1. ## Low pass filter

With R = 1k ohm, C = 3.1827x10^-8, Find 6db cutoff freq.

How do i get the eqn?

2. The simple answer to your question is that you have to write the equations of the circuit according to Kirchhoff's circuit laws, and then use the relation of the change of voltage across the capacitor and the current flowing through it:

$I(t) = C\frac{{dV(t)}}
{{dt}}$

3. If i am using voltage divider rule?

4. R = R
C = 1/jwC

5. ok, but do you have to find the cutoff frequency of Vout or Vout1?

6. Vout

7. $Z_{out1} = \left( {R + \frac{1}
{{jwc}}} \right)||\frac{1}
{{jwc}} = \frac{{1 + Rjwc}}
{{2jwc - Rw^2 c^2 }}$

now we use voltage divider:

$
V_{out1} = V_{in} \frac{{Z_{out1} }}
{{Z_{out1} + R}}
$

now we'll use the voltage divider again:

$
V_{out} = V_{out1} \frac{{\frac{1}
{{jwc}}}}
{{R + \frac{1}
{{jwc}}}} = V_{out1} \frac{1}
{{1 + Rjwc}} =
$

$
= V_{in} \frac{{Z_{out1} }}
{{Z_{out1} + R}}\frac{1}
{{1 + Rjwc}}
$

$
H_{out} (jw) = \frac{{Z_{out1} }}
{{Z_{out1} + R}}\frac{1}
{{1 + Rjwc}}
$

8. hi,

your Vout is the voltage across Zout

but my Vout is the voltage across the 2nd Cap(RHS)

are both the same?

9. Originally Posted by MilK
hi,

your Vout is the voltage across Zout

but my Vout is the voltage across the 2nd Cap(RHS)

are both the same?
A little blunder on my part, but I've fixed it, check my post again.

10. I see, thks

If i am told to find the 6db cutoff freq,

how do i translate this statement into your eqn?

i got this eqn:

Vout[ (-1/jw^3c^3) + R^3 + (4R^2/jwc) - (4R/w^2c^2) ] = Vin[ (-R/w^2c^2) - (1/jw^3c^3)]

how do i solve for f?

11. you can easily verify that the passband gain is 0db thus we are interested in finding the frequency in which the gain of the transfer function is -6db:

$
20\log _{10} \left| {H_{out} \left( {jw} \right)} \right| = - 6$

solve for w...

12. I see...

But how can i simplifiy my eqn:

Vout[ (-1/jw^3c^3) + R^3 + (4R^2/jwc) - (4R/w^2c^2) ] = Vin[ (-R/w^2c^2) - (1/jw^3c^3)]

to get the magnitude? struck at here.

13. Originally Posted by Peritus
$Z_{out1} = \left( {R + \frac{1}
{{jwc}}} \right)||\frac{1}
{{jwc}} = \frac{{1 + Rjwc}}
{{2jwc - Rw^2 c^2 }}$

now we use voltage divider:

$
V_{out1} = V_{in} \frac{{Z_{out1} }}
{{Z_{out1} + R}}
$

now we'll use the voltage divider again:

$
V_{out} = V_{out1} \frac{{\frac{1}
{{jwc}}}}
{{R + \frac{1}
{{jwc}}}} = V_{out1} \frac{1}
{{1 + Rjwc}} =
$

$
= V_{in} \frac{{Z_{out1} }}
{{Z_{out1} + R}}\frac{1}
{{1 + Rjwc}}
$

$
H_{out} (jw) = \frac{{Z_{out1} }}
{{Z_{out1} + R}}\frac{1}
{{1 + Rjwc}}
$
after substituting Zout1 we get:

$
\left| {H_{out} \left( {jw} \right)} \right| = \frac{1}
{{\left[ {1 - \left( {Rwc} \right)^2 } \right]^2 + \left( {3wc} \right)^2 }}

$

14. Hi,

I got struck after simplify to this eqn:

(-2.03(10^-9))w^2 - (1.026(10^-18))w^4 -1 =0

15. just substitute: x = w^2 and solve the quadratic equation...

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