Flippy

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16 years, 321 days

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These are answers submitted by Flippy

It's in the exponent!
The exponent in my original picture reads:
exp( [ i * 2 * Pi * sqrt(...) ] / L )

You placed the L correctly though so there's no problem.

 

Anyway, I have now confirmed that what I did in my last post was actually right. Simply substracting 50 orso dB's from the total plot and then manually editing the axes to read 50..90 dB (for example) is apparently exactly the same as plotting it from 50 to 90 dB only.

So I guess while my particular problem has been solved, the actual problem of how to plot polar coords in different ranges is not.

However I don't need this anymore right now so I guess it's solved now!

Thanks!!

Actually there is an ell in the image, it just looks like a 1 because for some reason it's not italic.

Now, I now noticed p0 is probably a bit too small, you might want to change it to about 0.1 or something like that.
The central spike (at a = 90*) should be around 90 to 105 dB's.

If you do that, you will see that the overall shape of the graph will get more 'puffed up' as can be seen in my images.


I want this plot to be constrained from around 50 to 90 dB's.

So I don't want to change the formula (changing the values of p0 or R will also change the shape to the way I want it) but I just want to limit the plotting from 50 to 90 dB.

I have done this now by changing the value of p0 and then manually editing the text on the axes.
I know this is probably "not done" but I'm pretty sure that the graph I'm getting will look the same when I actually only plot it from 60 to 90 dB... or similar.
Here's a polar graph I made where I edited the text manually (note, the actual axes are hidden and are still only going from 0 to 30 dB's!)

 

Hope this helps.

Thanks for your replies!

I think you are close yes, but I'm not that advanced with Maple and I cannot understand everything you said. I understand most of it though.

For reasonal parameters, first of all you are missing a "l" in the exp. The "i * 2 * Pi * q(i)" should be divided by l, which is the wavelength of the sound. ("l" is the lowercase letter L everywhere in this post, not the complex i !!)

Then, try something like R = 100, l = 0.12, d = 0.06 (this is ESSENTIAL! d should be 1/2 * l ), p0 = 0.05.

As long as you keep d = 1/2 * l and R >> l then you can change p0. Changing p0 will change the shape of the graph dramatically which is presumably due to the log...

 

I know that when you substract some reasonable value from the function, it appears as though you have actually cut out a hole in the graph and shrunk it back to close the hole again. However, this way, the axes will still show 0..30 dB instead of 60..90 for example.

The problem here is that I'm not sure if it's "allowed" to simply substract say 40 dB from 90 dB, since dB's are not linear.

In other words, 90dB - 60 dB is NOT equal to 60 dB - 30 dB (if they were linear both would yield 30 dB but they don't)

I now realise you are probably confusing the "l" (lowercase L) with the "I" (uppercase i ). The font on this forum makes them look the same... You should be able to see what I mean in the first picture in my post which shows the formula I use.

 

Also, I see you are getting negative dB's in your last plot. If possible I would like to prevent those because they are well below the human treshold of hearing and it wouldn't make alot of sense plotting those when we are trying to compare this theoretical graph with measurements we made.

Sorry I see the pictures aren't working. You can copy paste the url in your browser to see them.

Right, from the start, this time with pictures:


The function I have is the following:
[IMG]http://i28.tinypic.com/23lyyol.jpg[/IMG]

It is used to sum a row of sound sources while keeping in mind the difference in distance between the point where you are measuring and all the sources. The square roots are in essence the distance from some point P to speaker number i. The distance d is the distance between two speakers. l is the wavelength of the sound and a is the angle under which you are measuring. R is the distance from point P to the middle of the row of speakers.

So basically, while you go around in a circle (changing a) with radius R, the distance from you to each speaker will vary, which will in turn give the sound at your position some interference. If you know a little physics you will understand that a difference in distance between two sources that emit sound coherently will result in a phase difference between the two waves at a certain distance away from the speakers. I am trying to graph this phase difference.

 

Now, when the distance d between two speakers is exactly half of the wavelength l then the sound will interfere destructively at the sides causing a minimum in sound level, while it will interfere constructively in front of the row of speakers (a = Pi/2) causing a maximum in sound level. Basically, we have build a sound directing device.

 

Now, the formule above gives me a sound pressure. I have to compute the sound level from that which is a logaritmic function

20*log[10](p/p_ref)

where p_ref is some reference value, 20*10^-6.

 

When I graph that in polar coordinates, this is what I get, depending on the value of p0:
[IMG]http://i25.tinypic.com/4lfsww.jpg[/IMG]


What I want to do now is to cut out a circle in the middle of the graph, and stick the endpoints back in the middle.

It SHOULD be equivalent to the following graph, which I got using a (much) smaller value for p0. The only difference is that the following graph is plotted from 0 to around 30 dB, while my graph should be plotted from roughly 40 to 100 dB. (40 is the minimum value, I want the minimum value to be in the center so to speak)
[IMG]http://i30.tinypic.com/142r7m1.jpg[/IMG]

 

Is there any way I can do this without manually adjusting the numbers on the axes? I don't know if it's even allowed to do that, it probably changes the whole graph...

 

Hope this helps...!

No, I don't want to connect the endpoints by a line.

 

It's very hard for me to explain this (my main language is not even english :p) but what I mean can be visualed like this:

You graph a certain function in polar coordinates.

You then 'draw' a circle (r= some value in polar coordinates) in the same graph

You take some scissors and cut out the circle

You then manipulate the 'paper' (i know, you cant do that, but it would be possible on a computer lol) to shrink everything back until the hole is filled again.

Maybe an illustration would do justice but I dont have the time to make one right now, maybe tomorrow.


again, I dont know how to describe this... It's probably not even mathematical so that's also why I wondered if it would even be possible...

Ok thanks. So while you can 'hide' the plot so to speak in a certain range, there is no way to represent this as a new polar plot that starts from 2 at the center?

There is no way to 'take the endpoints and squeeze them together in the middle of the axis' so to speak..?

If not, there is really no point in changing the scale in my problem.

Thanks for your reply though!

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