Need help with some antenna maths

"PčéJäy" <wiseguyatiinetnetau> wrote in message
news:3f979462$1_1@news.sydney.pipenetworks.com...
Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j
Click to expand...
Hahahahahahaha... I'm laughing my head off after the holier than thou
Calculus thread that was on here

First of all....

this function is a dipole pattern function, specifically a Full Wave Dipole.

a = amplitude factor
x = angle from 0 to 360 degrees

The whole lot is squared but dont worry about that. (Its simply an intensity
calculation, hence the S)

Of course if you know this you know the shape is a figure 8 (for dipoles
under 1 lambda) and you also know the minimums are at 0 and 180 degrees.

So plot it by hand (Matlab or graph paper) - let a = whatever its not
important to shape.
Then if you want to solve its minimums quickly use an iterative method such
as the newton raphson method.


I'm laughing because the original Math test post from EW (prick) chastised
recent grads for often not knowing this type of stuff... and here we are...

In my experience, (I'm a mildly experienced PhD grad) I've sat in a heap of
interviews in this country and some of the half witted engineers and
managers makes me wonder how many of them are in business. If I sound like a
cheased off engineer its because I am.... basically I've been in the market
for a decent job in this country for over a year... guess what.. nada.. In
this time I've been offered several overseas... one just yesterday and
now.... since I'm running out of options (and money) am in the position
where I am either going to set up my own firm or leave our shores. I've
applied for bench tech jobs through to design roles..... The one recurring
theme in all of them is the lack of real know-how when it comes to design,
and their present hack and try mentality. Experience counts for little in my
opinion, you can either do it or you can't. Most talk about their 'unique
experience' but I've yet to meet one who can do it properly outside a
government lab.

So whats a guy like me to do? I may have less than 5 years work
experience, but I've been nibbling on the electronics teet for 20 years. I'm
the kid who was building quad arrays and living in the radio shed as a
kid.... I became a tech... then an engineer.. .and now a doc... and what do
I see? No jobs, underpaid jobs and a bunch of idiots in business.

To leave or not to leave...
 
hahaha.. too busy laughing and crying... its actually a 3/2 wavelength
dipole... as well as the 0,180 nulls there will be 4? extras... should be
around the 85,95 mark and corresponding angles the opposite side.
 
"Jeff" <noflames@flame.org> wrote in message
news:3f9e35b5_1@news.iprimus.com.au...
"PčéJäy" <wiseguyatiinetnetau> wrote in message
news:3f979462$1_1@news.sydney.pipenetworks.com...
Hi all,

Could some kind soul point me in the right direction to solve or
simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j



Hahahahahahaha... I'm laughing my head off after the holier than thou
Calculus thread that was on here

First of all....

this function is a dipole pattern function, specifically a Full Wave
Dipole.

a = amplitude factor
x = angle from 0 to 360 degrees

The whole lot is squared but dont worry about that. (Its simply an
intensity
calculation, hence the S)

Of course if you know this you know the shape is a figure 8 (for dipoles
under 1 lambda) and you also know the minimums are at 0 and 180 degrees.

So plot it by hand (Matlab or graph paper) - let a = whatever its not
important to shape.
Then if you want to solve its minimums quickly use an iterative method
such
as the newton raphson method.


I'm laughing because the original Math test post from EW (prick) chastised
recent grads for often not knowing this type of stuff... and here we
are...

In my experience, (I'm a mildly experienced PhD grad) I've sat in a heap
of
interviews in this country and some of the half witted engineers and
managers makes me wonder how many of them are in business. If I sound like
a
cheased off engineer its because I am.... basically I've been in the
market
for a decent job in this country for over a year... guess what.. nada.. In
this time I've been offered several overseas... one just yesterday and
now.... since I'm running out of options (and money) am in the position
where I am either going to set up my own firm or leave our shores. I've
applied for bench tech jobs through to design roles..... The one recurring
theme in all of them is the lack of real know-how when it comes to design,
and their present hack and try mentality. Experience counts for little in
my
opinion, you can either do it or you can't. Most talk about their 'unique
experience' but I've yet to meet one who can do it properly outside a
government lab.

So whats a guy like me to do? I may have less than 5 years work
experience, but I've been nibbling on the electronics teet for 20 years.
I'm
the kid who was building quad arrays and living in the radio shed as a
kid.... I became a tech... then an engineer.. .and now a doc... and what
do
I see? No jobs, underpaid jobs and a bunch of idiots in business.

To leave or not to leave...
Click to expand...


** The former sounds most suitable for you.




............ Phil
 
On Tue, 28 Oct 2003 19:44:49 +1000, "Jeff" <noflames@flame.org> put
finger to keyboard and composed:

hahaha.. too busy laughing and crying... its actually a 3/2 wavelength
dipole... as well as the 0,180 nulls there will be 4? extras... should be
around the 85,95 mark and corresponding angles the opposite side.
Click to expand...
You weren't paying attention. Very early in this thread I stated that
there was a null where "x = arc cos(1/3) = 70.53 deg". I also stated
that "as S(x) is symmetrical, a solution for x will be repeated
at pi-x, pi+x, and -x radians". And BTW, the OP asked for *maxima*,
not *minima*. So what is *your* answer?


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.
 
On Tue, 28 Oct 2003 19:22:23 +1000, "Jeff" <noflames@flame.org> put
finger to keyboard and composed:

Then if you want to solve its minimums quickly use an iterative method such
as the newton raphson method.
Click to expand...
Nobody is interested in the *minima*. We all know what they are and
where they occur.

As for the *maxima*, we would need to solve the equation S'(x)=0. The
Newton-Raphson method would require iterations of the form

x2 = x1 - S'(x)/S"(x)

Sure, this is a nice method, and I *did* consider using it, but S"(x)
gets to be very complicated. It was much easier to knock out some
simple Qbasic code to solve for S'(x)=0.

CLS
pi = 3.141592654#

FOR i = 40 TO 45 STEP .1

x = i / 180 * pi

dsdx = 1.5 * pi * SIN(x) * SIN(x) * SIN(3 * pi * COS(x)) - (1 + COS(3
* pi * COS(x))) * COS(x)

PRINT i, dsdx
NEXT i

You can progressively edit the FOR statement after each run by
changing the range of angles and step sizes. For example, on the
second pass, the range of angles could be reduced to 42.5 - 42.6 and
the step size could be changed to .01.

A similar technique could be implemented with a spreadsheet.


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.
 
One way is to use Excel.

Type the function in a cell. Define x as another cell. Make a guess for x.

Then use "Goal Seek" to solve for x. If there's more than one solution
you'll need to have different starting guesses.




"PčéJäy" <wiseguyatiinetnetau> wrote in message
news:3f979462$1_1@news.sydney.pipenetworks.com...
Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j
Click to expand...
 
Well, to be honest I wont sit around solving problems on this newsgroup...
I'm quite happy to talk about it ...but its not important to my life to pull
out the pen.... this problem is simple however.

My solution? Find all points where the first derivative is zero. This will
include maxima and minima of course.

Again unless you can "see" (if you plotted it you will see some) the answer
the easiest method is to use an interative method. ie. Conjugate gradient.

"Franc Zabkar" <fzabkar@optussnet.com.au> wrote in message
news:rhftpvg8mcsofbtsdh8vmhsh4qg3blbuv6@4ax.com...
On Tue, 28 Oct 2003 19:44:49 +1000, "Jeff" <noflames@flame.org> put
finger to keyboard and composed:

hahaha.. too busy laughing and crying... its actually a 3/2 wavelength
dipole... as well as the 0,180 nulls there will be 4? extras... should be
around the 85,95 mark and corresponding angles the opposite side.

You weren't paying attention. Very early in this thread I stated that
there was a null where "x = arc cos(1/3) = 70.53 deg". I also stated
that "as S(x) is symmetrical, a solution for x will be repeated
at pi-x, pi+x, and -x radians". And BTW, the OP asked for *maxima*,
not *minima*. So what is *your* answer?


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.
Click to expand...
 
"Jeff" <

Well, to be honest I wont sit around solving problems on this newsgroup...
I'm quite happy to talk about it ...but its not important to my life to
pull
out the pen.... this problem is simple however.

My solution? Find all points where the first derivative is zero. This will
include maxima and minima of course.

Again unless you can "see" (if you plotted it you will see some) the
answer
the easiest method is to use an interative method. ie. Conjugate gradient.
Click to expand...


** Which is almost the same as the answer I posted on the 24th.

And I ain't got no highfalutin PhD - like the disgruntled Jeff




........... Phil
 
And I ain't got no highfalutin PhD - like the disgruntled Jeff




.......... Phil
Click to expand...
Well.. plotting a curve and finding the maximum and minima is high school
work.. I certainly hope the original poster is not a degree student however
since this was an intensity eqn rather than a pattern eqn I suspect he was.

I must appologise for my shit winge... I'm finding it hard to remain
positive in the current climate...
 
On Wed, 29 Oct 2003 20:09:19 +1000, "Jeff" <noflames@flame.org> put
finger to keyboard and composed:

Well, to be honest I wont sit around solving problems on this newsgroup...
I'm quite happy to talk about it ...but its not important to my life to pull
out the pen.... this problem is simple however.
Click to expand...
My understanding is that the OP requires a "non-empirical" solution.
This makes the actual problem quite difficult, and possibly
unsolvable.

BTW, I'm waiting for the OP's tutor, or real Andy, to provide the
actual answer, ... if there is one. ;-)

My solution? Find all points where the first derivative is zero. This will
include maxima and minima of course.

Again unless you can "see" (if you plotted it you will see some) the answer
the easiest method is to use an interative method. ie. Conjugate gradient.
Click to expand...
Although it may give us little intellectual satisfaction, the most
expedient method presented in this thread would probably involve
Mathematica or GnuCalc. The next easiest would require a spreadsheet,
eg Excel.

"Franc Zabkar" <fzabkar@optussnet.com.au> wrote in message
news:rhftpvg8mcsofbtsdh8vmhsh4qg3blbuv6@4ax.com...
On Tue, 28 Oct 2003 19:44:49 +1000, "Jeff" <noflames@flame.org> put
finger to keyboard and composed:

hahaha.. too busy laughing and crying... its actually a 3/2 wavelength
dipole... as well as the 0,180 nulls there will be 4? extras... should be
around the 85,95 mark and corresponding angles the opposite side.

You weren't paying attention. Very early in this thread I stated that
there was a null where "x = arc cos(1/3) = 70.53 deg". I also stated
that "as S(x) is symmetrical, a solution for x will be repeated
at pi-x, pi+x, and -x radians". And BTW, the OP asked for *maxima*,
not *minima*. So what is *your* answer?
Click to expand...

- Franc Zabkar
--
Please remove one 's' from my address when replying by email.
 
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