# 2nd order low pass fillter

Discussion in 'Homework Help' started by Jaywin, Apr 22, 2008.

1. ### Jaywin Thread Starter New Member

Apr 20, 2008
9
0
For the 2nd order low pass filter
is the input impedence calculate like this:
Zin = 1/jwC1 + R2||(R2+1/jwC2)
if freq-> inf then the Zin is just R1||R2.
Am i correct? That is wave generated using Pspice.^^

2. ### Caveman Senior Member

Apr 15, 2008
471
1
Actually as f->infinity. Zin = R1. Assuming perfect opamp of course.

3. ### Caveman Senior Member

Apr 15, 2008
471
1
Otherwise, you just analyze the circuit to get the voltage at the node between R1 and R2, Va.

Zin = (Vin-Va)/R1.

There might be an easier way, but I don't know it.

4. ### Jaywin Thread Starter New Member

Apr 20, 2008
9
0
Zin = (Vin-Va)/R1?

(Vin-Va)/R1 isn't we wil obtain current? by the way, Va is same as Vc1 right?
Actually, the problem is to fulfill the specification which our
Zin >= 10kOhm
So if i put the R1 = 10kohm is also satisfy the specification right?

Zin = 1/jwC1 + R2||(R2+1/jwC2) is wrong?

5. ### Caveman Senior Member

Apr 15, 2008
471
1
Whoops,

Zin = R * Vin/(Vin-Va)

But yeah, R1 =10k should solve it, I think.
I frankly haven't had a chance to check your Zin since it is fairly complex to quickly calculate. I might look at it tonight.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,720
496
Is the input impedance your only specification? Don't you also have to meet a frequency response spec?

Go back to the Spice plot of Zin in your first post, and look carefully at the region between 3000 Hz and 4000 Hz. This may be a trick question.

If you make R1 = 10k, you also might want to make R2 = 10k, and make C1 and C2 bigger to meet the frequency response spec.

But if you do that, plot Zin at about 7000 Hz and look carefully at its behavior.

I think Zin has a second order numerator.

7. ### Jaywin Thread Starter New Member

Apr 20, 2008
9
0
The cutoff frequency is given f0 = 1kHz
If change the R then C also have to change also using f0 = 1/2paiRC
which RC = 159.15us right?
Why have to observe certian frequency to know the Zin whether meet the
>= 10k specificatin?
Is it just look at the red-circled region if it is >=10k then we meet the specification. Am i right?

R1=R2=10k

R1= R2=20k

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,720
496
Adjust your plot so that the maximum vertical axis magnitude is 25k instead of 300k or 500k.

9. ### Caveman Senior Member

Apr 15, 2008
471
1
Man, this is nasty, but here goes:

Zin = R1/(1-H*(1+jwC2*R2)/G)

In this case H is the transfer function of the 2-pole filter. Look at wikipedia.
G is the DC gain = (3.5k + 6k)/3.5k.

That's as far as I got.