[Maths Class Notes] on Laplace Transform Pdf for Exam

Laplace Transform Examples

The differential equation of time-domain form is hard to solve hence, it is converted into an algebraic equation of the frequency-domain form and solved. The result of the algebraic form is again transformed back to the original differential equation. In other words, a form of a question is transformed into another form of a question just to make it easier to solve. This transformation is done by using the Laplace Transform.

Laplace Transform is an integral part of Mathematics and finds its applications in simplifying the calculations of system modelling, electrical and electronic circuits, nuclear physics, etc. The communication medium like cell phones, FM/AM radios, etc which has now become a basic necessity of life also uses Laplace transform to convert time-varying waves to frequency function. 

The Laplace Transform Formula:

[Fleft( p right)intlimits_L^{} {fleft( z right){e^{ – pz}}dz,} ]

With the help of the Laplace formula, we can convert a function of z to the function of a complex variable (p). Here, z stands for time and is always greater than or equal to 0.

The integration in the aforementioned formula is carried out over a contour. Here, the contour (L) is in the z complex plane between a function F(z) and analytical function F(p). F(p) is a function of p which is a complex variable containing real numbers like σ and ω.

[p{text{ }} = sigma {text{ }} + {text{ }}iomega .]

Laplace Transform Definition Examples:

Example 1:

Find the Laplace transform of f(t) = [_{0, otherwise}^{2,  t epsilon [0,2]}]

Ans:

[F(s) = L{f(t)} = int_{0-}^{infty} f(t) e^{-at} dt = int_{0}^{2} 2e^{-at} dt + int_{2}^{infty} 0e^{-at} dt = frac{2}{s} – frac{2}{s} e^{-2s}]

Laplace Transform can be One – Sided or Two – Sided.

One-sided Laplace Transform – Unilateral Laplace Transform

Two-sided Laplace Transform – Bilateral Laplace Transform

 

Laplace Transform is closely connected to Fourier Transform. One-sided transform is a transform of the function which is equal to only if the value of t is between zero to infinity. The Fourier Transform of Unilateral Transform of the function is equal to zero if the value of t is from minus infinity to zero. Whereas the two-sided Laplace Transform is the Fourier Transform of.

1) Unilateral Laplace Transform

Unilateral Laplace Transform of an arbitrary signal can be defined as:

[Fleft( p right) = Lleft[ f right]left( p right) = intlimits_0^infty  {fleft( t right){e^{ – pt}}dt,} ]

Though the Unilateral Laplace Transform is different from the Bilateral Laplace transform, the Unilateral Transform of any signal is found to be identical to its Bilateral Transform.

2) Bilateral Laplace Transform

Laplace transform can be one-sided or two-sided. Bilateral Transform is a two-sided Laplace Transform with extending limits of integrations. A common unilateral transform can also be changed to bilateral Laplace transform by extending its limits. The formula for bilateral transform is given below: 

[Fleft( p right) = Lleft[ f right]left( p right) = intlimits_0^infty  {fleft( t right){e^{ – pt}}dt,} ]               

 

3) Inverse Laplace Transform

The Inverse Laplace Transform is the transformation into a function of time. In the Laplace inverse formula  F(s) is the Transform of F(t) whereas in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). This Laplace transform formula can be written as follows:

[fleft( t right) = {L^{ – 1}}{ F} left( t right) = frac{1}{{2pi i}}mathop {lim }limits_{T to infty } oint {_{_{gamma  – iT}}^{gamma  + {text{i}}{rm T}}} {e^{stFleft( s right)ds}}]

If integrable functions differ on Lebesgue measure then integrable functions can have the same Laplace transform. Thus, there is an inverse transform on the range of transform. The inverse of a complex function F(s) to produce a real-valued function f(t) is an inverse Laplace transformation of the function. If a unique function is continuous on 0 to ∞ limit and has the property of Laplace Transform. This function is an exponentially restricted real function. 

Methods of Using Laplace Theorem

Laplace transform is carried out in different functions to study its function in control system engineering. Mostly higher order differential equations are solved by using the Laplace equation. In order to solve a differential equation of a time function, multiply the function of time (t) by e-st, where, s = σ + j ω. The resulting product is integrated with respect to time and keeping zero and infinity as its limit. The result of this integration is called Laplace Transform.

Laplace Transform Table

S.no

f(t)

L(f(t)) = F(s)

S.no

f(t)

L(f(t)) = F(s)

1

1

[1/s]

11

[eleft( {at} right)]

[frac{1}{{left( {s{text{ }} – {text{ }}a} right)}}]

2

[{t_n}{text{ }}at{text{ }}t{text{ }} = {text{ }}1,2,3, ldots ]

[frac{{n!}}{{sleft( {n + 1} right)}}]

12

[tp,{text{ }}at{text{ }}p >  – 1]

[Gamma left( {p + 1} right)/sleft( {p + 1} right)]

3

√(t)

[frac{{sqrt Pi  }}{{2sleft( {3/2} right)}}]

13

[tleft( {frac{{n – 1}}{2}} right)at{text{ }}n{text{ }} = {text{ }}1,2,..]

[left( {1.3.5 ldots frac{{left( {2n – 1} right)surd pi }}{{(2n{text{ }}sleft( {frac{{n + 1}}{2}} right)}}} right)]

4

sin(at)

[frac{a}{{(s^{2} + a^{2})}}]

14

cos(at)

[frac{a}{{(s^{2} + a^{2})}}]

5

t sin(at)

[frac{{2as}}{{{{(s^{2} + a^{2})}^2}}}]

15

t cos(at)

[frac{{(s^{2} – a^{2})}}{{{{(s^{2} + a^{2})}^2}}}]

6

sin(at+b)

[frac{{(s{text{ }}sinleft( b right) + {text{ }}a{text{ }}cosleft( b right)}}{{(s^{2} + a^{2})}}]

16

cos(at+b)

[frac{{(s{text{ }}cosleft( b right) – a{text{ }}sinleft( b right)}}{{(s^{2} + a^{2})}}]

7

sinh(at)

[frac{a}{{(s^{2} – a^{2})}}]

17

cosh(at)

[frac{a}{{(s^{2} – a^{2})}}]

8

e(at)sin(bt)

[frac{b}{{left( {left( {s – a} right)2 + b^{2}} right)}}]

18

e(at)cos(bt)

[frac{{left( {s – a} right)}}{{left( {left( {s – a} right)2 + b^{2}} right)}}]

9

e(ct)f(t)

F(s-c)

19

tnf(t) at n = 1,2,3..

(-1)n Fn s

10

f'(t)

sF(s) – f(0)

20

f”(t)

[s_{2}F(s) – sF(0) – f'(0)]

Solved Examples

Example 1) Find the Laplace transform of [fleft( t right) = 1 – 2{e^{ – 2t}}]

Answer: [Fleft( s right) = Lleft{ {fleft( t right)} right} = fleft( t right) = intlimits_{0 – }^infty  {fleft( t right)} {e^{ – at}}{text{ }}dt = int_0^infty  {left( {1 – 2{e^{ – 2t}}} right)} {e^{ – at}}dt = intlimits_0^infty  {{e^{ – at}}} dt – 2intlimits_0^infty  {{e^{ – 2t}}} {e^{ – at}}dt = frac{1}{s} – frac{2}{{s + 2}}]

Example 2) Find the Laplace transform of [fleft( t right) = {t^2}{e^{ – 2x}}cos left( {3t} right)]

Answer: For convenience, let us define the following functions

g(t) = cos(3t)

[h(t) = e^{-2x} cos(3t) = e^{-2x} g(t)]

then, f(t) = [t^{2} h(t)]

Let G(s) = L{g(t)}, H(s) = L{h(t)}, F(s) = L{f(t)}

[G(s) = frac{s}{s^{2} + 9}] from Table

[H(s) = frac{s + 2}{(s + 2)^{2} + 9}]

[F(s) = -frac{d}{ds}[-frac{d}{ds} H(s)] = frac{2(s + 2)(s^{2} + 4s – 23)}{(s^{2} + 4s + 13)^{3}}]

Fun Facts

  • Laplace is also used in determining the structure of astronomical objects from the spectrum.

  • The communication medium like cell phones, FM/AM radios, also uses Laplace transform to convert time-varying waves to frequency function.

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