Ch 03 solution methods for dynamic models

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien §2.Response Types and Stability • Steady-state: the part of the response that remains with time • Transient: the part of

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3 Solution Methods for

Dynamic Models

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

System Dynamics 3.01 Solution Methods for Dynamic Models

§1.Differential Equations

- Anordinary differential equation(ODE): an equation containing

ordinary, but not partial, derivatives of the dependent variable

- The subject of system dynamics is time-dependent behavior,the independent variable in our ODEs will be time𝑡

2.Classification of Differential Equations

- Linear differential equation

- The order of differential equation: the order of thehighestderivativeof the dependent variable in the equation

3 𝑥+ 7 𝑥 + 2𝑥 = 5: second order differential equation

- A model can consist of more than one equation

3 𝑥1+ 5𝑥1− 7𝑥2= 5 𝑥

0

𝑡

𝑔 𝑡 𝑑𝑡

- Example 3.1.1 Separation of Variables for a Linear Equation

Use separation of variables to solve the following problem for

§1.Differential Equations4.Trial Solution Method

Consider 𝑥 + 𝑎𝑥 = 𝑏, 𝑡 ≥ 0

⟹ Find the solution in the form 𝑥 𝑡 = 𝐶 + 𝐷𝑒𝑠𝑡

Subtituiting the solution into the equation

𝑥 + 𝑎𝑥 = 𝑠𝐷𝑒𝑠𝑡+ 𝑎 𝐶 + 𝐷𝑒𝑠𝑡 ⟹ 𝑠 + 𝑎 𝐷𝑒𝑠𝑡+ 𝑎𝐶 = 𝑏Find𝑠, 𝐶

𝑠 + 𝑎 = 0

𝑎𝐶 = 𝑏⟹

𝑠 = −𝑎

𝐶 = 𝑏/𝑎Find𝐷 from the initial condition

𝑥 0 = 𝐶 + 𝐷𝑒0= 𝐶 + 𝐷 ⟹ 𝐷 = 𝑥 0 − 𝐶Therefore

𝑥 𝑡 =𝑏

𝑎+ 𝑥 0 −

𝑏

𝑎 𝑒−𝑎𝑡

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- Example 3.1.2 Two Distinct, Real Roots

Use trial solution method to solve the following problem for

There are two solutions⟹ the trial form is not suitable

Need an additional term with an arbitrary constant⟹ The

appropriate trial-solution form

𝑥 𝑡 = 𝐶 + 𝐷1𝑒𝑠 1 𝑡+𝐷2𝑒𝑠 2 𝑡

⟹

𝑠1= −2

𝑠2= −5

𝐶 = 2The solution

- Example 3.1.3 Two Repeated, Real Roots

𝑡 ≥ 0: 5 𝑥 + 20 𝑥 + 20𝑥 = 28, 𝑥 0 = 5, 𝑥 0 = 8

Solution

Trial solution form 𝑥 𝑡 = 𝐶 + 𝐷𝑒𝑠𝑡

Substituting this form into the ODE

5𝑠2+ 20𝑠 + 20 𝐷𝑒𝑠𝑡+ 20𝐶 = 28

⟹ 5𝑠2+ 20𝑠 + 20 = 0

20𝐶 = 28

⟹ 𝑠 = −2, 𝑠 = −2, 𝐶 = 1.4

Need an additional term with an arbitrary constant⟹ The

appropriate trial-solution form

𝑥 𝑡 = 𝐶 + (𝐷1+ 𝐷2𝑡)𝑒𝑠 1 𝑡

⟹ 𝑠1= −2, 𝐶 = 1.4The solution

- Example 3.1.4 Two Imaginary Roots

𝑡 ≥ 0: 𝑥 + 16𝑥 = 144, 𝑥 0 = 5, 𝑥 0 = 12

Solution

Trial solution form 𝑥 𝑡 = 𝐶 + 𝐷𝑒𝑠𝑡

Substituting this form into the ODE

𝑥 𝑡 = 𝐶 + 𝐷1+ 𝐷2𝑐𝑜𝑠4𝑡 + 𝑗 𝐷1− 𝐷2𝑠𝑖𝑛4𝑡

or 𝑥 𝑡 = 𝐶 + 𝐵1𝑐𝑜𝑠4𝑡 + 𝐵2𝑠𝑖𝑛4𝑡Evaluating𝑥(𝑡) and 𝑥(𝑡) at 𝑡 = 0

𝑥 0 = 𝐶 + 𝐵1= 5

𝑥 0 = 4𝐵2= 12 ⟹

𝐵1= −4

𝐵2= +3The solution

𝑥 𝑡 = 9 + 3𝑠𝑖𝑛4𝑡 − 4𝑐𝑜𝑠4𝑡

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- Example 3.1.5 Motion of a Robot-Arm Link

The equation of motion0.23 + 0.5𝑚𝐿2 𝜃

= 𝑇𝑚− 4.9𝑚𝐿𝑠𝑖𝑛𝜃Solve the equation for the case

𝑇𝑚= 0.5𝑁𝑚,𝑚 = 10𝑘𝑔,𝐿 = 0.3𝑚 Assume the system starts fromrest at 𝜃 = 0 and the angle 𝜃remains small

- Example 3.1.6 Two Complex Roots

𝑡 ≥ 0: 𝑥 + 6 𝑥 + 34𝑥 = 68, 𝑥 0 = 5, 𝑥 0 = 7Solution

Subtituting𝑥 𝑡 = 𝐶 + 𝐷𝑒𝑠𝑡into the ODE

𝑠2+ 6𝑠 + 34 𝐷𝑒𝑠𝑡+ 34𝐶 = 68

⟹ 𝑠2+ 6𝑠 + 34 = 0

34𝐶 = 68⟹ 𝑠 = −3 ± 5𝑗, 𝐶 = 2Therefore

𝑥 𝑡 = 𝐶 + 𝐷1𝑒(−3+5𝑗)𝑡+ 𝐷2𝑒(−3−5𝑗)𝑡

⟹ 𝑥 𝑡 = 𝐶 + 𝑒−3𝑡(𝐷1𝑒5𝑗𝑡+ 𝐷2𝑒−5𝑗𝑡)

⟹ 𝑥 𝑡 = 𝐶 + 𝑒−3𝑡(𝐵1𝑐𝑜𝑠5𝑡 + 𝐵2𝑠𝑖𝑛5𝑡)With the initial conditions𝑥 𝑡 = 2 + 𝑒−3𝑡 3𝑐𝑜𝑠5𝑡 +16

5𝑠𝑖𝑛5𝑡

5.Sumary of the trial solution method

Ordinary Differential Equation Solution form

§1.Differential Equations6.Assessment of Solution Behavior

- The characteristic equation can be quickly identified from ODE

by replacing 𝑥 with 𝑠, 𝑥 with 𝑠2, and so forth

- Example

3 𝑥 + 30 𝑥 + 222𝑥 = 148The characteristic equation

𝑠2+ 10𝑠 + 74 = 0

⟹ 𝑠 − −5 − 7𝑗 𝑠 − −5 + 7𝑗 = 0The solution in the form

𝑥 𝑡 = 𝑒−5𝑡 𝐶1𝑠𝑖𝑛7𝑡 + 𝐶2𝑐𝑜𝑠7𝑡 +2

3

𝑎 ≠ 0, 𝑎 2 < 4𝑏 : 𝑠 = 𝜎 ± 𝑗𝜔, 𝜎 = −𝑎/2, 𝜔 = 4𝑏 − 𝑎 2 /2 𝑥 𝑡 = 𝑒 𝜎𝑡 𝐶1𝑠𝑖𝑛𝜔𝑡 + 𝐶2𝑐𝑜𝑠𝜔𝑡 +𝑐

§2.Response Types and Stability

• Steady-state: the part of the response that remains with time

• Transient: the part of the response that disappears with time

• Free: the part of the response that depends on the initial conditions

• Forced: the part of the response due to the forcing function

𝑥 + 𝑎𝑥 = 𝐵, 𝑎 ≠ 0 𝑥 𝑡 =𝑏

𝑎 + 𝐶𝑒 −𝑎𝑡 System Dynamics 3.17 Solution Methods for Dynamic Models

1.The time constantThe 1storder model

𝑥 + 𝑎𝑥 = 𝑏

⟹ 𝑥 +1

𝜏𝑥 = 𝑏, 𝜏 ≡ 1/𝑎give the free response

𝑥 𝑡 = 𝑥 0 𝑒−𝑎𝑡

⟹ 𝑥 𝑡 = 𝑥(0)𝑒−𝜏𝑡

𝜏:time constant

• measure of the exponential decay curve

• estimate how long it will take for the transient response todisappear

𝑥 + 𝑎𝑥 = 𝑏, 𝑎 ≠ 0 𝑥 𝑡 =𝑏𝑎+ 𝐶𝑒 −𝑎𝑡 System Dynamics 3.18 Solution Methods for Dynamic Models

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From figure

• after 𝑡 = 𝜏, 𝑥(𝑡) has decayed to 37% of its initial value

• after𝑡 = 4𝜏, 𝑥(𝑡) has decayed to02%of its initial value

The time constant is useful also for analyzing the responsewhen the forcing function is a constant

The total response in terms of𝜏 by substituting 𝑎 = 1/𝜏

𝑥 𝑡 = 𝑏𝜏

steady state

+ 𝑥 0 − 𝑏𝜏 𝑒−𝑡𝜏 transient state

= 𝑥𝑠𝑠+ [𝑥 0 − 𝑥𝑠𝑠]𝑒−𝑡𝜏

The response approaches the constant value𝑏𝜏 as 𝑡 → ∞

𝑥𝑠𝑠= lim

𝑡→∞𝑥(𝑡) = 𝑏𝜏

The forced response (for which𝑥(0) = 0) is plotted in the figure

• at 𝑡 = 𝜏, the response is 63% of the steady-state value

• at𝑡 = 4𝜏, the response is98%of the steady-state value

• at 𝑡 = 5𝜏, the response is 99% of the steady-state value

For most engineering purposes:𝑥(𝑡) reaches steady-state at 𝑡 = 4𝜏

- Example 3.2.1 Responses for Second-Order, Distinct Roots

Identify the responses of 𝑥 + 7 𝑥 + 10𝑥 = 𝑐 (roots: −2, −5)Solution

2.The Dominant-Root Approximation

- The time constant concept is not limited to first-order models

It can also be used to estimate the response time of

higher-order models

- Example 3.2.2Responses for Second-Order, Complex Roots

Identify the responses of the equation

The coefficients𝐵1and𝐵2can be found in terms of arbitraryinitial conditions in the usual way

𝐵1= 𝑥 0 − 𝑐

34, 𝐵2=

34 𝑥 0 + 102𝑥 0 − 3𝑐170

Model’s time constant: τ = 1/3 ⟹ the response is essentially

at steady state for 𝑡 > 4τ = 4/3

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3.Time Constants and Complex Roots

- Example 3.2.3 Responses for Second-Order, Imaginary Roots

Identify the responses of 𝑥 + 16𝑥 = 𝑐

Solution

The characteristic roots𝑠 = ±4𝑗, the solution form

𝑥 𝑡 = 𝑐

16+ 𝐵1𝑐𝑜𝑠4𝑡 + 𝐵2𝑠𝑖𝑛4𝑡

no terms that disappear as𝑡 → ∞ ⟹ no transient response

The free and forced responses

§2.Response Types and Stability4.Stability

- Unstable: the free response approaches∞ as 𝑡 → ∞

- Stable: the free response approaches0

- Neutral stability: the borderline between stable and unstable

The free response does not approach both∞ and 0

- The stability properties of a linear model are determined from

its characteristics roots

- The first-order model 𝑥 + 𝑎𝑥 = 𝑓(𝑡)

The characteristic equation 𝑠 + 𝑎 = 0

The free response 𝑥 𝑡 = 𝑥(0)𝑒−𝑎𝑡

𝑠 = −𝑎 < 0: 𝑡 ⟶ ∞, 𝑥(𝑡) ⟶ 0: the model is stable

𝑠 = −𝑎 > 0: 𝑡 ⟶ ∞, 𝑥(𝑡) ⟶ ∞ : the model is unstable

𝑠 = 𝑎 = 0: 𝑡 ⟶ ∞, 𝑥(𝑡) ↛ 0, ∞: the model is neural stability

- The second-order models with the same initial conditions

- Stability Test for Linear Constant-Coefficient Models

• A constant-coefficient linear model isstableif and only if all

of its characteristic roots have negative real parts

• The model isneutrally stableif one or more roots have a

zero real part, and the remaining roots have negative real

parts

• The model isunstableif any root has a positive real part

§2.Response Types and Stability5.A Physical Example

- Pendulum motion equation (𝜃 ≈ 0)

𝑚𝐿2 𝜃 + 𝑐 𝜃 + 𝑚𝑔𝐿𝜃 = 0

- Equilibrium point: 𝜃 = 0

𝑠 =−𝑐 ± 𝑐

2− 4𝑚2𝐿3𝑔2𝑚𝐿2

• 2𝑚𝐿 𝐿𝑔 > 𝑐 > 0: damped oscillating

• 𝑐 > 2𝑚𝐿 𝐿𝑔: no oscillating

• 𝑐 = 0: oscillating about𝜃 = 0

Roots

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- Pendulum motion equation (𝜃 ≈ 0)

𝑚𝐿2 𝜃 + 𝑐 𝜃 + 𝑚𝑔𝐿𝜃 = 0

- Equilibrium point: 𝜃 = 𝜋

𝑠 =−𝑐 ± 𝑐

2+ 4𝑚2𝐿3𝑔2𝑚𝐿2

- Pendulum will not oscillate about𝜃 = 𝜋 but will

continue to fall away if disturbed⟹ the system

is unstable

- The model is based on the assumption that𝜃 ≈

0 ⟹ cannot draw any conclusions from the

model regarding the behavior when 𝜃 is not

near𝜋

System Dynamics 3.33 Fluid and Thermal Systems

7.Stability and Equilibrium

- Anequilibrium: a state of no change

The pendulum in the figure is in equilibrium at

𝜃 = 0 and when perfectly balanced at 𝜃 = 𝜋

• the equilibrium at 𝜃 = 0: stable

• the equilibrium at 𝜃 = 𝜋: unstable

⟹ the same physical system can have different

stability characteristics at different equilibria

- Stability is not a property of the system alone, but is a

property of a specific equilibrium of the system

- When we speak of the stability properties of a model, we are

actually speaking of the stability properties of the specific

equilibrium on which the model is based

- The figure shows a ball on a surface that has a valley and a hill

- The bottom of the valley is an equilibrium, and if the ball isdisplaced slightly from this position, it will

• oscillate forever about the bottom if there is no friction:

neutrally stable

• return to the bottom if friction is present:stable

- If displace the ball so much to the left that it lies outside thevalley, it will never return:locally stablebutglobally unstable

- If the system returns to its equilibrium for any initialdisplacement:globally stable

- The equilibrium on the hilltop isglobally unstable

§3.The Laplace Transform Method

- The Laplace transforms the problem in time-domain to

problem in s-domain, then applying the solution in s-domain,

and finally using inverse transform to converse the solution

back to the time-domain

- The Laplace transformℒ{𝑥 𝑡 } of a function 𝑥(𝑡) is defined as

ℒ 𝑥 𝑡 = lim

𝑇→∞ 0

𝑇

𝑥 𝑡 𝑒−𝑠𝑡𝑑𝑡but is usually expressed more compactly as

ℒ 𝑥 𝑡 =

0

∞

𝑥 𝑡 𝑒−𝑠𝑡𝑑𝑡

- Notation for the Laplace and inverse Laplace transforms

𝑋 𝑠 = ℒ 𝑥 𝑡

𝑥 𝑡 = ℒ−1{𝑋(𝑠)}

-Table of Laplace transform pairs

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§3.The Laplace Transform Method1.Transforms of Common Functions

- Example 3.3.1 Transform of a Constant

Suppose 𝑥(𝑡) = 𝑐 , a constant, for 𝑡 ≥ 0 Determine itsLaplace transform

SolutionFrom the transform definition

- The step function

Unit-step function

𝑢𝑠𝑡 = 0 𝑡 < 0

1 𝑡 > 0Step function

𝑥(𝑡) = 𝑀𝑢𝑠(𝑡)

𝑋 𝑠 = ℒ 𝑀𝑢𝑠𝑡 = 𝑀/𝑠

- Example 3.3.2 The Exponential Function

Derive the Laplace transform of the exponential function𝑥(𝑡) = 𝑒−𝑎𝑡, where𝑎 is a constant

SolutionFrom the transform definition

System Dynamics 341 Fluid and Thermal Systems

2.Properties of the Laplace Transform

- Example 3.3.3 The Sine and Cosine Functions

Derive the Laplace transforms of𝑒−𝑎𝑡𝑠𝑖𝑛𝜔𝑡 and 𝑒−𝑎𝑡𝑐𝑜𝑠𝜔𝑡,where𝑎 and 𝜔 are constants

SolutionRecall: - the Euler identity

𝑒𝑗𝜃= 𝑐𝑜𝑠𝜃 + 𝑗𝑠𝑖𝑛𝜃, with 𝜃 = 𝜔𝑡

- the relation1

𝑥 − 𝑗𝑦=

𝑥 + 𝑗𝑦(𝑥 − 𝑗𝑦)(𝑥 + 𝑗𝑦)=

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Comparing the above two equation, the Laplace transforms

of the exponentially decaying sine and cosine functions are

ℒ 𝑒−𝑎𝑡𝑐𝑜𝑠𝜔𝑡 = 𝑠 + 𝑎

(𝑠 + 𝑎)2+𝜔2

ℒ 𝑒−𝑎𝑡𝑠𝑖𝑛𝜔𝑡 = 𝜔

(𝑠 + 𝑎)2+𝜔2

-Shifting along the 𝑠-axisor multiplication by an exponential

- Example 3.3.4 The Function𝑡𝑒−𝑎𝑡

Derive the Laplace transform of the function𝑡𝑒−𝑎𝑡

= 1(𝑠 + 𝑎)2

- Example 3.3.5 The Function𝑡𝑐𝑜𝑠𝜔𝑡

Derive the Laplace transform of the function𝑡𝑐𝑜𝑠𝜔𝑡

- Example 3.3.6 The Shifted Step Function

If the discontinuity in the unit-step functionoccurs at𝑡 = 𝐷

b From the time-shifting property

The pulse= A unit-step + A shifted, negative unit-step

⟹ℒ 𝑃 𝑡 =1

𝑠 1 − 𝑒

−𝑠𝐷

§3.The Laplace Transform Method3.The Derivative Property

Applying integration by parts to the definition of the transform

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4.The Initial Value Theorem

Use to find the value of the function 𝑥(𝑡) at 𝑡 = 0+(a time

infinitesimally greater than0) with given the transform 𝑋(𝑠)

§3.The Laplace Transform Method5.The Final Value Theorem

Use to find the limit of the function𝑥(𝑡) as 𝑡 → ∞

𝑓 ∞ = lim

𝑡→∞𝑓(𝑡) = lim

𝑠→0𝑠𝐹(𝑠)Example

𝑋 𝑠 = 7(𝑠 + 4)2+49

⟹ 𝑥 ∞ = lim

𝑠→0𝑠𝑋(𝑠) = lim

𝑠→0

7𝑠(𝑠 + 4)2+49= 0This is confirmed by evaluating the inverse transform

𝑥 𝑡 = ℒ−1𝑋 𝑠 = 𝑒−4𝑡sin7𝑡

⟹ 𝑥 ∞ = 𝑒−(4×0)sin(7 × 0) = 0

The final value theorem does not apply to a periodic function

6.Solving Equations with the Laplace Transform

Consider the linear first-order equation

𝑥 + 𝑎𝑥 = 𝑓(𝑡)

𝑓(𝑡): the input 𝑎: a constant

Multiply both sides of the equation by𝑒−𝑠𝑡and then integrate

- Example 3.3.8 Step Response of a First-Order Equation

Suppose that the input𝑓(𝑡) of the equation 𝑥 + 𝑎𝑥 = 𝑓(𝑡) is astep function of magnitude𝑀 whose transform is 𝐹(𝑠) = 𝑀/𝑠

Obtain the expression for the complete responseSolution

The forced response is obtained from

- Example 3.3.9 Ramp Response of a First-Order Equation

Determine the complete response of the following model,

which has a ramp input 𝑥 + 3𝑥 = 5𝑡, 𝑥 0 = 10

Comparing the numerators

𝐶2+ 𝐶3= 0

𝐶1+ 3𝐶2= 03𝐶1= 5

⟹

𝐶1= 5/3

𝐶2= −5/9

𝐶3= 5/9The forced response

by𝑥(𝑡) = 5𝑡/3 − 5/9

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- Example 3.3.10 Transform Inversion for Complex Factors

Invert the following transform

⟹ 𝑥 𝑡 = 8𝑒−2𝑡𝑐𝑜𝑠7𝑡 −3

7𝑒

−2𝑡𝑠𝑖𝑛7𝑡

- Example 3.3.11 Step Response of a Second-Order Equation

Obtain the complete response of the following model

𝑥 + 4 𝑥 + 53𝑥 = 15𝑢𝑠𝑡 𝑥 0 = 8, 𝑥 0 = −19Solution

Transforming the equation gives

𝑠2𝑋 𝑠 − 𝑠𝑥 0 − 𝑥 0 + 4 𝑠𝑋 𝑠 − 𝑥 0 + 53𝑋 𝑠 =15

𝑠Solve for 𝑋(𝑠) using the given initial conditions

𝑋 𝑠 =𝑥 0 𝑠 + 𝑥 0 + 4𝑥(0)

𝑠2+ 4𝑠 + 53 +

15𝑠(𝑠2+ 4𝑠 + 53)

= 8𝑠 + 13

𝑠2+ 4𝑠 + 53+

15𝑠(𝑠2+ 4𝑠 + 53)The first term on the right of eq.(1) corresponds to the freeresponse8𝑒−2𝑡𝑐𝑜𝑠7𝑡 − (3/7)𝑒−2𝑡𝑠𝑖𝑛7𝑡 (Ex 3.3.10)

(1)

(2)

The second term on the right of eq.(1) corresponds to the

forced response It can be expressed as follows

=(𝐶1+ 𝐶2) 𝑠 + 2

2+ (4𝐶1+2𝐶2+ 7𝐶3)𝑠 +53𝐶1𝑠[ 𝑠 + 22+ 72]

Comparing numerators on the left and right sides

The complete response is the sum of the free and forcedresponses given by equations

(3)

Alternatively, combine the terms on the right side of eq.(1)

𝑋 𝑠 =𝑥 0 𝑠

2+ 𝑥 0 + 4𝑥 0 𝑠 + 15𝑠(𝑠2+ 4𝑠 + 53) =

8𝑠2+ 13𝑠 + 15𝑠[ 𝑠 + 22+ 72]

𝐶1+ 𝐶2= 84𝐶1+ 2𝐶2+ 7𝐶3= 13

§3.The Laplace Transform MethodStep response of a second-order equation with complex roots

Using a transform of the following form

𝐴𝑠2+ 𝐵𝑠 + 𝐶𝑠[ 𝑠 + 𝑎2+ 𝑏2]= 𝐶1

𝑥 𝑡 = 𝐶1+ 𝐶2𝑒−𝑎𝑡𝑐𝑜𝑠𝑏𝑡 + 𝐶3𝑒−𝑎𝑡𝑠𝑖𝑛𝑏𝑡

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§4.Transfer Function

- Response of a linear system= Free response + Forced response

• To focus the analysis on the effects of the input only by

taking the initial conditions to be zero temporarily

• Then, the free response due to any nonzero initial conditions

can be added to the result

- Consider the model 𝑥 + 𝑎𝑥 = 𝑓(𝑡)

Transforming both sides of the equation

𝑠𝑋 𝑠 + 𝑎𝑋 𝑠 = 𝐹 𝑠obtain thetransfer function

𝑇 𝑠 =𝑋(𝑠)𝐹(𝑠)=

1

𝑠 + 𝑎

- A Transfer Function is the ratio of the output of a system to the

input of a system, in the Laplace domain considering its initial

conditions and equilibrium point to be zero

The TF concept is extremely useful for several reasons

-Transfer Functions and Software

• Matlab accepts a description based on the TF

• Simulink uses TF as the basis graphical system descriptioncalled the block diagram

-ODE Equivalence

The transfer function⟺ The ODE

-The Transfer Function and Characteristic Roots

• The characteristic polynomial is the denominator of the TF

• The roots of characteristics equation give the informationabout the stability of the system

- Example 3.4.1 Two Inputs and One Output

Obtain the transfer functions𝑋(𝑠)/𝐹(𝑠) and 𝑋(𝑠)/𝐺(𝑠) for the

temporarily setting the other inputs equal to zero

𝑋(𝑠)

𝐹(𝑠)=

65𝑠2+ 30𝑠 + 40,

𝑋(𝑠)𝐺(𝑠)=

205𝑠2+ 30𝑠 + 40

- Example 3.4.2 A System of Equations

a.Obtain the transfer functions𝑋(𝑠)/𝑉(𝑠) and 𝑌(𝑠)/𝑉(𝑠) of thefollowing system of equations

𝑥 = −3𝑥 + 2𝑦

𝑦 = −9𝑦 − 4𝑥 + 3𝑣(𝑡)b.Obtain the forced response for𝑥(𝑡) and 𝑦(𝑡) if the input is𝑣(𝑡) = 5𝑢𝑠(𝑡)

𝑠 + 32

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