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<blockquote data-quote="Phil Graham" data-source="post: 35059" data-attributes="member: 430"><p>Re: Some math for you on voltage vs. excursion</p><p></p><p><span style="color: #222222"></span></p><p><span style="color: #222222"></span></p><p><span style="color: #222222">Langston,</span></p><p><span style="color: #222222"></span></p><p><span style="color: #222222">The following is a gross oversimplification, as it considers only a sinusoidally driven mass. A real system also has to consider the suspension compliance and resistive losses, AND the equivalent behaviors of the box air (i.e. mass, compliance, and loss), the port, etc.</span></p><p><span style="color: #222222"></span></p><p><span style="color: #222222">That said, the following should be enough to get you thinking. I make no attempt at exposition, as I'm pretty sure you know how to do the math. Hopefully there are no mistakes, as I wrote it out quickly. This is for the 1D case: <ol> <li data-xf-list-type="ol">Voltage to amplifier through the voice coil impedance Z (which is Rdc + vc inductance)</li> <li data-xf-list-type="ol">At low frequencies Rdc dominates, so V=IZ is (mostly) V=I(Rdc) and V and I are therefore mostly in phase</li> <li data-xf-list-type="ol">F(t)=Bl x I(t) (x ==cross product)</li> <li data-xf-list-type="ol">F=ma - where mass would be the inertial mass of the system (cone mass)</li> <li data-xf-list-type="ol">F(x,t)=m(d2x/dt2) - acceleration is the second derivative of displacement wrt to time</li> <li data-xf-list-type="ol">V(t)=A*sin(wt + p) where wt is angular frequency and p is phase</li> <li data-xf-list-type="ol">I(t) is then nearly (A*sin(wt + p))/Z</li> <li data-xf-list-type="ol">F(t)= Bl x [(A*sin(wt + p))/Z]</li> <li data-xf-list-type="ol">Ignoring all the constants, and the starting phase: sin(wt) α (d2x/dt2) - where α means proportional to</li> <li data-xf-list-type="ol">Integrate both sides twice with respect to t to find the displacement: x=∫(∫sin(wt) dt)dt</li> <li data-xf-list-type="ol">The first integral is: -1/w*cos(wt) + C</li> <li data-xf-list-type="ol">Second integral, ignoring the constant, is: (-1/w)∫cos(wt) = (-1/w)(1/w)sin(wt) + C</li> <li data-xf-list-type="ol">In the end, again ignoring the constant of integration: x(t) α -1/(w^2)sin(wt) - where x(t) is displacement as a function of time.</li> </ol><p>So, in the <em>simplest</em> case of an undamped mass being driven by a sinusoidally varying voltage source, the displacement of the mass has the <em>opposite</em> sign as the original driving voltage. Also the amplitude is inversely proportional to the square of the angular frequency. That means the lower the frequency, the higher the amplitude. For the spacing of an octave (i.e. freq. of 1 vs freq. of 2) it is straightforward to see where the 12dB/octave comes from:</span></p><p><span style="color: #222222">20log(2^2/1^2) == 20log(4/1)=12.04dB</span></p><p><span style="color: #222222"></span></p><p><span style="color: #222222">As an example of how this gets more complicated: For periodic driving of an under-damped simple harmonic system, which is closer to a loudspeaker in free air, the form of solution to the differential equation is:</span></p><p><span style="color: #222222">[ATTACH]149922[/ATTACH]</span></p><p><span style="color: #222222"></span></p><p><span style="color: #222222">Where the first term is the transient solution that depends on the initial conditions of system motion, and the second term is dependent on the driving force, and not the initial system motion. This solution is courtesy of the excellent <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html" target="_blank">GSU hyperphysics page</a>. </span></p><p><span style="color: #222222"></span></p><p><span style="color: #222222">The amplifier voltage is not a simple expression of the driver displacement.</span></p></blockquote><p></p>
[QUOTE="Phil Graham, post: 35059, member: 430"] Re: Some math for you on voltage vs. excursion [COLOR=#222222] Langston, The following is a gross oversimplification, as it considers only a sinusoidally driven mass. A real system also has to consider the suspension compliance and resistive losses, AND the equivalent behaviors of the box air (i.e. mass, compliance, and loss), the port, etc. That said, the following should be enough to get you thinking. I make no attempt at exposition, as I'm pretty sure you know how to do the math. Hopefully there are no mistakes, as I wrote it out quickly. This is for the 1D case:[LIST=1] [*]Voltage to amplifier through the voice coil impedance Z (which is Rdc + vc inductance) [*]At low frequencies Rdc dominates, so V=IZ is (mostly) V=I(Rdc) and V and I are therefore mostly in phase [*]F(t)=Bl x I(t) (x ==cross product) [*]F=ma - where mass would be the inertial mass of the system (cone mass) [*]F(x,t)=m(d2x/dt2) - acceleration is the second derivative of displacement wrt to time [*]V(t)=A*sin(wt + p) where wt is angular frequency and p is phase [*]I(t) is then nearly (A*sin(wt + p))/Z [*]F(t)= Bl x [(A*sin(wt + p))/Z] [*]Ignoring all the constants, and the starting phase: sin(wt) α (d2x/dt2) - where α means proportional to [*]Integrate both sides twice with respect to t to find the displacement: x=∫(∫sin(wt) dt)dt [*]The first integral is: -1/w*cos(wt) + C [*]Second integral, ignoring the constant, is: (-1/w)∫cos(wt) = (-1/w)(1/w)sin(wt) + C [*]In the end, again ignoring the constant of integration: x(t) α -1/(w^2)sin(wt) - where x(t) is displacement as a function of time. [/LIST]So, in the [I]simplest[/I] case of an undamped mass being driven by a sinusoidally varying voltage source, the displacement of the mass has the [I]opposite[/I] sign as the original driving voltage. Also the amplitude is inversely proportional to the square of the angular frequency. That means the lower the frequency, the higher the amplitude. For the spacing of an octave (i.e. freq. of 1 vs freq. of 2) it is straightforward to see where the 12dB/octave comes from: 20log(2^2/1^2) == 20log(4/1)=12.04dB As an example of how this gets more complicated: For periodic driving of an under-damped simple harmonic system, which is closer to a loudspeaker in free air, the form of solution to the differential equation is: [ATTACH=CONFIG]149922.vB5-legacyid=2041[/ATTACH] Where the first term is the transient solution that depends on the initial conditions of system motion, and the second term is dependent on the driving force, and not the initial system motion. This solution is courtesy of the excellent [URL="http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html"]GSU hyperphysics page[/URL]. The amplifier voltage is not a simple expression of the driver displacement.[/COLOR] [/QUOTE]
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