Zero-Forcing Equalizer

[Ahmed Alaa]

Hi all,

I want to know whether it is possible that a ZF equalizer enhances SNR instead of deteriorating it ? When can this happen ?

Thanks.

 

[zorro]

Hi Ahmed Alaa,

The equalizer enhances the ratio signal/(noise+ISI), with a criterion that is different for the ZF than for the LMS.
Adding other signal samples to the current sample with appropriate weights in order to reduce the ISI, it adds noise too. SNR ratio impairs.
(Maybe for some type of colored noise it could be different.)
But what is of interest and is really observed is the signal/(noise+ISI) ratio, that is improved by the equalizer.
Regards

Z

 

[Ahmed Alaa]

do you mean that SNR can accidentally increase or decrease after equalization ? i.e. Noise enhancement/ supression can take place for AWGN ?

 

[zorro]

Hi Ahmed,

signal-to-(noise+ISI) ratio increases after equalization.
Normally, signal-to-noise decreases after equalization. The equalizer reduces (ideally removes) ISI, but as a side effect it increases the noise.

Nevertheless, I can imagine a situation in which the noise too is decreased after equalization:
The equalizer is a filter that modifies the signal i such a way that it is near to satisfy the Nyquist criterion removing ISI. Doing that, some frequencies are attenuated or amplified more that others, following the spectrum of the sampled signal. For a particular noise spectrum before equalization, it is possible that this filter attenuates the total noise more than the useful signal. In that way SNR is increased.
But this example is not the normal case. For white noise at the input, SNR is decreased.

Regards


Z

 

[Ahmed Alaa]

I know that if the transversal filters coefficents used for equalization are: alpha i, then the SNR decreases by a factor = summation(alpha i ^ 2), what if this summation is less than 1 ? i.e. what if the filter coefficients used to equalize are too small ?

Logically, the frequency domain interpretation tells that SNR enhancement for white noise is not possible, but that is what I'm obtaining !!

 

[FvM]

In my opinion, the discussion lacks of a clear communication channel definition. If we assume a frequency selective channel with defined signal power, added white noise at the receiver input and succeeding equalizer, SNR will be decreased for all possible equalization filters.

 

[Ahmed Alaa]

FvM, yes you got the point, a flat fading channel would increase the SNR if the channel gain is greater than one, I was stricting my mind to frequency selective fading channels.

 

[zorro]

I know that if the transversal filters coefficents used for equalization are: alpha i, then the SNR decreases by a factor = summation(alpha i ^ 2), what if this summation is less than 1 ? i.e. what if the filter coefficients used to equalize are too small ?

This is true if the noise samples are independent. This happens is input noise is white (as normally is assumed).
If the filter coefficients used to equalize are too small, it is clear that the signal is attenuated too. Then it is necessary to take into account not only the attenuation or enhancement of noise, but that oif the signal too. So, SNR is always less at the equalizer output.

Let's take an approximation. If I'm not wrong, the amplitude of the signal sample at the equalizer output is increased by approximately [summation(alpha_i)] if ISI is not strong. So the power gain for the signal is [summation(alpha i)]^2 in that case.
Then, for AWGN at the input, approx.:

SNR(out) ~ SNR(in) * [summation(alpha_i)]^2 / summation(alpha_i ^ 2)

The factor by which SNR(in) is multiplied is always less than 1.
But, let me say again, the important point is the reduction of ISI.
Regards

Z

P.D.: please see post #14 for correction

 

[Ahmed Alaa]

Thanks zorro, but let's think in the issue in the frequency domain:

assume the signal is X and the channel is H, the received signal in AWGN is then:

Y = H*X + N,

the ZF equalizer divides by the transfer function H, thus Y = X + N/H, noise enhancement occurs for frequency selective channels because H has nulls that boost the noise power significantly, but if H is just a constant for a flat channel, then if H > 1, SNR increases and if H<1 the SNR decreases, This is the same idea of preamplifiers.

 

[zorro]

Hi again, Ahmed

Let's consider that H is just a constant for a flat channel.
Then SNR is the same at the input of the equalizer and the output, because both signal and noise are multiplied (or divided) by the same constant.
From your equations:

Input: signal is H*X and noise is N
Output: signal is X and noise is N/H

Or perhaps I'm misunderstanding you? :???:
Regards

Z

 

[Ahmed Alaa]

Yes you're right because noise is added after applying the channel, the input SNR is ~ H*X/ N and not X/N. I'll check my code again

 

[FvM]

SNR is the same at the input of the equalizer and the output, because both signal and noise are multiplied (or divided) by the same constant.

Depends on your definition of SNR. Signal and noise are spectral distributions, but SNR may be defined as scalar quantity, ratio of signal power total power to noise total power.

 

[Ahmed Alaa]

Depends on your definition of SNR. Signal and noise are spectral distributions, but SNR may be defined as scalar quantity, ratio of signal power total power to noise total power.

He defines SNR as the signal * transfer function power over the noise power, which is logical. I am comparing the BER of conventional BPSK in AWGN only (SNR = X/N) with BER of BPSK after applying a channel (SNR at input of equalizer = H*X/N and at the output is X/(N/H)), this means that my SNR will be enhanced -compared to the X/N in AWGN BPSK- if the power in (N/H) is less than the power in X ?? This is really confusing me

---------- Post added at 09:15 ---------- Previous post was at 09:13 ----------

Btw, I am calculating summation(alpha i) -> (the sum of filter coefficients squared which represents the amount of SNR loss) and for some channels it is +ve (in dB) which means that I'll loose SNR, and for other channels it is -ve (which means I'll gain SNR), and what I calculate is what I get on the curves whether loss or gain ! This is an observation that I can't interpret.

 

[zorro]

In post #8:

If I'm not wrong, the amplitude of the signal sample at the equalizer output is increased by approximately [summation(alpha_i)] if ISI is not strong.

I was wrong. The amplitude of the signal sample involves the samples of the response of the channel too: it is summation(alpha_i*h_-i). (h_i are the samples of the impulse response of the channel). Sorry.

Btw, I am calculating summation(alpha i) -> (the sum of filter coefficients squared which represents the amount of SNR loss) and for some channels it is +ve (in dB) which means that I'll loose SNR, and for other channels it is -ve (which means I'll gain SNR)

It is right that noise power increases as summation(alpha_i^2). But signal changes too:
Before equalization signal power is summation(h_i^2), but at the output of the equalizer it is summation(alpha_i*h_-i) if all the ISI is removed.

Regards

Z

 

[zorro]

Hi,

Let me add something.
If the distortion of the channel is in phase only, then an infinite length equalizer can ideally remove all the ISI (invert the channel characteristic) without noise impairment.
If the channel has distortion in its amplitude characteristic, then there will be an impairment in SNR (but a net gain because of ISI removal).
N.B.:"Channel" includes sampling, i.e. folded frequency characteristic.
Regards

Z


By the way, today is my 10th anniversary in Edaboard (formerly Elektroda)!!! :-D

 

[Ahmed Alaa]

Congratulations for your 10th anniversary

---------- Post added at 09:25 ---------- Previous post was at 09:24 ----------

So to conclude: SNR can't be enhanced by an equalizer