# Homework Help: High Pass filter

1. Nov 12, 2008

### gimini75

whats the transfer function of this simple High Pass filtre circut:

if R = 10000 ohm

C = 220 nF

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2. Nov 12, 2008

### Staff: Mentor

Are you sure about the transfer function you wrote in the document? I get something different. Can you show the steps you went through to calculate that transfer function?

3. Nov 12, 2008

### gimini75

Sorry, what did you got? Iam not sure about the transfer function and I don't know how to find it, if you found it, please let me know?

4. Nov 12, 2008

### Staff: Mentor

Er, no. Per the PF Rules, we don't give out answers here to homework/coursework questions. We can offer tutorial help, as long as you show your work and do the bulk of the work.

How are transfer functions defined in general? If the circuit were just two resistors instead of a resistor and capacitor, what would be the transfer function? When you change the input resistor to the capacitor (like in your document), what changes in the transfer function? Show us your attempt to do that math, and we can offer hints if you are missing something.

5. Nov 12, 2008

### gimini75

I found that the Transfer functon is:

Ts = 0.0022 / (0.0022s + 1)

Is it wright?

6. Nov 12, 2008

### Staff: Mentor

No, I don't think so. Your transfer function fot the HPF in the document needs to be a function of frequency. At low frequencies, the transfer function will be much less than 1, and at high frequencies, it should get very close to 1.

What is the "s" in your equation above? Please show each step of your math. Show us the case for 2 resistors first, and then show us how the math changes when the input component goes from a resistor to a capacitor.

7. Nov 12, 2008

### gimini75

This what i have done step by step:
V(s) = R I (s)
V(s)/I(s) = R =ZR
V(s)/I(s) = 1/Cs = ZC

Transfer function = Vo/Vi = ZR / (ZR + Zc) = R / (ZR + 1/Cs)

= [R / (R + Zc)] x Zc/Zc
= RC / (RCs + 1)

R (resistance) = 10 kilo ohm
C (capacitance) = 220 nF

Therefore the transfer function for the High Pass will be:

0.0022 / (0.0022S + 1)

8. Nov 12, 2008

### Staff: Mentor

I think you have it correct until you try to move the fraction out of the denominator. Look at the fraction again, and be careful moving it out of the denominator:

$$Z(s) = Z(j\omega) = \frac{R}{R + \frac{1}{j\omega C}}$$

9. Nov 12, 2008

### gimini75

Yes but it gives the same answer as mine if u put the value of R and C, I have tried this transfer function in MATLAB but it is incorrect transfer function?

10. Nov 12, 2008

### Staff: Mentor

Hint: there is something missing from your numerator. Remember, the transfer function Z(s) = Z(jw) should go to 1 as the frequency w goes to infinity....

11. Nov 13, 2008

### gimini75

Hi

Can you please discuss the features of Low pass and High pass filters for me and suggest their application areas?

Thanks

12. Nov 13, 2008

### Staff: Mentor

That sounds like another homework question, so no, we won't give you the answer to it. Tell us your thoughts instead....

13. Nov 13, 2008

### gimini75

Thanks

I think the features of HPF and LPF are in the simpleist circuit is when Vo is applied to a HPF the high frequency can pass easily but in a LPF the low frequency can pass easily that's what I think I don't know if Iam wrong?

14. Nov 13, 2008

### Staff: Mentor

Mostly correct. The transfer funtion Z = Vo/Vi is basically the transfer from Vi --> Vo. for the HPF, low frequencies are attenuated, and high frequencies are passed with little attenuation. The LPF's behavior is just the opposite in its Vi --> Vo behavior.

Now tell us when you would typically use each one. What would be some typical applications of HPFs and LPFs?

15. Nov 14, 2008

### gimini75

I think the typical application of HPFs and LPFs is in the audio sound system?

16. Nov 14, 2008

### Staff: Mentor

That's a good example. Think of the treble and bass adjustments on your stereo...