Content

• Preliminary information
• Steps
• Part 1 of 3: The Basics
• Part 2 of 3: Properties of the Laplace transform
• Part 3 of 3: Finding the Laplace Transform by Series Expansion

The Laplace transform is an integral transform that is used to solve differential equations with constant coefficients. This transformation is widely used in physics and engineering.

While you can use the appropriate tables, it is helpful to understand the Laplace transform so that you can do it yourself if needed.

Preliminary information

• Given a function ${ displaystyle f (t)}$defined for ${ displaystyle t geq 0.}$ Then Laplace transform functions ${ displaystyle f (t)}$ is the next function of each value ${ displaystyle s}$, at which the integral converges:
  • ${ displaystyle F (s) = { mathcal {L}} {f (t) } = int _ {0} ^ { infty} f (t) e ^ {- st} mathrm {d} t}$
• The Laplace transform takes the function from the t-region (time scale) to the s-region (transformation region), where ${ displaystyle F (s)}$ is a complex function of a complex variable. It allows you to move the function to an area where a solution can be found more easily.
• Obviously, the Laplace transform is a linear operator, so if we are dealing with a sum of terms, each integral can be calculated separately.
  • ${ displaystyle int _ {0} ^ { infty} [af (t) + bg (t)] e ^ {- st} mathrm {d} t = a int _ {0} ^ { infty} f (t) e ^ {- st} mathrm {d} t + b int _ {0} ^ { infty} g (t) e ^ {- st} mathrm {d} t}$
• Remember that the Laplace transform only works if the integral converges. If the function ${ displaystyle f (t)}$ has discontinuities, it is necessary to be careful and correctly set the limits of integration in order to avoid uncertainty.

Steps

Part 1 of 3: The Basics

1. 1 Substitute the function into the Laplace transform formula. In theory, the Laplace transform of a function is very easy to calculate. As an example, consider the function ${ displaystyle f (t) = e ^ {at}}$, where ${ displaystyle a}$ is a complex constant with ${ displaystyle operatorname {Re} (s) operatorname {Re} (a).}$
   • ${ displaystyle { mathcal {L}} {e ^ {at} } = int _ {0} ^ { infty} e ^ {at} e ^ {- st} mathrm {d} t}$
2. 2 Estimate the integral using the available methods. In our example, the estimation is very simple and you can get by with simple calculations. In more complex cases, more complex methods may be needed, for example, integration by parts or differentiation under the integral sign. Constraint condition ${ displaystyle operatorname {Re} (s) operatorname {Re} (a)}$ means that the integral converges, that is, its value tends to 0 as ${ displaystyle t to infty.}$
   • ${ displaystyle { begin {aligned} { mathcal {L}} {e ^ {at} } & = int _ {0} ^ { infty} e ^ {(as) t} mathrm {d } t & = { frac {e ^ {(as) t}} {as}} Bigg _ {0} ^ { infty} & = { frac {1} {sa}} end {aligned}}}$
   • Note that this gives us two types of Laplace transform, with sine and cosine, since according to Euler's formula ${ displaystyle e ^ {iat}}$... In this case, in the denominator we get ${ displaystyle s-ia,}$ and it only remains to determine the real and imaginary parts. You can also evaluate the result directly, but that would take a little longer.
     • ${ displaystyle { mathcal {L}} { cos at } = operatorname {Re} left ({ frac {1} {s-ia}} right) = { frac {s} {s ^ {2} + a ^ {2}}}}$
     • ${ displaystyle { mathcal {L}} { sin at } = operatorname {Im} left ({ frac {1} {s-ia}} right) = { frac {a} {s ^ {2} + a ^ {2}}}}$
3. 3 Consider the Laplace transform of a power function. First, you should define the transformation of the power function, since the linearity property allows you to find the transformation for of all polynomials. The power function is a function of the form ${ displaystyle t ^ {n},}$ where ${ displaystyle n}$ - any positive integer. Can be integrated piece by piece to define a recursive rule.
   • ${ displaystyle { mathcal {L}} {t ^ {n} } = int _ {0} ^ { infty} t ^ {n} e ^ {- st} mathrm {d} t = { frac {n} {s}} { mathcal {L}} {t ^ {n-1} }}$
   • This result is expressed implicitly, but if you substitute several values ${ displaystyle n,}$ you can establish a certain pattern (try to do it yourself), which allows you to get the following result:
     • ${ displaystyle { mathcal {L}} {t ^ {n} } = { frac {n!} {s ^ {n + 1}}}}$
   • You can also define the Laplace transform of fractional powers using the gamma function. For example, in this way you can find the transformation of a function such as ${ displaystyle f (t) = { sqrt {t}}.}$
     • ${ displaystyle { mathcal {L}} {t ^ {n} } = { frac { Gamma (n + 1)} {s ^ {n + 1}}}}$
     • ${ displaystyle { mathcal {L}} {t ^ {1/2} } = { frac { Gamma (3/2)} {s ^ {3/2}}} = { frac { sqrt { pi}} {2s { sqrt {s}}}}}$
   • Although functions with fractional powers must have cuts (remember, any complex numbers ${ displaystyle z}$ and ${ displaystyle alpha}$ can be written as ${ displaystyle z ^ { alpha}}$, because the ${ displaystyle e ^ { alpha operatorname {Log} z}}$), they can always be defined in such a way that the cuts lie in the left half-plane, and thus avoid problems with analyticity.

Part 2 of 3: Properties of the Laplace transform

1. 1 Let us find the Laplace transform of the function multiplied by eat{ displaystyle e ^ {at}}. The results obtained in the previous section allowed us to find out some interesting properties of the Laplace transform. The Laplace transform of functions such as cosine, sine, and exponential function seems to be simpler than the power function transform. Multiplication by ${ displaystyle e ^ {at}}$ in the t-region corresponds to shift in the s-region:
   • ${ displaystyle { mathcal {L}} {e ^ {at} f (t) } = int _ {0} ^ { infty} f (t) e ^ {- (sa) t} mathrm {d} t = F (sa)}$
   • This property immediately allows you to find the transformation of functions such as ${ displaystyle f (t) = e ^ {3t} sin 2t}$, without having to calculate the integral:
     • ${ displaystyle { mathcal {L}} {e ^ {3t} sin 2t } = { frac {2} {(s-3) ^ {2} +4}}}$
2. 2 Let us find the Laplace transform of the function multiplied by tn{ displaystyle t ^ {n}}. First, consider multiplication by ${ displaystyle t}$... By definition, one can differentiate a function under an integral and get a surprisingly simple result:
   • ${ displaystyle { begin {aligned} { mathcal {L}} {tf (t) } & = int _ {0} ^ { infty} tf (t) e ^ {- st} mathrm { d} t & = - int _ {0} ^ { infty} f (t) { frac { partial} { partial s}} e ^ {- st} mathrm {d} t & = - { frac { mathrm {d}} { mathrm {d} s}} int _ {0} ^ { infty} f (t) e ^ {- st} mathrm {d} t & = - { frac { mathrm {d} F} { mathrm {d} s}} end {aligned}}}$
   • Repeating this operation, we get the final result:
     • ${ displaystyle { mathcal {L}} {t ^ {n} f (t) } = (- 1) ^ {n} { frac { mathrm {d} ^ {n} F} { mathrm {d} s ^ {n}}}}$
   • Although the rearrangement of the operators of integration and differentiation requires some additional justification, we will not present it here, but only note that this operation is correct if the final result makes sense. You can also take into account the fact that the variables ${ displaystyle s}$ and ${ displaystyle t}$ do not depend on each other.
   • Using this rule, it is easy to find the transformation of functions such as ${ displaystyle t ^ {2} cos 2t}$, without re-integration by parts:
     • ${ displaystyle { mathcal {L}} {t ^ {2} cos 2t } = { frac { mathrm {d} ^ {2}} { mathrm {d} s ^ {2}}} { frac {s} {s ^ {2} +4}} = { frac {2s ^ {3} -24s} {(s ^ {2} +4) ^ {3}}}}$
3. 3 Find the Laplace transform of the function f(at){ displaystyle f (at)}. This can be easily done by replacing the variable with u using the definition of a transform:
   • ${ displaystyle { begin {aligned} { mathcal {L}} {f (at) } & = int _ {0} ^ { infty} f (at) e ^ {- st} mathrm { d} t, u = at & = { frac {1} {a}} int _ {0} ^ { infty} f (u) e ^ {- su / a} mathrm {d } u & = { frac {1} {a}} F left ({ frac {s} {a}} right) end {aligned}}}$
   • Above, we found the Laplace transform of functions ${ displaystyle sin at}$ and ${ displaystyle cos at}$ directly from the exponential function. With this property, you can get the same result if you find the real and imaginary parts ${ displaystyle { mathcal {L}} {e ^ {it} } = { frac {1} {s-i}}}$.
4. 4 Find the Laplace transform of the derivative f′(t){ displaystyle f ^ { prime} (t)}. Unlike the previous examples, in this case have to integrate piece by piece:
   • ${ displaystyle { begin {aligned} { mathcal {L}} {f ^ { prime} (t) } & = int _ {0} ^ { infty} f ^ { prime} (t ) e ^ {- st} mathrm {d} t, u = e ^ {- st}, mathrm {d} v = f ^ { prime} (t) mathrm {d} t & = f (t) e ^ {- st} Big _ {0} ^ { infty} + s int _ {0} ^ { infty} f (t) e ^ {- st} mathrm {d } t & = sF (s) -f (0) end {aligned}}}$
   • Since the second derivative occurs in many physical problems, we find the Laplace transform for it as well:
     • ${ displaystyle { mathcal {L}} {f ^ { prime prime} (t) } = s ^ {2} F (s) -sf (0) -f ^ { prime} (0) }$
   • In the general case, the Laplace transform of the nth order derivative is defined as follows (this allows solving differential equations using the Laplace transform):
     • ${ displaystyle { mathcal {L}} {f ^ {(n)} (t) } = s ^ {n} F (s) - sum _ {k = 0} ^ {n-1} s ^ {nk-1} f ^ {(k)} (0)}$

Part 3 of 3: Finding the Laplace Transform by Series Expansion

1. 1 Let us find the Laplace transform for a periodic function. The periodic function satisfies the condition ${ displaystyle f (t) = f (t + nT),}$ where ${ displaystyle T}$ is the period of the function, and ${ displaystyle n}$ is a positive integer. Periodic functions are widely used in many applications, including signal processing and electrical engineering. Using simple transformations, we get the following result:
   • ${ displaystyle { begin {aligned} { mathcal {L}} {f (t) } & = int _ {0} ^ { infty} f (t) e ^ {- st} mathrm { d} t & = sum _ {n = 0} ^ { infty} int _ {nT} ^ {(n + 1) T} f (t) e ^ {- st} mathrm {d} t & = sum _ {n = 0} ^ { infty} int _ {0} ^ {T} f (t + nT) e ^ {- s (t + nT)} mathrm {d} t & = sum _ {n = 0} ^ { infty} e ^ {- snT} int _ {0} ^ {T} f (t) e ^ {- st} mathrm {d} t & = { frac {1} {1-e ^ {- sT}}} int _ {0} ^ {T} f (t) e ^ {- st} mathrm {d} t end { aligned}}}$
   • As you can see, in the case of a periodic function, it is sufficient to perform the Laplace transform for one period.
2. 2 Perform the Laplace transform for the natural logarithm. In this case, the integral cannot be expressed in the form of elementary functions. Using the gamma function and its series expansion allows you to estimate the natural logarithm and its degrees. The presence of the Euler-Mascheroni constant ${ displaystyle gamma}$ shows that to estimate this integral, it is necessary to use a series expansion.
   • ${ displaystyle { mathcal {L}} { ln t } = - { frac { gamma + ln s} {s}}}$
3. 3 Consider the Laplace transform of the unnormalized sinc function. Function ${ displaystyle operatorname {sinc} (t) = { frac { sin t} {t}}}$ widely used for signal processing, in differential equations it is equivalent to the spherical Bessel function of the first kind and zero order ${ displaystyle j_ {0} (x).}$ The Laplace transform of this function also cannot be calculated by standard methods. In this case, the transformation of individual members of the series, which are power functions, is carried out, so their transformations necessarily converge on a given interval.
   • First, we write the expansion of the function in a Taylor series:
     • ${ displaystyle { frac { sin t} {t}} = sum _ {n = 0} ^ { infty} { frac {(-1) ^ {n} t ^ {2n}} {(2n +1)!}}}$
   • Now we use the already known Laplace transform of a power function. The factorials are canceled, and as a result we get the Taylor expansion for the arctangent, that is, an alternating series that resembles the Taylor series for the sine, but without factorials:
     • ${ displaystyle { begin {aligned} { mathcal {L}} left {{ frac { sin t} {t}} right } & = sum _ {n = 0} ^ { infty } { frac {(-1) ^ {n} (2n)!} {(2n + 1)!}} { frac {1} {s ^ {2n + 1}}} & = sum _ {n = 0} ^ { infty} { frac {(-1) ^ {n}} {2n + 1}} { frac {1} {s ^ {2n + 1}}} & = tan ^ {- 1} { frac {1} {s}} end {aligned}}}$