Electronics Forums > frequency of oscillation in a twin t oscillator

# frequency of oscillation in a twin t oscillator

Junior Member
Join Date: Nov 2011
Posts: 18

 11-17-2012, 12:47 PM
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?

Senior Member
Join Date: Jan 2011
Location: Derbyshire. UK
Posts: 2,508

 11-17-2012, 01:16 PM
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.

Senior Member
Join Date: Jan 2011
Location: Derbyshire. UK
Posts: 2,508

 11-17-2012, 01:33 PM
According to my book, the frequency of the twin T notch is

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

Where is the square root?

Junior Member
Join Date: Nov 2011
Posts: 18

 11-20-2012, 10:45 AM
Quote:
 Originally Posted by duke37 According to my book, the frequency of the twin T notch is F=1/(2*pi*R*C) Where is the square root?

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)
Attached Thumbnails

Senior Member
Join Date: Jan 2011
Location: Derbyshire. UK
Posts: 2,508

 11-20-2012, 12:28 PM
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?

Junior Member
Join Date: Nov 2011
Posts: 18

 11-20-2012, 04:57 PM
Cheers duke. Thats great,thanks for taking the time

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