# Really Simple Control Theory Question?

At least, I'm sure it'll be simple to some of the folks on here. I am looking at pole-zero diagrams: I am in particular looking
at the pole-zero diagram for a first order low pass RC filter.
It's open loop, unloaded, so obtaining the transfer function is trivial,
T.F. = (1/CR)/ ((1/CR) +S)
Nothing controversial there. I understand S is the complex variable,
S = (sigma + jw)
which is all to do with Laplace and stuff. We plot the T.F. on the complex s-plane and obtain a single pole at sigma = -(1/CR): the zero is at infinity.
My problem is what physical interpretation can we put on this pole? I understand that the T.F. along the jw axis is a back-to-back Bode plot, i.e. response versus frequency, and I understand that the sigma (real) axis is an indicator of exponential growth or decay, i.e exp-(sigma*t).
I cannot for the life of me though figure out how the transfer function ( Vout/Vin ) of an RC low-pass filter develops an infinite gain ( pole ) at what is zero-frequency on the -sigma axis ( for jw=0 ). Not even negative resistances explains the shape of the graph to me. Anyone have a physical interpretation as to the meaning of this pole at sigma = -(1/CR)?????
Apologies if this is not the sort of stuff to post here, I am unfamiliar with this group,
Andy.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
andrewpreece wrote:

trivial,
complex
plot,
(real)
exp-(sigma*t).
function
pole )

at
unfamiliar
It's a good question actually! For this case, you should think of frequency not in the usual sinusoidal sense, but in terms of the "frequency" of an exponential signal. If you give the system an input of the form exp(-at) where a is at the same radian frwquency of the filter pole, see what happens to the reponse.
Fred.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
andrewpreece wrote:

Maybe I have not understood the question but the pole is located (using your notation) at S=-(1/CR), not at sigma=-(1/CR) and the frequency response of your filter is obtained with sigma=0 in the T.F. (1/CR)/ ((1/CR) +S)
josu
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

complex
No, since S is a complex number ( sigma +jw ), if the pole is at S=-(1/CR), then since 1/CR is real then so must S be: jw is not real so the pole must be on the real axis, the sigma axis. This is not guesswork by me anyway, look in any text on pole/zero diagrams showing a first order RC lowpass filter and you will see the pole is on the real (sigma) axis. The frequency response along the jw axis does not show a pole, and anyway I understand the shape of the transfer function along the jw axis. It is the shape along the sigma axis ( which does not correspond with frequency, rather some exponential property ) that baffles me. The previous poster actually suggested an interesting line of thought to follow, though I've not satisfied myself yet that I understand the transfer function profile along the real ( sigma ) axis. I need to get thinking!
Andy.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
andrewpreece wrote:

If you derive the frequency response without Laplace transforms, simply solving the differential equation, you will find the response to be
1 1 ------- , where T, the time constant, = ---. 1 + jwT RC
If you make the same substitution -- T = 1/RC -- and multiply numerator and denominator of your expression by T, you will get
1 ------ . The similarity of form is not coincidental. 1 + sT
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
andrewpreece wrote:

I believe there are 2 different constructs here - one is the steady-state response to a sinusoid which can be formally obtained by evaluating the transfer function H(s) at s = jw.
The other construct is the "transfer function" H(s) which is the result of utilizing the Laplace Transform. By definition in order for the Laplace transform to converge, sigma must be > (in your case) the real pole (region of convergence stuff, etc). H(s) is not "analytic" at its pole.
At least, that's what I remember from 40 years ago.
Harvey
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

trivial,
pole )

unfamiliar
S=-(1/CR),
the
along
pole.
I don't disagree with some of the comments made, but none seem to answer my basic question: what is the physical meaning of a pole on the real (sigma ) axis, i.e when jw=0. Poles in themselves are nothing to be afraid of - I understand an integrator can be represented by a pole at the origin, understandable because with jw=0, i.e. a dc input, the output voltage climbs away to infinity, hence the Transfer Function is infinite. Also an LC resonant circuit with zero damping would have a pole ( probably two ) on the jw axis, one at the positive jw resonant frequency, and one at the negative jw resonant frequency. Again, easy to visualise, so why can I not figure out what the physical meaning of a pole on the real (sigma ) axis, at jw=0 means?
Andy
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
andrewpreece wrote:

variable,
the
at
pole? I

Bode plot,

(real)
exp-(sigma*t).
function
(
even
pole at

(using
frequency
be on

look in

filter
frequency
understand
shape
some
suggested an

that I

to get

by
result
real
its

(sigma )

an
with
Transfer
a pole

one at

figure out

Think of the s plane as an energy dissipation diagram. The imaginary axis corresponds to imaginary power flow, so complex poles come in conjugate pairs as power oscillates back and forth between two energy storage elements in the system (because the system is second order for two complex poles).
The distance along the real axis is the rate of real energy flow (associated with real power loss in resistors etc). Negative real pole parts mean that energy is leaving the system at a rate given by how far to the left the pole(s) is(are) located. Positive real components mean that energy flow is increasing and therefore the buildup means that it is unstable. The units of the sigma and jw axes are both radians per second so they represent rates of energy flow.
Understanding the physical meaning of poles and zeros is easier with complex conjugates as sinusoids are understood easily by anyone. Take a system with complex conjugate poles or zeros, put in a signal at those frequencies and you will see that the signal is either blocked (for zeros) or grows (for poles). I'll leave the maths for you to do.
Fred.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Hello Fred, I see where you are coming from. I like to think in terms of physical processes, so I tend to think of the negative real pole as representing real resistances ( as in 1/CR, taking my elctrical RC filter example ) because resistances dissipate energy, UNLESS they are negative resistances when they produce energy, leading to unstable behaviour, though in that case why isn't the whole positive real part of the pole-zero TF diagram infinite? I had noticed that both jw and sigma axes had the same units, radians/second as you say, but radians are dimensionless, and 'per second' has units of 1/time, so it can't be units of energy, as that would be joules/second, whatever the dimensions of joules turn out to be. I think both axes simply have units of frequency.

I shall plod on! I have noticed two things. First, a pole is drawn as the MAGNITUDE of the transfer function, but it turns out for the pole I'm looking at the sign goes to negative at infinity, so from the side , if you plotted sign as well as magnitude, a pole would look more like a tan function than a mountain. The second thing is that these axes, jw and sigma, they are the powers of the exponential function, e. Now 1/(t+(-t)) = 1/0 =infinity, but 1/[exp(t).exp(-t)] =1/[exp(0)] = 1. I am wondering if this pole has been created by the way the Laplace transform's involvement gives us a function of t in terms of a function of s. None of which probably makes any sense, but may be avenues to explore, as are your ideas on exponential 'frequency' and energy dissipation/flow.
Andy.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
A pole is a point where the denominator of the transfer function is equal to zero. Therefore, the value of the transfer function at that poiint is infinite.
Forget all the c%^p about pole-zero diagrams. A transfer function is an expression of the gain of a system element. If an element has infinite gain, it can generate an output with no associated stimulus. If an element has a pole corresponding to a term containing jw, it will exhibit a sinusoidal response at a frequency of w radians /sec when distrubed in any way - even if the dusturbance is specifically not sinusoidal. Similarly, a pole at a "frequency" of sigma will exhibit an exponential response with a time constant of 1/sigma seconds as part of its reaction to any disturbance.
Think in terms of a differential equation and the Characteristic Equation (which is the unforced solution, corresponding to the poles) and Particular Integral (wwhich is the response to a specific forcing or driving function). HTH, Bruce.
andrewpreece wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
I don't agree. My H(s) function does not have infinite gain at any frequency, but the transfer function blows up at the pole value. Also, it does not generate anything unless stimulated, so I think the pole-zero diagram stuff is still valid. The idea of poles and zeros is a little more subtle than many think.
Fred.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
fred snipped-for-privacy@hotmail.com wrote:

Your transfer function has infinite gain at the complex frequency that makes sT = -1. That's why it has a pole at sT + 1 = 0. The response to real frequencies is given by replacing s with jw. Your equation describes the open-circuit voltage at the tap of a a voltage divider which consists on a resistor R connected to a zero-impedance source and a capacitor C to ground. Elementary AC circuit analysis easily gives the frequency response without resort to Laplace.
Is this homework? Although I'd like to clear away confusion, I don't want to give away what it's intended you get for yourself.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
We are saying precisely the same thing, but frequency in this context is not confined to be sinusoidal - we need to consider inputs of the general form e^(st), not just e^(jwt). The Laplace transform method gives the response to many different input signal types.
No, formal homework finished decades ago, but problems still remain!
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
wrote:

Nope, not homework, I'm way too old for that. I understand the behaviour of first order RC lowpass filters like the back of my hand. I can find the frequency and phase response. I can solve the associated first order differential equation both by classical and Laplacian methods. Despite the fact that I can do all this stuff, I refuse to learn stuff parrot fashion. I do not completely understand the shape of the resulting pole-zero plot in the s-plane. I understand the response along the jw axis, but not the real axis. I can do the maths which gives me that shape as easily as the next man, but what does it mean?
Andy.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
hi The value of the T.F. becomes infinite at the value of the pole, this means that the system is in a state of saturation, we generally try and use the system in its active region (see transistors), even Vout\Vin infinity means that we have a saturated output for even very small values of the input. As you say that the system has a pole at -1/RC it surely ought to be on the real axis, the system in this case may be stable for all values of gain(mathematically).
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Hello, You cannot saturate an RC filter - it doesn't have a power supply, and it is a linear system, by definition then, saturation cannot occur. I agree that an infinite trasfer function (pole) can mean one of two things though: either a finite input causes an infinite output, or zero input creates some finite output. Both conditions would look like infinite gain. Neither condition can occur with an RC lowpass filter, but I am beginning to believe that the behaviour of the pole-zero plot along the sigma axis for a low pass RC filter is more a mathematical construct than to do with any observable property of a lowpass filter transfer function.
Fred-Stevens is the only respondent who is truly understanding what my question is getting at, I think.Thanks anyway,
Andy.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

How do you reconcile this view that the system behavior is entirely determined by the locations of the poles and zeros?
Scott
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Scott Seidman wrote:

filter
Returning to my first post. Take a system with transfer function H(s) = a/(s+a) Supply the system with an input u(t) = e^(-at) (which is a signal at the same frequency as that of the pole of H(s)). The output response will be y(t) = te^(-at). If you plot this you will see that y(t) rises up to a peak from zero before it decays back down again. The output RISES when given a DECAYING input at its pole frequency.
[Here is a Matlab script:
t=0:0.001:10; a=input('Enter the value of a: '); x=exp(-a*t); y=t.*x; plot(t,x,t,y,'r'),grid
]
So, although the function in the Laplace domain goes to infinity at the pole, this does not mean that the time response will do so. Remember from the definitions of Fourier and Laplace transforms that the time and frequency domains are related via infinite sums in the form of integrals, so if an integral from 0 to infinity gives an infinite answer, this does not mean that the integrand has to be infinite.
I hope the above helps.
Fred.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Andy,
Sorry to cause confusion about the units. What I meant was that it is helpful to think of the s plane as representing energy flow. I did not mean to imply that the units actually were Joules/sec etc. Similar to what is done in communication systems where normalization to a 1 Ohm resistor gives power, you can do a similar thing here to figure out the correct units. The analogy is what's important.
Fred.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
fred snipped-for-privacy@hotmail.com wrote in

I don't understand your point. First, the input is not a simple decaying exponential-- it is a decaying exponential times a step. I wouldn't call that a "decaying" input. ANY input of that form, regardless of the time constant, is going to make the output rise, then fall.
Scott