# very basic maths help

Discussion in 'Math' started by harry99932, Jan 2, 2011.

1. ### harry99932 Thread Starter Member

Dec 30, 2010
38
2
Hi guys can someone help me out im a verrrrrrry un mathematically minded person slowly learning lol Im trying to expand out this equation but not sure how to go about it.

Vout= Vin* [98+((2/k)*Vin)]

I think im right so far in thinking il end up with;

Vin*98 + Vin*(2/k) + Vin*Vin

Now if thats right, then for the 3rd part of the equation were

Vin= 0.01 sin(2pi800)

i get an equation of the type

(sinx)*(siny)= cos(x-y) - cos(x+y)

How do i apply this to 0.001 sin (2pi800) ? i.e what do i do with the two 0.001's etc?

2. ### edgetrigger Member

Dec 19, 2010
133
19

just this => Vin*98 + Vin*(2/k) * Vin

3. ### edgetrigger Member

Dec 19, 2010
133
19
correct formula is

sinx * siny = 1/2 * [cos(x-y) - cos(x+y)]

4. ### edgetrigger Member

Dec 19, 2010
133
19
i could not get what you want to do with Vin.

If Vin is a sine wave the equation would be Vin = Vmax* Sin(θ+phy)

[ i did not get the symbol for phy in ]

could you pls let me know what you want to do with Vin

5. ### harry99932 Thread Starter Member

Dec 30, 2010
38
2
Hi edgetrigger,

The 0.01 sin(2*pi*800) is definatley correct. Peak sine wave voltage is 0.01 at 800hz working from equation; Vpeak sine (ωτ) were t is irrelevant for now and 2*pi*800 is hertz to angular velocity

the equation is the input to output expression for a theoretical amplifier, were v out = the equation given below, the answer should be several (3 we think) voltages a dc component, fundamental frequency and a harmonic frequency.

The previous example we are given is

Vo= 0.001 + 9.962Vi + 0.1Vi^2 with Vin at 0.01 sin (2000*pi*t) we get a result of

0.001005 + 0.09962sin(2000*pi*t) - 5x10^-6cos(4000*pi*t)

0.001005 volts dc
0.09962 volts ac at 1000hz fundamental frequency
5x10^-6 volts ac at 2000hz second harmonic

Any ideas??

6. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
The answer is delayed, but, hey, I deserve some christmass free time.

Let me try this one:

$
V_{out}=V_{in}(98+\frac2kV_{in})\\
=98V_{in}+\frac2kV_{in}V_{in}\\
=9.8\sin (1600\pi t) +\frac2k 0.0001 \sin^2 (1600 \pi t)\\
9.8 \sin(1600 \pi t) + \frac2k 0.0001 \left( \frac12 (1- \cos (3200 \pi t)) \right )\\
=\frac{0.0001}{k}+9.8 \sin (1600 \pi t)- \frac{0.0002}{k} \cos (3200 \pi t)
$

So, I guess your assumptions where correct.

I used the formula proposed by edgetrigger.

As usual, any corrections are welcome.

7. ### harry99932 Thread Starter Member

Dec 30, 2010
38
2
thanks for that georacer and edge trigger, i got there in the end with a lot of head banging but your workings make a lot more sense than mine so nice to see how i got there
Really appreciate the help, its nice people like you that will eventually help us mere mortals kill maths once and for good

Happy new year