# problem math bessel function

Started by December 3, 2004
```All,

I have a series of other questions, but before i get to those ones,
i'm stuck on one part of an equation that appears on page 304 of
Digital Communications 4th Ed, Proakis.  It's equation 5.4-18 and i
can't seem to understand how the equality between the integral of the
exponential function and the zeroth ordered mod bessel function of the
first kind works.  The equation looks something like

1/2*pi int^{2*pi}_{0}  exp(2E(a*cos(th)+b*sin(th))) d(th)       eqn(1)

where int = integral and th = theta, representing phase
and z = a+sqrt(-1)*b

and it says that this equals

Io(2E*sqrt(|z|))                  eqn(2)

where Io is the mod bessel function

Now, I'm of the understanding that the bessel function is equal to

Io(z) = 1/2*pi int^{2*pi}_{0} exp(z*cos(th)) d(th)

but i can't see how he has made the leap from eqn(1) to eqn(2) above.
Perhaps i've been looking at it for too long?  Can anyone help?

CW
```
```CW wrote:
> All,
>
> I have a series of other questions, but before i get to those ones,
> i'm stuck on one part of an equation that appears on page 304 of
> Digital Communications 4th Ed, Proakis.  It's equation 5.4-18 and i
> can't seem to understand how the equality between the integral of the
> exponential function and the zeroth ordered mod bessel function of the
> first kind works.  The equation looks something like
>
> 1/2*pi int^{2*pi}_{0}  exp(2E(a*cos(th)+b*sin(th))) d(th)       eqn(1)

Hopefully, I won't make any dumb errors

a*cos(th) + b*sin(th)  can be expressed in the form  c*cos(th+phi),
where you can look up c and phi from the trig identities.

You can substitute that back into the integral.

If you make the change of variable theta=th+phi  the limits on the
integrals go from 0 to 2pi  to  -phi to 2pi-phi which you can break up
into two integrals

-phi to 0   and  0 to 2pi-phi

since the interval is periodic you can add 2pi to both the lower and
upper terms of the first integral without changing its value, which
gives you for the lower limit, 2pi-phi and the upper limit of 2pi. You
can now combine the limits with the second integral to get back to  a
single integral from 0 to 2pi.

>
> where int = integral and th = theta, representing phase
> and z = a+sqrt(-1)*b
>
> and it says that this equals
>
> Io(2E*sqrt(|z|))                  eqn(2)
>
> where Io is the mod bessel function
>
> Now, I'm of the understanding that the bessel function is equal to
>
> Io(z) = 1/2*pi int^{2*pi}_{0} exp(z*cos(th)) d(th)
>
> but i can't see how he has made the leap from eqn(1) to eqn(2) above.
> Perhaps i've been looking at it for too long?  Can anyone help?
>
> CW
```
```prada_white@yahoo.ca (CW) wrote

> I have a series of other questions, but before i get to those ones,
> i'm stuck on one part of an equation that appears on page 304 of
> Digital Communications 4th Ed, Proakis.  It's equation 5.4-18 and i
> can't seem to understand how the equality between the integral of the
> exponential function and the zeroth ordered mod bessel function of the
> first kind works.  The equation looks something like
>
> 1/2*pi int^{2*pi}_{0}  exp(2E(a*cos(th)+b*sin(th))) d(th)       eqn(1)
>
> where int = integral and th = theta, representing phase
> and z = a+sqrt(-1)*b
>
> and it says that this equals
>
> Io(2E*sqrt(|z|))                  eqn(2)
>
> where Io is the mod bessel function
>
> Now, I'm of the understanding that the bessel function is equal to
>
> Io(z) = 1/2*pi int^{2*pi}_{0} exp(z*cos(th)) d(th)
>
> but i can't see how he has made the leap from eqn(1) to eqn(2) above.
> Perhaps i've been looking at it for too long?  Can anyone help?

Write:

a*cos(th) + b*sin(th) = |z|*cos(al)*cos(th) + |z|*sin(al)*sin(th)
= |z|*cos(th-al)

sub into eqn(1):

1/2*pi int^{2*pi}_{0}  exp(2E*|z|*cos(th-al)) d(th)       eqn(A)

and that yields

Io(2E*|z|)

which isn't quite what you want.

I don't have that edition of Proakis, so I find the equation you're
referring to.  Are you sure that eqn(2) doesn't read:

Io(2E*sqrt(||z||))                  eqn(B)

i.e. there is the two-norm of z instead of the magnitude?

Ciao,

Peter K.
```
```Peter, Stan thanks.

but that's not a big issue, because it was more the route that i wanted
help with and that's what both you and Peter gave me.  Now i'm
beginning to see the light at the tunnels end.

Peter, you're correct, I threw the email together quite quickly and
used the variable 'z' too much, but in any case, the result is correct
and it should read, from Proakis:

1/2*pi int^{2*pi}_{0}  exp(2E*|z|*cos(th-al)) d(th) = Io(2E |z|)
eqn(1)

I'm happy with this bit, but i haven't got the connect between the
relationship

Io(A) = 1/2*pi int^{2*pi}_{0} exp(A*cos(th)) d(th)       eqn(2)

so can we still make the equivalence of

cos(th-al) = cos(th)
2*E*abs(z) = A

which yields

Io(2E abs(z))

I don't understand what's so special about the bessel function so that
you can do this?  Does it mean that i can equate

cos(th)  = cos(u+v+w+x) where u,v,w and x are given variables?
or any number of cosine arguments?

Thanks

CW

```
```CW,

The thing is, when you have:

int*{2*pi}_{0} f( cos(th) ) dth

it really doesn't matter what you're adding to "th" in the argument of
cos(.),
because you're integrating over one complete period of the cos
function.
Whether the period is over [0,2*pi) or [-al,2*pi-al) doesn't matter.

cos(th-al) = cos(th)

is certainly incorrect (unless al = 2*pi*k with k an integer).
Ciao,

Peter K.

```
```Ah yes, of course.  We're integrating the cosine function.  I forgot
about that, i was too busy trying to spot the black magic that was
happening between the arguments of the exponent and the bessel
function.

Thanks again.

CW

```
```Ah yes, of course.  We're integrating the cosine function.  I forgot
about that, i was too busy trying to spot the black magic that was
happening between the arguments of the exponent and the bessel
function.

Thanks again.

CW

```
```Just to follow up on my line of questions now.  What I was looking at
originally - prior to being sidetracked into understanding the bessel
function translation in Proakis - was a few steps in the math from a
journal paper on iterative demodulation and decoding.  The authors get
as far as:

= ... exp( A (exp(-j*th) int r(t) s*(t) dt + exp(j*th) int r*(t) s(t)
dt ))    eqn1

= ... exp( 2A cos(th) int Re{ r*(t) s(t) } dt )    eqn2

where th = phase, r(t) = known pulse shape, s(t) = rx signal and r*(t)
= conjugate of r(t).

and go on to say that the phase {th} is uniformly distributed over
0-2*pi and hence

= ... Io( 2A abs( int conj(r(t)) s(t) ) )   eqn3

where i am now thoroughly and wretchedly lost!

Note: there are a few more terms in the equation, but for brevity the
bit i don't understand is the steps above.

My problem is that the argument for the exp in eqn1 can be written as
follows

A ( C * exp(-j*th) + D * exp(j*th) )

where C and D are the integrals or matched filter outputs if you
prefer.  Which they conveniently express in the form of:

2 * A * cos(th) Re{D}
so how do we remove the C and D from  the above?

Thanks

CW

```
```Hi CW,

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

= F*G + conj(F)*conj(G)

= 2 Re{F*G}

= 2 cos(th) Re{G}

where F = exp(-j*th) and G = int r(t) s*(t) dt.

Then, along with the assumption of the uniform distribution of th,
they must be taking an expectation over th.
That should get you to eqn3.

Ciao,

Peter K.

```
```Hi CW,

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

= F*G + conj(F)*conj(G)

= 2 Re{F*G}

= 2 cos(th) Re{G}

where F = exp(-j*th) and G = int r(t) s*(t) dt.

Then, along with the assumption of the uniform distribution of th,
they must be taking an expectation over th.
That should get you to eqn3.

Ciao,

Peter K.

```