adv math[unit 2]

Advanced Engineering Mathematics

Laplace and Inverse Laplace Transform

LAPLACE AND INVERSE LAPLACE TRANSFORM

The Laplace transforms and its inverse has many important applications in mathematics, physics,

economics, engineering, and probability theory. Pierre Simon Marquis de Laplace originally developed the

integral transform for his work on the probability theory. However, the powerful, practical Laplace transform

techniques were developed only a century later by Oliver Heaviside.

The Laplace transform, because of its properties, makes solving ordinary differential equations much easier

by making operations in calculus algebraic. Moreover, compared to other transform techniques (such as

Fourier transform), the Laplace transform virtually exists to almost all kinds of functions, subject to certain

restrictions. It is then ideal to use Laplace transform to analyze systems modeled by differential equations

and are causal in nature.

2.1 Definition, Existence and Uniqueness of the Laplace Transform

Let f�t� be a causal function, so f�t� = 0 for t < 0. The Laplace transform F�s� of f�t� is the complex function defined for s ∈ ℂ by

F�s� = e��f�t�dt�� (2.1)

provided that the integral exists.

The Laplace transform basically maps the function f�t�, which is in t-domain (in applications, time-domain) into s-domain (which we will call later as the complex frequency domain). We shall use the following

notation

ℒ�f� = F�s� (2.2a)

read as the Laplace transform of the function f�t� is the function F�s�, and ℒ��F� = f�t� (2.2b)

which denotes the inverse Laplace transform of the function F�s� is the function f�t�. Also take note that the original functions f�t� depend on t and their transforms F�s� depend on s. We shall use lowercase letter for the original functions and the same letter in upper case to denote their transforms.

In Eq. 2.1 we need to evaluate the integral from zero to infinity; such integrals are called improper integrals.

Improper integrals are evaluated according to the rule

e��f�t�dt�� = lim�→� e��f�t�dt�

� (2.3)



Note that the function f�t� must be causal, that is its value is f�t� when t � 0 and 0 when t < 0. Example 2.1 Using the Laplace integral, find the Laplace transform of the following: (a) f�t� = 1 (b) f�t� = e��, a is a constant (c) f�t� = t (d) f�t� = cosωt, ω is a constant (e) f�t� = sinωt, ω is a constant Answers:

(a) F�s� = �� (b) F�s� = �� (c) F�s� = ��$ (d) F�s� = ��$%&$ (e) F�s� = &�$%&$

A function f�t� has a Laplace transform if it does not grow too fast, say, if for all t ≧ 0 and some constants M and k it satisfies the growth restriction |f�t�| ≦ Me,� (2.4)

The function f�t� also need not to be continuous in the whole interval, rather it should be piecewise continuous. A function is piecewise continuous on a finite interval a - t - b where f is defined, if this interval can be divided into finitely many subintervals in each of which f is continuous and has finite limit as t approaches either endpoint of such a subinterval from the interior.. Figure 2.1 illustrates a piecewise continuous function as an example.

Figure 2.1. A Piecewise Continuous Function

We can now state the existence of the Laplace transform in the following manner: if f�t� is defined and piecewise continuous on every finite interval on the semi-axis t � 0 and satisfies the growth restriction (Eq. 2.4) for all t � 0 and some constants M and k then the Laplace transform ℒ�f� exists for all s / k.



It can also be seen that when the Laplace transform of a given function exists, it is uniquely determined. In

the same manner, the inverse of a given transform is essentially unique.

2.2 General Properties of Laplace Transform: Linearity, s-shifting. Table of Laplace Transform. Because of some basic properties and the uniqueness of Laplace transform, we can deviate from using the

Laplace integral to all functions. All we need is to derive the Laplace transform of basic functions then apply

the properties. At the end of this section, a table of Laplace transform is presented.

Linearity Just as differentiation and integration are linear, Laplace transform is also a linear operation, that

is for any functions f�t� and g�t� whose transforms exist and any constants a and b, the transform of af�t� + bg�t� exists and ℒ2af�t� + bg�t�3 = aℒ2f�t�3 + bℒ2g�t�3 = aF�s� + bG�s� (2.5)

Example 2.2 Using the linearity theorem, and the previously obtained Laplace transform pairs, find the Laplace transforms of (a) cosh at (b) sinh at (c) cosωt (d) sin ωt

First shifting theorem: Shifting in the s-domain The Laplace transform has the very useful property that

if we know the transform of f�t�, we can immediately get that of e��f�t�, as follows: If f�t� has the transform F�s� (where s / k for some k), then e��f�t� has the transform F�s − a� (where s − a / k for some k). Thus,

ℒ2e��f�t�3 = F�s − a� e��f�t� = ℒ��2F�s − a�3 (2.6a) (2.6b)

Example 2.3 Apply the shifting theorem and the previously obtained Laplace transform pairs to obtain the Laplace transform of the following: (a) e�� cosωt (b) e�� sinωt



The table below shows the Laplace transform pairs we have obtained in the previous examples.

Figure 2.2 Table of Laplace Transform Pairs

We haven’t proved formulas 3, 4 and 5 yet, however, they can be obtained from formula 2 by induction.

Formulas 1 thru 3 are special cases of formula 4 (note that 0! = 1, which applies for formula 1). Example 2.4 Find the Laplace transform of the following functions using the table (variables other than t are considered constants)

(a) t8 − 2t (b) cos 2πt (c) e8� cosh t (d) e;��8<� (e) cos�ωt + θ�



Answers

(a) 8�> − 8�$

(b) ��$%?@$

(c) ��8��8�$��

(d) A>B�%8<

(e) � CD�E�&�FGE�$%&$

Example 2.5 Find the inverse Laplace transform of the following functions using the table (variables other than s are constants)

(a) ?��;@�$%@$

(b) �H�;�$%�8�I

(c) �J�$%?�%8K

(d) L�$%?�

(e) �M��√;OM�%√JO

Answers: (a) 4 cosπt − 3 sin πt (b) 1 −;8 t8 +�8 t? (c) 3e�8� sin 5t (d) 2 − 2e�?� (e)

AS√>�ATS√I√;%√J



Drill Problem 2.1

Find the Laplace transform of the following functions. Variables other than t are constants. 1. �t8 − 3�8 2. sin84t 3. e�� sinh 5t 4. sin U3t −�8V 5. −8 sin 0.2t 6. sin t cos t 7. �t + 1�; 8. 3.8te8.?� 9. −3t?e��.J� 10. 5e�� sin ωt

Find the inverse Laplace transform of the following functions. Variables other than s are constants. 11.

��$%J − ��%J

12. 8�%�Y�$��Y

13. ��8�%√8

14. G@ZZ$�$%G$@$

15. 8��%��%;�

16. �L��8K�$��

17. ��%��%<�

18. ?��8�$�Y�%�L

19. @�$%��@�%8?@$

20. 8��JY�$�?��8



2.3 Transforms of Derivatives and Integrals. Solutions of Ordinary Differential Equations

The power of Laplace transform underlies in the fact that it can transform calculus operations into algebraic

operations. This is due to its differentiation and integration properties. Thus, it is a very useful tool in solving

initial value problems in ordinary differential equations.

Differentiation Property The Laplace transform of the first derivative of f�t� is ℒ2f′�t�3 = sF�s� − f�0� (2.7a)

provided f�t� is continuous for all t � 0 and satisfies the growth restriction and f′�t� is piecewise continuous on every finite interval on the semi-axis t � 0. The Laplace transform of the second derivative of f�t� is

ℒ2f′′�t�3 = s8F�s� − sf�0� − f′�0� (2.7b)

provided f�t� and f′�t� are continuous for all t � 0 and satisfy the growth restriction and f′′�t� is piecewise continuous on every finite interval on the semi-axis t � 0. By induction, we can then have the Laplace transform of the n-th derivative of f�t� as

ℒ\f �G��t�] = sGF�s� − sG��f�0� − sG�8f ^�0�− ⋯−f �G��0� (2.7c)

provided f up to f �G�� are continuous for all t � 0 and satisfy the growth restriction and f �G� be piecewise continuous on every finite interval on the semi-axis t � 0. Example 2.6 Find the Laplace transform of the following using the differentiation property:

(a) te,� (b) t sinωt (c) sin8ωt Answers:

(a) ��,�$

(b) 8&��$%&$�$

(c) 8&$

��$%?&$�



Integration Property For a function f�t� that is piecewise continuous for t � 0and satisfies the growth restriction, then for s / 0, s / k, and t / 0,

ℒ ` f�τ�dτ�� b = 1s F�s� (2.8)

where F�s� is the Laplace transform of f�t�. Example 2.7 Find the inverse Laplace transform of the following using the integration property.

(a) ��$%&$�

(b) ��$��$%&$�

Answers:

(a) �&$ �1 − cosωt�

(b) �&$ − �FG&�&>

Take note that multiplication by s in s-domain is differentiation in t-domain and division by s in s-domain is integration in t-domain. This is the very useful property of the Laplace transform that makes it an ideal tool

in solving differential equations and initial value problems.

In illustrating how Laplace transform can solve initial value problem, consider the examples below:

Example 2.8 Find the general solution of the differential equation

(a) y^^ + 2y^ + 2y = 0 for y�0� = 1 and y^�0� = −3 (b) y^^ − y = t for y�0� = y^�0� = 1 Answer:

(a) y = e��cos t − 2 sin t� (b) y = e� + sinh t − t

The Laplace transform method of solving differential equations has the following advantages:

• Solving a nonhomogenous ODE does not require first solving the homogenous ODE.

• Initial values are automatically taken care of.

• Complicated R�t� (right sides of linear ODEs) can be handled very efficiently.



Drill Problems 2.2 A. Use the differentiation property to find the Laplace transform of the following: 1. t cos 5t 2. cos8πt 3. sinh8at 4. cosh8 �8 t 5. sin?t B. Use the integration property to find the inverse Laplace transform of the following:

6. ��>�@�$

7. ��H%�$

8. J�>�J�

9. ��H�?�$

10. 8�>%K�

C. Solve the following differential equations using Laplace transforms:

11. y^ + 4y = 0, y�0� = 2.8 12. y^ +�8 y = 17 sin 2t, y�0� = −1 13. y^^ − y^ − 6y = 0, y�0� = 6, y^�0� = 13 14. y^^ − 4y^ + 4y = 0, y�0� = 2.1, y^�0� = 3.9 15. y^^ + ky^ − 2k8y = 0, y�0� = 2, y^�0� = 2k

2.4. Unit Step Function. t-shifting. The unit step function or the Heaviside function u�t − a� is 0 for t < a, has a jump size of 1 at t = a, and is 1 for t / a or in a formula,

u�t − a� = i0, t < a1, t / a (2.9)

The special case of step function u�t� for which a = 0 and the general case u�t − a� for an arbitrary positive constant a is shown in the figures below.



Figure 2.3. The unit step function u(t) and its shifted version u(t-a)

The transform of u�t − a� can be found using the Laplace transform integral and is given as ℒ2u�t − a�3 = e��

s

(2.10)

The unit step function is typical engineering functions made to measure for engineering applications which

often involve functions that are either off or on. Multiplying functions f�t� with u�t − a� can produce all sorts of effects, as in examples illustrated below.

Figure 2.4. Illustration of the effect of multiplication of the unit step function to a given function

The second-shifting theorem: t-shifting If f�t� has the transform F�s� , then the shifted function f�t − a�u�t − a� has the transform e��F�s�, that is ℒ2f�t − a�u�t − a�3 = e��F�s�

(2.11)

If the conversion of f�t� to f�t − a� is inconvenient, replace it by ℒ2f�t�u�t − a�3 = e��ℒ2f�t + a�3

(2.12)



Example 2.8 Write the following function using unit step functions and find its transform

f�t� =klmln 2,0 < t < 112 t8,1 < t < 12πcos t t / 12π

Figure 2.5. Graph for the function in Example 2.8

Answer:

ℒ�f� = 2s − 2e��s +o 1s; + 1s8 + 12sp e�� − q 1s; + π2s8 +π8

8sr e�@�/8 − 1s8 + 1e�@�/8

Example 2.9 Find the inverse Laplace transform of

F�s� = e�� +e�8�s8 +π8 + e�;��s + 2�8

Answer

f�t� =klmln0 0 < t < 1− sin πtπ 1 < t < 20 1 < t < 2�t − 3�e�8��;� t / 3

Figure 2.6. Graph of the function of the answer in Example 2.9



Drill Problems 2.3 A. Sketch or graph the given function (assumed to be zero outside the given interval). Represent it using unit step function. Find its transform

1. t�0 < t < 1� 2. sin 3t�0 < t < π� 3. t8�t / 3� 4. 1 − e��0 < t < π� 5. sinωt�t / 6π/ω�

B. Find and sketch or graph f�t� if F�s� equals

6. �ATt

�$%&$

7. s�8 −�s�8 +s��e��

8. ATut

�$%8�%8

9. ��ATtvw

��,

10. 2.5 AT>.xt�AT$.yt�

2.5 Short Impulses. Dirac’s Delta Function

Phenomena of an impulsive nature, such as the action of forces or voltages over short intervals of time,

such as the action of forces or voltages over short intervals of time, arise in various applications. Such

impulses can be modeled using the Dirac’s delta function and can be solved very efficiently using Laplace

transform.

Consider the function

f,�t − a� = z1k , a - t - a + k0,otherwise (2.13)

This function represents, for instance, a force of magnitude 1/k acting from t = a to t = a + k, where k is positive and small. In mechanics, the integral of a force acting over a time interval a - t - a + k is called the impulse of the force. The area bounded by this function (which is its integral with respect to t) is unity.



Figure 2.7. The function fk(t-a)

Thus, from Figure 2.7,

I, = f,�t − a�dt�� = 1k dt�%,

� = 1 (2.14)

We take the limit of 2.13 as k → 0, denoted by δ�t − a�, that is δ�t − a� = lim,→� f,�t − a� = i∞, t = a0,otherwise (2.15)

which is called the Dirac delta function or the unit impulse function. The Laplace transform of δ�t − a� is given as

L2δ�t − a�3 = e�� (2.16)

The unit step and unit impulse functions can model several situations in electric and mechanical systems,

thus, knowing its Laplace transform is of great value to us.



Example 2.10 Determine the response of a system described by the differential equation y^^ + 3y^ + 2y = r�t� for y�0� = y^�0� = 0 and inputs r�t�. (a) r�t� = u�t − 1� − u�t − 2�

Figure 2.8. Input and Output Response for Problem (a) (b) r�t� = δ�t − 1� Answers: (a)

y�t� =klmln0 0 < t < 112 − e�� +12 e�8��8�1 < t < 2−e�� + e��8� +12 e�8�� − 12 e�8��8�t / 2

(b) y�t� = i00 < t < 1e�� − e�8�� t / 1

Figure 2.9. Output Response for Problem (b)



Drill Problems 2.4 Showing the details, find and graph the solution. 1. y^^ + y = δ�t − 2π� y�0� = 10, y′�0� = 0 2. y^^ + 2y^ + 2y = e�� + 5δ�t − 2�, y�0� = 0, y^�0� = 1 3. y^^ − y = 10δUt − �8V − 100δ�t − 1� y�0� = 10, y^�0� = 1 4. y^^ + 3y^ + 2y = 10Msin t + δ�t − 1�O y�0� = 1, y^�0� = −1 5. y^^ + 4y^ + 5y = 21 − u�t − 10�3e� − e��δ�t − 10� y�0� = 0, y^�0� = 1 6. y^^ + 2y^ − 3y = 100δ�t − 2� + 100δ�t − 3� y�0� = 1, y^�0� = 0 7. y^^ + 2y^ + 10y = 1021 − u�t − 4�3 − 10δ�t − 5� y�0� = 1, y^�0� = 1 8. y^^ + 5y^ + 6y = δUt − �8πV + u�t − π� cos t y�0� = y^�0� = 0 9. y^^ + 2y^ + 5y = 25t − 100δ�t − π� y�0� = −2, y^�0� = 5 10. y^^ + 5y = 25t − 100δ�t − π� y�0� = −2, y^�0� = 5

2.6 Application: Vibration of Spring

Consider a steel spring attached to a support and hanging downward.

Figure 2.10. Illustration of Mass-Spring System

The spring, within certain elastic limits, will obey the Hooke’s law: If a spring is stretched or compressed, its

change in length will be proportional to the force exerted upon it and when this force is removed, the spring

will return to its original position with its length and other physical properties unchanged. Thus, the force

that will cause the spring to stretch or compress x unit of length will be F� = kx (2.17)

where k is a spring constant which is related to the ratio of the force applied per unit length of displacement.



Let a body of weight w be attached at the lower end of the spring, and brought to the point of equilibrium where it can remain at rest. Once the weight is moved from the point of equilibrium, its motion will be

determined by a differential equation and associated initial conditions.

We have the following assumptions to simplify our analysis of the system. First the motion takes place

entirely in a vertical line, so that this will result in a linear equation. Spring motion taking place in two or

three dimensions results in nonlinear equations. Second, the displacement x of the object is measured positive downward and negative upward.

In addition to the spring force (Hooke’s law), there will, in general be a retarding force caused by the

resistance of the medium in which the motion takes place or by friction. This retarding force, or drag, will be

assumed to be proportional to the velocity of the object. We assumed it is that way so that we make our

differential equation linear, as a drag force proportional to the square or cube of the velocity leads to

nonlinear equation.

Thus, the motion of the object is determined by four forces:

• The force due to the mass, which is proportional to the acceleration of the object (by Newton’s

second law of motion,

F� = wg x^^�t� (2.18a)

• The retarding force, or the drag force, whether applied or caused by the medium, which opposes

the motion of the object, and is proportional to the velocity of the object (the constant b is the constant of proportionality characterizing the medium),

F� = bx^�t� (2.18b)

• The force applied on the spring by the object when it is moving, determined by Hooke’s law

F� = kx�t� (2.18c)

• A time varying external force which is proportional to the acceleration F��t� that it alone would impart on the object,

F� = wg F��t� (2.18d)

From these, the differential equation that describes the motion of the object can be written as



wg x^^�t� + bx^�t� + kx�t� = wg F��t� (2.19)

Multiplying 2.19 by �� and letting 2γ = <�� and β8 = ,�� , we have

x^^�t� + 2γx^�t� + β8x�t� = F��t� (2.19)

which is a second-order, nonhomogenous linear differential equation. The initial conditions required for this

problem, x�0� and x^�0� refer to the initial position of the object with respect to the equilibrium point and the initial velocity of the object respectively. As previously done, equations of this type can be solved using

Laplace transform techniques.

Undamped Motion. When the parameter γ = 0 the differential equation of 2.19 becomes x^^�t� + β8x�t� = F��t� (2.20)

Example 2.11 Find the response of a mass-spring system without damping to the following inputs: (a) hammerblow input βδ�t�at t = 0, zero initial conditions (b) no input, but with non-zero initial conditions (x�0� = x^�0� ≠ 0). (c) sinusoidal driving force Asinωt with ω ≠ β, x�0� = x^�0� ≠ 0. (d) sinusoidal driving force Asinωt with ω = β, x�0� = x^�0� ≠ 0 Answers: (a) x�t� = sin βt (b) x�t� = x�0� cosβt + �� sin βt (c) x�t� = x�0� cosβt + �� sin βt − �&��$�&$� sin βt + ��$�&$ sinωt (d) x�t� = x�0� cosβt + �� sin βt + �8�$ �sin βt − βt cos βt�



Example 2.12 1. A spring is such that a 5-lb weight stretches it 6 in. The 5-lb weight is attached, the spring reaches equilibrium, then the weight is pulled down 3 in below the equilibrium point and started off with an upward velocity of 6 ft/sec. Find an equation giving the position of the weight at all subsequent times. 2. (a) A spring is stretched 1.5 in by a 2-lb weight. Let the weight be pushed up 3 in above the equilibrium point and then released. Describe the motion. (b) For the same mass-spring system, let the weight be pulled down 4 in below the equilibrium point and given a downward initial velocity of 8 ft/sec. Describe the motion. Answers:

1. x�t� = �? �cos8t − 3 sin 8t� 2. (a) The equation of motion is x�t� = �? cos8t. That means, the amplitude of vibration is 3 in above and below the equilibrium point and the frequency of vibration is f = ?@ Hz.

(b) The equation of motion is x�t� = �; cos 8t + �8 sin 8t. The amplitude of vibration is 2√13 in above and below the equilibrium point and the frequency of vibration is still f = ?@ Hz.

The parameter γ is a factor related to damping; it represents the energy lost by the object because of retarding force while moving.

Damped Motion. When the parameter γ is non-zero, the motion that results is a damped one. Thus from the equation

x^^�t� + 2γx^�t� + β8x�t� = F��t� (2.19)

we can identify three possible scenarios for the motion.

Case 1: Underdamped Motion. This motion is characterized by an oscillation but eventually dies out as a

result of damping. When the parameter β / γ, the quantity β8 − γ8 = ω8 is a positive number and in general, it will have a solution of the form

x�t� = e��c� cosωt + c8 sinωt� +ϕ��t� (2.20)



Figure 2.11. Underdamped Oscillation

The term ϕ��t� represents the particular solution for the external input F��t�. Case 2: Critically Damped Motion. This motion is characterized by decay of motion without oscillation.

This happens when β = γ, and the general solution of this motion is x�t� = e��c� + c8t� +ϕ8�t� (2.21)

Figure 2.12. Critically Damped Oscillation

Case 3: Overdamped Motion. This motion is characterized by a slower decay of motion without

oscillating. When β / γ, the quantity β8 − γ8 =σ8 is a negative number, and the general solution of the differential equation becomes

x�t� = e��%�� + e�� +ϕ;�t� (2.23)

As the equation indicates, this motion decays slower than the critical damped motion case.



Example 2.13 Derive the general solutions of Equation 2.19 for the three cases mentioned above.

Example 2.14 An iron ball whose weight is 98 N stretches a spring 1.09 m. Determine the equation of motion of the object when it is pulled down 16 cm from its equilibrium and for the following damping parameters: (a) b = 0 (b) b = 10kg/sec (c) b = 60kg/sec (d) b = 100kg/sec Determine what type of motion the object undergoes. Answers (a) x�t� = 0.16 cos3t; undamped motion (b) x�t� = e��.J��0.16 cos2.96t + 0.027 sin 2.96t�; underdamped motion (c) x�t� = �0.16 + 0.48t�e�;�; critically damped motion (d) x�t� = 0.18e�� − 0.02e�K�; overdamped motion

Drill Problem 2.5 Solve each of the problems completely using Laplace transform.

1. A spring is such that a 4-lb weight stretches it 6 in. An impressed force �8 cos 8t is acting on the spring. If

the 4-lb weight is started from the equilibrium point with an imparted upward velocity of 4 ft/sec, determine the position of the weight as a function of time. 2. A spring is such that it is stretched 6 in by a 12-lb weight. The 12-lb weight is pulled down 3 in below the equilibrium point and then released. If there is an impressed force of magnitude 9 sin 4t, describe the motion.

3. A spring is such that a 2-lb weight stretches it �8 ft. An impressed force �? sin 8t is acting upon the spring.

If the 2-lb weight is released from a point 3 in below the equilibrium point, determine the equation of motion. 4. A spring is such that a 16-lb weight stretches it 1.5 in. The weight is pulled down to a point 4 in below the

equilibrium point and given an initial downward velocity of 4 ft/sec. An impressed force of 360 cos4t lb is applied. Find the position and velocity of the weight at time t = π/8 sec. 5. A 20-lb weight stretches a certain spring 10 in. Let the spring first be compressed 4 in, and then the 20-lb weight and then the 20-lb weight attached and given an initial downward velocity of 8 ft/sec. Find how far the weight would drop. 6. Consider an underdamped motion of a body of mass m = 2 kg. If the time between two consecutive

maxima is 2 sec and the maximum amplitude decreases to �? of its initial value after 15 cycles, what is the



damping constant of the system? 7. A certain straight line motion is determined by the differential equation d8xdt8 + 2γdxdt + 169x = 0 and the conditions that when t = 0, x = 0 and x^ = 8ft/sec. (a) Find the value of γ that leads to critical damping and determine x in terms of t. (b) Use γ = 12. Find x in terms of t. (c) Use γ = 14. Find x in terms of t.

8. A spring is such that a 2-lb weight stretches it �8 ft. An impressed force �? sin 8t and a damping force of

magnitude |F�| = |v| (v is the velocity of the object) are both acting on the spring. The weight starts �? ft below the equilibrium point with an imparted upward velocity of 3ft/sec. Find a formula for the position of the weight at time t. 9. A spring is such that a 4-lb weight stretches the spring 0.4ft. The 4-lb weight is attached to the spring (suspended from a fixed support) and the system is allowed to reach equilibrium. Then the weight is started from the equilibrium position with an imparted upward velocity of 2 ft/sec. Assume that the motion takes place in a medium that furnishes a retarding force of magnitude numerically equal to the speed in feet per second of the moving weight. Determine the position of the weight as a function of time. 10. A particle is moving along the x-axis according to the law d8xdt8 + 6dxdt + 25x = 0 If the particle started at x = 0 with an initial velocity of 12 ft/sec to the left, determine (a)x in terms of t, (b) times at which stops occur, and (c) the ratio between the numerical values of x at successive stops.

2.7 Application: Electric Circuits

An inductor is a passive electrical component that can store energy in a magnetic field created by the

electric current passing through it. An inductor’s ability to store magnetic energy is measured by its

inductance, in units of henries (henry in singular).

The effect of an inductor in a circuit is to oppose changes in current through it by developing a voltage

across it proportional to the rate of change of current. Thus, the time-varying voltage across the inductor is

vZ = L didt (2.24)

A capacitor is a passive electrical component consisting of a pair of conductors separated by a dielectric.

When there is a potential difference (voltage) across the conductors a static electric field develops in the

dielectric that stores energy. An ideal capacitor is characterized by a single constant value, capacitance,

measured in farad.



The time-varying voltage across the capacitor is given as

v� = q�t�C = 1C i�τ�dτ�� + v�t�� (2.25)

Series RLC circuit. Consider a series RLC circuit as shown below

Figure 2.13. Series RLC circuit

By Kirchhoff’s voltage law (KVL), we can derive an expression for current at any time t. We first consider the case when the input voltage is a dc source (a battery).

Example 2.14 Find an expression for the current at any time t for the series RLC circuit shown in Figure 2.13 when the input voltage is E volts dc if �Z� − �$

?Z$ (a) is positive. (b) is zero. (c) is negative.

Let �Z� − �$

?Z$ = β8 and �Z = 2γ. Answers:

(a) i�t� = ��Z e�� sin βt

(b) i�t� = �Z te��

(c) i�t� = �8�Z Me�� − e��%��O



In this example, notice that the answers are also consistent with the results of the damped mass-spring

system with constant input force. The first case is an underdamped system, the second one is a critically

damped system, and the last one is an overdamped case.

Note that for a series RLC circuit with a constant voltage source, the steady-state output (or response)

current, that is the output as time approaches infinity, is zero. This is due to the fact the capacitor becomes

open as time increases.

In the second example, a sinusoidal source is used as an input.

Example 2.15

Find the current i�t� in a series RLC circuit with R = 11Ω, L = 0.1H, C = 10�8F which is connected to a source voltage E�t� = 100 sin 400t. Assume that current and charge are zero when t = 0. Answer: i�t� = −0.2776e�� + 2.6144e�� − 2.3368cos 400t + 0.6467sin 400t

The transient response is the output that tapers off as time goes to infinity. In the example above, the first

two terms represent the transient response, while the last two terms represent the steady-state response.

Note that the steady state response of the circuit is a harmonic oscillation. The last two terms can be

written as

i��t� = 2.4246 sin�400t − 1.3008� Thus, the oscillation has an amplitude of 2.4246 A and frequency of

8��@ Hz. The Laplace transform is very useful in the analysis of various circuits including filters, transistor networks,

operational amplifiers, etc.

One of the interesting properties of the models for the mass-spring system and the electric circuits is their

similarity. In fact, if one will observe, the mass-spring model

wg x^^�t� + bx^�t� + kx�t� = wg F��t� (2.19)

and the series RLC equation

Li^^�t� + Ri^�t� + 1C i�t� = E′�t� (2.26)

has the same form, and a quantitative analogy can be deduced from such similarity. The following table

summarizes that analogy.



Figure 2.14. Analogy of Electrical and Mechanical Quantities

Drill Problems 2.6 1. Find the current of a series LC circuit when L = 0.2H, C = 0.05F and E = sin t �, assuming zero initial current and charge. 2. What are the conditions for an RLC circuit to be overdamped, critically damped and underdamped? 3. Find the steady-state current in the RLC circuit for the given data: a. R = 8Ω, L = 0.5H, C = 0.1F, E = 100 sin 2t b. R = 1Ω, L = 0.25H, C = 5 � 10�JF, E = 110 c. R = 2Ω, L = 1H, C = 0.05F, E = �J K sin 3t 4. Find the transient current in the RLC circuit for the given data: a. R = 6Ω, L = 0.2H, C = 0.025F, E = 110sin 10t b. R = 0.2Ω, L = 0.1H, C = 2F, E = 754 sin 0.5t c. R = 1/10Ω, L = 1/2H, C = 100/13F, E = e�?� U1.932 cos �8 t + 0.246 sin �8 tV 4. Find an expression for the current at any time t in the RLC circuit for the given data:

a. R = 4Ω, L = 0.1H, C = 0.025F, E = 10sin 10t b. R = 6Ω, L = 1H, C = 0.04F, E = 600�cos t + 4 sin t� c. R = 3.6Ω, L = 0.2H, C = 0.0625F, E = 164cos 10t

adv math[unit 2]

Engineering

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