I don't see a step, perhaps an impulse followed by a decaying
exponential?

I don't see a step, perhaps an impulse followed by a decaying
exponential?

The transfer function 1/(s+a) pairs with exp(-at)u(t) (or -exp(-at)u(-t)!!!). The step is built in. By starting your simulation at t=0, the step is sort of assumed. Otherwise, exp(-at) is unbounded on the left. Think of it as an unbounded exponential being windowed in the time domain by a step function.

Scott

A step has a DC component - it has a final nonzero value, otherwise it
would be pulse or something else. My input signal has no DC value.

fred snipped-for-privacy@hotmail.com wrote in

Your input signal has no steady state value, but it certainly has a mean, and it shows up as a non-zero value at the 0-frequency point of the fft.

exp(-at)u(t)= 0 for t<0 = exp(-at) for t>=0

This is unlike exp(-at), which is simply exp(-at) at all values of t

Scott

Your input signal has no steady state value, but it certainly has a mean, and it shows up as a non-zero value at the 0-frequency point of the fft.

exp(-at)u(t)= 0 for t<0 = exp(-at) for t>=0

This is unlike exp(-at), which is simply exp(-at) at all values of t

Scott

Ah, yes, I've just said something along the lines of your last para to Scott, in that I was looking for some explicit infinity at the pole, when in fact it may be more subtle to discover, though I came at it from a completely different direction. As to your conjecture about a decaying exponential input actually causing the filter o/p to rise for a short time before decaying again, I am familiar with that kind of waveform. The question is whether the o/p of the filter, whilst also decaying ever becomes infinitely more than the decaying input waveform. I shall now have to look at the rates at which exponentials approach zero, etc or possibly even their integrands. I wish I was a natural at maths! I have several approaches to work on now vis a vis this problem and see if I can knock them down with logic/maths.

Andy.

Well, I was rather speculating out loud there. What I was trying to say is that using maths it is possible to end up with something that is derived from an original function but rather different in appearance.

For instance, at first glance one would not associate a differential equation with its Laplace transform, which is an algebraic equation, but the two describe the same system but in different ways.

Sometimes mathematicans construct ways of avoiding infinities. I believe Riemann or someone developed a way of transforming a function from a plane onto a sphere. The thing was, on the plane the function would start at 0,0 then shoot to infinity in all directions, whereas on the sphere the function ranged between 0 and 1. I was simply wondering ( I am not a mathematician and find the whole area rather difficult ) whether the Laplace Transformation and its representation on a pole zero plot, had taken a function which naturally varies from 0 to 1, and transformed it so it varies between 0 and infinity, so rather baffling my brain which was looking for a property of an RC filter which could actually achieve infinity, when I should perhaps be looking for a gain of 1!!

All of which is probably the worst kind of twaddle but at least you can see what was going through my head when I wrote the pararaph you quoted.

Andy.

The transformation z=e^(st) actually transforms the infinite left half
of the s plane into a unit circle and DSP and control guys use it a lot.

This is what they call the Z-Plane?

Andy.

Yes. However there are many other transforms that do similar things.
The Z plane is related to Z transforms as the Laplace transform is
related to the s plane.

212.67.96.135:

Saturation can occur if the poles are on the wrong side of the plane. For simple RC, this can't happen. For negative values of C and R (and negative capacitance amplifiers are fairly common), it certianly can.

Perhaps you should consider the Region of Convergence of the Laplace Transform. The pole of an RC filter actually lies outside of the ROC

Scott

Saturation can occur if the poles are on the wrong side of the plane. For simple RC, this can't happen. For negative values of C and R (and negative capacitance amplifiers are fairly common), it certianly can.

Perhaps you should consider the Region of Convergence of the Laplace Transform. The pole of an RC filter actually lies outside of the ROC

Scott

negative

I shall look into that: I will have to look up what the Region of Convergence of the Laplace Transform is first! ( Opens up heavy copy of Kreysig's Advanced Engineering Mathematics and looks baffled ) :0)

Andy.

On Sun, 3 Apr 2005 21:21:12 +0100, andrewpreece wrote

I may be misinterpreting your problem but in short I think what is puzzling you is that something goes to infinity when S = -1/CR, (i.e. when S is purely real) but the physical filter does not have infinite output for DC input.

What goes to infinity at S = -1/CR is ONLY the S domain transfer function:

Vo(s) 1/CR ----- = ---------- Vi(s) S + 1/CR

But what happens in the S domain should not be confused with what happens in the time domain. As you well know, you have to invert the expression for Vo(s) in order to determine Vo(t), or alternatively apply the final value theorem if you only want to know the steady state output. In both cases the steady state output of the filter is equal to Vi, as expected.

I'm sorry if this doesn't address your true difficulty, but again I would stress that it is only the S domain function that has a pole and that one shouldn't confuse that with real world (time domain) behaviour. That is why it is imperative to express the T.F. as Vo(s)/Vi(s), not simply Vo/Vi.

Alan

I may be misinterpreting your problem but in short I think what is puzzling you is that something goes to infinity when S = -1/CR, (i.e. when S is purely real) but the physical filter does not have infinite output for DC input.

What goes to infinity at S = -1/CR is ONLY the S domain transfer function:

Vo(s) 1/CR ----- = ---------- Vi(s) S + 1/CR

But what happens in the S domain should not be confused with what happens in the time domain. As you well know, you have to invert the expression for Vo(s) in order to determine Vo(t), or alternatively apply the final value theorem if you only want to know the steady state output. In both cases the steady state output of the filter is equal to Vi, as expected.

I'm sorry if this doesn't address your true difficulty, but again I would stress that it is only the S domain function that has a pole and that one shouldn't confuse that with real world (time domain) behaviour. That is why it is imperative to express the T.F. as Vo(s)/Vi(s), not simply Vo/Vi.

Alan

complex

puzzling

in

the

why

Hello Alan,

You understand my problem perfectly! I think I have been at least partially guilty of confusing time domain and s-domain stuff ( I'm aware there's a difference but get confused ). The thing is, the pole-zero plot in the s-domain makes perfect sense if you look at its profile along the jw axis - you get back to back Bode plots of the frequency response of the RC filter. Time no longer is a part of the jw axis ( nor of the sigma axis, it has been integrated out ), yet I still recognise back-to-back Bode plots when I see them, and it is obvious this filter has a gain of 1 when w=0 ( or jw=0 if you like ), falling to a gain of zero at +/- infinity. However, when I look along the real ( or sigma ) axis, I can't expainthat profile, with its pole at sigma =(1/CR) in the same way that I can explain the profile along the jw axis.

I am struggling a bit with the maths of Laplace and the concepts behind it, with the help of a few kind souls on this newgroup, so that I can look at the profile of the real axis in a pole zero plot in future, and say, "that shape means the filter will have a response like this" just in the same way that I can already scan the jw axis and tell you what its frequency response is merely by observation.

Andy.

I don't understand what you mean here. The frequency response can be
determined graphically from the pole-zero diagram by calculating the
magnitude and phase contributions of each pole and zero while "walking"
up the jw axis. In effect you are graphically evaluating the effects of
residues in a portion of the Nyquist D-contour.

I don't know what you mean by "However, when I look along the real ( or sigma ) axis, I can't expainthat profile, with its pole at sigma =(1/CR) in the same way that I can explain the profile along the jw axis. ".

Fred.

I don't know what you mean by "However, when I look along the real ( or sigma ) axis, I can't expainthat profile, with its pole at sigma =(1/CR) in the same way that I can explain the profile along the jw axis. ".

Fred.

I don't quite understand your last sentence! Never heard of a D-contour, though I think residues are something vaguely mathematical!

I'll give it one last go, though without diagrams it is hard. Imagine a pole-zero plot in the s-plane: there is one pole at sigma = -(1/CR) for the particular example of a first order low pass RC filter.

Imagine that instead of looking at the three-dimensional surface of the contours of the transfer-function pole-zero plot H(s) or whatever you wish to call it, you look only at the transect directly above the jw axis. This will yield a profile symmetrical about the 0 point on the jw axis, with a value of 1 at jw=0, falling off in both the +jw and -jw directions, to zero at infinity. As an electrical engineer, I recognise this shape as a Bode plot, or rather two Bode plots back-to-back, one for positive frequencies, one for negative. This ties in with the reality I know, where if I inject a sinusoid into the input of this filter, and vary the frequency w, the output amplitude will fall off in the same way as the Bode plot. All this is easy for me to grasp.

If we switch our perspective around 90 degrees when looking at the pole-zero plot, the jw axis disappears since we're looking at it head on, and we instead see the real ( sigma ) axis running in a left-to-right fashion as it were. If we look only at the transect of the sigma axis makes with the pole-zero contour floating above it, we see a line that falls off to an amplitude of zero if you go to infinity along the positive or negative direction of the real axis. At the value of -(1/CR) on the sigma axis the magnitude curve (pole-zero contour ) goes to infinity. OK, it's a pole......

The thing is, I recognised what the shape of the magnitude contour directly above the jw axis meant - frequency response, Bode plot in fact. When I look at the contour , complete with pole, directly over the sigma axis, it means nothing to me. It is not a Bode diagram. What properties do I associate with it? Is it phase? Is it something I haven't heard of?

That is my problem. I recognise the sort of animal a Bode diagram is, but turn the plot 90 degrees and I cannot recognise what kind of animal that plot with an off centre hummock represents. Properties I associate with a filter are phase, gain, frequency response.

I'm afraid I've spawned a monster with this thread, I actually have benefitted from some new leads to look into from the folks on this thread, so what I need to do now is get a lot of scrap paper and start to try out a few ideas. My great problem ( apart from being thick ) is that I cannot settle for being able to apply a technique parrot-fashion, I need to understand what's-what and why,

cheers,

Andy.

On Wed, 6 Apr 2005 21:38:41 +0100, andrewpreece wrote

Andy,

Ah, now I see what you mean. I guess I should have spent more time reading all the posts in this thread then the penny might have dropped earlier.

Yes, its a good question as to why F(s) evaluated along the imaginary axis gives something meaningful like the frequency response but evaluating it along the real axis doesn't yield anything of obvious physical significance.

I can't give a quantitative answer as to why this is, but the following way of looking at it may shed some light:

Let F(s) = 1/S+A, i.e. a pole on the real axis at - A. Now if you plot |F(s)| on the Z axis (i.e. an axis orthoganal to the sigma and omega axes) you get something that looks like a solid hill of increasing steepness rising to infinity above the pole. Now take a sharp knife, hold it above the jw axis and slice the hill in a downward vertical direction. If you look at the curved top of the cut surface it is, of course, |F(jw)| i.e. the frequency response by definition. Similarly, if you cut the hill along the real axis, the cut surface represents |F(sigma)|. OK, you may say, I knew that and it doesn't answer the problem. That's true, but I think it does help to show that the first cut corresponds to something we all know because its a useful tool, i.e. the frequency response, whereas the second cut yields something which we don't recognize because is just not used in practice. So maybe its just a mistake to expect that every cross section should have practical utility ?

Alan

Andy,

Ah, now I see what you mean. I guess I should have spent more time reading all the posts in this thread then the penny might have dropped earlier.

Yes, its a good question as to why F(s) evaluated along the imaginary axis gives something meaningful like the frequency response but evaluating it along the real axis doesn't yield anything of obvious physical significance.

I can't give a quantitative answer as to why this is, but the following way of looking at it may shed some light:

Let F(s) = 1/S+A, i.e. a pole on the real axis at - A. Now if you plot |F(s)| on the Z axis (i.e. an axis orthoganal to the sigma and omega axes) you get something that looks like a solid hill of increasing steepness rising to infinity above the pole. Now take a sharp knife, hold it above the jw axis and slice the hill in a downward vertical direction. If you look at the curved top of the cut surface it is, of course, |F(jw)| i.e. the frequency response by definition. Similarly, if you cut the hill along the real axis, the cut surface represents |F(sigma)|. OK, you may say, I knew that and it doesn't answer the problem. That's true, but I think it does help to show that the first cut corresponds to something we all know because its a useful tool, i.e. the frequency response, whereas the second cut yields something which we don't recognize because is just not used in practice. So maybe its just a mistake to expect that every cross section should have practical utility ?

Alan

trivial,

plot,

exp-(sigma*t).

function

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at

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is

DC

function:

happens

for

value

cases

would

one

is

difference

of

plots

sigma )

same

it,

the

like

significance.

way

rising

the

axis,

it

Thanks for the reply Alan, you understand what I was getting at. You may well be right about the real axis cross section having no practical physical significance, but I had to assume that it did, and the fault was mine for being too stupid to grasp what it was!

In an earlier thread, Fred suggested that since we apply sinusoids to the jw axis, then we should ask ourselves what happens when we apply exponentials to the real axis ( since the sigma axis is a transformed verion of exp(sigma), just as the w axis is a transformed version of exp(jw).

Now, exp(sigma) is an exponential decay or growth, depending on the sign of sigma. What I did was to solve the linear differential equation for the RC low pass filter to get the time solution. This circuit has a 'pole' or time constant, of -(1/CR). Then I mathematically applied different exponential forcing functions to the circuit ( equivalent to injecting exponential input voltages of different time constants ), and solved for the output voltage as a function of time. Finally, I divided the output voltage by the input voltage to get the transfer function ( in time ).

Forcing Function Vout/Vin ( Transfer Function )

Exp(2t/cr) 1/3[1-exp(-3t/cr)] Exp(t/cr) 1/2[1-exp(-2t/cr)] Exp(0) 1[1-exp(-t/cr)] Exp(-t/cr) t/cr Exp(-2t/cr) 1[1-exp(t/cr)] * Exp(-3t/cr) 1/2[1-exp(-2t/cr)] * Exp(-4t/cr) 1/3[1-exp(-3t/cr)] *

* These results multiplied by -1 as on a pole zero plot to give a positive result.

All a bit confusing, but look at the coefficients of the Transfer Function, 1/3, 1/2, 1, etc. These are the values of the transfer function along the sigma axis in the s-domain. Note that the TF goes to infinity at sigma = -1/cr, and the results above give t/cr, but at t=infinity then that goes to infinity also. I'm thinking that the physical explanation of these results, as Fred suggested, is that as t>infinity, the ratio of output voltage over input voltage ( for an applied exponential voltage ) tends towards the coefficients of the TF I've given above. For instance. for this particular circuit, if we inject an exponential voltage of exp(t/cr), then the ratio of Vout/Vin tends to 1/2 at t=infinity, etc.

Of what practical import this is I cannot say, can't say I can think of any use for that info, but I think it may explain the shape of the transfer function directly above the real axis. When we get an infinite response at an input forcing function of exp(-t/cr), we do not get an infinite output anywhere, it simply happens that Vout falls more slowly than Vin: both decay towards zero at t=infinity, but nevertheless, Vout manages to be infinitely more than the infinitely small amount that Vin becomes!

Well, that's my take on things anyway, not quite the earth-shattering lifting of the veil I'd hoped for but I reckon that'll have to do!

Andy.

There seems to be a bit of confusion here (on my side too!). As far as
I understand things, the frequency response (Bode and Nyquist plots) is
the result of a sinusoidal input signal to a system. Thus, we put s=jw
and calculate the magnitude and phase. The sigma part of s is not
relevant in this case, even though the real poles contribute to the
frequency response. The frequency response is obtained by moving up the
jw axis and at each frequency w, the magnitude and phase contributions
of each pole, including those on the real axis, are added together.
Thus we get vectors from the w value on the jw axis to each pole and we
sum the magnitudes of each to get the resultant at frequency w.
Similarly, we add the angle contributions to get the resultant phase
angle at w. We don't need to consider the real axis itself in this
procedure. However, our discussion was about real poles and the effect
on the response. This is different from calculating the Bode plot. For
the above reasons, I don't understand what you are doing with the real
axis when developing a Bode plot.

I wasn't developing a Bode plot from the real axis Fred, it isn't possible of course as the real axis does not have a sinusoidal frequency associated with it like the jw axis. I was merely observing that that if we slice along the jw axis of the pole zero plot with a knife ( as it were ) held vertically, then we reveal a profile of the transfer function that is actually a Bode plot, gain ( vertical ) versus frequency.

Now if we slice through the transfer function pole zero plot with a vertically held knife along the real (sigma ) axis, we get a profile that is ...what? It certainly isn't a Bode plot. Does that profile have a special name? Does anyone use that profile for analysis in the same way they might use a Bode plot?

I posted a reply to another chap today, on this same thread, you should look at that, I followed your recommendations about theoretically applying exponential input signals ( forcing functions ) to the system, and seeing how the transfer function behaved. I was able to reproduce the profile of the response above the real axis. I can't say that it looks like useful info though, so the profile along the real axis seems not to contain immediately useful info as the profile along the jw axis does(who wants to know how a circuit responds to input exponentials? Sinusoids, yes, all the time, but not exponentials.

Andy.

Sorry for being so thick, but I just don't see how slicing along the jw
axis in the way you describe shows anything about gain versus
frequency. For example, a complex pole pair positioned with different
damping factors will give different gain values at the same frequency
(if wn is held fixed). How can this be seen from the sliced jw axis?

Fred.

Fred.

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