In Signal Processing and Communications, one often has to deal with unknown signals. For example, the wireless channel (air medium) distorts the transmitted signal in so many ways (multipath fading, ISI, etc.) that the receiver has to use its own methods to recover the signal. Generally, such signals are modeled as stochastic random processes and statistical methods are used to "demystify" the received waveform. The air medium can be considered to be a time-varying FIR filter and the receiver unit must estimate the properties of this wireless medium. Such techniques are called Channel Estimation methods.

I have begun a series of articles, in which I'll explain the various methods of channel estimation. In this post, I'll explain a method called as Maximum Likelihood Estimate. I used LaTeX and converted it to pdf. You can download the PDF here.

Please comment if you have anything to ask !

1. LTI Channel Model

For simplicity, let us assume that the channel is linear and time-invariant. Later, we’ll extend the model to time-varying channel.

Let the input sequence of information bits be s[n]. Let v[n] be additive noise. The output y[n] at the receiver can be modelled as

L∑−1 y[n] = h [k]s[n − k] + v [n ] k=0
(1)

where n = 0, 1,…,N − 1. We have assumed that the channel impulse response is h[k].

The channel estimator is defined as any static (known function of observable random variables that is itself a random variable) whose values are used to estimate τ(𝜃), where τ(⋅) is some funtion of an unknown parameter 𝜃.

2. Maximum Likelihood Estimator

The channel vector is given by

h = {h[0],h [1],⋅⋅⋅ ,h [L − 1]}T
(2)

where {⋅}T is the transpose of a vector. Suppose that we received N samples of the output

y = {y[0 ],y[1],⋅⋅⋅ ,y[N − 1]}T
(3)

we can rewrite the Eq(1) as

y = Sh + v
(4)

The matrix S is an N × L Toeplitz matrix consisting of the samples of the input sequence {s[n],n = 0,1,...,N − 1} given by

⌊ ⌋ s[0 ] s [N − 1] ⋅⋅⋅ s[N − L + 1] || s[1 ] s[0] ⋅⋅⋅ s[N − L + 2] || S = | .. .. .. .. | ⌈ . . . . ⌉ s[N − 1] s [N − 2] ⋅⋅⋅ s[0]
(5)

Let 𝜃 be the vector of unknown parameters that may contain the channel vector h and possibly the entire or part of the input vector s. Assume that the joint probability distribution of noise vector v and the input vector s is known. We can then obtain the probability density function (pdf) of the observation vector y.

The joint pdf of the observation fy(y; 𝜃) is called as the likelihood function. For simplicity, assume that 𝜃 is a single parameter. If the observation samples y[0],y[1],…,y[N − 1] are independent of each other, the likelihood function can be written as the product of individual densities:

fy(y;𝜃) = f(y[0],𝜃)f(y[1],𝜃)⋅⋅⋅f(y[N − 1],𝜃 )
(6)

Thus, the maximum likelihood estimator is the solution of the equation

dfy(y;𝜃)- d𝜃 = 0
(7)

Generally, it is easier to find the maximum of the logarithm of the likelihood, ln fy(y; 𝜃) because both fy(y; 𝜃) and ln fy(y; 𝜃) have their maximums at the same value of 𝜃. If, instead of a single unknown 𝜃, there are multiple unknowns, the normal derivative d∕d𝜃 in Eq(7) is replaced by the partial derivative ∂∕∂𝜃.

2.1 Example

Assume that the channel impulse response h = h[0],h[1],…,h[N − 1] is a Gaussian random process, that is

[ ] 1 1 2 fh[n](h[n];μ,σ) = --√----exp − --2(h [n] − μ) σ 2π 2σ
(8)

with mean −∞ < μ < ∞ and variance σ2,σ > 0. Find the MLE ˆμ and ˆ2 σ of μ and σ2 respectively.

It is reasonable to assume that the individual samples h[0],h[1],…,h[N − 1] are independent of each other. In such a case, the likelihood function is given by

N− 1 [ ] ∏ --1--- -1-- 2 LF (h[n ];μ, σ) = σ √2-π exp − 2σ2(h[n] − μ) n=0 [ ] ( 1 )N ∕2 1 N∑ −1 = ----2 exp − ---2 (h [n ] − μ )2 2πσ 2σ n=0
(9)

The logarithm of the likelihood function is given by

LLF = lnLF (h [n ];μ, σ) N−1 = − N--ln(2π ) − N-ln σ2 − -1--∑ (h[n] − μ)2 2 2 2σ2 n=0
(10)

At the maximum (or minimum) of LLF, the first derivative of LLF is 0. However, at the maximum, the second derivative of LLF is negative. In other words, we need to solve for μ = ˆμ such that

∂ ---LLF = 0 ∂μ ∂2-- ∂μ2 LLF < 0
(11)

Similarly, for σ2 = ˆσ2

∂-- ∂ϕ LLF = 0 2 ∂---LLF < 0 ∂ϕ2
(12)

where ϕ = σ2.

Wait until next week for the solution of above equations ! See you !

3. License

Copyright © 2010, Satish BD. Some rights reserved. All the content in this article is licensed under the Creative Commons BY-NC-SA 2.5 India License. The original location of this document can be found at my site: http://www.bdsatish.in. You can e-mail me at: mail@bdsatish.in

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