# 2nd order system

Status
Not open for further replies.

Hi

Regards
PG

#### Attachments

• 372.6 KB Views: 107

#### MrAl

##### Well-Known Member
Hi there PG,

To compute the poles you set the denominator equal to zero, then solve for the roots.

If you end up with complex roots then you have complex poles.

My question to you is:
Did you ever find a complex solution to a quadratic equation?
Did you ever plot a pole zero constellation?

#### Winterstone

##### Banned
PG1995,

A denominator D(s) of second order can always be written in the following form:

D(s)=K*(s-z1)*(s-z2) where z1 and z2 are the zeros (real or complex) of D(s), identical to the poles of the transfer function.

Of course, the poles can be complex - however a stable system requires that the real part of the poles is always negative (left hand part of the complex s-plane).

#### steveB

##### Well-Known Member
Q1: Poles can be complex because they exist in the complex frequency s-plane that we have talked about in other threads. Complex poles, when they exist in a system, are always the result of a quadratic that occurs in the denominator of the transfer function. Real systems, that have complex poles, always have complex roots as complex conjugate pairs. This directly results from the form of the roots/poles in your Q2.

Q2: This form is a direct result of your normal formula for finding roots of a quadratic. From your algebra, you know that ax^2+bx+c has solution (-b±sqrt(b^2-4ac))/2a. This can also be written as -b/2a±sqrt((b/2a)^2-c/a). Then, as Winterstone mentions, these become z1 and z2 in D(s)=K*(s-z1)*(s-z2).

PG1995,

#### PG1995

##### Active Member
Hi WinterStone

In general, Steve and MrAl are very patient persons or perhaps I'm just lucky! I have seen a lot of members here and on other forums who, in my opinion, don't even try to put forward their queries properly. For example, going through the replies by Steve and MrAl in the given threads substantiate my point. I think both Steve and MrAl would agree with me that I try to facilitate my helper as much as I can and try to make query as simple as possible.

My sincerest apologies to you and MrAl if you thought I wasn't being thankful and not valuing your time. I have seen instructors who in spite of being paid aren't helpful and don't really try to help the way they are supposed to but you people help out of goodwill and practically it actually costs you to help others.

Best regards
PG

#### PG1995

##### Active Member
Hi

Re: Q1
I understand it now. So far, quite strangely I have never used complex poles (or, perhaps I'm just wrong!) therefore this wrongly led to believe me otherwise. You are right that poles can be complex because only requirement for stability is that the poles should lie in left half plane.

Re: Q2
I believe I understand it too.

Thank you.

Regards
PG

#### Attachments

• 162.6 KB Views: 97
Status
Not open for further replies.