Discussion in 'General Electronics Chat' started by dileepchacko, Sep 1, 2009.

1. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
Dear All

Please calculate the cutoff frequency of this circuit. I think the R2 will not depend on the cutoff frequency, but it depends on the other parameter such as pass band gain and quality factor.

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2. ### t06afre AAC Fanatic!

May 11, 2009
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Can you tell more about why you struggle. You have drawn your figure in some sort of ecad system. Do this tool have a simulator you can use for simulation?

3. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
Yes I have drown this circuit in LT SPICE. I have the option for simulation, I can get the cutoff frequency by simulating the circuit, but I need the mathematical steps to solve this. I know the Simple LPF cutoff frequency ( 1/(2*pi*R*C)). How R2 is depends on this circuit characteristics.

4. ### t06afre AAC Fanatic!

May 11, 2009
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You have to find the transfer function. This is equal to Vout/Vin. In this case I think nodal analysis is the essayist way to solve this problem. Are you familiar with this form of circuit analysis?

5. ### rjenkins AAC Fanatic!

Nov 6, 2005
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A clue: The value of R2 will drastically affect the cutoff frequency, it is as important as R1.

6. ### Wendy Moderator

Mar 24, 2008
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Are you planning on using the classic 3db level? This is typically ½ the input signal voltage. This question is more important than it looks, because you have a voltage divider whose voltage out is entirely dependent on frequency, and you need to set a point where this magic level (3db) happens.

7. ### Electronworks New Member

Sep 1, 2009
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Combined impedance of C1 and R2 is product/sum, thus:

(R2/jwC)/(R2+1/jwC) = R2/(1+jwCR2).

Then use resistive divider equation to find transfer function:

[R2/(1+jwCR2)] / ([R2/(1+jwCR2)] + R1) = R2/(R2+R1(1+jwCR2))

cut off frequency is when denominator real = denominator imaginary

ie when (R2 + R1) = jwCR1R2 (expanding out the denominator)

If R1 = R2 = 1k and C = 1uF

2k = 2pi x f x 1uF x 1k x 1k (w = 2pi x f)

318 Hz

As a rough check, your 2 1k resistors are in parallel with a voltage source, so the cut off freq is the same as a single RC filter with a 500 Ohm resistor and a 1uF cap - 318 Hz according to my Casio..

is this what Linear tech say...?

Interested to know..

Bill Naylor

<snip>
Electronic Kits for Education and Fun

Last edited by a moderator: Sep 1, 2009
8. ### t06afre AAC Fanatic!

May 11, 2009
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Is not the 3 Db level equal to 1/square root(2) or approximately 0.707

9. ### Audioguru New Member

Dec 20, 2007
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-3dB is half the output power or 0.707 times the voltage.

10. ### Wendy Moderator

Mar 24, 2008
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Sheesh. I stepped in that one.

11. ### Tronster Member

Dec 29, 2008
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An LT Spice IV simulation of this circuit agrees precisely with your estimate.

12. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
Yes I need 3dB bandwidth. The 3dB bandwidth cutoff frequency is my requirement.

13. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
If the circuit is in cutoff frequency how we can write (R1+R2) = 2Pi*R1R2*f ?

Can you please explain me , Bill Naylor

14. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
The LT-SPICE simulation tool is showing the Cutoff frequency is = 314.534Hz. this value is based on the 3dB reference point. I have attached the file for the reference. It shows different value than 318hz.

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15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The cut-off frequency is equal
Vo/Vin=1/√2=0.7071067812=-3.010299957dB
And for this circuit is equal
Fc=1/ ( 2*PI*(R1||R2*C) ) = 318.3098862Hz
And simulation show:
Fc=318.313Hz

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16. ### Ron H AAC Fanatic!

Apr 14, 2005
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As Jony130 pointed out, the cutoff frequency occurs at -3.010299957dB. Another, easier to remember point, is the phase shift. It will be exactly -45.0000 degrees.