# Second order filter - quadratic formula and standard normalized form

#### Jonathan56

Jan 3, 2017
3
Hi,

I have a question regarding corner frequency for second order filter.

I don't understand how I can have two roots in the denominator which will indicate two corner frequencies each of them at different frequency and with 20db/decade and the same transfer function express in standard normalized form (express with the quality factor) indicate only one corner frequency (power of two <=> 40db/decade).
2 corner frequency or one?

Example with an LCR filter - two-pole low-pass filter
s = jw and w = 2 x pi x F
G(s) = 1 / (1 + sL/R + s²LC) = 1 / (1 + sA + s²B) with A = L/R and B = LC
G(s) = 1 / [ ( 1 - s/s1 ) (1 - s/s2) ]
with s1/2 = -A/(2B) x [1 -/+ rsqt(1 - 4B/A)]

==> This indicates that there is two corner frequencies

standard normalized form
G(s) = 1 / [1 + s(Qw0) + (s/w0)²]

==> This indicates that there is only one corner frequency

The example above is taken from the book:
Fundamentals of Power Electronics SECOND EDITION
Robert W. Erickson and Dragan Maksimovic
page 282

Jonathan

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,828
This is not really a contradiction.
You can build a filter where the two corner frequencies are different, one for each order of the filter.
You can also build a filter where the two corner frequencies are identical, looking like a single corner frequency.

Think of a second order filter as a series connection of two first order filters where you can define the corner frequency of each filter individually.

#### Jonathan56

Jan 3, 2017
3

I am talking about one same filter. LRC (L in serial with RC in parallel) Vout taken at RC

according the standard normalized form
Corner frequency = 1/ [2 x pi x sqrt (LC)] and Q = R x sqrt (C/L)
according the factorized form with the two roots (using the quadratic formula)
Corner frequency 1 = 1/(4 x pi x RLC) x [1 - sqrt(1 - 4R²C/L)]
Corner frequency 2 = 1/(4 x pi x RLC) x [1 + sqrt(1 - 4R²C/L)]

What is the corner frequency of the LRC filter?

#### LvW

Apr 12, 2014
604
Jonathan - why do you think that you would have two corner frequencies?
Just because you have found two roots?
Thats not logical.
Corner frequencies are defined for the magnitude response of a second order filter.
Now - you have to distinguish between two cases:

(a) Two REAL roots (poles of the function): In this case, you can (can !!) realize the filter with two first-order sections having two different corner frequencies. However, also a realization in one 2nd-order filter block is possible.

(b) When you decrease the degree of damping for the filter (smaller resistors) the two real poles approach each other until they form a double pole (Quality factor 0.5). If you further decrease damping the poles split up and form a conjugate-complex pole pair (equal real part and imag. part with different sign).
This gives a second-order filter with rather good selectivity (Butterworth or Chebyshev response) and a pole quality factor Qp>0.5. Such a filter has only one single cut-off frequency (but a pair of conjugate-complex poles).

#### Ratch

Mar 10, 2013
1,099
Jonathan - why do you think that you would have two corner frequencies?
Just because you have found two roots?
Thats not logical.
Corner frequencies are defined for the magnitude response of a second order filter.
Now - you have to distinguish between two cases:

(a) Two REAL roots (poles of the function): In this case, you can (can !!) realize the filter with two first-order sections having two different corner frequencies. However, also a realization in one 2nd-order filter block is possible.

(b) When you decrease the degree of damping for the filter (smaller resistors) the two real poles approach each other until they form a double pole (Quality factor 0.5). If you further decrease damping the poles split up and form a conjugate-complex pole pair (equal real part and imag. part with different sign).
This gives a second-order filter with rather good selectivity (Butterworth or Chebyshev response) and a pole quality factor Qp>0.5. Such a filter has only one single cut-off frequency (but a pair of conjugate-complex poles).

What makes you think that the "standard normalized form" has only one frequency?

Ratch

#### LvW

Apr 12, 2014
604
What makes you think that the "standard normalized form" has only one frequency?
Ratch - is this question directed to me?
Are you saying that a second-order function (always) has TWO corner frequencies?
It has (for Qp>0.5) two poles (conjugate-complex) and ONE SINGLE corner frequeny.

#### Ratch

Mar 10, 2013
1,099
Ratch - is this question directed to me?
Are you saying that a second-order function (always) has TWO corner frequencies?
It has (for Qp>0.5) two poles (conjugate-complex) and ONE SINGLE corner frequeny.

Sorry, I meant for the reply to go to Jonathon56. I am saying that a quadratic equation has two solutions, not one as he stated.

Ratch

#### Jonathan56

Jan 3, 2017
3
Hi LvW and Ratch.

Thanks for your answers it makes me realise my mistake in my analyse.
To answer to the question of Ratch, indeed you are right, I was just to focus on the assunptotic behaviour to sketch the magnitude bode plot (but without the magnitude of the transfert function...).
I guess I was trying to take to much shortcut in my analyse. Lesson learned!

Thanks.

#### robertrichardr

Dec 2, 2022
1
An algebraic expression to the power of 2 is called a square formula. The general form of a quadratic formula calculator is ax2 + bx + c = 0, where a, b and c are also real numbers, also called "numerical coefficients," and ≠ 0. Below, x is the unknown variable we need to find the solution. We know that the quadratic formula is used to find the solution (or root) of the quadratic formula ax2 + bx + c = 0.

Moderator
Nov 17, 2011
13,828

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