Ch.02 Modeling in Frequency Domain

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Chapter Objectives After completing this chapter, the student will be able to • Find the Laplace transform of time functi

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02 Modeling in Frequency

Domain

System Dynamics and Control 2.01 Modeling in Frequency Domain

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

Chapter Objectives

After completing this chapter, the student will be able to

• Find the Laplace transform of time functions and the inverseLaplace transform

• Find the transfer function (TF) from a differential equation andsolve the differential equation using the transfer function

• Find the transfer function for linear, time-invariant electrical networks

• Find the TF for linear, time-invariant translational mechanical systems

• Find the TF for linear, time-invariant rotational mechanical systems

• Find the TF for gear systems with no loss and for gear systemswith loss

• Find the TF for linear, time-invariant electromechanical systems

• Produce analogous electrical and mechanical circuits

• Linearize a nonlinear system in order to find the TF

§1.Introduction

- Mathematical models from schematics of physical systems

• transfer functions in the frequency domain

• state equations in the time domain

§2.Laplace Transform Review

- Transforms: a mathematical conversion from one way ofthinking to another to make a problem easier to solve

- The Laplace transform the problem in time-domain to problem in𝑠-domain, then applying the solution in 𝑠-domain, and finally usinginverse transform to converse the solution back to the time-domain

- The Laplace transform is named in honor of mathematician andastronomer Pierre-Simon Laplace (1749-1827)

- Others: Fourier transform, z-transform, wavelet transform,…

- The Laplace transform of the function𝑓(𝑡) for 𝑡 > 0 is defined

by the following relationship

𝑠 : complex frequency variable,𝑠 = 𝜎 + 𝑗𝜔 with 𝑠, 𝜔 are

real numbers,𝑠 ∈ 𝐶 for which makes 𝐹 𝑠 convergent

ℒ : Laplace transform

𝐹(𝑠): a complex-valued function of complex numbers

- The inverse Laplace transform of the function𝐹(𝑠) for 𝑡 > 0 is

defined by the following relationship

𝑢(𝑡) : the unit step function, 𝑢 𝑡 = 1 𝑖𝑓 𝑡 > 00 𝑖𝑓 𝑡 < 0

- The Laplace transform table

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- Ex.2.1 Laplace Transform of a Time Function

Find the Laplace transform of𝑓 𝑡 = 𝐴𝑒−𝑎𝑡𝑢(𝑡)

- Ex.2.2 Inverse Laplace Transform

Find the inverse Laplace transform of𝐹1𝑠 = 1/(𝑠 + 3)2

𝐹(𝑠)

Case 1 Roots of the Denominator of𝐹(𝑠) are Real and Distinct

In general, given an𝐹(𝑠) whose denominator has real anddistinct roots, a partial-fraction expansion

𝐹 𝑠 =𝑁(𝑠)𝐷(𝑠)

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ℒ𝑑𝑓𝑑𝑡 = 𝑠𝐹 𝑠 − 𝑓(0 − ) (Table 2.2 – 7)

ℒ𝑑2𝑓

- Ex.2.3 Laplace Transform Solution of a Differential Equation

Given the following differential equation, solve for𝑦(𝑡) if all

initial conditions are zero Use the Laplace transform

𝑑2𝑦

𝑑𝑡2+ 12𝑑𝑦

𝑑𝑡+ 32𝑦 = 32𝑢(𝑡)Solution

𝑌 𝑠 = 32𝑠(𝑠 + 4)(𝑠 + 8)=

𝐾1= 32(𝑠 + 4)(𝑠 + 8)𝑠→0= 1

𝐾2= 32𝑠(𝑠 + 8)𝑠→−4= −2

𝐾2= 32𝑠(𝑠 + 4)𝑠→−8= 1

𝑦 𝑡 = 1 − 2𝑒−4𝑡+ 𝑒−8𝑡𝑢(𝑡)

The𝑢(𝑡) in (2.20) shows that the response is zero until 𝑡 = 0

Unless otherwise specified, all inputs to systems in the text

will not start until𝑡 = 0 Thus, output responses will also be

zero until𝑡 = 0

For convenience, the𝑢(𝑡) notation will be eliminated from now

on Accordingly, the output response

𝑦 𝑡 = 1 − 2𝑒−4𝑡+ 𝑒−8𝑡 (2.21)

Run ch2p1 through ch2p8 in Appendix BLearn how to use MATLAB to

• represent polynomials

• find roots of polynomials

• multiply polynomials, and

• find partial-fraction expansions

§2.Laplace Transform ReviewCase 2 Roots of the Denominator of𝐹(𝑠) are Real and Repeated

𝐹 𝑠 = 2(𝑠 + 1)(𝑠 + 2)2

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𝑦 𝑡 = 2𝑒−𝑡− 2𝑡𝑒−2𝑡− 2𝑒−2𝑡 (2.26)

(2.23)

(2.24)

𝐹 𝑠 = 2(𝑠 + 1)(𝑠 + 2)2

Matlab F=zpk([], [-1 -2 -2],2)

Result F =

2 -(s+1) (s+2)^2Continuous-time zero/pole/gain model

(2.22)

In general, given an𝐹(𝑠) whose denominator has real anddistinct roots, a partial-fraction expansion

𝐹 𝑠 =𝑁(𝑠)𝐷(𝑠)

𝑖 = 1,2, … , 𝑟; 0! = 1

Case 3 Roots of the Denominator of𝐹(𝑠) are Complex or Imaginary

𝑠→0[(2.31) × 𝑠] ⟹ 𝐾1= 3/5

First multiplying (2.31) by the lowest common denominator,

𝑠(𝑠2+ 2𝑠 + 5), and clearing the fraction

𝐹 𝑠 = 3

𝑠(𝑠2+ 2𝑠 + 5)=

35

1

𝑠−

35

ℒ 𝐴𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 =𝐵 𝑠+𝑎 +𝐵𝜔

(𝑠+𝑎) 2 +𝜔 2 (Table 2.1 – 6&7)

𝐹 𝑠 =35

1

𝑠−

35

𝑠 + 2

𝑠2+ 2𝑠 + 5

=35

1

𝑠−

35

𝑠 + 1 + 1/2 2(𝑠 + 1)2+22

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𝐹 𝑠 = 2(𝑠 + 1)(𝑠 + 2)2

𝐹 𝑠 = 3𝑠(𝑠2+ 2𝑠 + 5)

Matlab F=tf([3],[1 2 5 0])

Result F =

3 -s^3 + 2 s^2 + 5 sContinuous-time transfer function

(2.30)

In general, given an𝐹(𝑠) whose denominator has complex or

purely imaginary roots, a partial-fraction expansion

To find𝐾𝑖

• the 𝐾𝑖’s in (2.42) are found through balancing the

coefficients of the equation after clearing fractions

• put (𝐾2𝑠 + 𝐾3)/(𝑠2+ 𝑎𝑠 + 𝑏) in to the form

𝐵 𝑠 + 𝑎 + 𝐵𝜔

(𝑠 + 𝑎)2+𝜔2

(2.42)

𝐹 𝑠 = 3𝑠(𝑠2+ 2𝑠 + 5)

𝐹 𝑠 =3/5

𝑠 −

320

2 + 𝑗1

𝑠 + 1 + 𝑗2+

2 − 𝑗1

𝑠 + 1 + 𝑗2Matlab numf=3; denf=[1 2 5 0]; [r,p,k]=residue(numf,denf)

Result r=[-0.3+0.15i; -0.3-0.15i; 0.6]; p=[-1+2i; -1-2i; 0]; k=[]

(2.30)

(2.47)

Run ch2sp1 and ch2sp2 in Appendix FLearn how to use the Symbolic Math Toolbox to

• construct symbolic objects

• find the inverse Laplace transforms of frequencyfunctions

• find the Laplace of time functions

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§2.Laplace Transform ReviewSkill-Assessment Ex.2.2

Problem Find the inverse Laplace transform of

𝐹 𝑠 = 10𝑠(𝑠 + 2)(𝑠 + 3)2

Solution Expanding𝐹(𝑠) by partial fractions

𝑠 + 3

𝐴 = 10(𝑠 + 2)(𝑠 + 3)2

𝑠→0

=5

9, 𝐵 =

10𝑠(𝑠 + 3)2 𝑠→−2

= −5

𝐶 = 10𝑠(𝑠 + 2)𝑠→−3=

⟹ 𝐹 𝑠 =59

1(𝑠 + 3)2+40

9

1

𝑠 + 3Taking the inverse Laplace transform

Matlab syms s; f=ilaplace(10/(s*(s+2)*(s+3)^2)); pretty(f)

Result exp(-3 t) 40 t exp(-3 t) 10 5

exp(2 t) 5 + +

§3.The Transfer Function

- The transfer function of a component is the quotient of theLaplace transform of the output divided by the Laplacetransform of the input, with all initial conditions assumed to bezero

- Transfer functions are defined only for linear time invariant systems

- The input-output relationship of a control system𝐺 𝑠 𝐶(𝑠)

- Ex.2.4 Transfer Function for a Differential Equation

Find the transfer function represented by

𝑑𝑐(𝑡)

𝑑𝑡 + 2𝑐(𝑡) = 𝑟(𝑡)Solution

Taking the Laplace transform with zero initial conditions

𝑠𝐶 𝑠 + 2𝐶(𝑠) = 𝑅(𝑠)The transfer function

𝐺 𝑠 =𝐶(𝑠)𝑅(𝑠)=

1

𝑠 + 2

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Run ch2p9 through ch2p12 in Appendix B

Learn how to use MATLAB to

• create transfer functions with numerators and

denominators in polynomial or factored form

• convert between polynomial and factored forms

• plot time functions

Run ch2sp3 in Appendix FLearn how to use the Symbolic Math Toolbox to

• simplify the input of complicated transfer functions aswell as improve readability

• enter a symbolic transfer function and convert it to alinear time-invariant (LTI) object as presented inAppendix B, ch2p9

- Ex.2.5 System Response from the Transfer Function

Given𝐺 𝑠 = 1/(𝑠 + 2), find the response, 𝑐(𝑡) to an input,

𝑟 𝑡 = 𝑢(𝑡), a unit step, assuming zero initial conditions

1

𝑠 + 2Taking the inverse Laplace transform

𝑐 𝑡 = 0.5 − 0.5𝑒−2𝑡 (2.60)

𝐺 𝑠 = 1

𝑠 + 2

𝑅 𝑠 =1𝑠

Result C =1/2 - exp(-2*t)/2

§3.The Transfer FunctionSkill-Assessment Ex.2.3

Problem Find the transfer function,𝐺 𝑠 = 𝐶(𝑠)/𝑅(𝑠), corresponding

to the differential equation

𝑠3𝐶 𝑠 + 3𝑠2𝐶 𝑠 + 7𝑠𝐶 𝑠 + 5𝐶 𝑠

= 𝑠2𝑅 𝑠 + 4𝑠𝑅 𝑠 + 3𝑅 𝑠Collecting terms

𝑠3+ 3𝑠2+ 7𝑠 + 5 𝐶 𝑠 = 𝑠2+ 4𝑠 + 3 𝑅(𝑠)The transfer function

𝐺 𝑠 =𝐶(𝑠)𝑅(𝑠)=

𝑠2+ 4𝑠 + 3

𝑠3+ 3𝑠2+ 7𝑠 + 5

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𝐺 𝑠 =𝐶(𝑠)

𝑅(𝑠)=

2𝑠 + 1

𝑠2+ 6𝑠 + 2Cross multiplying

§3.The Transfer FunctionSkill-Assessment Ex.2.5

Problem Find the ramp response for a system whose transferfunction

𝐺 𝑠 = 𝑠(𝑠 + 4)(𝑠 + 8)Solution For a ramp response

𝐵 = 1

𝑠(𝑠 + 8)𝑠→−4= −

116

𝐶 = 1

𝑠(𝑠 + 4)𝑠→−8=

132

1

𝑠 + 4+

132

1

𝑠 + 8The ramp response

§4.Electrical Network Transfer Functions

Summarizes the components and the relationships betweenvoltage and current and between voltage and charge under zeroinitial conditions

Simple Circuits via Mesh Analysis

- Ex.2.6Transfer Function - Single Loop via the Differential Equation

Find the transfer function𝑉𝐶(𝑠)/𝑉(𝑠)

SolutionThe voltage loop

𝐿𝐶𝑑

2𝑣𝐶

𝑑𝑡2 + 𝑅𝐶𝑑𝑣𝐶

𝑑𝑡 + 𝑣𝐶= 𝑣(𝑡)Taking Laplace transform assuming zero initial conditions

𝐿𝐶𝑠2+ 𝑅𝐶𝑠 + 1 𝑉𝐶𝑠 = 𝑉(𝑠)Solving for the transfer function

𝑉𝐶(𝑠)𝑉(𝑠)=

1𝐿𝐶

𝑠2+𝑅𝐿𝑠 +𝐿𝐶1Block diagram of series RLC electrical network

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Impedance

- A resistance resists or “impedes” the flow of current The

corresponding relation is 𝑣/𝑖 = 𝑅 Capacitance and

inductance elements also impede the flow of current

- In electrical systems an impedance is a generalization of the

resistance concept and is defined as the ratio of a voltage

transform𝑉(𝑠) to a current transform 𝐼(𝑠) and thus implies a

current source

- Standard symbol for impedance

𝑍(𝑠) ≡𝑉(𝑠)

𝐼(𝑠)

-Kirchhoff’s voltage law to the transformed circuit

[ Sum of Impedances ] × 𝐼 𝑠 = [ Sum of Applied Voltages ](2.72)

(2.70)

- The impedance of a resistor is its resistance

𝑍 𝑠 = 1𝐶𝑠

- For an inductor

𝑣 𝑡 = 𝐿𝑑𝑖

𝑑𝑡⟹ 𝑉 𝑠 = 𝐿𝐼 𝑠 𝑠The impedance of a inductor

𝑍 𝑠 = 𝐿𝑠

Series and Parallel Impedances

- The concept of impedance is useful because the impedances

of individual elements can be combined with series and

parallel laws to find the impedance at any point in the system

- The laws for combining series or parallel impedances are

extensions to the dynamic case of the laws governing series

and parallel resistance elements

- Series Impedances

• Two impedances are in series if they have the same current.

If so, the total impedance is the sum of the individualimpedances𝑖 𝑣 𝑅 𝐶

⟹𝑉(𝑠)𝐼(𝑠)≡ 𝑍 𝑠 =

𝑅𝐶𝑠 + 1𝐶𝑠and the differential equation model is

𝐶𝑑𝑣

𝑑𝑡= 𝑅𝐶

𝑑𝑖

𝑑𝑡+ 𝑖(𝑡)

- Parallel Impedances

• Two impedances are in parallel if they have the same

voltage difference across them Their impedances combine

by the reciprocal rule

11/𝐶𝑠+

1

𝑅⟹

𝑉(𝑠)𝐼(𝑠)≡ 𝑍 𝑠 =

𝑅𝑅𝐶𝑠 + 1and the differential equation model is

𝑅𝐶𝑑𝑣

𝑑𝑡+ 𝑣 = 𝑅𝑖(𝑡)

§4.Electrical Network Transfer FunctionsAdmittance

𝑌 𝑠 ≡ 1𝑍(𝑠)=

𝐼(𝑠)𝑉(𝑠)

In general, admittance is complex

• The real part of admittance is calledconductance

𝐺 =1𝑅

• The imaginary part of admittance is calledsusceptance

When we take the reciprocal of resistance to obtain theadmittance, a purely real quantity results The reciprocal ofresistance is called conductance

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Instead of taking the Laplace transform of the differential

equation, we can draw the transformed circuit and obtain the

Laplace transform of the differential equation simply by

applyingKirchhoff’s voltage law to the transformed circuit

The steps are as follows

1.Redraw the original network showing all time variables, such

as 𝑣(𝑡), 𝑖(𝑡), and 𝑣𝐶(𝑡), as Laplace transforms 𝑉(𝑠), 𝐼(𝑠), and

𝑉𝐶(𝑠), respectively

2.Replace the component values with their impedance values

This replacement is similar to the case of dc circuits, where

we represent resistors with their resistance values

- Ex.2.7 Transfer Function - Single Loop via Transform Method

SolutionThe mess equation using impedances

𝐿𝑠 + 𝑅 +1

𝐶𝑠 𝐼 𝑠 = 𝑉(𝑠)

⟹𝐼(𝑠)𝑉(𝑠)=

1

𝐿𝑠 + 𝑅 +𝐶𝑠1The voltage across the capacitor

𝑉𝐶𝑠 = 𝐼(𝑠)1

𝐶𝑠

⟹ 𝑉𝐶(𝑠)/𝑉(𝑠)

Simple Circuits via Nodal Analysis

- Ex.2.8 Transfer Function - Single Node via Transform Method

SolutionThe transfer function can be obtained bysumming currents flowing out of the nodewhose voltage is𝑉𝐶(𝑠)

𝑉𝐶(𝑠)

1/𝐶𝑠+

𝑉𝐶𝑠 − 𝑉(𝑠)

𝑅 + 𝐿𝑠 = 0: the current flowing out of the node through the

capacitor

: the current flowing out of the node through the

series resistor and inductor

⟹𝑉𝐶(𝑠)𝑉(𝑠)

𝑉𝐶(𝑠)

𝐼/𝐶𝑠

𝑉𝐶𝑠 − 𝑉(𝑠)

𝑅 + 𝐿𝑠

§4.Electrical Network Transfer FunctionsComplex Circuits via Mesh Analysis

To solve complex electrical networks - those with multiple loopsand nodes– using mesh analysis

1.Replace passive element values with their impedances2.Replace all sources and time variables with their Laplacetransform

3.Assume a transform current and a current direction in eachmesh

4.WriteKirchhoff’s voltage law around each mesh5.Solve the simultaneous equations for the output6.Form the transfer function

- Ex.2.10 Transfer Function– Multiple Loops

Find the transfer function𝐼2(𝑠)/𝑉(𝑠)

SolutionConvert the network intoLaplace transformsSumming voltages aroundeach mesh through whichthe assumed currents flow

𝑅1𝐼1+ 𝐿𝑠𝐼1− 𝐿𝑠𝐼2= 𝑉𝐿𝑠𝐼2+ 𝑅2𝐼2+1

§4.Electrical Network Transfer FunctionsCramer’s Rule

Consider a system of 𝑛 linear equations for 𝑛 unknowns,represented in matrix multiplication form as follows

𝐴𝑥 = 𝑏

𝐴 : (𝑛 × 𝑛) matrix has a nonzero determinant

𝑥 : the column vector of the variables 𝑥 = (𝑥1, … , 𝑥𝑛)𝑇

𝑏 : the column vector of known parametersThe system has a unique solution, whose individual values forthe unknowns are given by

𝑥𝑖=det(𝐴𝑖)det(𝐴), 𝑖 = 1, … , 𝑛

𝐴𝑖: the matrix formed by replacing the𝑖thcolumn of𝐴 by thecolumn vector𝑏

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𝑅1+ 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2= 𝑉

−𝐿𝑠𝐼1+ 𝐿𝑠 + 𝑅2+1

𝐶𝑠 𝐼2= 0UsingCramer’s rule

𝐼2= 𝐿𝐶𝑠

2𝑉

𝑅1+ 𝑅2𝐿𝐶𝑠2+ 𝑅1𝑅2𝐶 + 𝐿 𝑠 + 𝑅1Forming the transfer function

𝐺 𝑠 =𝐼2𝑠𝑉(𝑠)

2

𝑅1+ 𝑅2𝐿𝐶𝑠2+ 𝑅1𝑅2𝐶 + 𝐿 𝑠 + 𝑅1

The network is now modeled as the transfer function of figure

(2.82)(2.81)

Note

𝑅1+ 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2= 𝑉

−𝐿𝑠𝐼1+ 𝐿𝑠 + 𝑅2+1

𝐶𝑠 𝐼2= 0

Run ch2sp4 in Appendix FLearn how to use the Symbolic Math Toolbox to

• solve simultaneous equations using Cramer’s rule

• solve for the transfer function in Eq (2.82) using Eq

(2.80)

Complex Circuits via Nodal Analysis

- Ex.2.11 Transfer Function– Multiple Nodes

SolutionSum of currents flowing fromthe nodes marked𝑉𝐿(𝑠) and

𝐺1+ 𝐺2+1

𝐿𝑠 𝑉𝐿 − 𝐺2𝑉𝐶= 𝑉𝐺1

−𝐺2𝑉𝐿+ 𝐺2+ 𝐶𝑠 𝑉𝐶= 0Solving for the transfer function

𝑉𝐶(𝑠)𝑉(𝑠)=

(2.86)

(2.87)

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- Another way to write node equations is to replace voltage

sources by current sources In order to handle multiple-node

electrical networks, we can perform the following steps

1.Replace passive element values with their admittances

2.Replace all sources and time variables with their Laplace

transform

3.Replace transformed voltage sources with transformed

current sources

4.WriteKirchhoff’s current law at each node

5.Solve the simultaneous equations for the output

6.Form the transfer function

Norton's Theorem Any collection of batteries and resistances with two terminals is electrically equivalent

to an ideal current source 𝑖 in parallel with a single resistor 𝑟 The value of 𝑟 is the same as that in the Thevenin equivalent and the current 𝑖 can be found by dividing the open circuit voltage by 𝑟

- Ex.2.12 Transfer Function– Multiple Nodes with Current Sources

SolutionConvert all impedances toadmittances and all voltagesources in series with animpedance to currentsources in parallel with anadmittance using Norton’stheorem

Using the general relationship𝐼 𝑠 =

𝑌 𝑠 𝑉(𝑠) and summing currents at thenode𝑉𝐿(𝑠)

A Problem-Solving Technique

In all of the previous examples, we have seen a repeating

pattern in the equations that we can use to our advantage If we

recognize this pattern, we need not write the equations

component by component; we can sum impedances around a

mesh in the case of mesh equations or sum admittances at a

node in the case of node equations

- Ex.2.13 Mesh Equations via Inspection

Write the mesh equations for thenetwork

SolutionThe mesh equations for loop 1+ 2𝑠 + 2 𝐼1− 2𝑠 + 1 𝐼2− 𝐼3= 𝑉

(2.94.a)

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The mesh equations for loop 2

− 2𝑠 + 2 𝐼1+ 9𝑠 + 1 𝐼2− 4𝑠𝐼3= 0

(2.94.b)

The mesh equations for loop 3

−𝐼1− 4𝑠𝐼2+ 4𝑠 + 1 +1

𝑠 𝐼3= 0(2.94.c)

4 3 2

#1 == 24 s + 30 s + 17 s + 16 s + 1

20𝑠 3 + 13𝑠 2 + 10𝑠 + 1 𝑉 24𝑠 4 + 30𝑠 3 + 17𝑠 2 + 16𝑠 + 1

8𝑠 3 + 10𝑠 2 + 3𝑠 + 1 𝑉 24𝑠 4 + 30𝑠 3 + 17𝑠 2 + 16𝑠 + 1

𝑠 8𝑠 2 + 13𝑠 + 1 𝑉 24𝑠 4 + 30𝑠 3 + 17𝑠 2 + 16𝑠 + 1

Operational Amplifiers

- An operational amplifier (op-amp) is an electronic amplifier

used as a basic building block to implement transfer functions

- Op-amp has the following characteristics

1.Differential input,𝑣2𝑡 − 𝑣1𝑡

2.High input impedance,𝑍𝑖= ∞ (ideal)

3.Low output impedance,𝑍𝑜= 0 (ideal)

4.High constant gain amplification,𝐴 = ∞ (ideal)

- The output,𝑣𝑜(𝑡), is given by

𝑣𝑜𝑡 = 𝐴[𝑣2𝑡 − 𝑣1𝑡 ] (2.95)

§4.Electrical Network Transfer FunctionsInverting Operational Amplifiers

- If 𝑣2(𝑡) is grounded, the amplifier is called an invertingoperational amplifier

- The output,𝑣𝑜(𝑡), is given by

𝑣𝑜𝑡 = −𝐴𝑣1𝑡 (2.96)

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Inverting Operational Amplifiers

- Ex.2.14 Transfer Function– Inverting Op-Amp Circuit

Find the transfer function𝑉𝑜(𝑠)/𝑉𝑖(𝑠)Solution

The impedances1

𝑍1= 11/𝐶1𝑠+

Noninverting Operational Amplifiers

- Ex.2.15 Transfer Function– Noninverting Op-Amp Circuit

Find the transfer function𝑉𝑜(𝑠)/𝑉𝑖(𝑠)Solution

Skill Assessment Ex.2.6

Problem Find𝐺 𝑠 = 𝑉𝐿(𝑠)/𝑉(𝑠) using mesh and nodal analysis

𝑠 + 1 𝐼1 − 𝑠𝐼2 − 𝐼3= 𝑉

−𝑠𝐼1+ 2𝑠 + 1 𝐼2 − 𝐼3= 0

−𝐼1 − 𝐼2+ 𝑠 + 2 𝐼3= 0Solving the mesh equation for𝐼2

The voltage across𝐿

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