR, N0 >> f0, so Jmin = 1
s
s
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/5.00
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00
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NdfNf
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s
T
Ts
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Slide 27Wireless Communications
Given the channel response h(t), the MLSE algorithm chooses the input sequence {dk} that maximizes the likelihood of the received signal w(t).
MLSE does not encode each symbol dk by itself but it estimated the most likely data sequence based on Veterbi algorithm.
The algorithm is very efficient but it has a very high computation complexity.
Maximum Likelihood Seq. Estimation
Slide 28Wireless Communications
The DFE consists of a feedforward filter B (z) with the received sequence as input (similar to the linear equalizer) followed by a feedback filter D (z) with the previously detected sequence as input.
Decision Feedback Equalization
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Slide 29Wireless Communications
The basic idea behind decision feedback equalization is that once the information symbol has been detected, its ISI that it induces on future symbols can be estimated and subtracted.
Slide 30Wireless Communications
The feedback approximate the baseband channel F(z), the resultant feedback is subtracted from the incoming signal.
The feedback filter must be strictly causal.
Assuming W (z) has N1 taps and V (z) has N2 taps, we can write the DFE output as
The typical criteria for selecting the coefficients for W(z) and V (z) are either zero-forcing (remove all ISI) or MMSE (minimize the expected MSE between the DFE output and the original symbol).
2
1 1
0ˆ
N
iiki
Niik dvknywd
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Slide 31Wireless Communications
When both W(z) and V (z) have infinite duration, it was shown by Price that the
optimal feedforward filter for a zero-forcing DFE is1/G∗m(1/z∗).
For the MMSE criterion, we wish to minimize
Let fn = f[n] denote the samples of f(t). Then this minimization implies that the coefficients of the feedforward filter must satisfy the following set of linear equations:
For
2ˆ
kk ddE
*0
1
lNi
ili fwq
0,,, 10
0 * NililNffqlj iljjli
Slide 32Wireless Communications
The coefficients of the feedback filter are then determined from the feedforward coefficients by
It was shown that the resulting minimum MSE is
0
1Niikik fwv
s
s
T
Ts dfNff
NTJ/5.0
/5.00
0min lnexp
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Slide 33Wireless Communications
Although MLSE is the optimal form of equalization, its complexity precludes its widespread use.
Most of these techniques either reduce the number of surviving sequences in the Viterbi algorithm or reduce the number of symbols spanned by the ISI through preprocessing or decision-feedback in the Viterbi detector.
These reduced complexity equalizers have better performance versus complexity tradeoffs than the other equalization techniques, and achieve performance close to that of the optimal MLSE with significantly less complexity.
Other Equalization Methods
Slide 34Wireless Communications
All of the equalizers described so far are designed based on a known value of the composite channel response h(t) = g(t) ∗ c(t).
Since in wireless channels c(t) = c(τ, t) will change over time, the system must periodically estimate the channel c(t) and update the equalizer coefficients accordingly.
This process is called equalizer training or adaptive equalization . The equalizer can also use the detected data to adjust the equalizer coefficients. This process is called equalizer tracking.
Blind equalizers do not use training: they learn the channel response via the detected data only
Adaptive Equalizers: training and tracking
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Slide 35Wireless Communications
During training, the coefficients of the equalizer are updated at time k based on a known training sequence [dk−M, . . . , dk] that has been sent over the channel.
The length M of the training sequence depends on the number of equalizer coefficients that must be determined and the convergence speed of the training algorithm.
If the training algorithm is slow relative to the channel coherence time then the channel may change before the equalizer can learn the channel.
we will choose the updated coefficient {w−L(k + 1), . . ., wL(k + 1)} as the coefficients that minimize the MSE between dk and d^k
Slide 36Wireless Communications
Recall that
The w coefficients can be found by
Where
L
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pRkwkwkw LL11,,11
2*
*1
21
*1
**1
2
LkLkLk
LkLkLkLkLk
LkLkLkLkLk
yyy
yyyyyyyyyy
R
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Slide 37Wireless Communications
And
Note that the optimal tap updates in this case requires a matrix inversion, which requires N 2 to N 3 multiply operations on each iteration (each symbol time Ts).
However, the convergence of this algorithm is very fast: it typically converges in around N symbol times for N the number
of equalizer tap weights
TLkLkk yydp
Slide 38Wireless Communications
If complexity is an issue then the large number of multiply operations needed to do MMSE training can be prohibitive. A simpler technique is the least mean square (LMS) algorithm .
In this algorithm the tap weight vector w(k + 1) is updated linearly as
where k = dk − ˆ dk is the error between the bit decisions and the training sequence and Δ is the step size of the algorithm, which is a parameter that can be chosen.
The choice of Δ dictates the convergence speed and stability of the algorithm.
**1 LkLkk yykwkw
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Slide 39Wireless Communications
However, the LMS algorithm exhibits significantly reduced complexity compared to the MMSE algorithm since the tap updates only require approximately 2N +1 multiply operations per iteration.
Slide 40Wireless Communications
Consider a 5 tap equalizer that must retrain every .5Tc, where Tc is the coherence time of the channel. Assume the transmitted signal is BPSK with a rate of 1 Mbps for both data and training sequence transmission.
Compare the length of training sequence required for the LMS equalizer versus the Fast Kalman DFE. For an 80 Hz Doppler, by how much is the data rate reduced in order to do periodic training for each of these equalizers.
How many operations does each require for this training
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Slide 41Wireless Communications
The equalizers must retrain every .5Tc = .5/Bd = .5/80 = 6.25 msec. From the table, for a data
rate of Rb = 1/Tb = 1 Mbps, the LMS algorithm requires 10NTb = 50 × 10−6 seconds to train, and the Fast
Kalman DFE requires NTb = 50 × 10−5 seconds to train. If training occurs every 6.25 msec, the fraction of time
the LMS algorithm uses for training is 50 × 10−6/6.25 × 10−3 = .008. Thus, the effective data rate becomes
(1 − .008)Rb=.992 Mbps. The fraction of time used by the Fast Kalman DFE for training is 50 × 10−5/6.25 ×
10−3 = .0008, resulting in an effective data rate of (1 − .0008)Rb=.9992 Mbps. The LMS algorithm requires
approximately 2N + 1 = 11 operations for training per training period, whereas the Fast Kalman DFE requires
20N + 5 = 105 operations, an order of magnitude more than the LMS algorithm. With processor technology
today, this is not a significant difference in terms of processor requiremen