# poles and zeros of Laplace transform

Discussion in 'Mathematics and Physics' started by PG1995, Mar 23, 2013.

1. ### PG1995Active Member

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Hi

Q1:
For a system to be stable all of its poles should lie to the left of imaginary axis of the s-plane; even if one pole lies to the right of the s-plane it would make the system unstable. If poles lie on the imaginary axis then the system is marginally stable. This is the criteria which is given in most places. But I have also been told that for a stable system the number of zeros and poles should be equal. I have found some sources which indirectly also support this. For instance, check out this source:

But the issue is that I have seen many functions which don't have equal number of zeros and poles but still are considered stable (please don't forget that I'm just a newbie to the Laplace). For example, s/(s+2)(s+3) has poles at s=-2, -3 and zero at s=0. It's considered stable in spite of the fact that number of poles doesn't equal number of zeros in this case. You can also check this source. This another source summarizes the stability this way.

The significance of zeros is explained after 10:25 in link #1 below.

Q2:
How do I interpret poles and zeros physically in context of circuits? I understand that mathematically poles are roots of denominator and zeros are roots of numerator of a transfer function, H(s) = N(s)/D(s) = Y(s)/X(s). My complete question is here.

Thanks a lot.

Regards
PG

1: http://www.youtube.com/watch?v=5jYr0QktWxE (very good for understanding poles and zeroes)
2: http://www.youtube.com/watch?v=ZGPtPkTft8g (laplace transform, good one)
3: http://www.youtube.com/watch?v=UFPhhS3MCiM (feedback control example, okay)
4: http://www.youtube.com/watch?v=KFbUr5Vy_cc (poles zeroes final value theorem, okay)
5: http://www.youtube.com/watch?v=CgQBfjD-4uk (stability and pole location, very good one)
6: http://www.youtube.com/watch?v=cQdIVwKqj2M (poles and zeroes, good one)
7: http://www.youtube.com/watch?v=pSN7t79RxC4 (MIT lecture, laplace transform, good one)
9: http://www.electro-tech-online.com/custompdfs/2013/03/PoleZero.pdf
10: http://en.wikipedia.org/wiki/Pole–zero_plot
11: http://en.wikipedia.org/wiki/Pole_(complex_analysis)

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2. ### steveBWell-Known MemberMost Helpful Member

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Q1: Stable systems have poles only in the left half plane - end of story.

The idea of number of poles and zeros being equal comes about if you count poles and zeros at infinity, which is a bit of a confusing (even questionable) notion. Your thought that you've seen systems with more poles (e.g. s/((s+2)(s+3)) ) comes about because we typically do not consider zeros at infinity. Let's forget about the fact that s=∞ is not a well defined point because s is complex. Well, the limit of s/((s+2)(s+3)) as s -> ∞ is 0. So, in a sense this function has a zero at s=∞ and if we count this as one of the zeros, this system has 2 zeros and 2 poles.

It's one of those mumbo-jumbo things that does not really matter.

Q2: is a difficult question to answer actually.

The significance of poles and zeros can depend on the system, although zeros relate to inputs, while poles relate to internal feedback. A zero blocks the path of an input signal basically. This is why only poles matter for stability. I'm sure those words mean little to you now, but in time it should make sense. The simplest answer says that poles are places where the transfer function blows up and zeros are places where the transfer function is zero. However, that's a cop out answer because it is too obvious and not very helpful for intuition. However, with experience, the pole-zero language is something you understand even when it's hard to explain it to others. So, don't worry too much about this. Just use the concepts and be patient that it will all make sense with experience.

However, here is a crude attempt to answer. The most significance can be seen in Bode plots because poles make the magnitude response decrease at 20 dB/dec and zeros make the response increase at 20 dB/dec and the effects are cumulative such that a pole and zero cancel and two poles or two zeros gives 40 dB/dec. The angle response picks up 90 lag for every pole an 90 lead for every pole. If you are not familiar with this, then wait till you start doing Bode plots and you will see it clearly.

The bottom line is that pole and zero frequencies in the complex domain give effects that we can see in the real frequency domain, which we typically display on a Bode plot. In a sense, the effect of the poles and zeros are indirectly manifesting themselves like shadows or reflections (speaking poetically now). Only when poles or zeros are actually on the jw line (i.e σ=0) do we directly see the pole or zero on the Bode plot, and then it is truly a zero response or a blow-up point, at that frequency w.

Last edited: Mar 24, 2013
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3. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

As Steve mentioned, the transfer function s/((s+2)*(s+3)) has another zero at s=infinity. This is easy to see if we look at only the powers of s. In the numerator we have s to the first power, and in the denominator we have s to the second power because we end up with at least s^2 in the denominator. This reduces it to:
s/s^2
which is clearly just 1/s, and as s goes to infinity 1/s goes to zero, so we get another zero.

That's the short view, and the right way to do this is to take the derivative of the top and the derivative of the bottom and then we get:
1/(2*s+5)

and then take the limit as s goes to infinity. As s goes to infinity (in the denominator) we see the '5' becomes insignificant, so we are left with 1/infinity which is zero.

Just thought i would point that out as you will run into this many times.

If you look at the time equation at t=0 you'll see it is equal to 1, and stable around t=0.

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5. ### PG1995Active Member

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Thank you very much, Steve, MrAl.

@Steve: I'm sorry about the Q2. I linked wrong attachment where I said "My complete question is here" at the end. Please have a look again and see if it's somewhat easier for you to help me to understand some meaning of poles and zeros in context of that example or some other related example.

@MrAl:
I think you are referring to L'Hopital's rule. Thanks.

Regards
PG

6. ### steveBWell-Known MemberMost Helpful Member

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I would still give the same answer for Q2, even with the more specific information in hand.

The one thing we can do now is apply the Bode plot rules to the specific system. This is a good example to use Matlab to speed up the learning process.

In Matlab, try the following command,

sys=tf([4 16 0],[2 12 1])

This creates a system called "sys" using the "tf" command which is the transfer function command. You can use the Matlab help command "help tf" to get the details about this command, but basically you are entering the polynomial coefficients for the numerator and denominator.

Once the system is defined try the following commands to get the poles and zeros of the system,

pole(sys)
zero(sys)

You will find the zeros are at 0 and -4, which you already knew, and the poles are at -5.92 -0.085. What we see here is that the pole at -5.92 is relatively close to the zero at -4, which means their effects cancel out somewhat. Hence a bode plot will look similar to a system with a zero at 0 and a pole at -0.085. This means that the phase will start out at +90 deg. and the magnitude response will increase at 20 dB/decade. Then at the pole frequency w=0.085, the response will flatten out and the phase will come back to zero at higher frequency.

Use the "bode" command i Matlab to get the exact plot as follows,

bode(sys)

I've attached the plot, but you should do this yourself to verify.

With experience, you can do approximate Bode plots just by knowing the locations of the polse and zeros. So, they tell you a lot about the frequency behavior of a circuit/system.

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7. ### PG1995Active Member

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Thanks a lot, Steve.

As you can see there is a typo so I thought I better confirm it. I think it would be "The angle response picks up 90 lag for every pole and 90 lead for every zero". Kindly let me know. Thank you.

Regards
PG

Last edited: Mar 24, 2013
8. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Just dont get too comfortable with that because for many second order systems you'll need a template of the curve
It still helps to understand this stuff though.

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9. ### steveBWell-Known MemberMost Helpful Member

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Yes, that is correct.

As MrAl is cautioning, these Bode plot rules are meant to indicate tendencies for the Bode plot, not the actual response. If poles and zeros are very far apart, you will see the (nearly) full angle develop in the plot, and with the magnitude rules, you will see the full 20dB/dec slope clearly.

When a pole and zero are exactly equal, they will perfectly cancel, but if they are merely close to each other, they will not perfectly cancel and you will see some small differences. Also, when two poles are very near (or equal, or complex conjugates of each other) the behavior is modified and the details (which relates to second order system properties: oscillatory, damped or critically damped) matter.

These rules (developed by Bode) were very important in the era before computers (or calculators) were readily available to engineers. They needed to make plots quickly for design and the rules (there are more rules by the way) allowed making quick plots, without having to manually find points with a slide rule. Those guys were fast with the slide rule, but doing complex math calculations for many points to make a plot is tedious and time consuming.

However, we are very lucky to have Matlab as a design tool to do the heavy labor for us, and it is critical for doing effective modern control design. So, make Matlab your very good friend as you learn this stuff.

Last edited: Mar 25, 2013
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10. ### PG1995Active Member

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Hi

In Laplace transform s=σ+jω where "σ" is a real number. When we say that s-->∞ or s=∞, what does it really mean? As you also implied that s=∞ is not a well defined point because s is complex, I also see it this way. In my view, we can't say that s=∞ unless "jω" is zero which then means that σ=∞. Please guide me here. Thank you.

Regards
PG

Last edited: Apr 30, 2013
11. ### steveBWell-Known MemberMost Helpful Member

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In my opinion, it does not have a precise meaning because it is vague. However, for our purposes, it is not really a critical problem. If either σ or ω go to infinity, those typical transfer functions go to zero. So, in a sense, there is a zero at infinity.

I'm not a mathematician, but I like to visualize this as a large infinite diameter circle, and there is a zero at all points on this circle. Yet still we are calling this one zero, as if it is one point. It's a bit of a strange concept, but it works well enough.

Honestly, this point from the book is not a critical idea at all in practical applications, but it is good to be exposed to it at least in a cursory way. I also learned this back in school, but I haven't thought about it in over 20 years until you brought up the question. This is one of those pretty looking fruits that you can only squeeze a tiny amount of juice out of, ... the juice being a metaphor for practical value, and the prettiness hinting at the conceptual symmetry implied by the mathematics.

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12. ### PG1995Active Member

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Thank you.

I have tried to understand your visualization of an infinite circle. Here is my attempt. Kindly guide me if it's too wrong. I understand you were talking about it generally but I thought of exploring it further out of curiosity. Thanks a lot.

Regards
PG

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13. ### steveBWell-Known MemberMost Helpful Member

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These are interesting thoughts. I think you are seeing why I mentioned this is a vague concept, especially the way it is presented in engineering books. I don't think mathematicians would like our approach.

You mentioned jω going to infinity, but instead think of ω going to infinity. This has a clearer meaning. The j is there in s=σ+jω, but σ and ω are the variables that would actually be set to a number, or be varied in a limiting process.

I like how you used the polar form s=r exp(jθ). This is useful to say that r goes to infinity. There is no real circle, as you seem to have pointed out. It's just a mental concept to think of the zero as if it exists at the infinite perimeter of the two dimensional space.

Again, I don't see a lot of usefulness in this. Perhaps the one place where you might care about this is when doing Bode plots. Then the concept of a limit as ω goes to infinity has a usefulness if you start at ω=∞ and work back to ω=0. However, I think most people prefer to start at ω=0 and think of how poles and zero's affect the magnitude and phase response. Poles and zeros at ω=0 are much easier to think about. Then, you never need to worry about those zeros (or even poles) at infinity. They are just out there to balance the scales, but otherwise, who cares?

Last edited: May 3, 2013
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14. ### PG1995Active Member

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bode plots, poles and zeros

Hi

Regards
PG

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15. ### steveBWell-Known MemberMost Helpful Member

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It's not clear why you think that. The math says the phase is zero at w=0, and then it goes to -90 at w=infinity.

To see this, just write out the voltage divider equation to find the transfer function. Remember that Xc=1/(jwC). Including the j is important. Then take the limit as w goes to zero and then again as w goes to infinity.

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16. ### PG1995Active Member

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Thank you.

Yes, the math says so and I'm not saying math is incorrect. But in some cases conceptual understanding should take precedent. If I can't make 'conceptual' sense of phase response of a RC circuit then I'm totally hopeless when it comes to more complex circuits. So, please help me. If it's still not clear to you where I'm having trouble then kindly let me know. Thank you.

Regards
PG

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17. ### steveBWell-Known MemberMost Helpful Member

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I think I see where you are getting confused.

The diagram you show in the last post is for a transfer function T=V/I, which is also the total impedance Z of the circuit. Here you would be correct that the phase is -90 at ω=0 and 0 at ω=∞.

But, this is not what your original question is referring to. The original question was for transfer function T=Vo/Vi. This is a different situation. Instead of comparing the phase of the voltage and current on the supply, we are comparing the phase between the supply voltage and the load voltage.

In this case, ω=0 means the supply is directly on the load and no drop is across the resistor. Well, if it's the same voltage, then obviously they are in phase. As a byproduct, the current is 90 degrees out of phase with the supply voltage because capacitors always have a 90 degree lag on current.

However as ω goes to ∞, most of the voltage goes on the resistor; hence the current is in phase with the supply. The capacitor is in series and has the same current, and the capacitor always has voltage and current out of phase by 90 degrees. Hence, when ω goes to infinity, the supply and the load voltages must be 90 degrees out of phase.

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18. ### PG1995Active Member

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Thank you, Steve.

I think I understand it now. I believe it will do even if there are some loopholes in my understanding at the moment.

Regards
PG

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19. ### steveBWell-Known MemberMost Helpful Member

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There are some misconceptions in your description. For example, it is not correct that a capacitor is an open circuit and that as soon as the switch is thrown Vc=Vi. A capacitor looks like an open circuit to DC, only a long time after the switch is thrown. When the switch is first thrown, the capacitor looks like a short circuit, and current flows as the capacitor is charged up.

You are correct to stress the point, "assume that the source is sinusoidal". That is a key point. But, it's important to remember that a sinusoidal signal must exist for a long period of time, so that transient effects can die off. Throwing a switch is a transient act, so here you don't need to think about what happens right when the switch is thrown. However, a long time after the switch is thrown, your way of thinking is probably good enough to get the point.

Another way to think about it is to note that a very slow sine wave gives the capacitor sufficient time to charge and discharge in order that the cap voltage can follow the input source voltage. This creates a very small phase angle.

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20. ### PG1995Active Member

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Thank you very much for the corrections.

Could you please also help me with these queries? Thanks.

Regards
PG

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21. ### steveBWell-Known MemberMost Helpful Member

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PG,

The "s" in question is indeed the normal s=σ+jω Laplace complex frequency parameter. The whole reason for considering s=jω is that this is how we obtain our Bode plots for the system. Whenever, you see Bode plots with magnitude and phase, they are obtained from the transfer function by substituting jω in place of s. The values are complex, so we need both magnitude and phase to represent the number in polar form.

Now, why do we care about jω? Basically, because this restricts the input functions to pure sinusoidal functions with frequency ω.

In the case you mentioned, you had a two poles at -1 and a zero at s=-2. These poles and this zero are not on the jω axis, so you will not see the bode plot blow up (poles blow up, right?) at ω=-1 and it won't go to zero at -2 (zeros are zero, right?). So, there is generally no need to worry about comparing pole/zero locations and s=jω substitutions.

Sometimes poles or zeros end up on the jw axis, and then you will see them on the Bode plots. For example, your zero at 0 will result in a Bode plot that goes to zero at DC. If you had an example with poles on the imaginary axis, then the Bode plot would go to infinity at the frequency corresponding to the pole.

Last edited: Jul 15, 2013
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