Low pass filter

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• Feb 26th 2008, 08:48 AM
MilK
Low pass filter
With R = 1k ohm, C = 3.1827x10^-8, Find 6db cutoff freq.

How do i get the eqn?
• Feb 26th 2008, 11:16 AM
Peritus
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:

$I(t) = C\frac{{dV(t)}}
{{dt}}$
• Feb 26th 2008, 02:27 PM
MilK
If i am using voltage divider rule?
• Feb 26th 2008, 02:28 PM
MilK
R = R
C = 1/jwC
• Feb 26th 2008, 02:44 PM
Peritus
ok, but do you have to find the cutoff frequency of Vout or Vout1?
• Feb 26th 2008, 05:24 PM
MilK
Vout
• Feb 27th 2008, 02:18 AM
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}}
$
• Feb 27th 2008, 06:08 AM
MilK
hi,

your Vout is the voltage across Zout

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

are both the same?
• Feb 27th 2008, 06:57 AM
Peritus
Quote:

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.
• Feb 27th 2008, 07:27 AM
MilK
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?
• Feb 27th 2008, 09:25 AM
Peritus
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...
• Feb 27th 2008, 09:38 AM
MilK
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. (Worried)
• Feb 27th 2008, 09:50 AM
Peritus
Quote:

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 }}

$
• Feb 27th 2008, 09:03 PM
MilK
Hi,

I got struck after simplify to this eqn:

(-2.03(10^-9))w^2 - (1.026(10^-18))w^4 -1 =0
• Feb 27th 2008, 10:38 PM
Peritus
just substitute: x = w^2 and solve the quadratic equation...
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