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Considering Fourier analysis and let's say I'm walking on the blue frequency axis in the below 3D plot from zero towards infinity:

enter image description here

So each time I encounter a non zero blue bar, I check the frequency at that point and measure the bar's height and say: Oh I found another sinusoidal component of the signal at this particular frequency and at this particular amplitude. This is how I grasp the idea of Fourier analysis.

But when I read about the s-plane I'm extremely confused to make it visual and link it to the actual signal as in the above fashion. I just cannot give a similar meaning what it is that really about.

Now imagine a system's output and input in time domain is given. Vout(t)/Vin(t) is transformed into s-domain and we call this ratio as H(s) which is the transfer function.

Here is a transfer function which is a surface on the s-plane:

enter image description here

Imagine I am walking on the s-plane. And imagine at a point on the s-plane I stop and the point s is a complex number and it is s = ωj + σ. Now if I look upwards towards the surface above me, obviously there will be a point on the surface. Lets call this point x. So what should it mean to me in terms of the signal or the transfer function?

Ok so at this particular point I found "what"? What is "x"? Is it a component of something? And my second question is if H(s) were a signal's transform not a transfer function what would x be in that case?

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  • $\begingroup$ If you walk the line $s = 0 + j\omega$, that is the same blue frequency axis as in your 3D plot, so $x$ is the magnitude of that radian frequency, in this special case of walking on the imaginary axis in the s-plane. $\endgroup$ – Andy Walls Apr 24 '17 at 12:46
  • $\begingroup$ For the special case of $s = 0 + j\omega$, $H(s = j\omega)$ is the Fourier Transform of your hypothetical signal, so $x = |H(j\omega)|$, the magnitude of the Fourier Transform of the signal. $\endgroup$ – Andy Walls Apr 24 '17 at 12:55
  • $\begingroup$ When I'm somewhere on the s-plane, if I look upwards towards to the surface above me, obviously there will be a point on the surface. There will be a height between me and the surface point on top of me. I called this point x in my question. So is x = H(s) and is x a real number? $\endgroup$ – floppy380 Apr 24 '17 at 13:44
  • $\begingroup$ The value of $H(s)$ for any $s$ is generally a complex number. So if you want to talk about a surface above the s-plane, you need to talk about $|H(s)|$ if you want to assign a real number at a point in the s-plane. That's why I deliberately wrote $x=|H(s)|$. $\endgroup$ – Andy Walls Apr 24 '17 at 13:56
  • $\begingroup$ If H(s) is the surface it would be H(s)=|H(s)| But you say H(s) is a complex number. How can we visualize H(s)? $\endgroup$ – floppy380 Apr 24 '17 at 14:11
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Using the bilateral Laplace transform, the interpretation of the transfer function $H(s)$ of a linear time-invariant (LTI) system is straightforward. Since the output signal is given by the convolution of the input signal and the system's impulse response

$$y(t)=\int_{-\infty}^{\infty}x(t-\tau)h(\tau)d\tau\tag{1}$$

with $x(t)$ the input signal and $h(t)$ the impulse response, we get for the specific input signal $x(t)=e^{s_0t}$, $s_0\in\mathbb{C}$,

$$y(t)=e^{s_0t}\int_{-\infty}^{\infty}e^{-s_0\tau}h(\tau)d\tau=e^{s_0t}H(s_0)\tag{2}$$

According to $(2)$, the transfer function evaluated at $s=s_0$ equals the the complex multiplication factor (eigenvalue) applied to an input signal of the form $x(t)=e^{s_0t}$ (eigenfunction).

In general I do not think that it is helpful to consider a physical interpretation of the Laplace transform of an arbitrary signal. The Laplace transform is rather a tool that simplifies certain operations, e.g. by transforming convolutions to multiplications, and differential equations to algebraic equations.

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  • $\begingroup$ When I'm somewhere on the s-plane, if I look upwards towards to the surface above me, obviously there will be a point on the surface. There will be a height between me and the surface point on top of me. I called this point x in my question. So is x = H(s) and is x a real number? $\endgroup$ – floppy380 Apr 24 '17 at 13:43
  • $\begingroup$ H(s) contains both magnitude and phase info, so is best treated as complex. $\endgroup$ – hotpaw2 Apr 24 '17 at 13:56
  • $\begingroup$ Calling H(s) as x is also confusing, as how do you now specify locations on the xy plane an which you are standing. $\endgroup$ – hotpaw2 Apr 24 '17 at 14:00
  • $\begingroup$ Obviously H(s) is not the same thing with s and it is not the same thing with |H(s)| as well as you say then. How can we visualize H(s) in this case? For s we say s = ωj + σ and it is obvious what we mean by s in that 3D plot. But can you give an example for H(s) and describe its location? $\endgroup$ – floppy380 Apr 24 '17 at 14:17
  • $\begingroup$ @doncarlos: What you show in your 3D plot is the absolute value $|H(s)|$. $\endgroup$ – Matt L. Apr 24 '17 at 15:12
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In the Laplace transform, being on the line $s = 0 \pm j\omega$ means you are looking at the pure sinusoids contained in the signal. This is exactly as you mentioned in your understanding of Fourier analysis. So this means that you are looking at the time-domain terms:

$$e^{j\omega{t}}$$

Which again represent the pure sinusoids composing your signal.

However once you are off this line and have some $s$ such that: $$s = \sigma \pm j\omega$$

Not only are you looking at the $e^{j\omega{t}}$ sinusoids terms, you are now looking at time domain terms: $$e^{\sigma{t}}e^{j\omega{t}}$$

When talking about a stable system, poles being on the left-hand-plane of the $s$-domain means that $\sigma < 0$ so the sinusoids you are considering contained in the signal now have a damping envelope. If $\sigma$ was positive, the poles would be placed in the right-hand-plane and the signal would explode to infinity as time went on. This is why we want poles to stay out of the right-hand-plane if we want a stable system.

The takeaway is that in Fourier analysis you are considering the pure sinusoids composing your signal. With the Laplace transform you are considering sinusoids with that exponential term attached. In the case for a stable or oscillating system, you are considering the damped sinusoids ($s = -\sigma \pm j\omega$), decaying exponentials ($s = -\sigma$), pure sinusoids ($s = \pm j\omega$), and constant amplitude ($s = 0$) components composing your signal.

This is what makes Fourier analysis appropriate for identifying frequency content and the Laplace transform appropriate for analyzing stability and performance parameters.

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  • $\begingroup$ I read your answer. So what I understand from your words is that: If Im at a point on the s plane call it s1=σ1+jω1 and if the |H(s)| for that point is zero it means the function does not have component for s=σ1+jω1. And if Im at a point on the s plane call it s2=σ2+jω2 and if I look upwards to the surface I measure |H(s2)|, so is this the amplitude of the s2=σ2+jω2 components of the signal?? Is that what you mean? $\endgroup$ – floppy380 Apr 28 '17 at 13:20
  • $\begingroup$ When looking at $|H(s)|$, yes that is correct. However understand that the actual value of $s$ determines what kind of component you're looking at, like mentioned above. $\endgroup$ – Envidia May 1 '17 at 15:52
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Given a process with an input signal, a transfer function and an output, it is important to note that the transfer function in and of itself doesn't tell you anything about the input signal. What the transfer function tells you is the relationship between the input and the output (i.e. what the process will do to ANY input). In this sense, the transfer function is independent of the input.

When you consider the poles of a transfer function, i.e. the roots of the characteristic polynomial (poignant though super fancy name for the denominator), what it tells you is the characteristics of the output signal.

Simply put, the presence of complex poles (and they always come in conjugate pairs) informs you of the presence of oscillations in the output. The larger the magnitude of the complex part, the larger the amplitude of the oscillation. The other terms inform you of other factors like the damping frequency.

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  • $\begingroup$ When I'm somewhere on the s-plane, if I look upwards towards to the surface above me, obviously there will be a point on the surface. There will be a height between me and the surface point on top of me. I called this point x in my question. So is x = H(s) and is x a real number? $\endgroup$ – floppy380 Apr 24 '17 at 13:44
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Consider just the unit line (x = 1) on the complex plane. For any point one this line, the gain at that frequency is the distance from all the zeros divided by the reciprocal of the distance to all the poles. The sum of (positive or negative) angles from all those things also gives a phase at that point.

So the locations of all the zeros and poles tells you everything important about the unit line. And, for an analytic function, the behavior of the unit line (if enough of it converges) determines the rest of the complex surface, so you can ignore all the rest as redundant stuff (but maybe picturesque).

Another view. If you walk up the unit line (x=1,y= from -inf to +inf), the height of the abs(H(s)) surface above your head is the magnitude of the Fourier transform of the transfer function.

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