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

# problem math bessel function

Started by ●December 3, 2004

Reply by ●December 3, 20042004-12-03

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

Reply by ●December 3, 20042004-12-03

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.

Reply by ●December 6, 20042004-12-06

Peter, Stan thanks. Stan i believe you have an incorrect sign in your th+phi derivation, 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) In eqn(1), using your notation, we have cos(th-al) instead of cos(th) 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

Reply by ●December 6, 20042004-12-06

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. Your equivalence: cos(th-al) = cos(th) is certainly incorrect (unless al = 2*pi*k with k an integer). Ciao, Peter K.

Reply by ●December 6, 20042004-12-06

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

Reply by ●December 6, 20042004-12-06

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

Reply by ●December 6, 20042004-12-06

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

Reply by ●December 6, 20042004-12-06

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.

Reply by ●December 6, 20042004-12-06

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.