# BER dependence on SINR

Started by September 15, 2004
```Hello,

In case of

r(t) = s(t) + i(t) + n(t)

s(t) - desired signal
i(t) - interference signal

then does the BER depend on Signal-to-{Interference + Noise} Ratio
(SINR) directly

For e.g. if s(t) is Binary PSK, can I use the Q-function directly by
replacing the SNR by SINR to estimate the BER.

Cheers,
Vimal
```
```I think (but I am just a beginner) the following:

if i(t) is gaussian white noise and uncorrelated with n(t), you can use the
SINR directly

> Hello,
>
> In case of
>
> r(t) = s(t) + i(t) + n(t)
>
> where r(t) - received signal
>       s(t) - desired signal
>       i(t) - interference signal
>       n(t) - additive Gaussian noise
>
> then does the BER depend on Signal-to-{Interference + Noise} Ratio
> (SINR) directly
>
> For e.g. if s(t) is Binary PSK, can I use the Q-function directly by
> replacing the SNR by SINR to estimate the BER.
>
> Cheers,
> Vimal

```
```Hint: Consider different types of interference.

Cheers,
Jake

vimal_bhatia2@yahoo.com (Vimal) wrote in message news:<b6fc6dda.0409151050.311b4aea@posting.google.com>...
> Hello,
>
> In case of
>
> r(t) = s(t) + i(t) + n(t)
>
> where r(t) - received signal
>       s(t) - desired signal
>       i(t) - interference signal
>       n(t) - additive Gaussian noise
>
> then does the BER depend on Signal-to-{Interference + Noise} Ratio
> (SINR) directly
>
> For e.g. if s(t) is Binary PSK, can I use the Q-function directly by
> replacing the SNR by SINR to estimate the BER.
>
> Cheers,
> Vimal
```
```> Hint: Consider different types of interference.
>
>    Cheers,
>    Jake
>

If the interference, is lets say binary antipodal +/- 1 ... then can we use SINR...
```
```If the interference is another BPSK modulated signal, and if the
desired user is employed with only a matched filter, then we cannot
use SINR. This is as follows:
After matched filtering, we have
r = A1*X1 + A2*X2 + n,
where A1 is the desired user's signal amplitude, X1 is the desired
user's BPSK signal, A2 is the interference amplitude, X2 is the
corresponding BPSK signal. It is now straightforward that, by first
conditioned on X2, one can find the average BER of the desired user,
which then has to be averaged over the statistics of X2.

For completeness, assuming the standard deviation of n as sigma and X1
and X2 are equally likely BPSK signals, the following is the final
expression for the BER:

BER = 0.5*Q((A1+A2)/sigma) + 0.5*Q((A1-A2)/sigma).

hope that helps
Ramesh

vimal_bhatia2@yahoo.com (Vimal) wrote in message news:<b6fc6dda.0409160248.6d1f88f3@posting.google.com>...
> > Hint: Consider different types of interference.
> >
> >    Cheers,
> >    Jake
> >
>
> If the interference, is lets say binary antipodal +/- 1 ... then can we use SINR...
```
```> For completeness, assuming the standard deviation of n as sigma and X1
> and X2 are equally likely BPSK signals, the following is the final
> expression for the BER:
>
> BER = 0.5*Q((A1+A2)/sigma) + 0.5*Q((A1-A2)/sigma).
>
> hope that helps
> Ramesh

For more complex example, when QPSK modulation is assumed for both desired
and interfered signals,
can we still use a similar equation together with the following additional
definitions?

> BER = 0.5*( 0.5*Q((A1+A2re)/sigma) + 0.5*Q((A1-A2re)/sigma)) +
0.5*( 0.5*Q((A1+A2im)/sigma) + 0.5*Q((A1-A2im)/sigma)) +

where A1 = w*h1, A2re = Re{w*h2}, A2im = Im{w*h2} while w =
conj{h1}/abs{h1},
in which h1 and h2 denote complex channel response and
Re{}, Im{}, conj{.}, and abs{.} stand for real, image, conjugate and
absolute, respectively.

--
Best regards,
James K. (txdiversity@hotmail.com)
- Any remarks, proposal and/or indicator to text would be greatly respected.
- Private opinions: These are not the opinions from my affiliation.
[Home] http://home.naver.com/txdiversity

```
```Hi Vimal : you can always use the SINR directly to estimate BER but you may
not get a very good estimate. If the interference is gaussian at the output
of your matched filter then you'll get a good estimate , else you may well
Best of luck - Mike

"Ramesh" <ecerams@rediffmail.com> wrote in message
> If the interference is another BPSK modulated signal, and if the
> desired user is employed with only a matched filter, then we cannot
> use SINR. This is as follows:
> After matched filtering, we have
> r = A1*X1 + A2*X2 + n,
> where A1 is the desired user's signal amplitude, X1 is the desired
> user's BPSK signal, A2 is the interference amplitude, X2 is the
> corresponding BPSK signal. It is now straightforward that, by first
> conditioned on X2, one can find the average BER of the desired user,
> which then has to be averaged over the statistics of X2.
>
> For completeness, assuming the standard deviation of n as sigma and X1
> and X2 are equally likely BPSK signals, the following is the final
> expression for the BER:
>
> BER = 0.5*Q((A1+A2)/sigma) + 0.5*Q((A1-A2)/sigma).
>
> hope that helps
> Ramesh
>
>
> vimal_bhatia2@yahoo.com (Vimal) wrote in message
> > > Hint: Consider different types of interference.
> > >
> > >    Cheers,
> > >    Jake
> > >
> >
> > If the interference, is lets say binary antipodal +/- 1 ... then can we
use SINR...

```
```On Mon, 20 Sep 2004 09:12:53 +0000 (UTC), "Mike Yarwood"
<mpyarwood@btopenworld.com> wrote:

>Hi Vimal : you can always use the SINR directly to estimate BER but you may
>not get a very good estimate. If the interference is gaussian at the output
>of your matched filter then you'll get a good estimate , else you may well
>Best of luck - Mike

Mike, do you mean good performance by noting good estimation?

To be more clear my question, I'd like show one example related to my
question. If considered SNR is high,

BER ~= 0.5*Q((A1-A2)/sigma) --> 0.5*Q(0) = 0.25

for good estimation where A1 = A2 is assumed, but for simple Gaussian
approximation case which is obviously not good estimation, it is

BER ~= Q(A1/(A2+sigma)) -> Q(1) = 0.16

In this case, the above result reveals that the good estimation
produces worse BER.

Thus, as long as my analysis is correct, the good estimation dose not
(directly) mean for the good performance, right?

BR,
------
James K. (txdiversity@hotmail.com)
[Home] http://home.naver.com/txdiversity
```
```James K. <txdiversity@hotmail.com> wrote in message news:<3ovjl0tu6ibhlaqib5mun7sg759m3sl13l@4ax.com>...
> On Mon, 20 Sep 2004 09:12:53 +0000 (UTC), "Mike Yarwood"
> <mpyarwood@btopenworld.com> wrote:
>
> >Hi Vimal : you can always use the SINR directly to estimate BER but you may
> >not get a very good estimate. If the interference is gaussian at the output
> >of your matched filter then you'll get a good estimate , else you may well
> >Best of luck - Mike
>
> Mike, do you mean good performance by noting good estimation?
>
> To be more clear my question, I'd like show one example related to my
> question. If considered SNR is high,
>
> BER ~= 0.5*Q((A1-A2)/sigma) --> 0.5*Q(0) = 0.25
>
> for good estimation where A1 = A2 is assumed, but for simple Gaussian
> approximation case which is obviously not good estimation, it is
>
> BER ~= Q(A1/(A2+sigma)) -> Q(1) = 0.16
>
> In this case, the above result reveals that the good estimation
> produces worse BER.
>
> Thus, as long as my analysis is correct, the good estimation dose not
> (directly) mean for the good performance, right?
>
> BR,
> ------
> James K. (txdiversity@hotmail.com)
> [Home] http://home.naver.com/txdiversity

Hi James,

> Mike, do you mean good performance by noting good estimation?

I don't think so. I was just talking about the quality of the
estimation, if it predicts low BER when it should predict high or high
when it should be low then its not as good as an estimator which gives
a better prediction.

> Thus, as long as my analysis is correct, the good estimation dose not
> (directly) mean for the good performance, right?

If your analysis is correct then the better method of estimation has
produced the most accurate estimate of the BER and that estimate of
the BER is higher than the estimate produced by the worse estimation
method , right.

Best of luck -Mike
```