Forums

ISI

Started by aamer February 28, 2007
Dear friends,

As we all know inter symbol interference(ISI) is , by which the
transmitted symbol smears or runs into the adjacent symbol. But, in one of
the IEEE paper I have seen the mathemetical definition as

ISI= sum(abs(h).^2)- max(abs(h).^2) / max(abs(h).^2)

where h= conv(c,w)

c = impulse response of channel
w = equalizer taps.

what I know is......to negotiate the effects of channel, the response of
equalizer in time domain should be inverse of channel response. In other
words, convolution of channel response and the equalizer response should
be delta like in time domain.

so, h given above is a vector which has one of its values as 1. But how
does it related to the intensity of symbol being smeared into the adjacent
one.

can some one please help me in explaining the definition of ISI through
the above equation.

kind regards
On Wed, 28 Feb 2007 07:52:01 -0000, aamer <raqeebhyd@yahoo.com> wrote:
> > Dear friends, > > As we all know inter symbol interference(ISI) is , by which the > transmitted symbol smears or runs into the adjacent symbol. But, in on=
e =
> of > the IEEE paper I have seen the mathemetical definition as > > ISI=3D sum(abs(h).^2)- max(abs(h).^2) / max(abs(h).^2)
I'm assuming that there should be some extra brackets, i.e.: ISI=3D (sum(abs(h).^2)- max(abs(h).^2)) / max(abs(h).^2)
> > where h=3D conv(c,w) > > c =3D impulse response of channel > w =3D equalizer taps. > > what I know is......to negotiate the effects of channel, the response =
of
> equalizer in time domain should be inverse of channel response. In oth=
er
> words, convolution of channel response and the equalizer response shou=
ld
> be delta like in time domain. >
If the equaliser is perfect, then yes, h will be an impulse. But then = you'd have zero ISI, so the equation above would be correct.
> so, h given above is a vector which has one of its values as 1. But ho=
w
> does it related to the intensity of symbol being smeared into the =
> adjacent > one. > > can some one please help me in explaining the definition of ISI throug=
h
> the above equation.
Imagine transmitting a single symbol through the channel and equaliser, = = with amplitude =3D 1, at time n=3D0. The resulting sequence will be h[n= ]. = Therefore, the total amount of energy available after the equaliser will= = be sum(abs(h).^2). If we assume for simplicity that the maximum value o= f = h[n] occurs at n=3D0, then the amount of energy that lands in the correc= t = position will be max(abs(h).^2). Therefore, the total amount of energy = = that is in the wrong position (i.e. ISI) will be the remainder, i.e. = (sum(abs(h).^2) - max(abs(h).^2)). For comparison purposes, this has be= en = normalised by dividing by the amount of non-ISI energy. -- = Oli
>On Wed, 28 Feb 2007 07:52:01 -0000, aamer <raqeebhyd@yahoo.com> wrote: >> >> Dear friends, >> >> As we all know inter symbol interference(ISI) is , by which the >> transmitted symbol smears or runs into the adjacent symbol. But, in
on=
>e = > >> of >> the IEEE paper I have seen the mathemetical definition as >> >> ISI=3D sum(abs(h).^2)- max(abs(h).^2) / max(abs(h).^2) > >I'm assuming that there should be some extra brackets, i.e.: > >ISI=3D (sum(abs(h).^2)- max(abs(h).^2)) / max(abs(h).^2) > >> >> where h=3D conv(c,w) >> >> c =3D impulse response of channel >> w =3D equalizer taps. >> >> what I know is......to negotiate the effects of channel, the response
=
>of >> equalizer in time domain should be inverse of channel response. In
oth=
>er >> words, convolution of channel response and the equalizer response
shou=
>ld >> be delta like in time domain. >> > >If the equaliser is perfect, then yes, h will be an impulse. But then
=
> >you'd have zero ISI, so the equation above would be correct. > > >> so, h given above is a vector which has one of its values as 1. But
ho=
>w >> does it related to the intensity of symbol being smeared into the = > >> adjacent >> one. >> >> can some one please help me in explaining the definition of ISI
throug=
>h >> the above equation. > >Imagine transmitting a single symbol through the channel and equaliser,
=
> = > >with amplitude =3D 1, at time n=3D0. The resulting sequence will be
h[n=
>]. = > >Therefore, the total amount of energy available after the equaliser
will=
> = > >be sum(abs(h).^2). If we assume for simplicity that the maximum value
o=
>f = > >h[n] occurs at n=3D0, then the amount of energy that lands in the
correc=
>t = > >position will be max(abs(h).^2). Therefore, the total amount of energy
=
> = > >that is in the wrong position (i.e. ISI) will be the remainder, i.e. = > >(sum(abs(h).^2) - max(abs(h).^2)). For comparison purposes, this has
be=
>en = > >normalised by dividing by the amount of non-ISI energy. > > > >-- = > >Oli >
Dear Oli, Yes, you are right. there must be extra bracket to seperate numerator and the denominator. I got the point. Thank you for your effort. But, I have one more question. As you said, when a symbol is transmitted through channel and equalizer the resulting sequence is h(n) which is mathematically given as (as I mentioned in my previous post) h(n)= conv(c,w)...........................(1) where c = channel impulse response and w = equalizer taps but I thought the resulting sequence is calculted when the input symbol is multiplied with the filter taps, given as y(n)= x*w ................................(2) where x = trasmitted symbol and w = equalizer taps. but i think the the equations (1) and (2) are not the same. They give different outputs. I think received sequence is y(n) and h(n) just checks whether the combined responses of channel and equalizer are delta like. Is it right??? or am I missing some point??????? Thanks in advance
On Feb 28, 6:01 am, "aamer" <raqeeb...@yahoo.com> wrote:
> >On Wed, 28 Feb 2007 07:52:01 -0000, aamer <raqeeb...@yahoo.com> wrote: > > >> Dear friends, > > >> As we all know inter symbol interference(ISI) is , by which the > >> transmitted symbol smears or runs into the adjacent symbol. But, in > on= > >e = > > >> of > >> the IEEE paper I have seen the mathemetical definition as > > >> ISI=3D sum(abs(h).^2)- max(abs(h).^2) / max(abs(h).^2) > > >I'm assuming that there should be some extra brackets, i.e.: > > >ISI=3D (sum(abs(h).^2)- max(abs(h).^2)) / max(abs(h).^2) > > >> where h=3D conv(c,w) > > >> c =3D impulse response of channel > >> w =3D equalizer taps. > > >> what I know is......to negotiate the effects of channel, the response > = > >of > >> equalizer in time domain should be inverse of channel response. In > oth= > >er > >> words, convolution of channel response and the equalizer response > shou= > >ld > >> be delta like in time domain. > > >If the equaliser is perfect, then yes, h will be an impulse. But then > = > > >you'd have zero ISI, so the equation above would be correct. > > >> so, h given above is a vector which has one of its values as 1. But > ho= > >w > >> does it related to the intensity of symbol being smeared into the = > > >> adjacent > >> one. > > >> can some one please help me in explaining the definition of ISI > throug= > >h > >> the above equation. > > >Imagine transmitting a single symbol through the channel and equaliser, > = > > = > > >with amplitude =3D 1, at time n=3D0. The resulting sequence will be > h[n= > >]. = > > >Therefore, the total amount of energy available after the equaliser > will= > > = > > >be sum(abs(h).^2). If we assume for simplicity that the maximum value > o= > >f = > > >h[n] occurs at n=3D0, then the amount of energy that lands in the > correc= > >t = > > >position will be max(abs(h).^2). Therefore, the total amount of energy > = > > = > > >that is in the wrong position (i.e. ISI) will be the remainder, i.e. = > > >(sum(abs(h).^2) - max(abs(h).^2)). For comparison purposes, this has > be= > >en = > > >normalised by dividing by the amount of non-ISI energy. > > >-- = > > >Oli > > Dear Oli, > > Yes, you are right. there must be extra bracket to seperate numerator and > the denominator. > > I got the point. Thank you for your effort. But, I have one more > question. > As you said, when a symbol is transmitted through channel and equalizer > the resulting sequence is h(n) which is mathematically given as (as I > mentioned in my previous post) > > h(n)= conv(c,w)...........................(1) > > where c = channel impulse response > and w = equalizer taps > > but I thought the resulting sequence is calculted when the input symbol is > multiplied with the filter taps, given as > > y(n)= x*w ................................(2) > > where x = trasmitted symbol > and w = equalizer taps. > > but i think the the equations (1) and (2) are not the same. They give > different outputs. > > I think received sequence is y(n) and h(n) just checks whether the > combined responses of channel and equalizer are delta like. Is it right??? > or am I missing some point??????? > > Thanks in advance
The system is structured like this: +----------+ +-----------+ | | | | x[n] =====>| Channel |=========>| Equalizer |=======> y[n] | c[n] | | w[n] | | | | | +----------+ +-----------+ The equalizer output is equal to the transmitted symbol x[n] convolved with the cascade of the channel and equalizer transfer functions. So, the h[n] you talked about before is equal to the cascaded transfer function, or c[n] * w[n] (* = convolution). So, you can say that: y[n] = x[n] * c[n] * w[n] or y[n] = x[n] * h[n], where h[n] = c[n] * w[n] Your receiver takes the channel's output (equal to x[n] * c[n]) and applies it to your (probably FIR) equalizer to get the best estimate you can of the input symbol x[n]. Ideally, h[n] would be an impulse (and y[n] would equal x[n], with some delay), but that's not likely unless you know the exact channel transfer function and can build a reasonable inverse filter. Jason
On Feb 28, 1:52 am, "aamer" <raqeeb...@yahoo.com> wrote:

> what I know is......to negotiate the effects of channel, the response of > equalizer in time domain should be inverse of channel response. In other > words, convolution of channel response and the equalizer response should > be delta like in time domain. >
This is not the optimal receiver. The optimal receiver is one that considers the effect of ISI as a convolutional code, and does sequence estimation at the receiver, such as using the Viterbi algorithm. What you are doing is a zero-forcing equalizer. It makes detection of the symbol easier, but in general it is sub-optimal. Think about it: in the frequency domain it amplifies the part of your signal that is attenuated, in effect amplifying the noise in that part of the frequency. The original paper is this one: G.D. Forney, Jr., "Maximum-likelihood sequence estimation of digital sequences in the presence of intersymbol interference," IEEE Trans. Inform. Theory, vol. 18, pp. 363--378, May 1972. The above is valid for when the noise is Gaussian. When it is not Gaussian, then a modification of the metrics in the Viterbi algorithm is required, and this has also been studied. Good communication theory textbooks (Proakis, Biglieri, etc) also cover this subject. Julius
>On Feb 28, 1:52 am, "aamer" <raqeeb...@yahoo.com> wrote: > >> what I know is......to negotiate the effects of channel, the response
of
>> equalizer in time domain should be inverse of channel response. In
other
>> words, convolution of channel response and the equalizer response
should
>> be delta like in time domain. >> > >This is not the optimal receiver. The optimal receiver is >one that considers the effect of ISI as a convolutional code, >and does sequence estimation at the receiver, such as using >the Viterbi algorithm. > >What you are doing is a zero-forcing equalizer. It makes >detection of the symbol easier, but in general it is >sub-optimal. Think about it: in the frequency domain >it amplifies the part of your signal that is attenuated, >in effect amplifying the noise in that part of the frequency. > >The original paper is this one: > >G.D. Forney, Jr., "Maximum-likelihood sequence estimation of digital >sequences in the presence of intersymbol interference," IEEE Trans. >Inform. Theory, vol. 18, pp. 363--378, May 1972. > >The above is valid for when the noise is Gaussian. When it >is not Gaussian, then a modification of the metrics in the >Viterbi algorithm is required, and this has also been studied. > >Good communication theory textbooks (Proakis, Biglieri, etc) >also cover this subject. > >Julius > >
Thanks alot for the efforts. I have understood, what I was looking for!!!
On 28 Feb 2007 06:15:27 -0800, "julius" <juliusk@gmail.com> wrote:
>G.D. Forney, Jr., "Maximum-likelihood sequence estimation of digital >sequences in the presence of intersymbol interference," IEEE Trans. >Inform. Theory, vol. 18, pp. 363--378, May 1972. > >The above is valid for when the noise is Gaussian. When it >is not Gaussian, then a modification of the metrics in the >Viterbi algorithm is required, and this has also been studied.
Hi Julius, Do you have any references for non-gaussian non-white noise case? Thanks.
On Mar 2, 10:14 am, mk <kal*@dspia.*comdelete> wrote:
> Hi Julius, > Do you have any references for non-gaussian non-white noise case? > > Thanks.
My knowledge in this is limited only to white, non-Gaussian noise. In the simplest modification, the metric cost function is changed from Euclidean distance to something else. If you google for: intersymbol interference non-gaussian You'll get a few papers. Hope that helps, Julius
On Mar 4, 8:32 pm, "julius" <juli...@gmail.com> wrote:
> On Mar 2, 10:14 am, mk <kal*@dspia.*comdelete> wrote: > > > Hi Julius, > > Do you have any references for non-gaussian non-white noise case? > > > Thanks. > > My knowledge in this is limited only to white, non-Gaussian noise. > In the simplest modification, the metric cost function is changed > from Euclidean distance to something else. > > If you google for: > > intersymbol interference non-gaussian > > You'll get a few papers. > > Hope that helps, > Julius
This has got to be tough to implement in practice. In addition to the problem of estimating the channel, the modified VA uses a parametric noise model whose parameters must also be estimated. Hopefully a training sequence is available for the channel estimates. As for the noise, I suppose one could observe a bunch of samples and try to fit something to it. I think fading, impulsive channels such as HF radio would be a popular application for this. John
On Mar 5, 4:43 am, sampson...@gmail.com wrote:
> This has got to be tough to implement in practice. In addition to the > problem of estimating the channel, the modified VA uses a parametric > noise model whose parameters must also be estimated. Hopefully a > training sequence is available for the channel estimates. As for the > noise, I suppose one could observe a bunch of samples and try to fit > something to it. > > I think fading, impulsive channels such as HF radio would be a popular > application for this. > > John
I agree. If knowledge of the noise statistics in not known, then it must be estimated. Even for the case of AWGN, it's not that easy to do. Remember that even in the simplest case of estimating the mean of a sequence of white Gaussian samples we already have to look at the Student-t distribution. In general if the correct metric is not Euclidean, it is very possible that computing the metric is a harder problem than the sequence estimation itself! Julius