# frequency of oscillation in a twin t oscillator

Discussion in 'Electronics Homework Help' started by bazzel, Nov 17, 2012.

1. ### bazzel

18
0
Nov 10, 2011
Hello all.I have a math question that hopefully someone can answer

I got so far in working out the frequency of a twin t but got stuck at this point

1/6.28 square root of 1/225x 0.0000000010

it should work out as 1059.9 Hz but i cant figure out how to get there, any thoughts?

2. ### duke37

5,364
772
Jan 9, 2011
I might be able to do this if you add brackets such as

1/(SQRT(1/225) * 1e-9)

You should have an even number of brackets.

3. ### duke37

5,364
772
Jan 9, 2011
According to my book, the frequency of the twin T notch is

F=1/(2*pi*R*C)

Where is the square root?

4. ### bazzel

18
0
Nov 10, 2011

Ive attatched a scanned copy of the formula if thats any help.
What Im struggling with is getting to the bottom answer in brackets (6666) from the sum above it (square rooot of 1/225 x 10minus 10)

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5. ### duke37

5,364
772
Jan 9, 2011
This should have been answered by an Australian, they are upside down!

So it is an asymetrical circuit with different resistances so you get SQRT(R1*R2)

F = 1/(2*pi)*SQRT(1/(R1*R2*C1*C2))

R1=15k
R2=1.5k
C1=100n = 1E-7
C2=10n = 1E-8

using the notation that E is 10 to the power of
Take the inner brackets
A= R1*R2*C1*C2 = 2.25E-6
1/A = 444444
SQRT (1/A) = 6666.7
F=6666.7/(2*pi) = 1061

Near enough?

6. ### bazzel

18
0
Nov 10, 2011
Cheers duke. Thats great,thanks for taking the time