# LMS and delta impulse response- Equalization concepts

I have been reading about equalization and channel estimation in order to apply to my application of estimating trajectory path for a two-wheeled robot. The application is quite similar to equalization. I am studying the Least Mean Square algorithm (LMS) and have some basic doubts regarding how to determine if equalization is proper or not i.e., if there is a measure to quantify how much of the signal has been estimated properly. I am following this MATLAB implementation.

• Doubt 1 :

The cost function for LMS is constructed using the input, so we are minimizing the error function. So, based on my understanding the input is known to the receiver. Hence, the LMS equalizer is not blind. Please correct me if wrong.

• Doubt2:

On running the code (using the same parameter settings in the code), I cannot understand what is the significance of the fourth tap having a magnitude one. Is this the delta impulse in the third subplot. Does having a delta function mean that equalization is correct? I had read that for equalization to happen, the convolution of the estimated channel coefficients (here these are the weights of the LMS) and the known channel impulse should be a delta function. Does this graph convey the same? Could somebody please explain this concept of delta impulse and the graph?

The delta function is also correct. Suppose that the channel impulse response is $h(t)$ and frequency response $H(f)$. Then the equalizer should ideally be the inverse of this function in frequency domain (remember the convolution in time domain and hence multiplication in frequency domain). $$H(f)\cdot \frac{1}{H(f)} = 1$$ and its inverse Fourier transform is $$F^{-1}(1) = \delta (t)$$
• Thank you for your reply. I still have doubts regarding the impulse response concept. Could you please clarify these points? (1)In the code, the input is filtered through a channel modeled as FIR :filter(c,1,u) where $c$ is the channel coefficients, $u$ is the input. The order is M=5. So, does this mean that there are 4 delays, hence the fourth tap =1? (2) In the first sub-plot, why is the third point higher? (3) What is the difference between the second sub-plot and third sub-plot? and lastly – Ria George Jun 11 '17 at 21:52
• (1) No. The 4th tap is 1 because the code is written in a way that the Rx time base is 4 taps behind this delta function. Remember there are 2 convolution, first of data with channel and second, of this Rx input with equalizer. So don't worry about the tap being 4 or not. Just remember that convolution of two signals with lengths $L_1$ and $L_2$ is $L_1+L_2-1$. – QMC Jun 12 '17 at 4:41