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[Explore Stack Internal](https://stackoverflow.co/internal/?utm_medium=referral&utm_source=math-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover) # [What exactly is Laplace transform?](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform) [Ask Question](https://math.stackexchange.com/questions/ask) Asked 13 years, 8 months ago Modified [10 years, 1 month ago](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform?lastactivity "2016-02-16 04:38:19Z") Viewed 138k times This question shows research effort; it is useful and clear 83 Save this question. Show activity on this post. 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) f ( t ) and L(f(t)) L ( f ( t ) ). For example:- let f(t)\=t f ( t ) \= t, L(t)\=1s2 L ( t ) \= 1 s 2 f(t) f ( t ) gives time view but how does 1s2 1 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? x¨\+ωnx\=f(t) x ¨ \+ ω n x \= f ( t ) - [laplace-transform](https://math.stackexchange.com/questions/tagged/laplace-transform "show questions tagged 'laplace-transform'") [Share](https://math.stackexchange.com/q/181160 "Short permalink to this question") Share a link to this question Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this question to receive notifications [edited Feb 16, 2016 at 4:38](https://math.stackexchange.com/posts/181160/revisions "show all edits to this post") [](https://math.stackexchange.com/users/1796/tomcuchta) [tomcuchta](https://math.stackexchange.com/users/1796/tomcuchta) 3,43511 gold badge1919 silver badges2525 bronze badges asked Aug 10, 2012 at 21:07 [](https://math.stackexchange.com/users/34351/hasexams) [hasExams](https://math.stackexchange.com/users/34351/hasexams) 2,38955 gold badges2626 silver badges3838 bronze badges 6 - 1 Have a look at this: [math.stackexchange.com/questions/6661/…](http://math.stackexchange.com/questions/6661/laplace-transformations-for-dummies "laplace transformations for dummies") M Turgeon – [M Turgeon](https://math.stackexchange.com/users/19379/m-turgeon "10,925 reputation") 2012-08-10 21:27:49 +00:00 [Commented Aug 10, 2012 at 21:27](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment417485_181160) - 2 [cambridge.org/us/features/chau/webnotes/chap2laplace.pdf](http://www.cambridge.org/us/features/chau/webnotes/chap2laplace.pdf) dato datuashvili – [dato datuashvili](https://math.stackexchange.com/users/3196/dato-datuashvili "1 reputation") 2012-08-10 21:51:29 +00:00 [Commented Aug 10, 2012 at 21:51](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment417495_181160) - 1 here is additional link [phy.duke.edu/~hx3/physics/FourierLaplace.pdf](http://www.phy.duke.edu/~hx3/physics/FourierLaplace.pdf) dato datuashvili – [dato datuashvili](https://math.stackexchange.com/users/3196/dato-datuashvili "1 reputation") 2012-08-10 22:09:32 +00:00 [Commented Aug 10, 2012 at 22:09](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment417509_181160) - 2 It's the same thing as with generating functions in combinatorics: It's a **very helpful** formal device, but for heaven's sake don't think that it has **any** intuitive physical, analytical, or geometrical meaning. Christian Blatter – [Christian Blatter](https://math.stackexchange.com/users/1303/christian-blatter "234,246 reputation") 2014-07-15 18:31:55 +00:00 [Commented Jul 15, 2014 at 18:31](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1790056_181160) - 1 [math.stackexchange.com/questions/428408/…](https://math.stackexchange.com/questions/428408/physical-interpretation-of-laplace-transforms/2156002#comment6701804_2156002 "physical interpretation of laplace transforms") D.R. – [D.R.](https://math.stackexchange.com/users/405572/d-r "11,427 reputation") 2019-11-23 20:49:03 +00:00 [Commented Nov 23, 2019 at 20:49](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment7088342_181160) \| [Show **1** more comment](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform "Expand to show all comments on this post") ## 5 Answers 5 Sorted by: [Reset to default](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform?answertab=scoredesc#tab-top) This answer is useful 57 Save this answer. Show activity on this post. 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 ('ddt d d t' or 'ddx d d x') is converted into multiplication by 's 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 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 ω ω in sinωt sin ⁡ ω t, think of it as a collection of points in C C that characterizes the behavior of f^\=Lf f ^ \= L f. For example, look at the Laplace transform of f(t)\=eαt f ( t ) \= e α t, which is f^(s)\=1s−α f ^ ( s ) \= 1 s − α. The single point α α (which may be complex) completely characterizes the time domain behavior. More generally, the poles and zeros of f^ f ^ characterize the time domain behavior of f f. Very loosely speaking, if f^ f ^ has poles p1,...,pn p 1 , . . . , p n, then we expect f f to have time-domain 'behaviors' of the forms ep1t,...,epnt e p 1 t , . . . , e p n t (the zeros, and pole multiplicities of f^ f ^ complicate this simplistic viewpoint somewhat). So, think of the frequencies (ie, poles & zeros) as characterizing the structure of f^ f ^. In your question, I think you meant the system x¨\+ω2nx\=f(t) x ¨ \+ ω n 2 x \= f ( t ). The unilateral transform gives s2x^(s)−sx(0)−x′(0)\+ω2nx^(s)\=f^(s), s 2 x ^ ( s ) − s x ( 0 ) − x ′ ( 0 ) \+ ω n 2 x ^ ( s ) \= f ^ ( s ) , where x^,f^ x ^ , f ^ are the Laplace transforms of x,f x , f respectively. This equation is typically written in the following form, which shows the relationship between the input f^ f ^ , the (time) initial conditions x(0),x′(0) x ( 0 ) , x ′ ( 0 ) , and the output x^ x ^ : x^(s)\=sx(0)\+x′(0)s2\+ω2n\+f^(s)s2\+ω2n. x ^ ( s ) \= s x ( 0 ) \+ x ′ ( 0 ) s 2 \+ ω n 2 \+ f ^ ( s ) s 2 \+ ω n 2 . We can see that the 1s2\+ω2n 1 s 2 \+ ω n 2 term 'contributes' two poles (at s\=±iωn s \= ± i ω n ) lying on the imaginary axis to x^ x ^ . So, we expect (at least) behaviors involving t↦sinωnt t ↦ sin ⁡ ω n t and t↦cosωnt t ↦ cos ⁡ ω n t . If we take f\=0 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)cosωnt\+x′(0)ωnsinωnt x ( t ) \= x ( 0 ) cos ⁡ ω n t \+ x ′ ( 0 ) ω n sin ⁡ ω 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 f^ f ^ in order to compute x^ x ^. If we take f(t)\=eiωt f ( t ) \= e i ω t (admittedly not real, but easier to compute), we have f^(s)\=1s−iω f ^ ( s ) \= 1 s − i ω, which gives x^(s)\=1(s−iω)(s2\+ω2n) x ^ ( s ) \= 1 ( s − i ω ) ( s 2 \+ ω n 2 ). If we take w≠wn w ≠ w n, then using a partial fraction expansion we can write x^(s)\=1ω2n−ω2(1s−iω−s\+iωs2\+ω2n) x ^ ( s ) \= 1 ω n 2 − ω 2 ( 1 s − i ω − s \+ i ω s 2 \+ ω n 2 ), which gives x(t)\=1ω2n−ω2(eiωt−cosωnt−iωωnsinωnt) x ( t ) \= 1 ω n 2 − ω 2 ( e i ω t − cos ⁡ ω n t − i ω ω n sin ⁡ ω n t ). If w\=wn w \= w n, then we obtain x^(s)\=i2ωn(1s2\+ω2n−1(s−iωn)2) x ^ ( s ) \= i 2 ω n ( 1 s 2 \+ ω n 2 − 1 ( s − i ω n ) 2 ), which corresponds to x(t)\=12w2nsinωnt−i2ωnteiωnt x ( t ) \= 1 2 w n 2 sin ⁡ ω n t − i 2 ω n t e i ω n t. Notice the response in this case is unbounded, even though the input is bounded. The 't t' term arises because of the pole of multiplicity 2 at s\=iωn s \= i ω n. [Share](https://math.stackexchange.com/a/181261 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this answer to receive notifications [edited Aug 11, 2012 at 8:30](https://math.stackexchange.com/posts/181261/revisions "show all edits to this post") answered Aug 11, 2012 at 7:35 [](https://math.stackexchange.com/users/27978/copper-hat) [copper.hat](https://math.stackexchange.com/users/27978/copper-hat) 179k1010 gold badges128128 silver badges275275 bronze badges [Add a comment](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| This answer is useful 16 Save this answer. Show activity on this post. 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](http://en.wikipedia.org/wiki/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) 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 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) h ( t ). Laplace-transform the impulse response, and you obtain the transfer function, H(s) 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) x ( t ), and obtain its Laplace transform X(s) X ( s ). What is X(s) X ( s )? It is the transfer function of the LTI system whose impulse response is x(t) 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](http://books.google.com/books/about/Mathematics_A_Very_Short_Introduction.html?id=DBxSM7TIq48C): "a mathematical object is what it does" ;-) [Share](https://math.stackexchange.com/a/181181 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this answer to receive notifications [edited Aug 10, 2012 at 22:46](https://math.stackexchange.com/posts/181181/revisions "show all edits to this post") answered Aug 10, 2012 at 22:26 [](https://math.stackexchange.com/users/19003/rod-carvalho) [Rod Carvalho](https://math.stackexchange.com/users/19003/rod-carvalho) 3,0691818 silver badges1313 bronze badges 3 - 1 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. Rod Carvalho – [Rod Carvalho](https://math.stackexchange.com/users/19003/rod-carvalho "3,069 reputation") 2012-08-10 22:35:27 +00:00 [Commented Aug 10, 2012 at 22:35](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment417519_181181) - 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. Matt L. – [Matt L.](https://math.stackexchange.com/users/70664/matt-l "10,994 reputation") 2014-04-17 10:51:41 +00:00 [Commented Apr 17, 2014 at 10:51](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1575702_181181) - @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 – [Val](https://math.stackexchange.com/users/39930/val "1,529 reputation") 2014-07-27 16:42:35 +00:00 [Commented Jul 27, 2014 at 16:42](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1815268_181181) [Add a comment](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| This answer is useful 4 Save this answer. Show activity on this post. use s\=iw s \= i w then the transformation becomes Fourier transform. Then you get the frequency as w\=2πf w \= 2 π 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 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 0 when f→∞ f → ∞ [Share](https://math.stackexchange.com/a/181166 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this answer to receive notifications answered Aug 10, 2012 at 21:35 [](https://math.stackexchange.com/users/29940/seyhmus-g%C3%BCng%C3%B6ren) [Seyhmus Güngören](https://math.stackexchange.com/users/29940/seyhmus-g%C3%BCng%C3%B6ren) 8,00033 gold badges2828 silver badges5050 bronze badges 5 - 1 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 s \= i w 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 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. Seyhmus Güngören – [Seyhmus Güngören](https://math.stackexchange.com/users/29940/seyhmus-g%C3%BCng%C3%B6ren "8,000 reputation") 2012-08-10 21:57:20 +00:00 [Commented Aug 10, 2012 at 21:57](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment417497_181166) - 1 I think it is important to point out that obtaining the Fourier transform from the Laplace transform by setting s\=iω s \= i ω only works if the ROC contains the imaginary axis. It definitely does not work like this in all cases. Matt L. – [Matt L.](https://math.stackexchange.com/users/70664/matt-l "10,994 reputation") 2014-04-17 10:56:18 +00:00 [Commented Apr 17, 2014 at 10:56](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1575710_181166) - @MattL. yes that is the reason why Laplace has a different name than Fourier. Seyhmus Güngören – [Seyhmus Güngören](https://math.stackexchange.com/users/29940/seyhmus-g%C3%BCng%C3%B6ren "8,000 reputation") 2014-04-17 11:00:55 +00:00 [Commented Apr 17, 2014 at 11:00](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1575714_181166) - Right, but you said "use s\=iω s \= i ω and the transformation becomes Fourier transform", which is not true in general, so your comment makes only limited sense. Matt L. – [Matt L.](https://math.stackexchange.com/users/70664/matt-l "10,994 reputation") 2014-04-17 11:04:15 +00:00 [Commented Apr 17, 2014 at 11:04](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1575720_181166) - @MattL.okay there is a misunderstanding here. I dont claim that whenever we insert s\=iw s \= i w 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 s \= i w , 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 s \= i w in the transformation and what do you see? it gives exactly the same formula with the Fourier transform. Seyhmus Güngören – [Seyhmus Güngören](https://math.stackexchange.com/users/29940/seyhmus-g%C3%BCng%C3%B6ren "8,000 reputation") 2014-04-17 11:15:15 +00:00 [Commented Apr 17, 2014 at 11:15](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1575734_181166) [Add a comment](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| This answer is useful 3 Save this answer. Show activity on this post. 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 a then time is 1a 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 y \= t, that is, for each sec(any appropriate time unit) same number of meteors were found in that region. For example at 10 10 secs 10 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 a and 1a 1 a graph. So we take each x x points as values of time and each y 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 x axis and values of frequency on y y axis. Take any value on time domain and find its y y values. You get corresponding frequency value. For time \=1 time \= 1 , frequency \=1 frequency \= 1. As you can see when time approaches 0 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. [Share](https://math.stackexchange.com/a/1214018 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this answer to receive notifications [edited Oct 15, 2015 at 8:28](https://math.stackexchange.com/posts/1214018/revisions "show all edits to this post") [](https://math.stackexchange.com/users/-1/community) [Community](https://math.stackexchange.com/users/-1/community)Bot 1 answered Mar 31, 2015 at 7:12 [](https://math.stackexchange.com/users/227692/max-hazard) [Max Hazard](https://math.stackexchange.com/users/227692/max-hazard) 3122 bronze badges [Add a comment](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| This answer is useful 1 Save this answer. Show activity on this post. Each point in the s-plane is to be evaluated in the Laplace Transform Eq, i.e, H(s)\=∫∞0h(t)e−stdt H ( s ) \= ∫ 0 ∞ h ( t ) e − s t d t Since t t goes from 0→∞ 0 → ∞ , then e−st e − s t can be thought of as an infinite sequence. Therefore the value s s of e−st e − s t provides the equation defining this sequence, and the integral of h(t) 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) h ( t ) \= exp ⁡ ( \+ k t ), then when s\=\+k s \= \+ k, i.e, H(s\=\+k)\=∫∞0e\+kte−stdt\=∞ H ( s \= \+ k ) \= ∫ 0 ∞ e \+ k t e − s t d t \= ∞ important? Yes, it will give the pole, but so will all other s s that are outside convergence? [Share](https://math.stackexchange.com/a/757596 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this answer to receive notifications [edited Apr 17, 2014 at 10:45](https://math.stackexchange.com/posts/757596/revisions "show all edits to this post") [](https://math.stackexchange.com/users/88595/user88595) [user88595](https://math.stackexchange.com/users/88595/user88595) 4,58911 gold badge2626 silver badges3131 bronze badges answered Apr 17, 2014 at 10:22 [](https://math.stackexchange.com/users/143776/user143776) [user143776](https://math.stackexchange.com/users/143776/user143776) 1111 bronze badge 1 - 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 – user164536 2014-07-15 17:46:14 +00:00 [Commented Jul 15, 2014 at 17:46](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform#comment1790765_757596) [Add a comment](https://math.stackexchange.com/questions/181160/what-exactly-is-laplace-transform "Use comments to ask for more information or suggest improvements. 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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 ('d d t' or 'd d x') 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 ω in sin ⁡ ω t, think of it as a collection of points in C that characterizes the behavior of f ^ \= L f. For example, look at the Laplace transform of f ( t ) \= e α t, which is f ^ ( s ) \= 1 s − α. The single point α (which may be complex) completely characterizes the time domain behavior. More generally, the poles and zeros of f ^ characterize the time domain behavior of f. Very loosely speaking, if 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 f ^ complicate this simplistic viewpoint somewhat). So, think of the frequencies (ie, poles & zeros) as characterizing the structure of f ^. In your question, I think you meant the system x ¨ \+ ω n 2 x \= f ( t ). The unilateral transform gives s 2 x ^ ( s ) − s x ( 0 ) − x ′ ( 0 ) \+ ω n 2 x ^ ( s ) \= f ^ ( s ) , where x ^ , 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 f ^, the (time) initial conditions x ( 0 ) , x ′ ( 0 ), and the output x ^: x ^ ( s ) \= s x ( 0 ) \+ x ′ ( 0 ) s 2 \+ ω n 2 \+ f ^ ( s ) s 2 \+ ω n 2 . We can see that the 1 s 2 \+ ω n 2 term 'contributes' two poles (at s \= ± i ω n) lying on the imaginary axis to x ^. So, we expect (at least) behaviors involving t ↦ sin ⁡ ω n t and t ↦ cos ⁡ ω 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 ⁡ ω n t \+ x ′ ( 0 ) ω n sin ⁡ ω 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 f ^ in order to compute x ^. If we take f ( t ) \= e i ω t (admittedly not real, but easier to compute), we have f ^ ( s ) \= 1 s − i ω, which gives x ^ ( s ) \= 1 ( s − i ω ) ( s 2 \+ ω n 2 ). If we take w ≠ w n, then using a partial fraction expansion we can write x ^ ( s ) \= 1 ω n 2 − ω 2 ( 1 s − i ω − s \+ i ω s 2 \+ ω n 2 ), which gives x ( t ) \= 1 ω n 2 − ω 2 ( e i ω t − cos ⁡ ω n t − i ω ω n sin ⁡ ω n t ). If w \= w n, then we obtain x ^ ( s ) \= i 2 ω n ( 1 s 2 \+ ω n 2 − 1 ( s − i ω n ) 2 ), which corresponds to x ( t ) \= 1 2 w n 2 sin ⁡ ω n t − i 2 ω n t e i ω 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 ω n.