Physical Significance of the Laplace Transform

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cnh1995

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I have used Laplace transform during my EE studies to solve differential equations and in control system analysis, but we were taught that as a tool kit to make the math easier. The physical meaning was never explained. I know basic time and frequency domain concepts (thanks to Fourier series), but I am having a hard time understanding the physical reality of the Laplace transform.

As per my understanding from my textbook and notes, the s in Laplace transform is called complex frequency and is given by s=σ+jω, where the real part σ tells how fast the signal decays(exponential part) and jw corresponds to the frequency of oscillations (sinusoidal part) of the signal.

So, if f(t)=1, F(s) is 1/s or 1/(σ+jω).
If σ= -1 and w= ±1, we get |F(s)|= 0.707 and phase angle Φ= -45°. This means the signal f(t)=1 contains a an exponentially decaying sinusoidal signal of time constant 1 whose peak magnitude is 0.707 and frequency of oscillations of this decaying signal is 1 Hz. Is this correct?
If yes, does this mean if I take "all" the possible values of σ and w and compute the magnitudes and phase angles for each 1/s term (God knows how ), and add all of them, will that addition result in f(t)=1?

Please correct me if I am wrong. Thanks a lot in advance.

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jim hardy

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but I am having a hard time understanding the physical reality of the Laplace transform.
i would love to grasp that concept myself.

One doesn't have to understand impulse and momentum to use a hammer
but it does enrich the experience.

I suspect you're on the right track
recall that the Laplace transform is integral of product of time function and e-st
so i think of it as "hammering" a function with all possible frequencies

but that only gives me confidence that it's a valid tool..

I use it to crack differential equations like a monkey would a hammer to crack a coconut - with no genuine understanding

old jim

anorlunda

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https://en.wikipedia.org/wiki/Laplace_transform said:
The Laplace transform is similar to the Fourier transform. While the Fourier transform of a function is a complex function of a realvariable (frequency), the Laplace transform of a function is a complex function of a complex variable. Laplace transforms are usually restricted to functions of t with t ≥ 0. A consequence of this restriction is that the Laplace transform of a function is a holomorphic function of the variable s. Unlike the Fourier transform, the Laplace transform of a distribution is generally a well-behaved function. Techniques of complex variables can also be used to directly study Laplace transforms. As a holomorphic function, the Laplace transform has a power series representation. This power series expresses a function as a linear superposition of moments of the function. This perspective has applications in probability theory.
I can't say that paragraph does for me except to say "The Laplace transform is similar to the Fourier transform. " That satisfies my curiosity, and it fits with the intuitive feel that it lets us work in the frequency domain, with all the advantages that come with that.

The Z transform is also a cousin and it can be very useful in digital systems where we work with samples and difference equations. A simulation engineer facile in t-s-z domains is well equipped.

Baluncore

The fourier transform converts between two orthogonal planes, time and frequency. In the frequency domain it cannot represent the time something happened, or how it varied during the time window.

A step impulse and the resulting exponential decaying sine wave cannot be represented in the frequency domain by the fourier transform. It takes complex frequency to represent the decaying signals. That is the domain of the Laplace transform.

You might think of Laplace transform space as a plane or a volume, at an angle part way between the time plane and the frequency plane.

jasonRF

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If you understand Fourier transforms then Laplace transforms can usually be thought of in the same way. So I will try to show how they are related, using three examples that illustrate the three general cases.

Let's start with the Fourier transform of a function $f(t)$
$$\hat{F}(\omega) = \int_{-\infty}^\infty f(t) \, e^{-j\omega t} \, dt$$
In electrical engineering we require the variable $\omega$ to be real. If you think of the functions as vectors and the integral as a dot product, then the Fourier transform is projecting the function $f(t)$ onto the space of exponentials $e^{j\omega t}$. The inverse transform then reconstructs the original function
$$f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{F}(\omega) \, e^{j\omega t} \, d\omega$$
So you are just adding up all of the frequency components with the correct amplitude and phase.

For the Laplace transform, I am assuming that you are using the unilateral Laplace transform, which is for functions that are zero for $t<0$. In this case, for a given function $f(t)$ the Laplace transform is
$$F(s) = \int_0^\infty f(t) \, e^{-st} dt$$
Here the variable $s$ is a complex variable, and the Laplace transform in general only converges for a right-half plane $Re(s)>\sigma_0$ (that is, the real part of $s$ larger than some real number $\sigma_0$). This region over which the transform converges is called the Region of Convergence (ROC). The Laplace transform is projecting the function $f(t)$ onto the space of exponentials $e^{st}$. The inverse Laplace transform reconstructs the original function, but is now a contour integral in the complex plane; the contour is a vertical line that is inside the ROC:
$$f(t) = \frac{1}{j2\pi}\int_{\sigma_1-j\infty}^{\sigma_1+j\infty} F(s) \, e^{s t} \, ds$$
for any $\sigma_1>\sigma_0$. So the weird thing is that when you reconstruct the original function, you only need to add up the components along $s=\sigma_1 + j\omega$ for a single value of $\sigma_1$, but all values of $\omega$.

Why use the Laplace transform at all, when we have the Fourier transform? Well, it allows us to analyze signals that grow exponentially, whereas Fourier does not. Let's look at a few examples. Below I will use the unit step function $u(t)$ which is $1$ for $t>0$ and 0 for $t<0$.

1. A damped exponential, $f(t) = u(t) e^{-t} e^{j\omega_0 t}$. In this case $F(s) = 1/(s + 1 - j\omega_0)$ with ROC $Re(s)>-1$. The Fourier transform is $\hat{F}(\omega) = 1/(j\omega - j\omega_0 + 1)$. Note that in this case $\hat{F}(\omega) = F(j\omega)$; this will always be true when the imaginary axis $Re(s)=0$ is inside the ROC. The Fourier and Laplace transforms are equivalent for this case.

2. A growing exponential, $f(t) = u(t) e^{t} e^{j\omega_0 t}$. In this case $F(s) = 1/(s - 1 - j\omega_0)$ with ROC $Re(s)>1$. Note that the Fourier transform does not exist, which is always the case when the imaginary axis $Re(s)=0$ is outside of the ROC. So for this case the Laplace transform is useful, while the Fourier transform is not.

3. Your example, $f(t) = u(t)$. In this case $F(s) = 1/s$ and the ROC is $Re(s)>0$. This case brings up another difference between the way electrical engineers usually work with Fourier transforms, in that we allow the Fourier transforms to be generalized functions. For your example, the Fourier transform is $\hat{F}(\omega) = 1/(j\omega) + \pi \delta(\omega)$, where $\delta(w)$ is the Dirac delta function. Note that
$\hat{F}(\omega) \neq F(j\omega)$ for this case, which is usually true when the imaginary axis $Re(s)=0$ is the boundary of the ROC. The Fourier transform usually includes generalized functions such as the delta function (or its derivatives) for this case. Clearly the Fourier and Laplace transforms are not equivalent in this case; you should use whichever is more useful for your task at hand. For controls, Laplace is more useful. For some signal analysis problems Fourier may be more useful.

So, if f(t)=1, F(s) is 1/s or 1/(σ+jω).
If σ= -1 and w= ±1, we get |F(s)|= 0.707 and phase angle Φ= -45°. This means the signal f(t)=1 contains a an exponentially decaying sinusoidal signal of time constant 1 whose peak magnitude is 0.707 and frequency of oscillations of this decaying signal is 1 Hz. Is this correct?
No. Since $s=-1+j\omega$ is outside the ROC, you cannot in general evaluate $F$ there and expect to get anything meaningful.

Hope that helped.

Jason

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• • TobilobaEinstein, Asymptotic, Delta2 and 4 others

cnh1995

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Thank you @jasonRF for your detailed explanation. Looks like I need to carefully study the concept of ROC. Will post here if I get stuck.

Thanks all for your inputs! phinds

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I use it to crack differential equations like a monkey would a hammer to crack a coconut - with no genuine understanding
Hey, at least you cracked them. When I started out in diff eq it was more like a monkey cracking ITSELF in the head with a hammer • • Asymptotic, PhanthomJay, jim hardy and 1 other person

eq1

Gold Member
You might want to check out this YouTube video. It basically just echos what JasonRF said, but it does it graphically. The MIT OCW diff eq class has a great lecture where the instructor derives Laplace as well, also using basically the same line of reasoning.

jasonRF

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I really like the video - Thanks! He is actually using the bilateral Laplace transform (so the function can be defined for all $t$, not just $t\geq 0$), so the ROC is a strip in the complex plane: $\sigma_a < Re(s) < \sigma_b$. The three cases I outlined stay the same - it just matters whether the imaginary axis is inside the ROC, outside the ROC, or a boundary of the ROC.

I would have loved taking a class from that professor. He is amazingly clear.

Jason

• Asymptotic and cnh1995

cnh1995

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does this mean if I take "all" the possible values of σ and w and compute the magnitudes and phase angles for each 1/s term (God knows how ), and add all of them, will that addition result in f(t)=1?
So this should only work in the ROC for F(s)=1/s, right?

In general, if I calculate the magnitude (M) and phase (Φ) of Laplace transform at a particular value of σ and w (in ROC), M represents the peak of the sinusoidal component, w is the frequency of that component and σ tells how fast that component is decaying (or growing, if σ is positive). Is this correct? Sorry if this sounds repetitive, I am trying to think Laplace quantities along the lines of Fourier series coefficients .

cnh1995

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So this should only work in the ROC for F(s)=1/s, right?

In general, if I calculate the magnitude (M) and phase (Φ) of Laplace transform at a particular value of σ and w (in ROC), M represents the peak of the sinusoidal component, w is the frequency of that component and σ tells how fast that component is decaying (or growing, if σ is positive). Is this correct? Sorry if this sounds repetitive, I am trying to think Laplace quantities along the lines of Fourier series coefficients .
I strongly feel I am missing something in this reasoning.

Also, Prof Douglas said in the video, "Any value of s that doesn't make the product of Impulse Response and e-st (for all t) either zero or barely infinite is some boring number that we don't really care about". Looks like I am asking about the physical significance of the same boring number in the post above.
So, in ROC, what is physical meaning of the magnitude of the Laplace transform at a particular location? At poles, it shoots up to infinity and at zeros, it drops down to zero. What about other values in the ROC?

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berkeman

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Oh dear, thread is closed temporarily for Moderation...

berkeman

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Okay, after a bit of cleanup, the thread is re-opened. Thanks for your patience.

• cnh1995

eq1

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So, in ROC, what is the physical meaning of the magnitude of the Laplace transform at a particular location?
It's just the numeric result of the prescribed computation. It's not a very satisfactory answer but it is the right answer. Personally, I prefer to think of the equation as a machine and rather than trying to assign physical meaning to the output of the machine (the magnitude of the Laplace transform at a particular location) assign physical meaning to each of the computational steps in the equation. Figure out why that step is there and what its doing.

Below is another youtube video to put you on the right track. It is for Fourier but, as established, Fourier is just Laplace with no decay. The video is not a strictly correct interpretation of Fourier but it will teach how to think about the equation.

So, for Laplace the real question is, what is the difference between $f(t) e^{-jwt}$ and $f(t) e^{-σt}e^{-jwt}=f(t) e^{-(σ+jw)t}=f(t) e^{-st}$? Answer that and then you'll have it.

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• boB_K7IQ and cnh1995

AVBs2Systems

Hi.

I would like to add another aspect which I believe might contribute a little more to the understanding.
$$\textbf{The role of the damping exponential} \,\,\,\, e^{-(\sigma t) }$$
The Laplace transform extends the set of signals for which the notion of a spectrum is available. It does this by adding an exponentially decaying part to the signal. In this post I deal with the one sided laplace transform.
$$\displaystyle \int_{0}^{\infty}{e^{- (\sigma t)}}_{\small \text{Damper} } \cdot f(t) \cdot e^{- j( \omega t) } \,\,\,\, \text{d}t$$
The role of the damping exponential is to ensure the signal decays to zero and does not blow up as the integral is taken to infinity.
We know that the fourier transform for many signals does not exist, but their laplace transform does. Three of these signals are:
$$\epsilon(t) , t, t^{n}$$

The laplace transform only exists for the values of $\sigma$ such that:
$$\displaystyle \int_{0}^{\infty}{e^{- (\sigma t)}} \cdot |f(t)| \,\,\,\, \text{d}t \in \mathbb{R}$$
Notice:
$$\displaystyle \int_{0}^{\infty}{e^{- (\sigma t)}}_{\small \text{Damper} } \cdot f(t) \cdot e^{- j( \omega t) } \,\,\,\, \text{d}t = \displaystyle \int_{0}^{\infty} f(t) \cdot e^{-t (\sigma +j \omega ) } \,\,\,\, \text{d}t$$
Now above, depending on $f(t)$ the value of $\sigma$ must be appropriately restricted, so that after multiplying $$- t \cdot \sigma$$ we still get a negative damping that is of a large enough magnitude to ensure the integral converges.

Look at this signal for example:
$$f(t) = e^{ (t)}$$
Now:
$$\displaystyle \int_{0}^{\infty}{e^{- (\sigma t)}} \cdot e^{( t)} \cdot e^{- j( \omega t) } \,\,\,\, \text{d}t = \displaystyle \int_{0}^{\infty} e^{ - t( \sigma + j \omega - 1)} \,\,\,\, \text{d}t$$
This:
$$\Re \bigg( \sigma + j \omega - 1 \bigg) \gt 0 \iff \sigma \gt 1$$
For this signal, the damping factor has to be greater than 1 for its laplace transform to exist, if its less, then there would be no damping. If sigma was negative, youd get the minus t multiplied by a negative sigma, which is positive giving a growing exponential, not a damped one. The range of values of the damping factor that are permitted, are what you specify when you determine the region of convergence, and there are some other subtleties here also.

$$\textbf{The fourier transform is a special case of the laplace transform when s} \,\,\, = j \omega$$
The fourier transform is simply a special case of the laplace transform when s = j w.
$$\displaystyle \int_{0}^{\infty} f(t) \cdot e^{- ( s t) } \,\,\,\, \text{d}t \,\,\, s= j \omega \iff \displaystyle \int_{0}^{\infty} f(t) \cdot e^{- j( \omega t) }$$
The fourier transform exists for BIBO stable systems and absolutely integrable signals only, and these are systems whose region of convergence includes the imaginary axis ( real part zero ).

The Laplace transform exists for a larger class of signals and systems, they do not have to be bibo stable or absolutely integrable.

Evaluating the systems transfer function at $s = j \omega$ gives the frequency domain representation. We mostly use the laplace transform for stability of the system, and finding the exact nature of stability (BIBO or not) and the impulse response. I cannot think how one should explain taking the magnitude of $F(s)$ physically.
$$\textbf{System stability and the pole zero map : }$$

I will cover the above in a later post, because I do need to re read this carefully to answer in a satisfactory depth.

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• jim hardy and cnh1995

AVBs2Systems

f(t)=1 contains a an exponentially decaying sinusoidal signal of time constant 1 whose peak magnitude is 0.707 and frequency of oscillations of this decaying signal is 1 Hz. Is this correct?
Hi
No unfortunately not, because the mathematical definition of $f(t)$ is:
$$f(t) = 1$$
Whereas an exponentially decaying sinusoidal signal of time constant 1 whose peak magnitude is 0.707 and frequency of oscillations of this decaying signal is 1 Hz is this:
$$g(t) = \dfrac{e^{-t} \cdot \cos(t )}{\sqrt{2}}$$
There is no point where they are equal that I can currently think of, but I did not add a phase to the sinusoid, either way there will be maximum one, or at best two points where the functions equal one, because the damping is there. :

I say $$\forall t \,\,\,\, f(t) \ne g(t)$$
I feel that if we are creative and smart. we can always add another phase shifted and equally exponentially growing sinusoid to always result in one. But this is not anything related to the spectrum or the laplace transform.

The alternate representations of signals in terms of base signals such as sinusoids, as you know, are only fourier series and the fourier transform.
So I think this line of thinking in terms of interpreting the $F(s)$ (the s domain) is not valid.

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• anorlunda and cnh1995

anorlunda

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The fourier transform is a special case of the laplace transform when s=jω​
Thank you. That's a useful way to view it.

• AVBs2Systems

Kyle Gonterwitz

One physical application of the laplace transform that helps to explain the phsical significance is in automatic flight control systems in aerospace applications.

In an oversimplified nutshell... Flight control surface responses can be measured and plotted in Bode Diagrams. Laplace transforms can be applied to the diagrams to break them down into algebraic formulae resulting in gain values apllied to control circuits and stability root locus plots.

Im not sure Im saying that all correctly but for more info look into Bode Diagrams, control systems theory, and root locus plots. The application of this mathematics to solving the physical problem in digital feedback control systems is one of the most elegant applications of mathematics to solve a physical problem imho.

rude man

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A step impulse and the resulting exponential decaying sine wave cannot be represented in the frequency domain by the fourier transform.
It can but the math is not as neat.
For example, the Fourier transform of the infinitely-extended sin(wt) is jw[δ(w+w0) - δ(w-w0)] with δ(x) the Dirac delta function (or "distribution" for the fusspots).

In general, any function not meeting the Dirichlet conditions can still be prepresented by a Fourier integral but the transform will always include the Dirac delta function:

As a practical matter the Laplace is usually the preferred transform for non-Dirichlet functions.

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• AVBs2Systems and berkeman

rude man

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I have used Laplace transform during my EE studies to solve differential equations and in control system analysis, but we were taught that as a tool kit to make the math easier. The physical meaning was never explained. I know basic time and frequency domain concepts (thanks to Fourier series), but I am having a hard time understanding the physical reality of the Laplace transform.
I recommend H.H. Skilling, Electrical Engineering Circuits, John Wiley & Sons, hopefully still in print or go to ebay, for a good explanation of how the integral for the coefficients of the complex Fourier series evolves to the Fourier integral to the Laplace integral. It's good to let an expert explain this rather difficult subject.

• cnh1995

AVBs2Systems

It can but the math is not as neat.
For example, the Fourier transform of the infinitely-extended sin(wt) is jwδ(w+w0) - δ(w-w0) with δ(x) the Dirac delta function (or "distribution" for the fusspots).

In general, any function not meeting the Dirichlet conditions can still be prepresented by a Fourier integral but the transform will always include the Dirac delta function:

As a practical matter the Laplace is usually the preferred transform for non-Dirichlet functions.
Hi.
I also agree with the post, because the fourier transform of the dirac impulse and the sine and cosine both exist.
If by step impulse it is meant that:
$$\delta(t)$$
Then the fourier transform exists and is a constant 1, which is the same as its laplace transform.
$$\delta(t) \iff X(j \omega) = 1 = X(s)$$
And for a sine:
$$\sin( \omega_{0} t) = \dfrac{1}{2 j } \cdot \big( e^{j(\omega_{0} t) } - e^{j(\omega_{0} t) } \big) \iff X(j \omega) = \dfrac{j \pi}{2} \delta( \omega + \omega_{0}) - \dfrac{j \pi}{2} \delta( \omega - \omega_{0} )$$
The fourier transform of the decaying sinusoid also exists:
$$f(t) = e^{-(\sigma t)} \cdot \cos(\omega_{0} t)$$
And is:
$$X(j \omega) = \dfrac{\sigma + j \omega }{-\omega^{2} + \omega_{0}^2 + a^{2} + j 2 \sigma \omega }$$
Though initially I would have thought to use convolution in the freqeuncy domain as the exponentially decaying sinusoid can be thought of as the product of two functions. I think the dirac pulse is the only function whose fourier and laplace transform are equal.

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rude man

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And BTW thanks for the link. A very good treatise on Fourier and apparently a lot of other stuff too.

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• AVBs2Systems

AVBs2Systems

And BTW thanks for the link. A very dood treatise on Fourier and apparently a lot of other stuff too.
It says not secure on google but it seems like a good site created by signal processing lovers like us out of altruism hehe and thank you you are welcome.

• rude man

rude man

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I think the dirac pulse is the only function whose fourier and laplace transform are equal.
Yes, but if you set s = jw they're all the same. This is of course the basis for the phasor transform.

One should distinguish between one-sided and two-sided Fourier transforms. You gave the 2-sided for sin(wt) and 1-sided for the damped sinusoid. You can get the one-sided sinusoid by setting a=0 in your damped sinusoid transform. For the two-sided sinusoid the Fourier transform would also have included delta functions. (But then I guess it wouldn't have been damped, would it? )

• AVBs2Systems

AVBs2Systems

Yes, but if you set s = jw they're all the same. This is of course the basis for the phasor transform.

One should distinguish between one-sided and two-sided Fourier transforms. You gave the 2-sided for sin(wt) and 1-sided for the damped sinusoid. You can get the one-sided sinusoid by setting a=0 in your damped sinusoid transform. For the two-sided sinusoid the Fourier transform would also have included delta functions. (But then I guess it wouldn't have been damped, would it? )
Correct!
Actually in our studies we always assumed the damped sinusoid begins at t = 0
So I should have redefined it as:
$$f(t) = \epsilon(t) \cdot e^{-(\sigma t)} \cdot \cos(\omega_{0} t)$$

Where $\epsilon(t)$is the unit step function, so the function is zero for all t less than zero. Which I have done now.

Now we use, double sided fourier transform, but one sided laplace transform. I believe the reason is to do with causality, because for a causal system the impulse response is zero for negative time.
I will have to ask them why because I am not sure. • rude man

"Physical Significance of the Laplace Transform"

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