# Proper method for generating Channel coefficients in MATLAB

I am trying to estimate the coefficients of a Moving average model using popular estimation method such as Least Squares (LS). For educational purposes, I am trying to see if LS can work well when the coefficients are complex valued or not. I have not seen any article or text book where LS is applied to complex parameter estimation. In many papers, I have seen that the input data is complex valued, channel is complex valued and the measurement noise is complex valued. When is this setting applicable?

1. What if only the channel coefficients are complex valued (input and noise are real valued), then where is this setting applicable?

2. What is the channel coefficients and the input are complex valued, then where is this applied?

• why did you generate h_R and h_I containing two elements ? What do h_R and h_I mean ? And why did you add the element 1 in h ? – AlexTP Jul 26 '17 at 11:47
• complex channel coefficient is just a way to represent the independent real coefficients. You just need to generate h = [hR_0, hR_1, hR_2] + 1i * [hI_0, hI_1, hI_2]. The independence/correlation between coefficients depend on your model. And if I were not wrong, the number of element of h is the order of your MA model. – AlexTP Jul 27 '17 at 8:06
• see my answer, it is much easier to format the text in answer part :) – AlexTP Jul 27 '17 at 8:20

• Complex channel coefficient is just a way to represent the independent real coefficients. You just need to generate h = [h_0, h_1, h_2] = [hR_0, hR_1, hR_2] + 1i * [hI_0, hI_1, hI_2]. The independence/correlation between coefficients depend on your model. And if I were not wrong, the number of element of h is the order of your MA model.
• The idea behind complex representation is that if the orthogonal set of vector space of baseband signal of bandwidth $[0,W]$ is $\left\lbrace \phi_k(t) \right\rbrace$, the orthogonal set of vector space of passband signal $[f_c-W,f_c+W]$ is $\left\lbrace \phi_k(t) \cos(2\pi f_ct), \phi_k(t) \sin(2\pi f_ct) \right\rbrace$. More details can be found at Read chapter 2. Furthermore, as real signals have symmetric spectrum, it is convenient to represent these signals by its Baseband Equivalent versions. Thus, a complex representation is used.

So, if you work in the baseband version of signal, channel coefficients are generally complex.

The continuos time white noise is a model of infinite-dimensional signal. When you project this signal to the orthogonal set to obtain the discrete time noise samples, you follow the dimension of the orthogonal set, thus the baseband noise samples are generally complex.

The input data can be complex or real depending on how you want to use available dimensions.

After all, yes, you can process your signal as you have written, regardless real or complex assumptions.

for k = 2:N y(k) = x(k) + 0.6*x(k-1) + 0.3*x(k-2);

• Thanks for the update. After the real and imaginary parts of the channel are generated, is it legal to do h=[1 sqrt(h_R.^2+h_I.^2)]; and then perform y(k) = x(k)+ h(2)*x(k-1) + 0.3*x(k-2)? If so, then why generate complex channel coefficients in the first place? I am trying to apply estimators such as Least Squares and Least Mean Squares to estimate the channel coefficients. If I perform convolution of the signal using h=[1 sqrt(h_R.^2+h_I.^2)]; upon estimation, I will get real valued channel coefficients and not complex (as the output is real valued). – Srishti M Jul 29 '17 at 6:26
• Should I not be doing convolution of the data with the complex channel parameters, and not with the square of the coefficients? What is the physical implication of doing the square and then doing convolution...the estimation results would be different in both the cases. These are not clear to me. Looking forward for your insights on what is the proper way to do. Thank you very much. – Srishti M Jul 29 '17 at 6:26
• I have no idea why you have square your complex coefficients. Physically, complex discrete coefficients model magnitudes and phases why the norms (square of absolute) model energy. Convolution operation models the passage of a signal over a linear system. It is up to you to choose which operation depending on your models. – AlexTP Jul 30 '17 at 6:39
• So, for complex valued coefficients, I should use the complex valued coefficients and not the square to do convolution with the data. Is this the correct method? – Srishti M Jul 31 '17 at 1:28
• Yes I think so ... – AlexTP Jul 31 '17 at 22:05