# What exactly is Laplace transform?

I've been working on Laplace transform for a while. I can carry it out on calculation and it's amazingly helpful. But I don't understand what exactly is it and how it works. I google and found out that it gives "less familiar" frequency view.

My question is how does Laplace Transform give frequency view?

I don't understand the connection between $f(t)$ and $\mathscr{L} (f(t))$. For example:- let $f(t) = t$, $\mathscr{L}(t)={1 \over s^2}$

$f(t)$ gives time view but how does $1 \over s^2$ give the frequency view? Somebody help me to understand what exactly is it. Thank you!!

Can anyone explain it in some physical phenomenon? Like harmonic oscillator? $$\ddot {x} + \omega_n x = f(t)$$

The Laplace transform is a useful tool for dealing with linear systems described by ODEs. As mentioned in another answer, the Laplace transform is defined for a larger class of functions than the related Fourier transform.

The 'big deal' is that the differential operator ('$\frac{d}{dt}$' or '$\frac{d}{dx}$') is converted into multiplication by '$s$', so differential equations become algebraic equations. In other words, convolution in the time or space-domain becomes multiplication in the s-domain. Another, often unspoken, 'big deal' is that the transform is unique in some sense (eg, if the transforms of two continuous functions agree, then the functions agree in the original domain). So if you can solve the problem in the s-domain, then you have solved it, in some sense, in the original domain. There is a formula for inversion, although tables are typically used for inversion. However, the inversion formula shows how the poles of the transformed functions manifest themselves in the time or space domain.

The Laplace transform comes in a few varieties; for engineering applications the most usual is the unilateral transform (behavior for $t<0$ is not relevant). Fourier transforms are often used to solve boundary value problems, Laplace transforms are often used to solve initial condition problems. Also, the Laplace transform succinctly captures input/output behavior or systems described by linear ODEs.

Regarding the 'frequency view'; instead of thinking of frequency as the $\omega$ in $\sin \omega t$, think of it as a collection of points in $\mathbb{C}$ that characterizes the behavior of $\hat{f} = \mathscr{L} f$. For example, look at the Laplace transform of $f(t) = e^{\alpha t}$, which is $\hat{f}(s) = \frac{1}{s - \alpha}$. The single point $\alpha$ (which may be complex) completely characterizes the time domain behavior. More generally, the poles and zeros of $\hat{f}$ characterize the time domain behavior of $f$. Very loosely speaking, if $\hat{f}$ has poles $p_1,...,p_n$, then we expect $f$ to have time-domain 'behaviors' of the forms $e^{p_1 t},...,e^{p_n t}$ (the zeros, and pole multiplicities of $\hat{f}$ complicate this simplistic viewpoint somewhat). So, think of the frequencies (ie, poles & zeros) as characterizing the structure of $\hat{f}$.

In your question, I think you meant the system $\ddot {x} + \omega_n^2 x = f(t)$. The unilateral transform gives $$s^2 \hat{x}(s) -s x(0) - x'(0)+ \omega_n^2 \hat{x}(s) = \hat{f}(s),$$ where $\hat{x}, \hat{f}$ are the Laplace transforms of $x,f$ respectively. This equation is typically written in the following form, which shows the relationship between the input $\hat{f}$, the (time) initial conditions $x(0), x'(0)$, and the output $\hat{x}$: $$\hat{x}(s) = \frac{s x(0) + x'(0)}{s^2 + \omega_n^2} + \frac{\hat{f}(s)}{s^2 + \omega_n^2}.$$ We can see that the $\frac{1}{s^2 + \omega_n^2}$ term 'contributes' two poles (at $s = \pm i \omega_n$) lying on the imaginary axis to $\hat{x}$. So, we expect (at least) behaviors involving $t \mapsto \sin \omega_n t$ and $t \mapsto \cos \omega_n t$.

If we take $f = 0$, you can see (meaning look up a table of transforms) that the initial conditions translate into a time function of the form $x(t) = x(0) \cos \omega_n t + \frac{x'(0)}{\omega_n} \sin \omega_n t$. So, in this particular problem, the initial conditions 'remain' forever.

If we take the system to be at rest initially (ie, take the initial conditions to be zero), then we need to know $\hat{f}$ in order to compute $\hat{x}$. If we take $f(t) = e^{i \omega t}$ (admittedly not real, but easier to compute), we have $\hat{f}(s) = \frac{1}{s-i\omega}$, which gives $\hat{x}(s) = \frac{1}{(s-i\omega)(s^2 + \omega_n^2)}$. If we take $w \neq w_n$, then using a partial fraction expansion we can write $\hat{x}(s) = \frac{1}{\omega_n^2 - \omega^2}(\frac{1}{s-i\omega} - \frac{s+i \omega}{s^2 + \omega_n^2})$, which gives $x(t) = \frac{1}{\omega_n^2 - \omega^2}(e^{i \omega t} - \cos \omega_n t - i \frac{\omega}{\omega_n} \sin \omega_n t)$. If $w = w_n$, then we obtain $\hat{x}(s) = \frac{i}{2 \omega_n}(\frac{1}{s^2+\omega_n^2} - \frac{1}{(s-i\omega_n)^2} )$, which corresponds to $x(t) = \frac{1}{2 w_n^2} \sin \omega_n t - \frac{i}{2 \omega_n} t e^{i \omega_n t}$. Notice the response in this case is unbounded, even though the input is bounded. The '$t$' term arises because of the pole of multiplicity 2 at $s = i \omega_n$.

What is your background? Are you a Mathematics major, or a Physics / Engineering major?

The purpose of the Laplace Transform is to transform ordinary differential equations (ODEs) into algebraic equations, which makes it easier to solve ODEs. However, the Laplace Transform gives one more than that: it also does provide qualitative information on the solution of the ODEs (the prime example is the famous final value theorem).

Do note that not all functions have a Fourier Transform. The Laplace Transform is a generalized Fourier Transform, since it allows one to obtain transforms of functions that have no Fourier Transforms. Does your function $f (t)$ grow exponentially with time? Then it has no FT. No problem, just multiply it by a decaying exponential that decays faster than $f$ grows, and you now have a function that has a Fourier Transform! The FT of that new function is the LT (evaluated on a line on the complex plane parallel to the imaginary axis).

If you are an engineering student who first encountered Laplace Transforms in your Signals & Systems class, then think about the name "signals & systems". Linear time-invariant (LTI) systems can be fully described by an impulse response, say $h (t)$. Laplace-transform the impulse response, and you obtain the transfer function, $H(s)$. What is the point? The point is that exponential functions (including complex exponential functions and, thus, sines and cosines) have simple Laplace Transforms. Thus, you can take a signal $x (t)$, and obtain its Laplace transform $X (s)$. What is $X (s)$? It is the transfer function of the LTI system whose impulse response is $x (t)$ itself! You have an LTI system that serves as a "signal generator", so to speak. Do you want to know how an LTI system responds to a sinusoid? Laplace-transform the sinusoid, Laplace-transform the system's impulse response, multiply the two (which corresponds to cascading the "signal generator" with the given system), and compute the inverse Laplace Transform to obtain the response. To summarize: the Laplace Transform allows one to view signals as the LTI systems that can generate them.

What is the Laplace Transform? Quoting Tim Gowers: "a mathematical object is what it does" ;-)

• Then think of the LT as a tool that simplifies the analysis of RLC circuits and mass-spring systems. The LT not only gives you the solutions of ODEs, it also helps you design a circuit / mechanical system such that it has some desired properties. In other words, the LT is useful not only for analysis, but also for synthesis. Aug 10, 2012 at 22:35
• Good answer. However, you seem to suggest that the Laplace transform is more general than the Fourier transform. This is not the case. There are many important function for which only the Fourier transform exists, but not the Laplace transform. Think of any periodic function, or impulse responses of ideal brick-wall filters such as ideal low-pass, band-pass etc. These functions can only be treated by the Fourier transform. Apr 17, 2014 at 10:51
• @RodCarvalho Doesn't transfer function imply dependence on input? If you say that your generator generates X, regardless if the input, then what is the transfer function?
– Val
Jul 27, 2014 at 16:42

use $s=iw$ then the transformation becomes Fourier transform. Then you get the frequency as $w=2\pi f$. Now you can analyze the signal at the transform domain. In the time domain you have a signal linearly increasing with time and at the transform domain the absolute value of the transform goes to $0$ when the frequency goes to infinity. It means in the signal there is always a change.. but it is not abrupt and it is not so big because asympototically you get $0$ when $f\rightarrow \infty$

• your oscillator is coming from the solution of a differential equation and the the function which satisfies it is a cosine with a particular frequency. When you get the transform you will see one specific frequency because with $s=iw$ Fourier transform is actually constructing the basis of exponentials with various frequencies. Since your signal has only one specific frequency you get only one $1$ at that specific frequency.It is a sort of projection from the time domain to the frequency domain with exponential basis. Whenever your signal has that frequency you get a dirac at that frequency. Aug 10, 2012 at 21:57
• I think it is important to point out that obtaining the Fourier transform from the Laplace transform by setting $s=i\omega$ only works if the ROC contains the imaginary axis. It definitely does not work like this in all cases. Apr 17, 2014 at 10:56
• @MattL. yes that is the reason why Laplace has a different name than Fourier. Apr 17, 2014 at 11:00
• Right, but you said "use $s=i\omega$ and the transformation becomes Fourier transform", which is not true in general, so your comment makes only limited sense. Apr 17, 2014 at 11:04
• @MattL.okay there is a misunderstanding here. I dont claim that whenever we insert $s=iw$ in the result of the Laplace transform we will get the fourier transform. This implicitely assumes that ROC is pure imaginary. I am saying the following "use $s=iw$, then the transformation becomes Fourier transform" I dont mean the result of transformation but the transformation itself. Write down the Laplace transform and put $s=iw$ in the transformation and what do you see? it gives exactly the same formula with the Fourier transform. Apr 17, 2014 at 11:15

Its easy when you question yourself how is time and frequency are related. You remember frequency is the inverse of time. So if frequency is $a$ then time is $\frac{1}{a}$. Now let us say you have a continuous set of time values that is, a continuous function representing "time" values. Lets say a high tech camera recorded numbers of meteors in a region in sky.Let us say that function is $y=t$, that is, for each sec(any appropriate time unit) same number of meteors were found in that region. For example at $10$ secs $10$ meteors were found and so on.

Now what is frequency? It is the inverse of time. So in our example we would say how many meteors were found in that region per seconds.But remember previously we plotted time vs number of meteors. This graph is fixed.Forget about that graph. We want a new graph that tells us what is the frequency corresponding to time values. We would like to plot time vs frequency. $a$ and $\frac{1}{a}$ graph. So we take each $x$ points as values of time and each $y$ point as values of frequency. What you asked can be plotted in one graph and understood easily. Now you see values of time on $x$ axis and values of frequency on $y$ axis. Take any value on time domain and find its $y$ values. You get corresponding frequency value. For $\text{time } = 1$ , $\text{ frequency } = 1$. As you can see when time approaches $0$ frequency approaches infinite. Don't ask where are the meteors in this graph, that was just for explaining what time values actually are. You can latter plot frequency vs number of meteors.

Each point in the s-plane is to be evaluated in the Laplace Transform Eq, i.e, $$H(s) = \int_0^\infty h(t)e^{-st}dt$$ Since $t$ goes from $0 \to \infty$, then $e^{-st}$ can be thought of as an infinite sequence.
Therefore the value $s$ of $e^{-st}$ provides the equation defining this sequence, and the integral of $h(t)$ to this sequence generates a measure of h(t) to this sequence.

What I still dont understand is e.g, if $h(t) = \exp(+kt)$, then when $s = +k$, i.e, $$H(s=+k) = \int_{0}^\infty e^{+kt}e^{-st} dt = \infty$$ important?
Yes, it will give the pole, but so will all other $s$ that are outside convergence?

• I am no expert -- just (re) learning this stuff myself. I think the answer to user143776's question about convergence and poles is that, if you look on a Laplace transform table, the transform of e^kt is 1/(s-k), but the right hand column of the table does point out that this is only true for "s>k" -- so I suppose this means the transform is undefined (infinite) for s >= k, just like your integral implies. This means that not only is H(s) infinite at the pole, but also when s is greater than the pole on the real axis.
– user164536
Jul 15, 2014 at 17:46