# problem math bessel function

Started by December 3, 2004
```Hmmm. Google News Beta barfed.

Ciao,

Peter K.

```
```Peter,

Thanks again.  When you write it out like that, it seems so straight
forward that i'm embarrassed to say that i've spent a couple of weeks
trying to understand what they had done!

One thing that still puzzles me though, is when you get to this stage

= exp( 2A cos(th) int Re{ r*(t) s(t) } dt )

and we assume a uniform phase {th}, does that mean that we really don't
care what the phase is (or in other words, we average over all possible
values of the phase), so therefore, we use the absolute value of abs (
int r*(t) s(t) ) instead of the Re{}?   This effectively removes the
dependence on the phase.

This then gives us eqn3

= Io( 2A abs( int r*(t) s(t) ) )

Is my line of thinking correct?  Or I should say, is my reasoning
correct?

Cheers

CW

```
```] One thing that still puzzles me though, is when you get to
] this stage
]
] = exp( 2A cos(th) int Re{ r*(t) s(t) } dt )
]
] and we assume a uniform phase {th}, does that mean that we
] really don't care what the phase is (or in other words, we
] average over all possible values of the phase),

Yes, that's what appears to be happening, though I haven't
seen the maths.

] so therefore, we use the absolute value of abs (
] int r*(t) s(t) ) instead of the Re{}?   This effectively
] removes the dependence on the phase.

??  You appear to be confusing the phase, th, with the phase
(argument, angle) of the expression   r*(t) s(t) .

These are two completely different things, as far as I'm
aware. ??

abs(x(t)) != Re{x(t)}
unless x(t) is positive and real-valued.

Ciao,

Peter K.

```
```] One thing that still puzzles me though, is when you get to
] this stage
]
] = exp( 2A cos(th) int Re{ r*(t) s(t) } dt )
]
] and we assume a uniform phase {th}, does that mean that we
] really don't care what the phase is (or in other words, we
] average over all possible values of the phase),

Yes, that's what appears to be happening, though I haven't
seen the maths.

] so therefore, we use the absolute value of abs (
] int r*(t) s(t) ) instead of the Re{}?   This effectively
] removes the dependence on the phase.

??  You appear to be confusing the phase, th, with the phase
(argument, angle) of the expression   r*(t) s(t) .

These are two completely different things, as far as I'm
aware. ??

abs(x(t)) != Re{x(t)}
unless x(t) is positive and real-valued.

Ciao,

Peter K.

```
```Peter,

Sorry to keep on labouring the point, but I want to make sure that i
have a solid understanding of the assumptions that are being made.
Perhaps if i elaborate a bit more on what it is that is happening, you
may be able to see where i'm heading.

The detector is phase noncoherent, and therefore, i'm trying to
determine what assumptions are to be made when the carrier phase (th)
is unknown at the receiver.  Perhaps i am indeed confusing the phase
{th} with the argument of the expression r*(t) s(t), but to me they are
the same thing, are they not?  {th} in my case, represents the unknown
channel phase shift, which i'm assuming remains constant over the
duration of my integral (int r*(t) exp(-j*th)* s(t) dt).  In other
words, I'm integrating over the rx symbol duration shifted by the
channel induced phase.  If the phase shift of the channel is zero, and
we have an impulse channel with relatively high SNR, then r(t) ~= s(t)
(i.e. what we tx is approx same as what we rx) the phase will therefore
cancel when i perform the matched filter receiver of ( int r*(t) s(t)
dt).

For the case of a nonzero channel phase shift, we simply perform the
same integral as above (int r*(t) exp(-j*th) s(t) dt) but we take the
exp{} term outside the integral thus exp(-j*th) int r*(t) s(t) dt.
Does this not mean that the channel phase {th} and the argument of the
expression int r*(t) s(t) dt are one and the same?

This seems to be turning into a long winded post, so i'll try and wrap
up by asking, given all of the above and assuming that the channel
phase {th} is uniformly distributed over 0-2*pi, what simplification
can we make to the expression

= ... exp( 2A cos(th) Re(int r*(t) s(t) dt) )
?

Thanks for your patience

CW

```
```Peter,

Sorry to keep on labouring the point, but I want to make sure that i
have a solid understanding of the assumptions that are being made.
Perhaps if i elaborate a bit more on what it is that is happening, you
may be able to see where i'm heading.

The detector is phase noncoherent, and therefore, i'm trying to
determine what assumptions are to be made when the carrier phase (th)
is unknown at the receiver.  Perhaps i am indeed confusing the phase
{th} with the argument of the expression r*(t) s(t), but to me they are
the same thing, are they not?  {th} in my case, represents the unknown
channel phase shift, which i'm assuming remains constant over the
duration of my integral (int r*(t) exp(-j*th)* s(t) dt).  In other
words, I'm integrating over the rx symbol duration shifted by the
channel induced phase.  If the phase shift of the channel is zero, and
we have an impulse channel with relatively high SNR, then r(t) ~= s(t)
(i.e. what we tx is approx same as what we rx) the phase will therefore
cancel when i perform the matched filter receiver of ( int r*(t) s(t)
dt).

For the case of a nonzero channel phase shift, we simply perform the
same integral as above (int r*(t) exp(-j*th) s(t) dt) but we take the
exp{} term outside the integral thus exp(-j*th) int r*(t) s(t) dt.
Does this not mean that the channel phase {th} and the argument of the
expression int r*(t) s(t) dt are one and the same?

This seems to be turning into a long winded post, so i'll try and wrap
up by asking, given all of the above and assuming that the channel
phase {th} is uniformly distributed over 0-2*pi, what simplification
can we make to the expression

= ... exp( 2A cos(th) Re(int r*(t) s(t) dt) )
?

Thanks for your patience

CW

```
```Peter,

Sorry to keep on labouring the point, but I want to make sure that i
have a solid understanding of the assumptions that are being made.
Perhaps if i elaborate a bit more on what it is that is happening, you
may be able to see where i'm heading.

The detector is phase noncoherent, and therefore, i'm trying to
determine what assumptions are to be made when the carrier phase (th)
is unknown at the receiver.  Perhaps i am indeed confusing the phase
{th} with the argument of the expression r*(t) s(t), but to me they are
the same thing, are they not?  {th} in my case, represents the unknown
channel phase shift, which i'm assuming remains constant over the
duration of my integral (int r*(t) exp(-j*th)* s(t) dt).  In other
words, I'm integrating over the rx symbol duration shifted by the
channel induced phase.  If the phase shift of the channel is zero, and
we have an impulse channel with relatively high SNR, then r(t) ~= s(t)
(i.e. what we tx is approx same as what we rx) the phase will therefore
cancel when i perform the matched filter receiver of ( int r*(t) s(t)
dt).

For the case of a nonzero channel phase shift, we simply perform the
same integral as above (int r*(t) exp(-j*th) s(t) dt) but we take the
exp{} term outside the integral thus exp(-j*th) int r*(t) s(t) dt.
Does this not mean that the channel phase {th} and the argument of the
expression int r*(t) s(t) dt are one and the same?

This seems to be turning into a long winded post, so i'll try and wrap
up by asking, given all of the above and assuming that the channel
phase {th} is uniformly distributed over 0-2*pi, what simplification
can we make to the expression

= ... exp( 2A cos(th) Re(int r*(t) s(t) dt) )
?

Thanks for your patience

CW

```
```Sorry for the multiple posts.  Google is playing tricks on me.

```
```Hi,

] Sorry to keep on labouring the point, but I want to make sure that i
] have a solid understanding of the assumptions that are being made.
] Perhaps if i elaborate a bit more on what it is that is happening,
] you may be able to see where i'm heading.

No problem.

] The detector is phase noncoherent, and therefore, i'm trying to
] determine what assumptions are to be made when the carrier phase (th)
] is unknown at the receiver. Perhaps i am indeed confusing the phase
] {th} with the argument of the expression r*(t) s(t), but to me they
] are the same thing, are they not?

I can't see that from what you've written.

So we have;

s(t) = sent signal
r(t) = received signal

r(t) =  f( s(t), th ) + noise

i.e. the received signal is the sent signal modified somehow [f( . )]
by the phase th --- and corrupted by noise.

] {th} in my case, represents the unknown
] channel phase shift, which i'm assuming remains constant over the
] duration of my integral (int r*(t) exp(-j*th)* s(t) dt).

OK does this say that

r(t) = s(t) exp(j*th) + noise

??

] In other words, I'm integrating over the rx symbol duration shifted
] by the channel induced phase. If the phase shift of the channel is
] zero, and we have an impulse channel with relatively high SNR,
] then r(t) ~= s(t) (i.e. what we tx is approx same as what we rx)
] the phase will therefore cancel when i perform the matched filter
] receiver of ( int r*(t) s(t) dt).

If th is zero, then you'll have r(t) ~= s(t) where the "~" is the
additive noise. That still makes your integral:

int (s*(t) + n(t))s(t) dt = int |s(t)|^2 dt + int n*(t)s(t) dt

which won't have zero phase, though the expectation [mean] of it
will probably be zero, depending on s(t) and n(t).

] For the case of a nonzero channel phase shift, we simply perform the
] same integral as above (int r*(t) exp(-j*th) s(t) dt) but we take the
] exp{} term outside the integral thus exp(-j*th) int r*(t) s(t) dt.
] Does this not mean that the channel phase {th} and the argument of
] the expression int r*(t) s(t) dt are one and the same?

Not at all. the phase {th} is one thing, the argument of
int r*(t) s(t) dt is another. For zero noise, the argument
of your integral will be zero, regardless of the value of {th}.

] This seems to be turning into a long winded post, so i'll try and
] wrap up by asking, given all of the above and assuming that the
] channel phase {th} is uniformly distributed over 0-2*pi, what
] simplification can we make to the expression
]
] = ... exp( 2A cos(th) Re(int r*(t) s(t) dt) )
] ?

Well, if you're trying to find the mean value of this with respect to
th, then just write:

E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] =
int exp( 2A cos(th) Re(int r*(t) s(t) dt) ) p_{th}(th) dth

where p_{th}(th) is the pdf of th. But p_{th}(th) =  1/(2*pi) so

E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] =
int exp( 2A cos(th) Re(int r*(t) s(t) dt) ) 1/(2*pi) dth

which will bring you to the

E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] = Io( ... )
expression of your previous post.

Ciao,

Peter K.

```
```Peter K. wrote:
> Hi,
>
> ] Sorry to keep on labouring the point, but I want to make sure that
i
> ] have a solid understanding of the assumptions that are being made.
> ] Perhaps if i elaborate a bit more on what it is that is happening,
> ] you may be able to see where i'm heading.
>
> No problem.
>
> ] The detector is phase noncoherent, and therefore, i'm trying to
> ] determine what assumptions are to be made when the carrier phase
(th)
> ] is unknown at the receiver. Perhaps i am indeed confusing the phase
> ] {th} with the argument of the expression r*(t) s(t), but to me they
> ] are the same thing, are they not?
>
> I can't see that from what you've written.
>
> So we have;
>
> s(t) = sent signal
> r(t) = received signal
>
> r(t) =  f( s(t), th ) + noise
>
> i.e. the received signal is the sent signal modified somehow [f( . )]
> by the phase th --- and corrupted by noise.
>
> ] {th} in my case, represents the unknown
> ] channel phase shift, which i'm assuming remains constant over the
> ] duration of my integral (int r*(t) exp(-j*th)* s(t) dt).
>
> OK does this say that
>
> r(t) = s(t) exp(j*th) + noise
>

Sorry.  Yes, that's exactly what i'm saying, in a convoluted way at
least.

> ??
>
> ] In other words, I'm integrating over the rx symbol duration shifted
> ] by the channel induced phase. If the phase shift of the channel is
> ] zero, and we have an impulse channel with relatively high SNR,
> ] then r(t) ~= s(t) (i.e. what we tx is approx same as what we rx)
> ] the phase will therefore cancel when i perform the matched filter
> ] receiver of ( int r*(t) s(t) dt).
>
> If th is zero, then you'll have r(t) ~= s(t) where the "~" is the
> additive noise. That still makes your integral:
>
> int (s*(t) + n(t))s(t) dt = int |s(t)|^2 dt + int n*(t)s(t) dt
>
> which won't have zero phase, though the expectation [mean] of it
> will probably be zero, depending on s(t) and n(t).
>

Yep.

> ] For the case of a nonzero channel phase shift, we simply perform
the
> ] same integral as above (int r*(t) exp(-j*th) s(t) dt) but we take
the
> ] exp{} term outside the integral thus exp(-j*th) int r*(t) s(t) dt.
> ] Does this not mean that the channel phase {th} and the argument of
> ] the expression int r*(t) s(t) dt are one and the same?
>
> Not at all. the phase {th} is one thing, the argument of
> int r*(t) s(t) dt is another. For zero noise, the argument
> of your integral will be zero, regardless of the value of {th}.
>

Okay, I agree with you here, I was trying to argue for the equivalence
of the unknown channel phase (th) and the result of the expression

exp(-j*th) int r*(t) s(t) dt

with high SNR.  I think we can move off this issue because i was really
flogging a dead four legged mammal.

> ] This seems to be turning into a long winded post, so i'll try and
> ] wrap up by asking, given all of the above and assuming that the
> ] channel phase {th} is uniformly distributed over 0-2*pi, what
> ] simplification can we make to the expression
> ]
> ] = ... exp( 2A cos(th) Re(int r*(t) s(t) dt) )
> ] ?
>
> Well, if you're trying to find the mean value of this with respect to
> th, then just write:
>
> E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] =
> int exp( 2A cos(th) Re(int r*(t) s(t) dt) ) p_{th}(th) dth
>
> where p_{th}(th) is the pdf of th. But p_{th}(th) =  1/(2*pi) so
>
> E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] =
> int exp( 2A cos(th) Re(int r*(t) s(t) dt) ) 1/(2*pi) dth
>

Sorry to sound thick once again, but the p(th) = 1/2pi is because we
have a 1/2*pi'th probability of selecting a given phase from the
uniform distribution?  Okay, you don't have to answer that, it's
correct when i think it through.

> which will bring you to the
>
> E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] = Io( ... )
> expression of your previous post.

Now, are you saying that the expression above is

E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] = Io( 2A * Re(int r*(t)
s(t) dt) ))

from my earlier post.  Or are you saying that it is

E[ exp( 2A cos(th) Re(int r*(t) s(t) dt) ) ] = Io( 2A * abs(int r*(t)
s(t) dt) ))

??

What i'm getting at is that i don't understand what effect taking the
Expectation of the above has on the argument of int r*(t) s(t) dt ? If
any?

I'm not trying to stretch your patience on this one, so don't give up
on me and my niggling questions just yet!

Thanks

CW

```