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- Thread starter gimini75
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- #2

berkeman

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

if R = 10000 ohm

C = 220 nF

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?

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- #4

berkeman

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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.

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I found that the Transfer functon is:

Ts = 0.0022 / (0.0022s + 1)

Is it wright?

Ts = 0.0022 / (0.0022s + 1)

Is it wright?

- #6

berkeman

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I found that the Transfer functon is:

Ts = 0.0022 / (0.0022s + 1)

Is it wright?

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

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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

berkeman

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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)

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:

[tex]Z(s) = Z(j\omega) = \frac{R}{R + \frac{1}{j\omega C}}[/tex]

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- #10

berkeman

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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....

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

Thanks

- #12

berkeman

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

Thanks

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

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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

berkeman

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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?

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?

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

- #16

berkeman

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

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

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