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LAPLACE TRANSFORMS

INTRODUCTION
LAPLACE TRANSFORM OF COMMONLY USED FUNCTIONS
SELECTED THEOREMS
PARTIAL FRACTION EXPANSION
SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS
SUMMARY


INTRODUCTION

To facilitate the solution of ordinary linear differential equations with constant coefficients, the unilateral or one-sided Laplace transform method is used extensively. This operational method has found wide acceptance in the field of circuit theory analysis. The advantages of using the Laplace transforms are listed below.

However there is a drawback in the usage of Laplace transforms. If the operational method is used mechanically without an understanding of theory behind it, the results obtained can at times be erroneous. In some cases, the differential equations can be solved more easily using the classical approach, outlined in the previous chapter.

eqn1_000

A time domain function and its corresponding function in Laplace transform have one to one correspondence. A time domain function is expressed with lower case letter, and the Laplace transform function is denoted by the capital letter.

eqn1_001

As stated by equation (1), the one sided transform is normally defined for the time range from zero to infinity. In order to get the Laplace transform of a time function, we should know whether it exists. The functions used in circuit analysis are Laplace transformable. Given that equation (2) is satisfied, the Laplace transform of a function in time domain is defined by equation (3). You need to remember this equation. In equation (3), L is the Laplace transform operator, and s is the Laplace variable. Normally, s is a real value. The inverse of a Laplace transform function is its corresponding time function, and the inverse operation is indicated, as shown by equation (4).

In order that a function f(t) can be transformed, it should satisfy certain conditions. The sufficient conditions for the existence of the Laplace transform fro a given function f(t) are given below.

If a function f(t) satisfies the above conditions, then the Laplace integral F(s) converges or in other words F(s) exists. The term ‘continuous’ means that the derivative of f(t) has a finite value at each instant of t within the time interval considered. A function is piecewise continuous in a finite number of sub-intervals if it is continuous within each sub-interval and if it has finite values at the start and end of each sub-interval. If a function f(t) is such that its value changes suddenly say at t = t1, then the function has a discontinuity at t = t1. At points of discontinuity, the value of derivative of f(t) tends to infinity.

Excitation functions used in circuit analysis satisfy the above conditions. We study next how we can get the Laplace Transform of commonly used functions.

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LAPLACE TRANSFORM OF COMMONLY USED FUNCTIONS

The time functions, for which we find the corresponding Laplace transform functions are the same functions, defined as the inputs to a system in the lecture on differential equations. They are:
 
 1. Unit step function.
 2. Impulse Function.
 3. Ramp function.
 4. Time t, raised to the power of n.
 5. Exponential function.
 6. Sinusoidal Function, and
 7. Complex Sinusoid function.

UNIT STEP FUNCTION

eqn1_002

eqn1_003

We take up the unit step function first. Equation (5) define the unit step function in time domain. The corresponding function in terms of the Laplace transform is defined by equation (6). The integral can be evaluated, as shown by equation (7). The Laplace transform of unit step function is 1/s. You need to remember this.

eqn1_004

stepFn1

Equation (8) expresses the Laplace transform of unit step function. It can be seen that when s is equal to zero, the value of F(s) is infinity. A value of s for which a Laplace transform function becomes infinite is called as the pole of the function. We have seen in the last lecture that the pole of an exponential function is at - a. The unit step function can be called an exponential function, with its pole at origin. The sketch in Figure 1 shows how the unit step function can be represented in both the time domain and the s domain. The cross at the origin denotes that the pole is at the origin.

IMPULSE FUNCTION

eqn1_005

impulse2

The Laplace transform of the impulse function is obtained, as shown by equation (10). It is equal to one. The impulse function is displayed in time domain, as shown in Fig. 2. It can be seen that an impulse function has no pole and no zero.

RAMP FUNCTION

ramp03

eqn1_006

The ramp function is expressed by equation (11). Its transform is defined by equation (12). Integrate by parts, as shown by equations (13) and (14). The transform of ramp function is expressed by equation (15). We find that the Laplace transform expression for ramp function contains two poles at the origin. The ramp function and the position of its poles are displayed by Fig. 3.

FUNCTION tn

eqn1_007

Normally we do not come across the function defined by equation (16). But using the results of this function, we can get the Laplace Transform of unit step function and Ramp function. To get the Laplace Transform of the time function in equation (16), we resort to integration by parts, as displayed by equations (17), (18) and (19). We can evaluate equation (19) using the Gamma Integral.

eqn1_008

Gamma integral is presented by equation (20). Using this integral, equation (21) presents the Laplace transform of the time function described by equation (16). We can resort to integration by parts, as shown by equations (22) and (23). We get factorial of n, by successively applying integration by parts. The Laplace transform of function, tn, is expressed by equation (24).

eqn1_009

In equation (36), if we let n be equal to 0, we get the Laplace Transform of unit-step function. In equation (24), if we let n be equal to 1, we get the Laplace Transform of ramp function.

EXPONENTIAL FUNCTION

eqn1_010

exp04

Equation (25) defines a decaying exponential function. Its Laplace Transform is obtained as shown by equation (26). Equation (25) and equation (26) form a unique pair. Given 1/(s + a), its inverse is the decaying exponential function in equation (25). Though the Table on Laplace transform pairs can be used to find the inverse time function, it is preferable to remember the Laplace Transform of frequently used functions. The decaying function and the plot of its pole position are displayed by Fig. 4. When the location of a moves to the origin, we have a unit step function.

SINUSOIDAL FUNCTION

eqn1_011

cos05

The cosine function is defined by equation (27). Its transform is obtained as shown by equation (28). We make use of Euler's identity in deriving the transform. The transform of the cosine function is displayed by equation (29). The cosine function and its transform form a unique pair. The plot of poles of a sinusoidal function is shown in Fig. 5. Both the poles are located on the imaginary axis in s plane. The Laplace transform of the cosine function has s as its numerator. We can state that the cosine function has zero at the origin. When s = 0, the Laplace transform of the cosine function is equal to zero.

eqn1_012

The sine function is defined by equation (30). Its transform is obtained as shown by equation (31). We make use of Euler's identity in deriving the transform. The transform of the sine function is displayed by equation (32). The sine function and its transform form a unique pair. You need to remember the transform expressions for the sinusoidal function.

COMPLEX SINUSOID FUNCTION

eqn1_013

comsine06

The complex sinusoid function, with a cosine function as its part, is defined by equation (33). Its transform is obtained as shown by equations (34), (35) and (36). We make use of Euler's identity in deriving the transform. It is better to remember the Laplace Transform of this complex sinusoid function.

eqn1_014

The complex sinusoid function, with a sine function as its part, is defined by equation (37). Its transform is obtained as shown by equations (38), (39) and (40). We make use of Euler's identity in deriving the transform.

A FUNCTION WITH MULTIPLE POLES

eqn1_015

In the chapter on differential equations, there has been a discussion on multiple poles, and how the assumed solution changes when the multiplicity of poles is 2. We can see the justification for the assumed solution in this sub-section. Equation (41) presents a function in time, similar to the assumed solution when the multiplicity of poles is 2. The Laplace transform of this function can be obtained, as shown by equations (42) and (43), wherein we have resorted to integration by parts. It is evident from the transform function in equation (43) that this function has two roots at - a, and the corresponding time function, defined by equation (41) contains the product of time t, and an exponential function. This sub-section serves to illustrate why we need to study the topic of Laplace transforms. We get additional insight into how we can obtain the solution of differential equations.

eqn1_016

The results obtained in the earlier section can be extended to an exponential function with an imaginary exponent. Based on the previous sub-section, the Laplace transform of the function defined by equation (44) is displayed by equation (45). Extending this example, we get the Laplace transforms of two more functions, defined by equations (46) and (47). We can multiply both the numerator and the denominator of the expression in equation (45) by the conjugate of the denominator. The imaginary part is the Laplace transform of the function in equation (45), whereas the real part is the Laplace transform of the function in equation (46). The next section describes the theorems relevant to this course.

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SELECTED THEOREMS

p04Thrms

In this section, some theorems relevant to circuit analysis are presented. The list of theorems that are described is listed above. When the Laplace transform of a repetitive function is to be obtained, we need to use shift theorem. It is also used in the derivation of convolution integral. The theorem on real differentiation forms the basis for applying Laplace transforms to solving linear differential equations. Theorem on complex differentiation is not used that often, but it can be applied where a system has multiple poles. The theorem on real integration is useful to highlight how the application of Laplace transforms converts differentiation and integration to algebraic operations. Given a Laplace transform function, the initial value theorem is useful in evaluating the initial value of a function in time domain, without finding the inverse transform. Complex integration theorem is not used that often, but is presented here to complement theorem on real integration. In the end, convolution integral is presented. We apply Laplace transforms to find the response of systems to different inputs. The justification for this approach stems from the validity of convolution integral, and hence it is necessary that the concept behind convolution integral is well understood. It is not essential to remember its proof.

The Shift Theorem or Translation in s domain

eqn1_017

Multiplication of a time function by an exponential function results in translation in s domain. Equation (48) presents this theorem. The evaluation of the transform is fairly straight forward. This theorem is very useful when the inverse transform of a given function is to be obtained.

eqn1_018

An example is presented to illustrate the use of this theorem. The task is to obtain the Laplace Transform of f(t), given g(t), as shown by equation (49). Function g(t), is specified by equation (50), and it is a cosine function. The Laplace transform of the cosine function is expressed by equation (51). Using the shift theorem, the Laplace transform of the complex sinusoid function can be obtained, as shown by equation (52).

eqn1_019

Another example is presented to illustrate the use of the shift theorem. Given the Laplace transform function, its inverse is to be obtained. The denominator of the given function can be presented, as shown by equations (54) and (55). Using the shift theorem, we can extract out the exponential function, and the remaining part is seen to be the Laplace transform of the sine function. Hence the inverse transform of the given transform function is obtained, as displayed by equation (56).

Translation in time domain

Fig07

In Figure 7, two functions are shown. They are identical in shape, but the second waveform is shifted from the first by a seconds. In other words, the second waveform lags the first waveform by a seconds. The second waveform is said to be shifted in time domain by a seconds with respect to the first waveform. For a signal, obtained by shifting in time domain, the theorem on translation in time domain can be applied, and we can get its Laplace transform function. We can extend this concept to a signal such as a square wave signal, and obtain its transform.

eqn1_020

The Laplace transform of the shifted signal is obtained as shown here. Equation (57) defines the Laplace transform. Since the time shifted signal has zero value from 0 to a seconds, the lower limit can be a instead of 0. We can let l be equal to ( t - a), as shown by equation (58). Then the transform of the shifted function can be obtained, as shown by equation (59). Equation (60) summarizes what this theorem means.

eqn1_021

An example is presented to illustrate the use of this theorem. The task is to obtain the Laplace Transform of g(t), given f(t), as shown by equation (61). Laplace transform of f(t), is expressed by equation (62). Using the theorem on Translation in Time domain, the Laplace transform of g(t) can be obtained as shown by equation (63).

Fig 08

Another example is presented to illustrate the use of the shift theorem. The task is to obtain the Laplace Transform of the function shown in Fig. 8. The waveform in Fig. 8 is a rectangular pulse, with an amplitude of K, and a period, equal to a. It can be visualized, as the sum of two signals. The first is a step function of magnitude K, applied at t = 0 ,equal to zero, and another signal of magnitude minus K, applied at t = a. The sum of two signals is the rectangular pulse, shown in Fig. 8.

eqn1_022

The signal in Fig. 8 can be represented by equation (61). Using the theorem on translation in time domain, we can obtain the Laplace transform of this signal, as shown by equation (62). We can extend the result to a train of pulses, as shown below.

Fig09

A square wave signal is shown in Fig. 9. It can be seen as the sum of a series of time-shifted signals. Then the Laplace transform of the square wave can be obtained.

eqn1_023

Equation (63) expresses the square wave signal as the sum of a series of time-shifted signals. Then the Laplace transform of the square wave can be obtained, as shown by equation (64). Using the binomial theorem, we can express the Laplace transform for the square wave signal, in a compact form, as shown by equation (64).

Real Differentiation

ole1_024

This theorem shows how to obtain Laplace Transform of a derivative of a time function. Equation (65) presents this theorem. This theorem states that the Laplace transform of the derivative of a function is obtained by multiplying the Laplace transform of that function by s, and then subtracting the initial value of that function. The initial value is evaluated at time, t = 0+. The lower limit can be zero, instead of 0 +, if the value of the function at time t = 0-, has not been changed by an impulse occurring at t = 0. Normally this may not be the case, and we can use the lower limit as zero, instead 0+. The Laplace Transform of second derivative is presented by equation (66), and we have made use of the result obtained in equation (65). Remember equations (65) and (66), and they are used repeatedly when the solution of a linear differential equation is obtained using Laplace transforms.

ole1_025

The usefulness of theorem on real differentiation is highlighted now. Since the unit step function is the derivative of the ramp function, we can obtain the Laplace transform of the unit step function from the Laplace transform of the ramp function, as shown by equations (67) and (68). Similarly, we know that the impulse function is the derivative of the unit step function, and hence we can obtain the Laplace transform of the impulse function from the Laplace transform of the unit step function, as shown by equation (69). We can conclude that differentiation in time domain corresponds to multiplying by s in the s domain. The process of differentiation becomes an algebraic operation when using Laplace transforms.

eqn1_026

An example is presented to illustrate the use of the theorem on real differentiation. The task is to find the Laplace transform of the function in equation (70). Equation (65) presents theorem on real differentiation. The cosine function can be expressed as the derivative of the sine function. Using the Laplace Transform of the sine function and the theorem on real differentiation, we can get the Laplace transform of the cosine function, as shown by equations (71), (72), (73) and (74). The purpose of presenting this example is to illustrate the use of theorem on real differentiation.

eqn1_027

The theorem on real differentiation states that differentiation in time domain is equivalent to multiplying by s in s domain. This theorem, the theorem on complex differentiation, puts forth the other case. Multiplication by time in time domain is equivalent to differentiation in the s domain. Since s can be a complex value, complex differentiation yields the stated result. Equation (75) presents the theorem on complex differentiation. The proof of this theorem is presented by equations (76) and (77).

eqn1_028

By making use of the theorem on complex differentiation, we can get the Laplace transform of the ramp function , given the Laplace transform of the unit step function, and this process is illustrated by equations (78) and (79).

ole1_029

By making use of the theorem on complex differentiation, we can prove a corollary of this theorem, stated by equation (80). Given equation (81), we get equations (82) and (83). We can prove the corollary by continuing along the same line.

eqn1_30

Example 6, presented by equation (84), shows how the theorem on complex differentiation can be applied.

eqn1_031

Example 7, presented by equation (85), shows how the theorem on complex differentiation can be applied. Two more examples are presented below.

eqn1_032

eqn1_033

eqn1_39

If the Laplace transform of f(t) is F(s), the Laplace transform of the integral of f(t) is expressed by equation (88) . To account for the effect of f(t), prior to t = 0, the theorem can be stated as shown by equation (89). Given an initial value of zero, it can be seen that integration in time domain, amounts to dividing by s in the s domain.

eqn1_035

Equation (90) presents the theorem on real integration. The Laplace transform of the integral of f(t) is expressed by equation (91). Equations (92) and (93) show the steps involved in integrating equation (91) by parts.

eqn1_036

Equation (94) presents proof for the theorem on real integration. The relationship between integration in time domain and division by s in s-domain is illustrated by an example.

eqn1_037

The Laplace transform of the ramp function can be obtained from the Laplace transform of the unit step function, since the ramp function is the integral of the unit step function. Equation (95) shows how the Laplace transform of the ramp function can be obtained. It is known that unit step function is of zero value, for t < 0.

eqn1_038

Examples 11 and 12 are presented to illustrate the application of the theorem on real integration. They are self-explanatory. It can be seen that integration in time domain corresponds to division by s in s-domain.

eqn1_039

This theorem enables us to find the value of function f(t), at t = 0+,, directly from the Laplace transform of f(t). It is to be noted that this theorem does not yield f(0), but yields only f(0+). Let both f(t), and its derivative be Laplace transformable. The theorem and its proof are presented by equations (98), (99), (100) and (101).

We get the value of f(0+), because we approach the origin through positive values of time t. This theorem states that the behaviour of f(t), in the neighbourhood of t = 0, is the same as the behaviour of sF(s), in the neighbourhood of s at infinity. Thus it is possible to obtain the value of f(0+) directly from sF(s), as s tends to infinity . For this theorem, there are no restrictions on the location of poles of sF(s).

eqn1_040

The theorem on complex integration is presented by equation (102). Let the Laplace transform of f(t) be F(s). As t approaches zero, let the ratio of f(t) over t be a finite value. The integral of F(s) is evaluated, as shown by equation (103). The order of integration can be interchanged, as shown by equation (103). Since the exponential function tends to zero as s tends to infinity, we get equation (104).

eqn1_041

An example is presented to illustrate the application of theorem on complex integration. Given the Laplace transform of the ramp function, the Laplace transform of the unit step function can be obtained, as shown by equation (106).

eqn1_042

This property forms the basis for applying Laplace transforms to linear systems. . Equation (107) defines the convolution integral. For causal systems, we can use equation (108), where the lower limit is zero. Equation (109) shows how the Laplace transform of the Convolution integral can be obtained.

eqn1_043

Interchanging the order of integration, as shown by equation (110). We can use variable p, to represent (t - t). We then get equations (112) and (113). It is not necessary to remember the proof. But remember the result presented by equation (113).

The set of commonly used functions and their transforms are presented by a table, shown below.


Table of Laplace Transforms: Table 1
t Function s Function
u(t) = unit step function ole00.gif

Linearity: ole01

ole02
Differentiation: ole03 ole04
ole05 ole06
Integration: ole07 ole08
Complex translation: ole09 ole10
Real translation: ole11 ole12
ole13 ole14
ole15 ole16
ole17 ole18
ole19 ole20
Convolution: ole21 ole22
Table 1: (Continued)
function f(t) Function F(s)
Impulse Function: ole25 1
ole26 ole27
ole28 ole29
ole30 ole31
ole32 ole33
ole34 ole35
ole36 ole37
ole38 ole39
ole40 ole41
ole42 ole43
ole44 ole45

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PARTIAL FRACTION EXPANSION

Usually a Laplace transform function has to be expanded into its partial fractions before the inverse can be determined. It is important and essential to master the technique of expanding a transform function into its partial fractions. In order to expand a function into partial fractions, it is essential that the function should be a rational function. A rational function, where the order of the numerator is less than the order of the denominator, can be expressed as the sum of partial fractions with unknown coefficients. There is correspondence between the pole of each partial fraction and the poles of the given function. The task here is to determine the unknown coefficients.

There are three cases that arise in expansion of a Laplace transform function into partial fractions. The given Transform function can have:

1. Distinct and real roots,
2. Multiple Roots, and
3. Complex Pair of roots.

The technique of partial fraction expansion is explained with the help of examples.

eqn1_044

Example 13 presents the first case. The given Laplace transform function has two poles and one zero. The order of the denominator is two and the order of the numerator is one. We can express the given Laplace transform function , as the sum of two partial fractions, one corresponding to the pole at - 1, and the other corresponding to the pole at - 2.

eqn1_045

The given function is expressed as the sum of two fractions, as shown by equation (115). The two coefficients, k1 and k2 , are determined using the Heaviside cover up method, which is the technique normally adopted. The first step is to multiply both sides by the denominator of the given function in equation (115). This results in equation (116).

eqn1_046

It can be seen from equation (116) that if we let s = - 2, we can evaluate coefficient k2. To evaluate k1, set s = - 1. The values of coefficients, k1 and k2, are shown by equation (117). By substituting these values into equation (116), we get equation (118). The solution can be verified as shown by equation (118), by assigning some other value to s, say one. It is necessary to verify the solution, since it is possible to commit a mistake in obtaining the solution.

eqn1_047

Example 14 presents a case with symbolic coefficients for a system with three poles and two zeros. It is seen that the order of the numerator is less than that of the denominator. The given function is presented by equation (119), and it is expressed as the sum of three fractions, with each fraction corresponding to a pole. As before, multiply both sides by the denominator of the given function, and obtain equation (120). It can be seen that the coefficients can be evaluated, as shown by equation (121). Next another example is presented to illustrate this technique.

eqn1_048

Example 15 presents the Laplace transform function of a system with three poles and two zeros. It is seen that the order of the numerator is less than that of the denominator. The given function is presented by equation (122), and it is expressed as the sum of three fractions, with each fraction corresponding to a pole. Multiply both sides by the denominator of the given function, and obtain equation (124). It can be seen that the coefficients can be evaluated, as shown by equation (125). To evaluate c1, s is set to be - 1. To evaluate c2, s is set to be - 2. To evaluate c3, s is set to be - 3. The solution is presented by equation (126). It can be seen that the technique is quite easy to understand.

eqn1_049

Example 16 presents a case with symbolic coefficients for a system with three poles and two zeros. In this case, the multiplicity of poles is three, and all the poles of the system are at the same location. The given function is presented by equation (127), and it is expressed as the sum of three fractions, with the order of each fraction varying by one. If there are three poles at the same location, three fractions are necessary, when the numerator contains one or more zeros. As before, multiply both sides by the denominator of the given function, and obtain equation (128). The value of coefficient, k1 can be evaluated as shown by equation (129), by letting s be equal to - p.

eqn1_050

When the multiplicity of poles is more than one, the normal technique is to use differentiation as a way of reducing the order. Both sides of equation (128) can be differentiated with respect to s, and we get equation (130). In equation (130), let s be equal to - p, then we get coefficient k2. Differentiate equation (130) and get equation (132). Let s be equal to - p, and get the value of coefficient k3. We follow up with a numerical example.

eqn1_051

Example 17 presents the Laplace transform function of a system with three poles and two zeros. In this case, the multiplicity of poles is three, and all the poles of the system are at -1. The given function is presented by equation (134), and it is expressed as the sum of three fractions, with the order of each fraction varying by one. If there are three poles at the same location, three fractions are necessary, when the numerator contains one or more zeros. As before, multiply both sides by the denominator of the given function, and obtain equation (135). The value of coefficient, k1 can be evaluated as shown by equation (136), by letting s = - 1.

eqn1_052

When the multiplicity of poles is more than one, the normal technique is to use differentiation as a way of reducing the order. Both sides of equation (135) can be differentiated with respect to s, and we get equation (137). In equation (137), let s = - 1, then we get coefficient k2. Differentiate equation (137) and get equation (139). Get the value of coefficient, k3, from equation (139). We follow up with another numerical example.

eqn1_053

Example 18 presents the Laplace transform function of a system with four poles, and three zeros. In this case, the multiplicity of one of the poles is three, and the multiple poles of the system are at minus 1. The given function is presented by equation (141), and it is expressed as the sum of four fractions, as shown by equation (141). As before, multiply both sides by the denominator of the given function, and obtain equation (142). The value of coefficient, k1 can be evaluated as shown by equation (143) by letting s = - 3.

The value of coefficient, k2 can be evaluated as shown by equation (144) by letting s = - 1.

eqn1_054
 
To get the coefficients of the other fractions for the multiple pole, the normal technique is to use differentiation as a way of reducing the order. Both sides of equation (142) can be differentiated with respect to s, and we get equation (145). In equation (145), let s = - 1, then we get coefficient k3 as shown by equation (146). Differentiate equation (145) and get equation (147). Let s = - 1, and get the value of coefficient, k4 as shown by equation (148). The answer is expressed by equation (149). This example is more complex than the expressions which we will come across later, when we apply the Laplace transform to differential equations.

eqn1_55

Example 19 presents the Laplace transform function of a system with one real pole, a complex pair of poles and two zeros. It can be expressed as the sum of two fractions. It is possible to express the partial fraction containing a quadratic term in the denominator, as the sum of two fractions, with the pole of the two fractions corresponding to either of the complex poles. That technique is illustrated in the next example. This problem adopts a different technique. As before, multiply both sides by the denominator of the given function, and obtain equation (151). The value of coefficient c1 can be evaluated as shown by equation (152) by letting s = - 1. The value of coefficient c3 can be evaluated as shown by equation (153) by letting s = 0.

The value of coefficient c2 can be evaluated as shown by equation (154), by letting s be equal to infinity. The answer is presented by equation (155).

eqn1_056

Example 20 presents the Laplace transform function of a system with one real zero, one set of complex pair of poles, and another pair of poles on the imaginary axis. It can be expressed as the sum of four fractions, as shown by equation (156). It is possible to express the given function as the sum of two fractions, with each fraction containing a second order term in the denominator. In such a case, each numerator will contain two unknown coefficients, and determining them by assigning values to s may not be that easy. We will end up with a set of four simultaneous equations. It is preferable to deal with evaluation of complex values than trying to solve a set of four simultaneous equations. With the technique adopted here, only two coefficients, A and B, have to be determined. We know how to obtain the conjugates and we can determine the other two coefficients. As before, multiply both sides by the denominator of the given function, and obtain equation (157). The value of coefficient, A can be evaluated as shown by equations (158) and (159), by letting s = j2.

The value of the conjugate of A, can be obtained as shown by equation (159).

eqn1_057

The value of coefficient B can be evaluated as shown by equation (160) by letting s = - 1 - j. The value of the conjugate of B can be obtained as shown by equation (160).

Then the given function can be expressed, as shown by equation (161). We can combine the four fractions into two fractions, to get rid of the imaginary parts. The result is shown by equation (162). It can be verified that the solution represents the given function.

eqn1_058

Example 21 presents the Laplace transform function of a system with one real zero and two sets of complex pair of poles. It can be expressed as the sum of four fractions, as shown by equation (163). It is possible to express the given function as the sum of two fractions, with each fraction containing a second order term in the denominator. In such a case, each numerator will contain two unknown coefficients, and determining them by assigning values to s may not be that easy. We will end up with a set of four simultaneous equations. It is preferable to deal with evaluation of complex values than trying to solve a set of four simultaneous equations. With the technique adopted here, only two coefficients, A and B, have to be determined. We know how to obtain the conjugates and we can determine the other two coefficients. As before, multiply both sides by the denominator of the given function, and obtain equation (164). The value of coefficient A can be evaluated as shown by equation (165) by letting s = - 1 - j2.

The value of the conjugate of A can be obtained as shown by equation (165).

eqn1_059

The value of coefficient B can be evaluated as shown by equation (166), by letting s = - 2 - j. The value of the conjugate of B can be obtained as shown by equation (166).

Then the given function can be expressed, as shown by equation (167). We can combine the four fractions into two fractions, to get rid of the imaginary parts. The result is shown by equation (168). It can be verified that the solution represents the given function. In the next section, we see how we can obtain the solution to linear differential equations.

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SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS

Homogeneous Response

ole60

Now it is shown how we can apply Laplace transforms to get the solution of linear differential equations. In the process, we also learn how we can get the inverse transform of Laplace transform functions. We need to use the table of transforms. Equation (169) presents the homogeneous differential equation of a first order system. We know how to get the Laplace transform of the derivative of a function. We first obtain the product of s, and the Laplace transform of the function, y(t). Subtract from this product y(0), and the result is the Laplace transform of the derivative of y(t). Equation (170) presents the Laplace transform of the given differential equation. Equation (171) presents the Laplace transform of Y(s), and we get its inverse, as shown by equation (172).

ole61

Equation (173) presents the homogeneous differential equation of a second order, over-damped system. The Laplace transform of the first derivative and the Laplace transform of the second derivative are obtained, as shown by equations (65) and (66).

ole62

Equation (174) presents the homogeneous differential equation of a second order, over-damped system. Equation (176) presents the Laplace transform of the given differential equation. Equation (177) presents the Laplace transform of Y(s).

It can be expanded into partial fractions, as shown by equation (177). We can obtain the values of coefficients, A and B, as follows.

ole63

Multiply both sides of equation (177) by the denominator of the Laplace transform function, and this process leads to equation (178). Let s = - 1. The value of A is obtained to be 18. Let s = - 3, and the value of B is obtained as - 8. Thus we get equation (180). The inverse transform is presented by equation (181).

ole64

Equation (182) presents the homogeneous differential equation of a second order, critically-damped system. The Laplace transform of the first derivative and the Laplace transform of the second derivative, are obtained, as shown by equations (65) and (66).

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Equation (182) presents the homogeneous differential equation of a second order, critically-damped system. Equation (183) presents the Laplace transform of the given differential equation. Equation (184) presents the Laplace transform of Y(s), and it can be expanded into partial fractions, as shown by the same equation. We can obtain the values of coefficients, A and B, as follows.

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Multiply both sides of equation (184) by the denominator of the Laplace transform function, and this process leads to equation (185). Let s = - 2. The value of A is obtained to be 26. Differentiate equation (185) with respect to s, and obtain equation (186). The value of B is obtained as 10. Thus we get equation (187). The inverse transform is presented by equation (188).

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Equation (189) presents the homogeneous differential equation of a second order, under-damped system. Equation (190) presents the Laplace transform of the given differential equation. Equation (191) presents the Laplace transform of Y(s), and it can be expanded into partial fractions, as shown by the same equation. We can obtain the values of coefficients, A and B, by comparing coefficients on either side. Equation (192) presents the inverse transform.

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Equation (193) presents the homogeneous differential equation of a second order, undamped system. Equation (194) presents the Laplace transform of the given differential equation. Equation (195) presents the Laplace transform of Y(s), and equation (196) presents the inverse transform.

Step Response

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In this sub-section, we find out how we can obtain the response of first and second order systems to a step input. Equation (197) presents the differential equation of a first order system, with a step input and a non-zero initial condition. We know how to get the Laplace transform of the derivative of a function. We first obtain the product of s and the Laplace transform of the function, y(t). Subtract from this product y(0), and the result is the Laplace transform of the derivative of y(t). Equation (198) presents the Laplace transform of the given differential equation. Equation (199) presents the Laplace transform of Y(s), and we can expand it as the sum of two partial fractions, as shown by equation (200).

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Multiply both sides of equation (200) by the denominator of the Laplace transform function, and this process leads to equation (201). Let s = 0. The value of A is obtained to be 2. Let s = - 5, and the value of B is obtained as 8. Thus we get equation (203). The inverse transform is presented by equation (204).

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Equation (205) presents the differential equation of a second order, over damped system, with zero initial conditions. From the definition of the Laplace transform of the first derivative, and the Laplace transform of the second derivative of a function, we get equation (206). Equation (207) presents the Laplace transform of Y(s), and we can expand it as the sum of three partial fractions, as shown by equation(208). Multiply both sides by the denominator of the Laplace transform of Y(s). By assigning s = 0, obtain the value of coefficient, A. By assigning s = - 3,obtain the value of coefficient, B. By assigning s = -2, obtain the value of coefficient, C. The inverse obtained is expressed by equation (209).

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Equation (210) presents the differential equation of a second order, critically-damped system, with non zero initial conditions. From the definition of the Laplace transform of the first derivative, and the Laplace transform of the second derivative of a function, we get equation (211). Equation (212) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of three partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (213).

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Equation (214) presents the differential equation of a second order, under-damped system, with zero initial conditions. From the definition of the Laplace transform of the first derivative and the Laplace transform of the second derivative of a function, we get equation (215).

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In an under damped system, the natural frequency of oscillation, the damped frequency of oscillation, and the damping factor, are related, as shown by equation (216). Using this relationship, we can express equation (215) in a different way. Expand the function as the sum of two partial fractions, and we get equation (217). The numerator of the second term can be split into two parts, as shown by equation (218). The inverse transform is expressed by equation (219). Spend some time over this problem, and try to get the solution on your own.

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Equation (220) presents the differential equation of a second order, under-damped system, with non zero initial conditions. This example has coefficients, which are real numbers, and it is easier than the previous example. The Laplace transform of the function is expressed by equation (221), and it can be expanded into the sum of partial fractions, as shown by equations (221) and (222). The inverse transform is expressed by equation (223).

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Equation (224) presents the differential equation of a second order, undamped system, with zero initial conditions. Equation (225) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of two partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (226).

Impulse Response

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In this sub-section, we find out how we can obtain the response of first and second order systems to an impulse input. Equation (227) presents the differential equation of a first order system, with an impulse input. From equation (228), equation (229) is obtained. Equation (229) presents the Laplace transform of the given differential equation. The inverse obtained is expressed by equation (230).

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Equation (231) presents the differential equation of a second order, over-damped system, with an impulse input. From the definition of the Laplace transform of the first derivative, and the Laplace transform of the second derivative of a function, we get equation (232). Equation (233) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of two partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (234).

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Equation (235) presents the differential equation of a second order, critically-damped system, with an impulse input. From the definition of the Laplace transform of the first derivative and the Laplace transform of the second derivative of a function, we get equation (236). Equation (237) presents the Laplace transform of the given differential equation. The inverse obtained is expressed by equation (238).

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Equation (239) presents the differential equation of a second order, under damped system, with an impulse input. From the definition of the Laplace transform of the first derivative, and the Laplace transform of the second derivative of a function, we get equation (240). Equation (241) presents the Laplace transform of the given differential equation. The inverse obtained is expressed by equation (242).

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Equation (243) presents the differential equation of a second order, undamped system, with an impulse input. Equation (244) presents the Laplace transform of the given differential equation. The inverse obtained is expressed by equation (245).

Ramp Response

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In this sub-section, we find out how first and second order systems respond a ramp input. Equation (246) presents the differential equation of a first order system, with a ramp input. Equation (247) presents how the Laplace transform of the given differential equation can be obtained. The Laplace transform of the given differential equation can be expanded into the sum of partial fractions, as shown by equation (248) . The inverse obtained is expressed by equation (249).

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Equation (250) presents the differential equation of a second order, over-damped system, with a ramp input. Express both sides of equation (250) in terms of the Laplace transforms, as shown by equation (251). The Laplace transform of the given differential equation can be expanded into the sum of partial fractions, as shown by equation (252) . The inverse obtained is expressed by equation (253).

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Equation (254) presents the differential equation of a second order, critically-damped system, with ramp input. Express both sides of equation (254) in terms of the Laplace transforms, as shown by equation (255). The Laplace transform of the given differential equation can be expanded into the sum of partial fractions, as shown by equation (256) . The values of coefficients, A, B, C, and D, can be obtained as shown below.

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Multiply both sides of equation (256) by the denominator of the Laplace transform of Y(s). The resultant equation is presented by equation (257). Equation (258) shows how the values of A and C, can be obtained. Obtain the value of coefficient A, by assigning - 2 to s. Let s = 0, and obtain the value of coefficient, C. Differentiate equation (257), and obtain equation (259). In equation (259), let s = 0, and then obtain the value of coefficient D, as shown by equation (260). Let s = 1, and obtain the value of coefficient B, as shown by the same equation. The inverse obtained is expressed by equation (262).

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Equation (263) presents the differential equation of a second order, under-damped system, with ramp input. Express both sides of equation (263) in terms of the Laplace transforms, as shown by equation (264). The Laplace transform of the given differential equation can be expanded into the sum of partial fractions, as shown by equation (265) . The values of coefficients, A, B, C and D, can be obtained as shown below.

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Multiply both sides of equation (265) by the denominator of the Laplace transform of Y(s). The resultant equation is presented by equation (266). Let s = 0, and then obtain the value of coefficient A, as shown by equation (267). Differentiate equation (266), and obtain equation (268). In equation (268), let s = 0, and then obtain the value of coefficient B, as shown by equation (269).

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With values of A and B, substituted into equation (268), we get equation (270).
 Differentiate equation (270), and obtain equation (271). Let s = 0, and then obtain the value of coefficient D, as shown by equation (272).

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With value of D, substituted into equation (271), we get equation (273). The only unknown, the coefficient C, can be determined by assigning a suitable value to S.

Let s be equal to 1, and obtain the value of coefficient C, as shown by equation (274). From the values of C and D, we get equation (275). The Laplace transform of the function is presented by equation (276), and the inverse of this function is presented by equation (277).

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Equation (278) presents the differential equation of a second order, undamped system, with ramp input. The Laplace transform of the given differential equation can be expanded into the sum of partial fractions, as shown by equation (279) . The inverse obtained is expressed by equation (280).

Exponential Response

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In this sub-section, we find out how we can obtain the response of first and second order systems to an exponential input. Equation (281) presents the differential equation of a first-order system with an exponential input. From equation (281), equation (282) is obtained. It is seen that the pole of the first order system is at - 3, whereas the pole of the excitation function is at - 2. Equation (283) presents the Laplace transform of the given differential equation. The inverse obtained is expressed by equation (284).

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Equation (285) presents the differential equation of a first-order system with an exponential input. From equation (285), equation (286) is obtained. It is seen that the pole of the first order system is at - 2, and the pole of the excitation function is also at - 2. Equation (286) presents the Laplace transform of the given differential equation. The inverse obtained is expressed by equation (287). When the multiplicity of poles is two, the solution is as shown by equation (287).

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Equation (288) presents the differential equation of a second order, over-damped system with an exponential input. By applying Laplace transform to both sides of equation (288), we get equation (289). Equation (290) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of three partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (291).

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Equation (292a) presents the differential equation of a second order, over -damped system with an exponential input. Equation (292b) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of three partial fractions, as shown by the same equation. By multiplying both sides of equation (292b) by the denominator of the Laplace transform function, we get equation (293). When s = - 2, we can determine the value of A, as shown by equation (294). Differentiate equation (293) and obtain equation (294). When s = - 2, we can determine the value of B, as shown by equation (295).

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With value of B, substituted into equation (295), we get equation (296). The only unknown, the coefficient C, can be determined by assigning a suitable value to s.

Let s be equal to 1, and then obtain the value of coefficient C, as shown by equation (297). The Laplace transform of the function is presented by equation (298), and the inverse of this function is presented by equation (299).

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Equation (300) presents the differential equation of a second order, critically-damped system with an exponential input. Equation (301) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of three partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (302).

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Equation (303) presents the differential equation of a second order, critically-damped system with an exponential input. Equation (304) presents the Laplace transform of the given differential equation. There are three poles at - 3. The inverse obtained is expressed by equation (305). When the multiplicity of poles is three, the solution is as shown by equation (305). It has to contain an exponential term, corresponding to the pole, and a second order term in time, t.

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Equation (306) presents the differential equation of a second order, under-damped system with an exponential input. By applying Laplace transform to both sides of equation (306), we get equation (307). Equation (308) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of two partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (309).

Sinusoidal Response

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This sub-section shows how the response of first and second order systems can be obtained, when the input is a sinusoidal signal. Equation (310) presents the differential equation of a first order system with a sinusoidal excitation function. By applying Laplace transform to both sides of equation (310), we get equation (311). Equation (311) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of three partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (312).

ole100

Equation (313) presents the differential equation of a first order system with a sinusoidal input. By applying Laplace transform to both sides of equation (313), we get equation (314). Equation (314) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of three partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (315).

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Equation (316) presents the differential equation of a second order, over-damped system with a sinusoidal input. By applying Laplace transform to both sides of equation (316), we get equation (317). Equation (317) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of four partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (318). When the input is sinusoidal, using the classical approach for solving the differential equation may be easier.

ole102

Equation (319) presents the differential equation of a second order, critically-damped system, with a sinusoidal input. By applying Laplace transform to both sides of equation (319), we get equation (320). Equation (320) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of four partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (321). When the input is sinusoidal, using the classical approach for solving the differential equation may be easier.

ole103

Equation (322) presents the differential equation of a second order, under-damped system, with a sinusoidal input. By applying Laplace transform to both sides of equation (322), we get equation (323). Equation (323) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of three partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (324). When the input is sinusoidal, using the classical approach for solving the differential equation may be easier.

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Equation (325) presents the differential equation of a second order, undamped system, with a sinusoidal excitation function. By applying Laplace transform to both sides of equation (325), we get equation (326). Equation (326) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of two partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (327).

Response to Complex Sinusoid Function

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This sub-section shows how the response of first and second order systems can be obtained, when the input is a complex sinusoid signal. Equation (328) presents the differential equation of a first order system with a complex sinusoid input. By applying Laplace transform to both sides of equation (328), we get equation (329). Equation (329) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of two partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (330). Perhaps it is easier to get the solution directly from the differential equation, without recourse to Laplace transforms.

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Equation (331) presents the differential equation of a second order, over-damped system with a complex sinusoid input. By applying Laplace transform to both sides of equation (331), we get equation (332). Equation (332) presents the Laplace transform of the given differential equation. It can be expanded, as the sum of two partial fractions, as shown by the same equation. The inverse obtained is expressed by equation (333). Perhaps it is easier to get the solution directly from the differential equation, without recourse to Laplace transforms.

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SUMMARY

This page has presented the definition of Laplace transforms at first. It has been shown how we can obtain the transform of the commonly used functions. The theorems relevant to this subject have been presented. The process of partial fraction expansion has been explained with the help of some examples. In the end, it is shown how we can solve differential equations using Laplace transforms.

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