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Topic: Differential equation of a microphone (Read 6030 times) previous topic - next topic
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Differential equation of a microphone

I have already described a speaker as an differential equation:

Tha amplifier sends out a certain voltage that makes a current flow in the coil of the speaker. The coil moves due to the magnetism (solid magnets + current trough coil), the membrane moves with it and the speaker produces sound.

electrical equation:
R.i + L di/dt = u - K.v

R = resistance of the coil
i = current in the coil
L di/dt = voltage at the coil
u = voltage from amplifier
v = velocity of the coil (B.l.v)

mechanical equation:
m.a + b.v + k.y = K.i

m = mass of membrane
b = friction-constant
y = displacement, v = velocity, a = acceleration (of membrane)
k = spring-constant (the membrane has a spring effect)
K.i = force on the coil of the magnetic field due to the current i (B.l.i)

These equations are correct.

As a microfone is the inverse of a speaker, it would also be very simple to design the differential equations of a microphone.

I think that the mechanical equation will stay the same and just the electrical equation will change.


I will redefine u:
As someone speaks to the membrane, there will come a voltage across the coil which will go to the amplifier = u (coil moves in the magnetic field of the solid magnets -> voltage) -> we get a microphone.

At first sight, I would only change the sign of u, making the equation:
R.i + L di/dt = - u - K.v
or should it be?
R.i + L di/dt = - u + K.v
Or maybe it is totaly wrong?

Can somebody help me to change the equations of the speaker, to suit a microphone?

Thanks in advance.

Differential equation of a microphone

Reply #1 –
It would seem to me that a microphone is in fact the exact same as a loudspeaker except in the values of current / voltage that are at its connectors.

In such a consideration, I think the formulae should stay the same, but redefine u as the voltage out of the microphone instead.

Edit : or have u=0?

Differential equation of a microphone

Reply #2 –
Quote
In such a consideration, I think the formulae should stay the same, but redefine u as the voltage out of the microphone instead.
[a href="index.php?act=findpost&pid=357900"][{POST_SNAPBACK}][/a]


I´m sorry, I edited my post so that you can see how u is redefined.

Differential equation of a microphone

Reply #3 –
Quote
I have already described a speaker as an differential equation:

Tha amplifier sends out a certain voltage that makes a current flow in the coil of the speaker. The coil moves due to the magnetism (solid magnets + current trough coil), the membrane moves with it and the speaker produces sound.

electrical equation:
R.i + L di/dt = u - K.v

R = resistance of the coil
i = current in the coil
L di/dt = voltage at the coil
u = voltage from amplifier
v = velocity of the coil (B.l.v)

mechanical equation:
m.a + b.v + k.y = K.i

m = mass of membrane
b = friction-constant
y = displacement, v = velocity, a = acceleration (of membrane)
k = spring-constant (the membrane has a spring effect)
K.i = force on the coil of the magnetic field due to the current i (B.l.i)

These equations are correct.

As a microfone is the inverse of a speaker, it would also be very simple to design the differential equations of a microphone.

I think that the mechanical equation will stay the same and just the electrical equation will change.


I will redefine u:
As someone speaks to the membrane, there will come a voltage across the coil = u (coil moves in the magnetic field of the solid magnets -> voltage) -> we get a microphone.

At first sight, I would only change the sign of u, making the equation:
R.i + L di/dt = - u - K.v
or should it be?
R.i + L di/dt = - u + K.v
Or maybe it is totaly wrong?

Can somebody help me to change the equations of the speaker, to suit a microphone?

Thanks in advance.
[a href="index.php?act=findpost&pid=357899"][{POST_SNAPBACK}][/a]


Wouldn't it also be wise to add, here, the equation that translates velocity to sound?
-----
J. D. (jj) Johnston

Differential equation of a microphone

Reply #4 –
In case any of you understand German, please consider this:
http://www.opus.ub.uni-erlangen.de/opus/vo...df/drarbeit.pdf

A friend and I were considering to develop a dynamic (observer/)controller to enhance audio reproduction. Then this guy came along and published something very similar so we dropped it. IIRC* the author develops a detailed dynamic model of a speaker.

*It's been a while, but from a quick look what I claim seems correct

Differential equation of a microphone

Reply #5 –
Quote
In case any of you understand German, please consider this:
http://www.opus.ub.uni-erlangen.de/opus/vo...df/drarbeit.pdf
[a href="index.php?act=findpost&pid=357915"][{POST_SNAPBACK}][/a]


Wow, thanks, that is a very advanced model and too advanced for my purposes. (making a space-state model of the form dx/dt = Ax + Bu, y = Cx + Du)

The differential equations in the first post are correct and suit my needs. I have just troubles in finding the model for the microfone. I think that only some signs have to change in the original equation.

Differential equation of a microphone

Reply #6 –
Maybe I'm swinging wildly wrong here, but does the name "Raes" ring a bell?

Differential equation of a microphone

Reply #7 –
Quote
Maybe I'm swinging wildly wrong here, but does the name "Raes" ring a bell?
[a href="index.php?act=findpost&pid=358004"][{POST_SNAPBACK}][/a]


Yes it does, but as there is a lot of discussion going on between the enginieering students about this matter, I think this is a better place to post my problem 

Differential equation of a microphone

Reply #8 –
Yes, this problem looked a bit too familiar.

Some "hints":

1) The assumption that loudspeaker == microphone is correct. That's not only true in theory - you can actually use a loudspeaker as a microphone with suitable amplification.

2) Mechanically, "obviously" nothing changes. Electrically, the flow is reversed. You got this far, too.

3) Figuring out the sign changes is most easily done by making a graph of the thing, rather than staring at the equations. I don't think it makes much sense to talk about the signs of the unknowns without a graph in the first place?

4) The problem is an illustration about state spaces to solve control systems engineering problems. It doesn't really have anything to do with audio, nor should you worry too much about that physical equation. Nobody is going to care if you got some sign reversed somewhere. But they will care if you can't manipulate the physical system into a state-space model.

5) If you're still stuck or unsure, mail the teacher and explain your problem or what you aren't sure of.