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Before asking my questions I would like to write down briefly about my big picture understanding so far. I will go with an example RLC circuit below:

enter image description here

Above circuit’s transfer function H(s) can be written as H(s)=Vout(s)/Vin(s)

Since all is in series the same current I(s) passes through the components, this circuit can be treated as a voltage divider. So by using relevant Laplace transforms, the voltage across R which is Vout can be written in terms of Vin as:

Vout(s) = Vin(s)*R / (sL+R+1/sC)

So the transfer function H(s) can be simplified as:

H(s) = sR / (s^2*L+sR+1/C)

Now that we have the transfer function, we can find the response of this circuit for a given input in s domain and then transform it by taking inverse Laplace transform to plot it in time domain. One can also find the location of poles and zeros on the s plane.

Before that, first I will start with a step response of this circuit in LTspice to estimate the impulse response in time domain which is the transfer function. Impulse response of an LTI system in time domain is the transfer function. Since I cannot create a real Dirac function I apply a very sharp rising edge step input and see the response in LTspice. I also do the same thing by using MATLAB in s domain.

Since step input X(s)=(1/s) in s domain, the output can be written as:

Y(s) = X(s)*H(s)

Y(s) = R / (s^2*L+sR+1/C)

So taking the inverse Laplace transform of Y(s), L-1[Y(s)] yields a time domain response Vout(t).

And below plots shows the same result is obtained by using LTspice and MATLAB:

enter image description here

Ok so if we look at the above plots we can see that the circuit response to a sharp step input is a decaying sinusoidal with f=50Hz(ω = 2*pi*f= 314), in other words damped oscillation at a particular frequency.

Now the reason for the question I’m going to ask is about some observations or the relation between the transfer function and the time domain response.

First of all here is the 3D plot of |H(s)| of this circuit above the s plane:

enter image description here

At this point, from the transfer function H(s) = sR / (s^2*L+sR+1/C) one can already determine the two poles and the zero locations on the s plane.

First I look at the |H(s)| vs imaginary axis as below:

enter image description here

My first observation of the transfer function of the plot above is that the location of the poles are where ω = 314 on the imaginary axis. This value of the ω is exactly the same when we observed the step response or natural frequency of the circuit where it damped with an oscillation at ω = 314. So I think what pole’s imaginary part indicates is the impulse response’s frequency. Is that correct?

Second is that, if we look at the real axis which is σ axis of s=ωj+ σ plane as follows:

enter image description here

And as seen above, σ for the point of the pole is around -100. I think this negative sign indicates the decay rate of the sinusoid as e^-100. Is this correct?

And here is the top view of the s plane:

enter image description here

Now regarding these observations so far,(if you agree with them) my questions are as follows:

  1. From the observations of the s plane, the locations of the poles turned out to be complex numbers which are carrying the knowledge of the time domain input response i.e the natural response. So as you see if it is true one can relate this particular pole points to something in time domain. But what about the other points on the s plane rather than a pole or a zero. Pick a random s point on the s plane and find out what |H(s)| for that point. What does that indicate in time domain?

  2. Another observation is that this RLC circuit’s response is decaying. Does that mean this system is stable? If so, is there a passive network example where the system would be unstable? Or do we need an active component to observe such phenomena?

  3. At the poles |H(s)| mathematically goes to infinity, not in MATLAB not in LTspice. Is that because applying a Dirac delta function is just a theoretical concept?

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  • \$\begingroup\$ If you choose a point on the positive jω axis, the vector to that point from a pole or zero tells you the gain and phase angle of the root (inverse of gain in the case of a pole) at that particular frequency. If you choose an arbitrary point in the s-plane and do something arbitrary with it you get arbitrary information. \$\endgroup\$ – Chu Apr 30 '17 at 23:31
  • \$\begingroup\$ Any transfer function with a non-trivial denominator will have at least one pole, i.e. goes to infinity at a particular finite value of s. That's how a pole is defined. If you plot the magnitude of H(s), then of course you'll get infinite spikes at the poles. \$\endgroup\$ – Chu Apr 30 '17 at 23:31
  • \$\begingroup\$ @Chu 1-)"choose a point on the positive jω axis, vector to that point from a zero tells the gain" But if I choose jω as j*100000, the vector becomes huge but the output is so small. You mean by gain Vout/Vin here?. Can you draw/link what you mean here. Would be great a figure would help visualize what you mean. 2-) For a pole p on s domain we have something to say in time domain. But for a complex point sx on s plane which has the magnitude of H(sx) what can we say in time domain? Nothing? It doesn't represent anything in time domain? Maybe Im looking for something doesn't have meaning here? \$\endgroup\$ – user16307 Apr 30 '17 at 23:36
  • \$\begingroup\$ if it's a pole, the gain is the inverse of vector length. If it's a zero the gain is the vector length. \$\endgroup\$ – Chu Apr 30 '17 at 23:41
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    \$\begingroup\$ Take, for example, \$H(s) = \large \frac{1}{a+s}\$, which has a pole at \$s=-a\$. Letting \$s=j\omega\$, \$(a+j\omega)\$ is a vector that starts \$-a\$ and ends at \$j\omega\$ (from \$-a\$ to the origin, then from the origin to \$j\omega\$). The length of the vector is \$\sqrt{a^2 +\omega^2}\$, and the inverse gives gain at \$\omega\$: \$\large\frac{1}{\sqrt{a^2+\omega^2}}\$. \$\endgroup\$ – Chu Apr 30 '17 at 23:51
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Yes you are basically correct with your observations.

So I think what pole’s imaginary part indicates is the impulse response’s frequency. Is that correct?

Yes. Laplace transform is a tool to solve the differential equation that arises from your system. If you put in an input function (also transformed into Laplace domain, in this case a step function, that corresponds to 1/s), then you obtain for your output signal in the s-domain. Do a backwards Laplace transform to see the actual time-domain function. Thereby the real part of the pole-location corresponds to the damping factor, the number in the exponential term. The imaginary part of the pole corresponds to the omega in the sine/cosine part of your time domain solution. So we have the answer to your next question:

So I think what pole’s imaginary part indicates is the impulse response’s frequency. Is that correct?

Yes, it is. Also:

I think this negative sign indicates the decay rate of the sinusoid as e^-100. Is this correct?

Yes.

The questions in the end have a slightly longer answer: 1. Again, put in a L-transformed input signal of arbitrary time-domain function and look at the solution. Transform back and you'll get a function that corresponds to the time-domain solution. In case you look at a system of only a sine/cosine function as input (that has been there very long, not only starts at t=0) you can rather use fourier transform. Then you will get three kinds of information: First, the output signal will have the exact same frequency as the input signal. Secondly, the signal amplitude is what you get and the third information is how much phase-shift you have. This means, you put in a sine and get a shifted sine by some fixed angle phi for a fixed input frequency.

  1. Passive circuits are always stable. You would need a negative resistance to get a positive value in the exponent. This is only possible with active components (although rarely interpreted as such as a negative resistance).

3.No, you're mixing stuff up here. The poles are not diracs. They are just values for s, where the transfer function has zeros in the denominator. Look at the table here, equation 19: the pole of this equation is here: $$ (s-a)^2+b^2=0 \Rightarrow (s-a)^2=-b^2 \Rightarrow (s-a)=ib \Rightarrow s=ib+a $$ This means, if the output function (the solution to your problem) is as eq 19, you get the exponential decaying sine as solution in the time domain.

The Laplace transform is just another way of solving a very special kind of ordinary differential equations (odes): linear differential equations with constant coefficients. Another presumption is that your input signal is 0 for t<0. If these criteria are met, you can solve an ode algebraicly, that means you only need addition, subtraction, division and multiplication. This is oftentimes easier than solving an ode with other methods (separation of variables, varying constants, arbitrarily complicated stuff). This means you transform your ode, use +-*/, and transform back.

Edit: Use your transfer function and multiply by 1/s. This will give you something like $$ \frac{1}{s^2+as+b} $$ Transform it back and you'll see that it is exactly as you expected.

Edit. To answer your question 2) in the comments. Am I right, you're not referring to "if I change the location of the poles, how does this affect the time domain solution.", right? Sorry if I get this wrong again, but I think you're asking something that cannot be answered. It is so, that a function in the s-domain corresponds exactly with one time-domain function. You always look at the function as a whole. And this corresponds to another function as a whole in the time domain. Only the poles and zeros of the s-domain function are used as "fingerprint" of the s-domain function that tells you a lot about the function.

It is not that you evaluate the s-function somewhere or calculate "what is F(s) at point s1 (somewhere), and what does this tell me?" That is not the question you ask. There is no thing like "the point s1 corresponds to the output value at t=1.2sec" or something. And this makes perfect sense: you are usually basically dealing with polynomials divided by other polynomials with respect to s. And the poles and zeros are all you need to fully describe the function. (Interestingly this rises the question, what happens if I have a function like log(s) or something even more crazy... it translates into something in the time domain, that does not consist of "normal" e,sin,cos functions, as far as I think about it now. In this case talking about poles and zeros wont be helpful anymore, because then those do not describe the s-function fully anymore). You always have something like:$$ \frac{1}{(s+1)} (transform sign) e^{-1t} $$ The notion $$ \frac{1}{(s_1+1)}(transform sign)? $$ does not make sense. Its always the function as a whole that is transformed. In case you're banging your head against the wall because I still don't seem to get what youre actually asking, I am sorry.

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  • \$\begingroup\$ I didn't write that poles are Dirac. In time domain if the input is Dirac delta you expect the impulse response but Dirac delta is impossible to create is that the reason poles don't go to infinity in s domain? And my first question still remains as mystery "Pick a random s point on the s plane and find out what |H(s)| for that point. What does that indicate in time domain?. And your edit is part of my question already done in MATLAB. So thanks for the answer to the 2nd question. \$\endgroup\$ – user16307 Apr 30 '17 at 22:11
  • \$\begingroup\$ No I think this is not what you mean. Transform a dirac to L-domain and it is 1. So your response (i.e. your output) is the backtransform of the transfer function itself. You can do this on your own, take the H(s) function and transform it back to the time-domain (without multiplying it with 1/s but simply with 1). This function is the function that will occur in time domain. What do you mean they do not go to infinity? Poles by definition go to infinity. Or do you mean that the location of the poles? This is a property of the circuit, not the input signal. \$\endgroup\$ – jjstcool Apr 30 '17 at 22:28
  • \$\begingroup\$ You should do the following: use your circuit and put in a dirac, a step, and a sine of arbitrary frequency. Then put the result in a form that you can transform back with a laplace transfrom table. Look what comes out. The one with the sine will also give you a sine of the frequency you started with, not necessarily the natural frequency of the system (maybe both, the system one decaying to zero). If you can't do it, Ill do the math for you tomorrow, give me a quick hint in case you find it too difficult. \$\endgroup\$ – jjstcool Apr 30 '17 at 22:31
  • \$\begingroup\$ Ok let me articulate another way. 1-) Call one of the pole on the s plane "p". Now the value |H(p)| in the plot does not go to infinity. Why? I already plotted what you say I multiplied H(s) by 1/s and took the inverse Laplace. That was not the question. 2-) Secondly I'm asking: Imagine you are standing on a random point s1 (not a pole not a zero) on the s plane. And you look at the surface above you and you measure the distance |H(s1)|. Ok what does this mean in time domain? \$\endgroup\$ – user16307 Apr 30 '17 at 22:40
  • \$\begingroup\$ 1) It does, only the plot is not showing it. If you look at it algebraically, it is obvious. Sorry, I just saw that you actually did the multiplication with 1/s and plotted it. I was in a hurry when first reading your text. For 2) see my edit \$\endgroup\$ – jjstcool May 1 '17 at 0:43

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