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Mike, <p>Check out the following code:<p> Private Sub Form_Load()<p>'Following is for the first combo box which is first color band on resistor Combo1.AddItem "Black" Combo1.AddItem "Brown" Combo1.AddItem "Red" Combo1.AddItem "Orange" 'add the other colors as ...

Ok,
Fillcolor will not work by itself as you must specify the fillstyle first.<p>shape1.fillstyle = 0 'for solid fill
shape1.fillcolor = vbBlack 'black color<p>
I will post the code.<p>greg

How is the color specified? Did you use vbColor or #XXXXXX?

You can make a subtraction circuit with a reistor network and an op amp.<p>I'll have to find the specs<p>greg

Well it's more of a substitution:<p> Solving For Sin y: <p>Rearrange EQN A, solve for cosy<p>e^jy = cosy + jsiny e^jy - jsiny = cosy<p>Substitute into EQN B<p>e^-jy = cosy - jsiny e^-jy = (e^jy - jsiny) - jsiny e^-jy - e^jy = -2jsiny * -1 (e^jy - e^-jy = 2jsiny )/2j siny = ½j(e^jy - e^-jy)<p> Solvin...

See if this makes sense....<p> Remember the relationship sin wt = ½j (e^jwt - e^-jwt) EQN C<p>***see below for this derivation*****<p> Also note that the laplace transform for e^jwt is 1/(s-jw) and the laplace transform for e^-jwt is 1/(s+jw). We will now substitute this into the above relationship....

Hey Matt,<p>Diving into the world of socket programming...fun stuff....<p>Check out these links:<p> University Of Ontario <p> Examples In C <p> Socket In Java <p> Intro <p> Also, these search engin results:<p> Google Results <p>Have fun<p>greg

Mike,<p>You've got a ground buzz. Created by a ground loop.<p> All you have to do is lift the ground on the source you have connected to you amp. i.e. remove the ground on the laptop power supply. Most commercial sound systems have a ground switch so you can easily remove the ground. Also, most mixe...

Yes, it would change the required inductance. Re-work the formula for yourself to see what the new inductance is.<p>As far as first order and second order goes, the "order" is the number of filter stages. I'd perfer a second order filter, which as you mentioned offers a higher (or steeper)...

I would try more capacitance on the legs of the crystal. The is a lot of capacitance present between the traces on bradboards, which is not present (in the same amount) on the pcb. <p>greg

Just wondering who is the current Editor In Chief for N&V magazine? I remember seeing a ad for the job on the site.<p>Is Larry gone? Anyone know?<p>greg

Yeah,<p>It's actually the "0x"

Will,<p>Your exactly right.<p>0x8000 is equal to 32768.<p>0x signifies that it is a hexadecimal value and 32768 is a decimal value.<p>greg

Mike,<p>Well for a Cross Over Frequency of 100 Hz, you would require an inductance of<p>R = 2*pi*F*L L = 8 Ohms/(2*3.14*100) L = 12.7 mH<p>Check out this link on winding inductors. Remember you can make an inductor by winding electrical wire around a bobbin. :p <p> Inductor Winding <p> Good Site Rea...

Mike,<p>The crossover frequency is given by<p>R = 2*pi*F*L<p>R can be speaker impedance<p>Solve for L with your required Crossover Frequency.<p>Also, check this site out:<p>Crossover Spec's