I'm studying for class and noticed something I'm not sure about. A simple series BP filter with a capacitor, an inductor, and a resistor. Is that a first-order or a second order filter, because from what I've read it has to do with the number of reactive components, which means it's a second-order filter, but when I simulated the circuit it had the roll-off of a first-order filter.

Whats the deal?

Thanks.

 

It all depends on the configuration. Can you attach a screenshot?

 

It all depends on the configuration. Can you attach a screenshot?

I found a pic in google images: https://goo.gl/images/3bz6FM

 

Basically because the transfer function is a second order function, the filter is a second order BPF.

 

View attachment 169381
Basically because the transfer function is a second order function, the filter is a second order BPF.

Then how come it's roll-off is that of a first-order? That being 20db/decade.

 

Also, what does the "s" denote in your drawing?

 

The order of a filter is determind by the order of the denominator of the transfer function.
Nevertheless, each 2nd-order bandpass has a first-order roll-off of 20dB/dec only.

 

Thanks. My professor is going to have to explain this to me.

 

Then how come it's roll-off is that of a first-order? That being 20db/decade.

Here's the intuition: we can think of a 2nd-order BPF as a 1st-order LPF followed by a 1st-order HPF in series. Each 1st-order filter has a 20 dB/decade roll-off, but since they act "in different directions" -- the LPF attenuates the low frequencies, while the HPF attenuates the high frequencies -- we don't get the expected 40 dB / decade roll-off of a 2nd-order filter.

 

Here's the intuition: we can think of a 2nd-order BPF as a 1st-order LPF followed by a 1st-order HPF in series. Each 1st-order filter has a 20 dB/decade roll-off, but since they act "in different directions" -- the LPF attenuates the low frequencies, while the HPF attenuates the high frequencies -- we don't get the expected 40 dB / decade roll-off of a 2nd-order filter.

Ahhh. I was close to guessing the reason. I thought the 20db per side was the 40db roll-off. Thanks for the explanation.

 

Ahhh. I was close to guessing the reason. I thought the 20db per side was the 40db roll-off. Thanks for the explanation.

Here is another -simple - explanation:

* For low frequencies (low s-values) , the transfer functions 1st-order numerator (+20dB/dec) governs (because the denominator is app. constant) and
* For high frequencies we have the sum of the 2-nd order denominator negative slope (-40db/dec) plus the numerators positive slope (+20dB(dec)

EDIT: Transfer function (2nd-order bandpass)

H(s)=s*T*wo²/(wo² + s*wo/Q + s²)

 

Hereis anotther -simple - explanation:

* For low frequencies the transfer functions 1st-order numerator (+20dB/dec) governs, and
* For high frequencies we have the sum of the 2-nd order denominator negative slope (-40db/dec) plus the numerators positive slope (+20dB(dec)-

I'm gonna have to look at this later after I've gotten some sleep. Thanks for the explanation.

 

Also, what does the "s" denote in your drawing?

The "s" is the standard Laplace Transform variable which represents any complex frequency. The reason we use the Laplace Transfor is to solve linear differential equations with algebra, which is among the simplest of all manipulation methods. Sadly it does not work for non-linear differential equations. The graph is misleading because it suggests that "s" is a real quantity. In fact it belongs to the set of Complex numbers.

 

s=jw

 

s=jw

It is also possible for s to have a non-zero real part -- is it not?

 

Thanks for the replies. I was never taught that. My professor probably wont get into that with me. I'll have to look into that on my own.

 

Really, the circuit with the inductance and capacitance in series as shown is a series resonant circuit, with the L and C reactances cancelling at some frequency, leaving only the resistance part of the reactance to limit the current. The detailed analysis does not appear in my head right now at 4AM, my local time.

 

Thanks for the replies. I was never taught that. My professor probably wont get into that with me. I'll have to look into that on my own.

In case it has not been said, s-domain analysis technique is introduced in more advanced courses. But it is so easy to use that everyone who had it, go straight to it.

 

Also, what does the "s" denote in your drawing?

The 's' denotes a complex variable. In analog circuits the default is 't' with an 's' LaPlace transform and in discrete time it's 'n' instead of 's' for the 'z' transform.

 

s=jw

s is a complex number and = σ +jω where σ is the transient term and jω is the steady-state frequency term.
We often ignore the transient term since only the steady-state frequency response is of interest, so we substitute jω for s in the Laplace transform.
Thus you often see only jω in an analog filter Laplace transfer function.