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Until now, we have only investigated Fourier exponential transformations as a type of integrated transformation.Fourier -transformer is useful in endless domains.Laplace -transformationIn their initial differential comparison class.DESSE transformations are defined in half of the infinite domains and are useful for solving initial value problems for common differential comparisons.
Remark
The Laplace transformation is named after Pierre-Simon de Laplace (\ (1749-1827 \)). Laplace has made great contributions, especially in heavenly mechanics, tidal analysis and probability.
Remark
Integrated transformation\ ([a, b] \)With regard to the integrated core,\ (K (x, k) \).
Fourier and Laplace transformations are examples of a broader class of transformations that are known asIntegrated transformation.For a function\ (f (x) \)defined with an interval\ ((a, b) \), we define the integrated transformation \ [f (k) = \ int_ {a}^{b} k (x, k) f (x) d x, \ not -number \] where\ (K (x, k) \)is a specified core of the transformation.\ (K (x, k) = e^{i k x} \).I table \ (\ page index {1} \) We show different types of integrated transformations.
| Laplace -transformation | \ (F (s) = \ int_ {0}^{\ innty} e^{-s x} f (x) d x \) |
|---|---|
| Fourier -transformation | OV |
| Fourier Cosine Transform | \ (F (k) = \ int_ {0}^{\ inny} \ cos (k x) f (x) d x \) |
| Fourier Sinus Transformation | \ (F (k) = \ int_ {0}^{\ infty} \ sin (k x) f (x) d x \) |
| Mellin transformation | \ (F (k) = \ int_ {0}^{\ innty} x^{k-1} f (x) d x \) |
| Hankel -Transformation | \ (F (k) = \ int_ {0}^{\ innty} x j_ {n} (k x) f (x) d x \) |
It should be noted that these integrated transformers inherit the linearity of integration.\ (h (x) = \ alpha f (x)+\ beta g (x) \), Where\ (\ alpha \)In\ (\ beta \)are constant.{B} k (x, k) (\ alpha f (x)+\ beta g (x)) d x, \ number \\ & = \ alpha \ int_ {a}^{b} k (x, k) f(x) d x+\ beta \ int_ {a}^{b} k (x, k) g (x) d x, \ number \\ & = \ alpha f (x)+\ beta g (x). \ label{equal: 1} \ end {align} \] That is why we have shown linearity of the integrated transformations.We have seen the linearity property that is used for Fourier transformations and we will use linearity in the study of Lalace transformations.
Remark
Laplace transformation of\ (f, f = \ Mathcal {l} [f] \).
We are now going to Laplace -transformations.laplace -transformation of a position\ (f (t) \)is defined as \ [f (s) = \ mathcal {l} [f] (s) = \ int_ {0}^{\ inny} f (t) e^{-s t} d t, \ quad s> 0.\ Label {EQur.
Laplace transformations have also proved to be useful for the technique for solving blood circulation problems and performing system analysis.i figure \ (\ page index {1} \) It seems that a signal\ (x (t) \)Supplied as input to a linear system that is specified with\ (h (t) \).Man is interested in system output,\ (y (t) \)which is given by the intervention of the input and system functions.\ (x (t) \)In\ (h (t) \), Transformation of Output is indicated as a product of the Laplace transformations in the S -domain.To do this in practice, we must of course know how we calculate Laplace transformations.
Properties and examples of Laplace transformations
Usually use Laplace transformations by referring to a table with transformation pair.Shown in Table \ (\ Page index {3} \), we can do many applications of the Laplace transformation to section, we show how these transformations can be used to summarize the endless series and to solve start -up value problems forCommon differential comparisons.
We start with some simple transformations.They are found by simply using the definition of the Laplace transformation.
Example \ (\ Pugandex {1} \)
Show it\ (\ Mathcal {l} [1] = \ frac {1} {\ varsigma} \).
Solution
For this example we are inserting\ (f (t) = 1 \)In the definition of the Laplace transformation: \ [\ Mathcal {l} [1] = \ int_ {0}^{\ Innty} e^{-s t} D t \ non-number \] This is an incorrect integrated to aTo introduce to an upper limit to be introduced for one and then leave\ (A \ rightarrow \ infty \)We will not always write this limit, but it will be understood that this is how such incorrect integral is calculated.{}^^{\ Innty} e^{-s t} D t} D t \ Nonumber \\ & = \ Lim} e^^{-s t}} \ to the right) _ {a}} \ Nonumber \\\ellGlue _ {a \ Rightarrow \ Innty}} Left (-\ frac {1} {s} E^{-s A}+} Frac {1} {s} \ to the right) = \ frac {1} {s}.\ Label {Equr.: 3} \ End {align} \]
So we have discovered that the Laplace transformation is of 1\ (\ frac {1} {\ frac {1} {s}} \).This result can be expanded to any constant\ (C \)the use of the linearity of transformation,\ (\ Mathcal {l} [c] = C \ Mathcal {l} [1] \).Derfor \ [\ Mathcal {l} [c] = \ frac {c} {s}.
Example \ (\ Pugandex {2} \)
Show that \ (\ mathcal {l} \ Left [E ^^ {a t} \ Right] = \ frac {1} {s-a} \), for \ (s> a \).
Solution
In this example we can easily calculate the transformation. \ Label {equr.: 4} \ end {align} \]
Note that the last limit was calculated as\).This is only true as a\ (-s <0 \), of\ (s> \)a. [In fact, one could be complex.In this case we would only be larger than the real part of\ (a, s> \ operatorName {re} (a) .1 \)
Example \ (\ Pugandex {3} \)
Vis at \ (\ no {fine {l} [\ cos a t] = \ frac {s}} + 2 ^ ^} + 2 ^ ^} +).
Solution
For these examples we could re -insert the trigonometric functions into the transformation and integrate.F.erigonometric function and the exponential function.However, there is a much easier way to calculate these transformations.
Remember that/ (Tie cairy to make me resty = k ọw &on.L} [\ It is a t]. \ noumer \] Transforming these complex exponential at the same time will offer the Laplace transformations to the sinus and kosine functions!
The transformation is simply calculated as \ [\ mathcal {l} \ left [e^{i a t}} right] = {0}^{\ Innty} e^} e ^^ {-S t} d t = \int_ {0}^{\ inny} e^{-(s-i a) t} d t = \ frac {1} {s-i a}. \ numer \] Note that we could easily use the result to transform aExponential that had already been proven.In this case\ (s> \ operatorName {re} (\ Mathrm {ia}) = 0 \).
We now extract the real and imaginary parts of the result using the complex conjugate of the name: \ [\ frac {1} {s-i a} = \ frac {1} {s-i a} \ frac {s+i a} {S+in a} = \ frac {s+i a} {2 {2}+a^{2}}. \ Non -\] Reading the real and imaginary parts, we find the searched transformations, \ Start{align} \ mathcal {l} [\ cos a t] & = \ frac {s} {S^ {2}+ a^ {2}}} \ Nonumber \\ \ Mathcal {l} [\ sin a t] & = \Frac {a} {S^{2}+a^{2}}. \ Label {Equr.: 5} \ End {Align} \]
Example \ (\ Pugandex {4} \)
Show it\ (\ (\ (\ Mathcal {l} [t] = \ svig {1} {s {2}} \).
Solution
For this example we evaluate \ [\ mathcal {l} [t] = \ int_ {}^{\ inny} t e^^} D t \ number \] This integrated can be evaluated using the integration method through the parts: OV0 IE}. \ Label {EQur.: 6} \ End {align} \]
Exemplay \ Pagandex {5} \)
Show it\)For Non -Negative Integaal\ (N \).
Solution
We have seen\ (n = 0 \)In\ (n = 1 \)Sager:\ (\ Walls {in} [௧] = \ Pick {௧} {s} \)In\ (\ (\ (\ Mathcal {l} [t] = \ svig {1} {s {2}} \)We now generalize these results according to non -negative entire troops,\ (n> 1 \), van\ (T \).We regards the integrated \ [\ mathcal {l} \ left [t^{n} \ right] = {0}^{\ innny} t^{n} e^{-s t} d t.After the previous example we are re -integrated with parts:\ (^{1} \)ov {-s t} \ til højre |_ {0}^{\ innty}+\ frac {n} {s} \ int_ {0}^{\ infty} t^{-n} e^{-s t} d t \ nummer \\ & = \ frac {n} {s} \ int_ {0}^{\ innty} t^{-n} e^{-s t} d t.\ label {ækv.: 7} \ end {align} \]
We can continue to integrate with parts until the final is calculated in full.\ (t^{n-1} \).So we can write the result as \ [\ mathcal {l} \ left [t^{n} \ right] = \ frac {n} {s} \ mathcal {l} \ left [t^{n-1}to the right]. \ no number \]
Remark
We calculate\ (\ int_ {0}^{\ inny} t^{n} e^{-s t} d t \)By transforming it into a starting value problem for a first -order difference comparison and finding the solution using an iterative method.
This is an example of a recursive definition of a sequence.In this case we have a series of integrals.angiver \ [i_ {n} = \ mathcal {l} \ left [t^} \ right] = \ int_ {}^{\ Innty} t^{n} e^{-S T} D t \ Nonumber \] and notes that\ (I_ {0} = \ mathcal {l} [1] = \ svig {1} {s} \)we have the following: \ [I_ {n} = \ frac {n} {s} I_ {n-1}, \ quad I_ {0} = \ frac {1} {s}. \ Label {equal: 8} \] This is also what a difference comparison is called.\ (I_ {0} \).
Finding the solution of this first order -difference comparison is easy to do with the help of simple iteration.Note, to replace\ (N \)met\ (n-1 \), We have \ [in {n-1} = \ frac {n-1} {s} in {n-2} \ text {.} \ noummer \]
Repeat the process we find \ [\ Start {align} I_ {n} & = \ frac {n} {s} I_ {n-1} \ Nonumber \\ & = \ frac {n} {s} \ left (\ frac {n-1} {s} I_ {n-2} \ after the right) \ Nonumber \\ & = \ frac {n (n-1)} {s^{2}} I_ {n-2} \ \Non-cumber \\ & = \ frac {n (N-1) (N-2)} {s^{3}} I_ {n
We can repeat this process until we\ (I_ {0} \), as we know.We must carefully count the number of iterations.We do this by Itere\ (k \)Time and then discover how many steps take us to the well -known starting value.-1} \ numer \\ & = \ frac {n (N-1; Nonumber \\ & = \ ldots \ Nonumber \\ & = \ frac {n (N-1) (N-2) \ ldots (N-K+1) N (N-k)} {s^{k}} I_ {n-k}.
Because we know that\ (I_ {0} = \ frac {1} {s} \), we choose to come by\ (k = n \)Reach the [I_ {n} = \ frac {n (N-1) (N-2) \ ldots (2) (1)} {s^{n}} I_ {0} = \ frac {n!} {S^{n+1}}. \ Number \] That's why we showed that\).
Such iterative techniques are useful for achieving a variety of integral, such as\)
As a last note, this result can be expanded to fallen in which\ (N \)Is not a whole number.Section 5.4.Husk, that the Gamma function is generalization of the factor function and is defined as \ [\ Gamma (x) = \ int_ {}^{\ inny} t^{x-1} e {-t} d t.\ (t^{x-1} \):] Nasty\ (x-1 \)A whole number and\ (S = 1 \), We have it \ [\ Gamma (x) = (x -1)!\ Mathcal {l} \ Left [t^{p} \ Right] = \ frac {\ gamma (p+1)} {s^{p+1}} \ non -cumber \] to\ (P> -1 \).
Now we are ready to introduce extra features of the Laplace transformation in Table \ (\ Page index {3} \).And the following sections.
| Laplace Transform -Properties |
|---|
| \ (\ Mathcal {l} [a f (t)+b g (t)] = a f (s)+b g (s) \) |
| \ (\ Mathcal {l} [t f (t)] =-\ frac {d} {d s} f (s) \) |
| \) |
| ovprime} (0) \) |
| \ (\ Mathcal {l} \ venstre [e^{a t} f (t) \ højre] = f (s-a) \) |
| \ (\ Mathcal {l} [H (t-a) f (t-a)] = e^{-a s} f (s) \) |
| \)s) g (s) \) |
Example \ (\ Pugandex {6} \)
Show it\).
Solution
We must \ [\ mathcal {l} \ left [\ frac {d f} {d t}} \ Right] = \ int_ {}^{{\ Innty} {d t} e}} d} dT. \ Number \] We can move the derivatus\ (F \)By integrating with parts.\ (u = e^{-s t} \)In\ (v = f (t) \)we have \ [\ start {align} \ mathcal {l} \ left [\ frac {d f} {d t}} \ Right] & = {0}^{\ infty} \ frac {d f} {d t} e^isse (t) e^{-s t} d t \ non-number \\ & = -f (0)+s f (s). \ Label {equ.12} \ end {align}}] here we accepted that\ (F (t) e ^ {- \ text {st}} \)overthrow\ (T \).
The final result is that \ [\ mathcal {l} \ left [\ frac {d f} {d t} \ right] = s f (s) -f (0). \ Numer \]
Example \ (\ Pugandex {7} \)
Show itovprime} (0) \).
Solution
We can calculate this Laplace transformation using two integrations after parts or we can use the latest result.\ (g (t) = \ frac {d f (t)} {d t} \), we hebben \ [\ Mathcal {l} \ links [\ frac {d^{2} f} {d t^{2}} \ rechts] = \ Mathcal {l} \ links [\ frac {d g} {d t t } eierstokken [\ frac {d f} {d t} \ rechts] = s f (s) -f (0) \ non -number \] so, \ [\ start {align} \ mathcal {l} \ left [\ frac { d^{2} f} {d t^{2}} \ naar rechts] & = s g (s) -f^{\ prime} (0) \ nonumber \\ & = s [s f (s) -f ( 0)] -f^{\ prime} (0) \ nonumber \\ & = s^{2} f (s) -s f (0) -f^{\ prime} (0).\ label {equ.13} \ end {aanpassen} \]
We return to the other properties in Table \ (\ Page index {3} \) After viewing some applications.
