Maths Help?

[pwdrive]

Never been good at maths, anyways here goes, two formulas from the link below can be found under the heading Operating Conditions:


http://www.r-type.org/articles/art-167.htm

I've also replicated them here, first formula......Pout = (Vin(RMS) * gm)squared * RL

2nd formula....Vin(RMS) = (1/gm) * (Pout/RL)0.5.

I've used the following figures for the 1st formula:

Vin(RMS)=7 Volts

gm = 7mA/V

RL = 2,400 Ohms.

Punching those into the 1st formula gives Pout as 5.76 Watts.

Entering 5.76 Watts into the second formula and still using 7mA/V and 2400 Ohms doesn't return a Vin of 7 Volts(RMS) Well not the way I'm doing it

Anyone know what's wrong or where I've gone wrong?

Also posted on UKVRR

Lawrence.

[ppppenguin]

The 2nd formula is shown in an odd way on that website. The 0.5 is a superscript, meaning raise the number to the power of 0.5. Or in common parlance, take the square root.

I think you may be multiplying by 0.5.

[pwdrive]

Yes I was multiplying by 0.5

Do I square root the whole lot or just Pout/RL?

'scuse my ignorance.

Lawrence.

[pwdrive]

Tried it with square root of Pout/RL, the result didn't twigg at first (brain lapse) but all is ok now....6.998 Volts which is near as damn it to 7 Volts.

Lawrence.

[ppppenguin]

If you're writing computer code or in other situations where special symbols and superscripts aren't possible a square root is usually expressed as:

SQRT(x)

or

x^0.5

[pwdrive]

Thanks.

Lawrence.

[pwdrive]

Another question, if gm is unknown but Vin, RL and Pout are known how would the formula be re arranged to obtain gm?

Lawrence.

[ppppenguin]

Start with:   Pout = (Vin(RMS) * gm)^2* RL

We need to get gm out on its own so move RL to the other side: Pout/RL = (Vin(RMS) * gm)^2

Get rid of the square by square rooting both sides: SQRT(Pout/RL) = Vin * gm

Finally move Vin to the other side: (SQRT(Pout/RL) )/ Vin = gm

Voila!

[pwdrive]

It all workie workie's, thanks again for that.

Lawrence.

[ppppenguin]

The mathematical pedants may wish to call me out for square rooting both sides without mentioning the problem that can bring.

What's the square root of 4?

Well 2 of course, as any fule kno.

But what is -2 squared?

4. Simples.

So what's the square root of 4?

There are 2 correct answers to the question, 2 and -2.

Not a probkem in pwdrive's example but just a warning to be careful when dealing with quadratic functions.

This fallacy isn't a quadratic problem, just a concealed division by zero:
http://www.quickanddirtytips.com/educati...e-that-1-2

Anyone for tackling imaginary numbers? (i for the maths bods, j for us engineers)
Reckon that Mark Hennessey would be a better tutor than me.