1. Class 3 Linear System Solution Using the Laplace Transform
System Dynamics
Class 3
Linear System Solution Using the Laplace Transform

2. Introduction
The Laplace transform provides a systematic and general method for solving linear ODEs, and is especially useful either for nonhomogeneous equations whose right-hand side is a function of time or for sets of equations.
Another advantage is that the transform converts linear differential equations into algebraic relations that can be handled easily
Solving a nonhomogeneous ODE does not require first solving the homogeneous ODE.
Initial values are automatically taken care of.
Complicated inputs (right sides of linear ODEs) can be handled mor efficiently.

3. Introduction
The Laplace transform L[x(t)] of a function x(t) is defined as follows:
The variable of integration, t, is arbitrary, and the transform is a function of the parameter s, which is a complex number. This definition is the so-called one-sided transform, which is used in applications where the variable x(t) is zero for t < 0. In our applications we will always assume that this is true.

4. Introduction
A simpler notation for the transform uses the uppercase symbol to represent the transform of the corresponding lowercase symbol; that is,
The process of determining the time function x(t) whose transform is X(s) is denoted by:
The transforms of common functions that have transforms have been derived and tabulated

5. Example: Transform of a Constant
Suppose x(t) = c, a constant, for t ≥ 0. and 0 t < 0 for Determine its Laplace transform, X(s)

6. Example: Transform of the Step Function
The unit step function us(t) is defined as:
Suppose that the function x(t) is zero for t < 0 and a nonzero constant, say M, for t > 0. Then we can express it as x(t)= Mus(t).
The value of M is the magnitude of the step function; if M = 1, the function is the unit step function. From the results of the previous slide, we can see that the Laplace transform of x(t)= Mus(t) is M/s.

7. Example: Transform of the Exponential Function
Derive the Laplace transform of the exponential function x(t)=e−at , where a is a constant:

8. Properties of the Laplace Transform

9. Properties of the Laplace Transform: The Linearity Property
The Laplace transform is a definite integral, and thus it has the properties of such integrals. For example, a multiplicative constant can be factored out of the integral, and the integral of a sum equals the sum of the integrals.
These facts lead to the linearity property of the transform; namely, for the functions f(t) and g(t), if a and b are constants, then:

10. Properties of the Laplace Transform: The Linearity Property
One use of the linearity property is to determine transforms of functions that are linear combinations of functions whose transforms are already known. For example, if x(t) = 6 + 4e−3t , its transform is:
The inverse transform also has the linearity property, so that:
We can often avoid the integration operations by using the linearity property and suitable identities, as shown in the example in the next slide.

11. Example: Laplace Transforms of the Sine and Cosine Functions
Use the linearity property and Euler identity, e jωt = cos ωt + j sin ωt, to derive the Laplace transforms of the exponentially decaying sine and cosine functions, e−at sin ωt and e−at cos ωt, where a and ω are constants
Using the linearity property we have:
Using Euler identity we have:
Using the transform of exponential, we have
To separate the real part from the imaginary part in the expression above, multiply the numerator and denominator by the complex conjugate of the denominator and use the relationship (x − j y)(x + j y) = x2 + y2 to get:
Comparing with the first equation above, we get
For a = 0, we get

12. Properties of the Laplace Transform: Multiplication by an Exponential ↔ Shifting along the s-axis
Another property of the Laplace transform is called shifting along the s-axis or multiplication by an exponential. This property states that:
To derive this property, note that

13. Properties of the Laplace Transform: Multiplication by an Exponential ↔ Shifting along the s-axis
Example: Laplace Transform of the function te-at
Use the multiplication by an exponential property to determine the Laplace transform of the function te-at
Solution: Letting x(t) = t, we have

14. Properties of the Laplace Transform: Multiplication by t ↔ Differentiation with respect to s
Another property is multiplication by t. It states that
To derive this property, note that

15. Properties of the Laplace Transform: Multiplication by t ↔ Shifting along the s-axis
Example: Laplace transform of the function t cos ωt:
Use the multiplication by t property to derive the Laplace transform of the function t cos ωt
Solution: Letting x(t) = cos ωt, we have
In arriving at the previous result, we used the following expression for the derivative of the division of two functions

16. Properties of the Laplace Transform: Shifting along the t axis ↔ Multiplication by an Exponential
This property states that if X(s) is the transform of x(t) then e-sDX(s) (D > 0) is the transform of the time shifted function:
To prove this property, note that
Let τ = t + D → dτ = dt and substitute in the above expression, we obtain:
Since the time shifted function defined above is zero for all t < D, the equation above may be written as:

17. Properties of the Laplace Transform: Shifting along the t axis ↔ Multiplication by an Exponential
Example: Laplace transform of a shifted unit step
Find the Laplace transform of the unit step function if the discontinuity occurs at t = D. The unit step function, denoted by us(t-D) as:
Solution

18. Properties of the Laplace Transform: Shifting along the t axis ↔ Multiplication by an Exponential
Example: Laplace transform of the rectangular pulse function
Find the Laplace transform of the rectangular pulse function P(t) using (i) the integration definition of the Laplace transform, and (ii) the time shift property :
Solution (i)
Solution (ii)

19. Properties of the Laplace Transform: Shifting along the t axis ↔ Multiplication by an Exponential
Example: Laplace transform of the rectangular pulse function
Find the Laplace transform of the rectangular pulse function P(t) using (i) the integration definition of the Laplace transform, and (ii) the time shift property.
Solution (i)
Solution (ii)

20. Properties of the Laplace Transform: The Derivative Property
The derivative property allows determining the Laplace transform of the derivative of a function x(t) as follows
Knowing the Laplace transform, X(s) = L [x(t)], of a function x(t), The Laplace transform L [ẋ(t)], of the derivative function dx(t)/dt can be found by applying integration by parts to the definition of the transform as follows

21. Properties of the Laplace Transform: The Derivative Property
Knowing the Laplace transform L [x(t)] = X(s) of a function x(t), the derivative property allows determining the Laplace transform the derivative of x(t), L [ẋ(t)] as follows
The derivative property can be derived by applying integration by parts as follows
The derivative relationship can be taken one step further as follows

22. Properties of the Laplace Transform: The Derivative Property
Example: Laplace transform of the cosine function
Knowing that the Laplace transform of the sine function
Find the Laplace transform of the cosine function using the derivative property

23. Properties of the Laplace Transform: The Derivative Property
Example: Using the derivative property to find the Laplace transform of known solutions to ODEs
Knowing that the function x(t) = e-at is the solution of the 1st order model ẋ + ax = 0, x(0) = 1, use the derivative property to find the Laplace transform of e-at L[e-at]
Solution

24. Properties of the Laplace Transform: The Initial Value Theorem
Sometimes we will need to find the value of the function x(t) at t =0+ (a time infinitesimally greater than 0), given the transform X(s). The answer can be obtained with the initial value theorem, which states that
Informal Proof:
Using the derivative property, we have

25. Properties of the Laplace Transform: The Final Value Theorem
The final value theorem, which states that
The theorem is true if the following conditions are satisfied:
The functions x(t) and dx/dt must possess Laplace transforms.
The function x(t) must approach a constant value as t → ∞. The latter condition will be satisfied if all the roots of the denominator of sX(s) have negative real parts.

26. Properties of the Laplace Transform: The Final Value Theorem
Example: Verify the validity of the final value theorem for the function x(t) = 1.
Solution: If x(t) = 1 then X(s) = 1/s, and the final value theorem is applied as follows:
which is correct.

27. Properties of the Laplace Transform: The Final Value Theorem
Example: Check if the final value theorem is valid for
Solution: the final value theorem is applied as follows:
The function x(t) corresponding to X(s) is x(t) = e−4t sin 7t, and approaches 0 as t → ∞, which is correct.

28. Solving Linear ODEs with the Laplace Transform
Differential Equation
Laplace Both Sides
Linearity Property
Derivative Property
Find X(s) Algebraically
Partial Fraction Expansion
Inverse Laplace to get x(t)

29. Solving Linear ODEs with the Laplace Transform
Example: Response of 1st Order System
Find the complete response of the 1st order ODE ẋ + ax = f(t) using the Laplace transform method
Differential Equation
Laplace Both Sides
Linearity Property
Derivative Property
Find X(s) Algebraically
Partial Fraction Expansion
Inverse Laplace to get x(t)

30. Partial Fraction Expansion: Numerator Conformance Method (NCM)
Example: Step Response of a 1st Order System
Find the complete response of the 1st order ODE ẋ + ax = f(t) whose f(t) is a step function of magnitude M
Solution: Using the results of the previous slide, we have
Since the Laplace transform of a step forcing function of magnitude M, is F(s) = M/s, the function X(s), and the corresponding x(t) are given by:
The second term in the above expression is evaluated as
The constants C1 and C2 that satisfy the above equation are found as follows:
The function x(t) is thus found from the Laplace transform tables to be

31. Partial Fraction Expansion: Numerator Conformance Method (NCM)
Example: Ramp Response of a First-Order System:
Obtain the expression for the complete response of the 1st order ODE ẋ + ax = 5t knowing that x(0) = 10.
Solution: Applying the Laplace transform to the equation, we obtain
Solving for X(s), we obtain
When there are repeated factors in the denominator of X(s), the partial fraction expansion must include one additional term for each extra repeated factor.
The forced response of the above expression may be written as:
Comparing the numerators we see that C2 + C3 = 0, C1 + 3C2 = 0, and 3C1 = 5. Thus, C1 = 5/3, C2 = −C1/3 = −5/9, and C3 = − C2 = 5/9. The complete response, which is the sum of the natural response and the forced response is therefore:

32. Partial Fraction Expansion: Numerator Conformance Method (NCM)
Complex Factors in the Denominator:
Complex factors in the denominator of the transform can be handled easily by using the fact that the complex conjugates s = −a ± jb correspond to the quadratic factor (s + a)2 + b2.

33. Partial Fraction Expansion: Numerator Conformance Method (NCM)
Example: Complex Factors in the Denominator
Invert the following transform by representing it as the sum of terms that appear in the Table of Laplace Transforms.
Solution: The roots of the denominator are s = −2 ± 7 j and so the transform can be expressed as:
Comparing numerators we see that
Solving the two resulting equations simultaneously, the constants are found to be C1 = 8 and C2 = -3/7. X(s) and the corresponding x(t) are thus given by:

34. Partial Fraction Expansion: Numerator Conformance Method (NCM)
Example: Step response of 2nd order system
Obtain the complete response of the following model:
Solution: Transforming the equation and solving for X(s) gives:
The first term on the right of the above expression corresponds to the free response and is the same transform inverted in the previous example. Thus the free response is:
The second term corresponds to the forced response. It can be expressed as
Comparing numerators on the left and right sides, we see that C1+C2 =0, 4C1+2C2+7C3 =0, and 53C1 =15. Thus, C1 =15/53, C2=−15/53, and C3=−30/371. The forced response and the complete response are thus given by:

35. Partial Fraction Expansion: Numerator Conformance Method (NCM)
Example: Step response of 2nd order system
The response has a radian frequency of 7, which corresponds to a period of 2π/7=0.897 s. The oscillations are difficult to see for t >2 because e−2t <0.02 for t >2.
So for most practical purposes we may say that the response is essentially constant with a value 15/53 for t >2.

36. Partial Fraction Expansion:
Step Response of 2nd Order System of Complex Roots
The step response of a second-order equation with complex roots results in a transform of the following form:
Using the procedure employed above, we can show that the coefficients are:
The response is

37. Partial Fraction Expansion The Limit Formula
To solve a differential equation by using the Laplace transform, we must be able to obtain a function x(t) from its transform X(s). This process is called inverting the transform. Unless the transform is a simple one appearing in the transform table, it will have to be represented as a combination of simple transforms.
In practice, we might encounter high-order system models or complicated inputs. Both situations require a general approach to obtaining the expansion, and this section develops such an approach.

38. Partial Fraction Expansion Coefficients The Limit Formula
Most transforms occur in the form of a ratio of two polynomials, such as:
In all of our examples, m ≤ n. If X(s) is of the form above, the method of partial fraction expansion can be used. Note that we assume that the coefficient an is unity. If not, divide the numerator and denominator by an. The first step is to solve for the n roots of the denominator. If the ai coefficients are real (as they will be for all our applications), any complex roots will occur in conjugate pairs.
There are two cases to be considered. The first is where all the roots are distinct; the second is where two or more roots are identical (repeated).

39. Partial Fraction Expansion: Distinct Roots
If all the roots are distinct, we can express X(s) in the expression of the previous slide in factored form as:
where the roots are s = −r1, −r2, , −rn. This form can be expanded as
where
Multiplying X(s) by the (s + ri) cancels that factor in the denominator of X(s) before the limit is taken. This is a good way of remembering the above formula. Each factor corresponds to an exponential function of time, and the inverse transform is:

40. Partial Fraction Expansion: Distinct Roots
Example: One Zero Root and One Negative Root
Obtain the inverse Laplace transform of
Solution
The denominator roots are s = 0 and s = −3, which are distinct and real. Thus the partial-fraction expansion has the form:
Using the formula in the previous slide, we obtain
The inverse transform is

41. Partial Fraction Expansion: Distinct Roots
Example: A Third Order System
Obtain the solution of the following problem using (i) The Laplace transform method and (ii) The trial method.

42. Partial Fraction Expansion: Distinct Roots
(i) Solution using the Laplace Transform
Laplace transform the system to obtain:
Solving for X(s) using the given initial values and noting that the roots of the resulting cubic in the denominator are s = -2, s = -3 and s = -5, we obtain:
Solving for the coefficients of the partial fraction expansion the solution is found to be:

43. Partial Fraction Expansion: Distinct Roots
(ii) Solution using the Trial Method
The characteristic equation for the homogeneous equation is:
whose roots are s = -2, s = -3 and s = Noting that the particular solution is xp(t) = 750/300 = 5/2, the general solution is found to be
The constants C1, C2 and C3 satisfying the initial conditions are found as shown:
The solution is:
The terms e−3t and e−5t die out faster than e−2t, so for t > 4/3, the response may be approximated by x(t) = 5/2 + (55/6)e−2t. For t > 2, the response is approximately constant at x = 5/2. The “hump” in the response is produced by the positive value of ẋ(0).

44. Partial Fraction Expansion: Distinct Roots
Using the trial method of the Laplace transformation method in solving ODEs is a personal choice. The Laplace method requires some algebra to handle the initial conditions, whereas the trial solution method requires that three simultaneous equations be solved.

45. Partial Fraction Expansion: Repeated Roots
Suppose that p of the roots have the same value s =−r1, and the remaining (n − p) roots are distinct and real. Then X(s) is of the form:
The expansion is:
and the coefficients in the expansion are found as:
The solution for the time function is thus:

46. Partial Fraction Expansion: Repeated Roots
Example: One Negative Root and Two Zero Roots
Obtain the solution of the following problem using (i) The Laplace transform method and (ii) The trial method.

47. The Impulse
The rectangular pulse function has a finite non-zero value for a non-zero time period. An impulse is similar to a pulse function but it models an input that is suddenly applied and removed after a very short time. It has an infinite magnitude for an infinitesimal time.
Consider the rectangular pulse function shown. Its transform was found (slide 18) to be M(1 − e−sD)/s. The area A under the pulse is A = MD is called the strength of the pulse. If we let this area remain constant at the value A and let the pulse duration D approach zero, we obtain the impulse. Because M = A/D, the transform F(s) is found by using the L’Hopital’s limit rule to be:
If the strength A=1, the function is called a unit impulse. In keeping with our interpretation of the initial conditions, we consider the impulse δ(t) to start at time t = 0 and finish at t = 0+, with its effects first felt at t = 0+.

48. The Impulse Example: Impulse Response of a Simple 1st Order Model
Obtain the unit-impulse response of the following model in two ways: (i) by separation of variables and (ii) with the Laplace transform. The initial condition is x(0) = 3. What is the value of x(0+)?

49. The Impulse Solution using Separation of Variables
Separate the variables and integrate both sides to obtain:
This is the solution for t > 0 but not for t = 0. Thus, x(0+) = 4 but x(0) = 3, so the impulse has changed x(t) instantaneously from 3 to 4.

50. The Impulse Solution using Laplace Transform
The transformed equation is
which gives a solution:
Again, this solution is for t > 0 but not for t = 0. Thus, x(0+) = 4 but x(0) = 3. This can also be verified by using the initial value theorem

51. The Impulse Example: Impulse Response of a Simple Second-Order Model
Obtain the unit-impulse response of the following model in two ways: (a) by separation of variables and (b) with the Laplace transform. The initial conditions are x(0) = 5 and ẋ(0) = 10. What are the values of x(0+) and ẋ(0+)?

52. The Impulse Solution using Separation of Variables
Let v = ẋ and separate the variables and integrate both sides to obtain:
Thus, ẋ(0+) = 11 and is not equal to x(0) = 3, so the impulse has changed ẋ(t) instantaneously from 10 to 11.
Now integrate ẋ = v = 11 to obtain
Thus, x(0+) = 5 and is equal to x(0). So for this model, the impulse has changed ẋ(t) from t = 0 to t = 0+ but does not change x.

53. The Impulse Solution using Laplace Transform
The transformed equation is
Note that the initial values used with the derivative property are the values at t = 0.
The initial value theorem gives:
Also, since L(ẋ) = sX(s) – x(0), we have
which is the result obtained before.