filter order

If you exclude all other devices that remember the past, it is true. If you allow crystals etc then the statement is false.

In this 'best of all possible worlds' you are correct. (As long as you count the stray inductance and capacitance in the circuit.)

George Herold

As long as the filter is an "all pole" filter, the statement is correct. However, some filters, such as the elliptic (Cauer) also have real zeros, which require more Ls and Cs. By definition, the order of the filter is the order of the denominator polynomial of its transfer function. Consider, for example, a 3 pole elliptic filter. This filter has 3 poles (because the order of the denominator polynomial is 3) and 2 zeros. The total number of reactive components (inductors and capicitors) is 4, even though the order of the filter is 3. The same problem exists with inverse Chebyshev filters, since these filters also have real zeros, as well as complex poles, so the number of reactive components exceeds the order of these filters. For the other classical filter types (Butterworth, Chebyshev, Legendre, Bessel) the number of reactive components equals the order of the filter. Pardon my long-windeness :-) Regards, Jon

So the original statement is also correct in this case i.e. the order (3) is not higher than four

The statement you make here differs from the one I posted originally and I think it is wrong to say the order equals the number of reactive components as for some networks e.g. the compensation network used for a 10:1 probe the order goes from 1 (for a network with two capacitors) to zero as the timeconstants are made equal

Thanks for that, your reasoning helps us gain insights

So the original statement is also correct in this case i.e. the order (3) is not higher than four

The statement you make here differs from the one I posted originally and I think it is wrong to say the order equals the number of reactive components as for some networks e.g. the compensation network used for a 10:1 probe the order goes from 1 (for a network with two capacitors) to zero as the timeconstants are made equal

Thanks for that, your reasoning helps us gain insights

Really? We physics types define the order of a rational function as the sum of the orders of the numerator and denominator polynomials.

Cheers

Phil Hobbs

HUH? I've never heard that before. Maybe you've had too much exposure to the leftists ?:-)

...Jim Thompson

So a low pass and a high pass with the same skirt behavior have different order??

...Jim Thompson

For instance, you can turn a polynomial approximation into a rational function approximation by using the orthogonality properties of Chebyshev polynomials: express the polynomial as a Chebyshev sum, set it equal to the (unknown) rational function (expressed as a ratio of Chebyshev sums), multiply through by the denominator, and apply the orthogonality relation. When you use all the information you have, the rational function has the same order as the polynomial if and only if 'order' is defined as the sum of the orders of the numerator and denominator.

Cheers

Phil Hobbs

It can easily be done lower then the number of Ls and Cs, but no higher.

It is true if LCR is constant, i.e. not varying with time, voltage, etc.

VLV

Bzzzzz! wrong answer. The order of the denominator is the order of the filter.

BTW, the original statement holds true even for eliptic filters. Look at it again:

"The order of a passive filter consisting of only Ls,Rs and Cs can never be higher than the number of Ls and Cs "

If you have some transmission zeros, it is true they add extra capacitors, but that isn't the issue. Even with the extra caps, the order of the filter won't be higher than the number of Ls and Cs. Now the sum of the Ls and Cs will not be the order, but that isn't what the theorem says.

Here is another theorem of sorts. The transfer function of any point along the ladder filter will have the same denominator. I think this falls out of Mason's rule.

Not that anyone does SCF these days, but if you did a SCF leapfrog filter, you generally got a better filter if you added transmission zeros. [This assumes you have filter software to build arbitrary responses.] The zeros didn't add any more op amps, so there was no noise penalty. Generally with any active filter, the attenuation of the input signal reaches the point where the signal falls in the noise floor. Often you could reduce the order of the filter using transmission zeros.

As examples learn more than rules, could you supply a trivial example of this?

regards, Candide

That's indeed what most textbooks tell us, but the same network can have a different order depending on the values of its components, I already mentioned a simple example of this:

r1 u_i ______

-----|______|---------- | |-------u_ o |_____||________| || | c1 - -------| | ------ | | _|_ | | | | | | __|__ r2 |_| _____ c2 | | | | | | ------ | _|_

when r1c1 r2c2 this is a first order network when r1c1=r2c2 this is a zero order network

Maximum of numerator and denominator orders, no?

Best regards, Spehro Pefhany

"The order of a passive filter consisting of only Ls,Rs and Cs can never be higher than the number of Ls and Cs "

Huh?

Consider the frequency mixer down, then a filter, then a frequency mixer up. The order of the filter is doubled. Think of lowpass to bandpass or bandstop transformation in the hardware.

Vladimir Vassilevsky DSP and Mixed Signal Consultant

the original post was about passive time invariant linear systems with Rs Ls and Cs, mixers don't fit in that picture

Mixers are passive. Nothing in the original post required the system to be linear and time invariant.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

Umm.. .how do you make a mixer with ideal lumped L, R, C only?

up.

A component is linear if its response is proportional to its input (double V, and the I doubles, like for R, L, C). A component is passive if it contains no amplifiers.

The usual "passive" components of catalogs, however, are exactly the linear components. The original statement is correct, but mixers with transformers and diodes, also "passive" components, are NOT linear, and are not built with only R, L, C.