# Bandpass Filter Question
**URL:** https://boards.straightdope.com/t/bandpass-filter-question/110059
**Category:** Factual Questions
**Created:** [May 21, 2002, 7:18am UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059 "2002-05-21T07:18:32Z")
**Posts on this page:** 11
**Page:** 1
### Author:  [@Phlip](https://boards.straightdope.com/u/Phlip)
#### Post date: [May 21, 2002, 7:18am UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/1 "2002-05-21T07:18:32Z")
I am working a Bandpass filter problem in my circuits class. The circuit is very basic. It is a cap in series with an inductor and resistor in parrellel.
If you work out the transfer function for this filter you will see that there is an s^2 in the numerator and nothing else.
My question is this: According to my book the standard form for a bandpass filter is (s x beta)/(s^2 + beta x s + wo). The transfer function for this circuit does not work out to have a beta on the top and the bottom. I am thinking that the circuit was set up wrong in the problem, but I am not sure.
Must beta be in both the numerator AND the denominator in a bandpassfilter transfer function? I am getting 1/RC in the beta position on the denominator, but nothing except s^2 in the numerator.
HELP!
---
### Author:  [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)
#### Post date: [May 21, 2002, 2:01pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/2 "2002-05-21T14:01:37Z")
I just worked it out by hand and got the following:
[s\*(R/L)] / [s^2 + s\*(R/L) + 1/(LC)]
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### Author:  [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)
#### Post date: [May 21, 2002, 2:47pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/3 "2002-05-21T14:47:14Z")
BTW: The above transfer function is for a cap and inductor in series, while the resistor is in parallel with V[sub]out[/sub].
If on the other hand the cap is in series, while the resistor, inductor and V[sub]out[/sub] are all in parallel, then you’ll get:
s^2 / [s^2 + s\*(1/RC) + 1/(LC)]
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### Author:  [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)
#### Post date: [May 21, 2002, 2:57pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/4 "2002-05-21T14:57:34Z")
Sorry to keep hammering this…
In my first post I assumed the cap and inductor were in series, and the resistor was in parallel with V[sub]out[/sub]. This gives the following transfer function:
[s\*(R/L)] / [s^2 + s\*(R/L) + 1/(LC)]
This is a bandpass filter.
In my second post I assumed the cap (alone) was is “in series,” while the resistor, inductor and V[sub]out[/sub] were all in parallel. This gives the following transfer function:
s^2 / [s^2 + s\*(1/RC) + 1/(LC)]
This is a **highpass** filter, _not_ a bandpass filter.
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### Author:  [@panamajack](https://boards.straightdope.com/u/panamajack)
#### Post date: [May 21, 2002, 5:23pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/5 "2002-05-21T17:23:03Z")
Maybe you knew this, but it’s not beta you need to worry about being missing, it’s the s. If you’re trying to build a bandpass filter and you don’t have any zeros at all ( i.e. no s in the numerator) you know something’s wrong. You need both poles & zeros to ‘cancel out’ for some portion of the spectrum in order to get a band.
Of course just having the zeros doesn’t mean you have a bandpass filter, as is obvious in the highpass filter given by **CrafterMan**.
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### Author:  [@erislover](https://boards.straightdope.com/u/erislover)
#### Post date: [May 21, 2002, 8:34pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/6 "2002-05-21T20:34:13Z")
I think the capactor and the resistor/inductor are in series with Vout.
```auto
-{{-
--||--< >--o **Vout**
-ww-
```
No? This is a bandpass filter, too, with an R there to limit (half, in principle, though without the values I can’t say for sure) the inductor’s inherent resistance due to the conductor length.
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### Author:  [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)
#### Post date: [May 21, 2002, 9:58pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/7 "2002-05-21T21:58:23Z")
> [@](#):
>
> \*Originally posted by erislover \*
> \*\*I think the capactor and the resistor/inductor are in series with Vout.
>
> ```auto
>
> -{{-
> --||--< >--o **Vout**
> -ww-
>
> ```
>
> No? This is a bandpass filter, too, with an R there to limit (half, in principle, though without the values I can’t say for sure) the inductor’s inherent resistance due to the conductor length. \*\*
I’m having a hard time understanding the topology of his filter.
Is he talking about a capacitor and inductor in series; with a resistor in parallel with V[sub]out[/sub]? That’s what I assumed in my first post.
Is he talking about a capacitor in series; with an inductor, resistor, and V[sub]out[/sub] all in parallel? That’s what I assumed in my second post.
Now you seem to being describing a third possibility.
Does your drawing show that _nothing_ is in parallel with V[sub]out[/sub]? If so, then it is not a filter, bandpass or otherwise, and cannot be analyzed as such. (Unless, of course, you include a finite load impedance.)
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### Author:  [@Phlip](https://boards.straightdope.com/u/Phlip)
#### Post date: [May 21, 2002, 10:08pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/8 "2002-05-21T22:08:39Z")
> [@](#):
>
> \*Originally posted by Crafter\_Man \*
> \*\*Sorry to keep hammering this…
>
> In my first post I assumed the cap and inductor were in series, and the resistor was in parallel with V[sub]out[/sub]. This gives the following transfer function:
>
> [s\*(R/L)] / [s^2 + s\*(R/L) + 1/(LC)]
>
> This is a bandpass filter.
> In my second post I assumed the cap (alone) was is “in series,” while the resistor, inductor and V[sub]out[/sub] were all in parallel. This gives the following transfer function:
>
> s^2 / [s^2 + s\*(1/RC) + 1/(LC)]
>
> This is a **highpass** filter, _not_ a bandpass filter. \*\*
Ok, I am confused as hell now. I got the exact same transfer function that you have above. s^2/[s^2+s blah blah
How is this a high pass filter? A high passfilter looks like such:
s/s+wo correct? What am I supposed to do with an s^2 term. On top of that, I have a quadratic in the denom that is not a perfect square. I have not seen a high pass that has a quadratic in the denom?!?
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### Author:  [@erislover](https://boards.straightdope.com/u/erislover)
#### Post date: [May 21, 2002, 10:34pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/9 "2002-05-21T22:34:24Z")
We’re just not clear on what the circuit looks like here, **Phlip**. Can you tell us which one exactly?
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### Author:  [@Chris\_Luongo](https://boards.straightdope.com/u/Chris_Luongo)
#### Post date: [May 21, 2002, 10:45pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/10 "2002-05-21T22:45:41Z")
I’d suggest that you visit the [Troubleshooting Forum](http://www.carsound.com/cgi-bin/ubbcgi/ultimatebb.cgi?ubb=forum&f=2&submit=Go) at [www.carsound.com](http://www.carsound.com). The moderator there, Dave Navone, is a physicist, and very smart; he’ll know.
---
### Author:  [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)
#### Post date: [May 21, 2002, 11:18pm UTC](https://boards.straightdope.com/t/bandpass-filter-question/110059/11 "2002-05-21T23:18:33Z")
> [@](#):
>
> \*Originally posted by Phlip \*
> \*\*
>
> Ok, I am confused as hell now. I got the exact same transfer function that you have above. s^2/[s^2+s blah blah
>
> How is this a high pass filter? A high passfilter looks like such:
>
> s/s+wo correct? What am I supposed to do with an s^2 term. On top of that, I have a quadratic in the denom that is not a perfect square. I have not seen a high pass that has a quadratic in the denom?!? \*\*
Welcome to the world of EE, **Phlip**. And _relax_; after we get this solved I’ll buy you a few pints of Guinness, O.K.?😉
As **s** goes to 0 the transfer function **H(s)** also goes to 0, and as **s** goes to infinity **H(s)** goes to 1.
Perhaps you’re confused because you’re not taking the limit correctly.
To take the limit, divide the numerator by **s** [sup]2[/sup] _and_ divide the denominator by **s** [sup]2[/sup]. **H(s)** will then be:
1 / [1 + 1/( **s** _R_C) + 1/(L_C_ **s** [sup]2[/sup])]
As **s** goes to 0 the second and third terms in the denominator explode toward infinity. This means the entire denominator goes toward infinity, and thus **H(s)** goes to 0.
As **s** goes to infinity the second and third terms in the denominator simply go to 0. This means **H(s)** = 1/1 = 1.
Do you believe it’s a (2[sup]nd[/sup] order) highpass filter now?
As far as it “not being a perfect square,” is there someone who says it has to be?
You also asked “what to do with the quadratic” in the numerator. All I can say is that it’s part of the transfer function.
Can I ask what you’re trying to solve for? Are you doing a Bode plot? Do you need to find the –3dB frequency? Are you trying to plot the magnitude over frequency? Or perhaps the phase?
Let me know…