LAPLACE TRANSFORM AND TRANSFER FUNCTION · 2019-10-22 · CHBE320 Process Dynamics and Control Korea University 5-1 CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor

5-1CHBE320 Process Dynamics and Control Korea University

CHBE320 LECTURE VLAPLACE TRANSFORM AND

TRANSFER FUNCTION

Professor Dae Ryook Yang

Fall 2019Dept. of Chemical and Biological Engineering

Korea University


PROCESS

Sensor

ActuatorController+-

Lectures IV to VII

Road Map of the Lecture V

• Laplace Transform and Transfer functions– Definition of Laplace transform– Properties of Laplace transform– Inverse Laplace transform– Definition of transfer function– How to get the transfer functions– Properties of transfer function


SOLUTION OF LINEAR ODE

• 1st-order linear ODE– Integrating factor:

• High-order linear ODE with constant coeffs.– Modes: roots of characteristic equation

– Depending on the roots, modes are• Distinct roots:• Double roots:• Imaginary roots:

• Many other techniques for different cases

For 𝑑𝑥𝑑𝑡 𝑎 𝑡 𝑥 𝑓 𝑥 , I.F. exp 𝑎 𝑡 𝑑𝑡

𝑥𝑒 𝑓 𝑡 𝑒 𝑥 𝑡 𝑓 𝑡 𝑒 𝑑𝑡 𝐶 𝑒

For 𝑎 𝑥 𝑎 𝑥 𝑎 𝑥 𝑓 𝑡 ,

𝑎 𝑝 𝑎 𝑝 𝑎 𝑎 𝑝 𝑝 𝑝 𝑝 0

𝑒 , 𝑒

𝑒 cos 𝛽 𝑡, 𝑒 sin 𝛽 𝑡

𝑒 , 𝑡𝑒

Solution is a linear combination of modes and the coefficients are

decided by the initial conditions.


LAPLACE TRANSFORM FOR LINEAR ODE AND PDE

• Laplace Transform – Not in time domain, rather in frequency domain– Derivatives and integral become some operators.– ODE is converted into algebraic equation– PDE is converted into ODE in spatial coordinate– Need inverse transform to recover time-domain solution

ODE or PDEu(t) y(t)

Transfer FunctionU(s) Y(s)

(Algebraic calculation)

(D.E. calculation)

𝔏 𝔏 𝔏 𝔏𝔏 𝔏


DEFINITION OF LAPLACE TRANSFORM

• Definition

– F(s) is called Laplace transform of f(t).– f(t) must be piecewise continuous.– F(s) contains no information on f(t) for t < 0.– The past information on f(t) (for t < 0) is irrelevant.– The s is a complex variable called “Laplace transform variable”

• Inverse Laplace transform

– and are linear.𝔏 𝔏 𝔏 𝑎𝑓 𝑡 𝑏𝑓 𝑡 𝑎𝐹 𝑠 𝑏𝐹 𝑠


LAPLACE TRANSFORM OF FUNCTIONS

• Constant function, a

• Step function, S(t)

• Exponential function, e-bt

𝔏 𝑎 𝑎𝑒 𝑑𝑡𝑎𝑠 𝑒 0

𝑎𝑠

𝑎𝑠

t

f(t)

a

t

f(t)

1

𝑓 𝑡 S 𝑡 1 for 𝑡 00 for 𝑡 0

𝔏 S 𝑡 𝑒 𝑑𝑡1𝑠 𝑒 0

1𝑠

1𝑠

𝔏 𝑒 𝑒 𝑒 𝑑𝑡1

𝑠 𝑏 𝑒1

𝑠 𝑏t

f(t)

1b>0

b<0


• Trigonometric functions– Euler’s Identity:

• Rectangular pulse, P(t)

𝑒 ≜ cos 𝜔 𝑡 𝑗 sin 𝜔 𝑡

cos 𝜔 𝑡12 𝑒 𝑒 sin 𝜔 𝑡

12𝑗 𝑒 𝑒

𝔏 sin 𝜔 𝑡 𝔏12𝑗 𝑒 𝔏

12𝑗 𝑒

12𝑗

1𝑠 𝑗𝜔

1𝑠 𝑗𝜔

𝜔𝑠 𝜔

𝔏 cos 𝜔 𝑡 𝔏12 𝑒 𝔏

12 𝑒

12

1𝑠 𝑗𝜔

1𝑠 𝑗𝜔

𝑠𝑠 𝜔

𝑓 𝑡 P 𝑡0 for 𝑡 𝑡

ℎ for 𝑡 𝑡 00 for 𝑡 0

𝔏 P 𝑡 ℎ𝑒 𝑑𝑡ℎ𝑠 𝑒

ℎ𝑠 1 𝑒

t

f(t)

h

tw

t

sin(t)1

2𝜋𝜔


• Impulse function,

• Ramp function, t

• Refer the Table 3.1 (Seborg et al.) for other functions

𝑓 𝑡 𝛿 𝑡 lim→

0 for 𝑡 𝑡 1/𝑡 for 𝑡 𝑡 0

0 for 𝑡 0

𝔏 𝛿 𝑡 lim→

1𝑡 𝑒 𝑑𝑡 lim

→

1𝑡 𝑠 1 𝑒 1

L′Hospital′s rule: lim→

𝑓 𝑡𝑔 𝑡 lim

→

𝑓 𝑡𝑔 𝑡

t

f(t)1/tw

tw

t

f(t)

1

1

𝛿 𝑡

𝔏 𝑡 𝑡𝑒 𝑑𝑡

𝑡𝑠 𝑒

𝑒𝑠 𝑑𝑡

1𝑠 𝑒 𝑑𝑡

1𝑠

Integration by part: 𝑓′ ⋅ 𝑔𝑑𝑡 𝑓 ⋅ 𝑔 𝑓 ⋅ 𝑔 𝑑𝑡




PROPERTIES OF LAPLACE TRANSFORM

• Differentiation𝔏

𝑑𝑓𝑑𝑡 𝑓 ⋅ 𝑒 𝑑𝑡 𝑓 𝑡 𝑒 𝑓 ⋅ 𝑠 𝑒 𝑑𝑡 by 𝑖. 𝑏. 𝑝.

𝑠 𝑓 ⋅ 𝑒 𝑑𝑡 𝑓 0 𝑠𝐹 𝑠 𝑓 0

𝔏𝑑 𝑓𝑑𝑡 𝑓 ⋅ 𝑒 𝑑𝑡 𝑓 𝑡 𝑒 𝑓 ⋅ 𝑠 𝑒 𝑑𝑡 𝑠 𝑓 ⋅ 𝑒 𝑑𝑡 𝑓 0

𝑠 𝑠𝐹 𝑠 𝑓 0 𝑓 0 𝑠 𝐹 𝑠 𝑠𝑓 0 𝑓 0

𝔏𝑑 𝑓𝑑𝑡 𝑓 ⋅ 𝑒 𝑑𝑡 𝑓 𝑡 𝑒 𝑓 ⋅ 𝑠 𝑒 𝑑𝑡

𝑠 𝑓 ⋅ 𝑒 𝑑𝑡 𝑓 0 𝑠 𝔏𝑑 𝑓𝑑𝑡 𝑓 0

𝑠 𝐹 𝑠 𝑠 𝑓 0 ⋯ 𝑠𝑓 0 𝑓 0

⋮


• If f (0) = f ’(0) = f ”(0) =…= f (n-1)(0) = 0,– Initial condition effects are vanished.– It is very convenient to use deviation

variables so that all the effects ofinitial condition vanish.

• Transforms of linear differential equations.

𝔏𝑑𝑓𝑑𝑡 𝑠𝐹 𝑠

𝔏𝑑 𝑓𝑑𝑡 𝑠 𝐹 𝑠

⋮

𝔏𝑑 𝑓𝑑𝑡 𝑠 𝐹 𝑠

𝑦 𝑡 𝔏

𝑌 𝑠 , 𝑢 𝑡 𝔏

𝑈 𝑠𝑑𝑦 𝑡

𝑑𝑡 𝔏

𝑠𝑌 𝑠 if 𝑦 0 0

𝜏𝑑𝑦 𝑡

𝑑𝑡 𝑦 𝑡 𝐾𝑢 𝑡 𝑦 0 0 𝔏

𝜏𝑠 1 𝑌 𝑠 𝐾𝑈 𝑠

𝜕𝑇𝜕𝑡 𝑣

𝜕𝑇𝜕𝑧

1𝜏 𝑇 𝑇

𝔏 𝜏 𝑣

𝜕𝑇 𝑠𝜕𝑧 𝜏 𝑠 1 𝑇 𝑠 𝑇 𝑠


• Integration

• Time delay (Translation in time)

• Derivative of Laplace transform

𝔏 𝑓 𝜉 𝑑𝜉 𝑓 𝜉 𝑑𝜉 𝑒 𝑑𝑡

𝑒𝑠 𝑓 𝜉 𝑑𝜉

1𝑠 𝑓 ⋅ 𝑒 𝑑𝑡

𝐹 𝑠𝑠 by 𝑖. 𝑏. 𝑝.

0

Leibniz rule: 𝑑𝑑𝑡 𝑓 𝜏 𝑑𝜏 𝑓 𝑏 𝑡

𝑑𝑏 𝑡𝑑𝑡 𝑓 𝑎 𝑡

𝑑𝑎 𝑡𝑑𝑡

𝑓 𝑡

𝑓 𝑡 𝜃 S 𝑡 𝜃

𝔏 𝑓 𝑡 𝜃 S 𝑡 𝜃 𝑓 𝑡 𝜃 𝑒 𝑑𝑡 𝑓 𝜏 𝑒 𝑑𝜏 let 𝜏 𝑡 𝜃

𝑒 𝑓 𝜏 𝑒 𝑑𝜏 𝑒 𝐹 𝑠

t

f(t)

𝜃

𝑑𝐹 𝑠𝑑𝑠

𝑑𝑑𝑠 𝑓 ⋅ 𝑒 𝑑𝑡 𝑓 ⋅

𝑑𝑑𝑠 𝑒 𝑑𝑡 𝑡 ⋅ 𝑓 𝑒 𝑑𝑡 𝔏 𝑡 ⋅ 𝑓 𝑡


• Final value theorem– From the LT of differentiation, as s approaches to zero

– Limitation: has to exist. If it diverges or oscillates, this theorem is not valid.

• Initial value theorem– From the LT of differentiation, as s approaches to infinity

𝑑𝑓𝑑𝑡 𝑑𝑡 𝑓 ∞ 𝑓 0 lim

→𝑠 𝐹 𝑠 𝑓 0 ⇒ 𝑓 ∞ lim

→𝑠 𝐹 𝑠

lim→

𝑑𝑓𝑑𝑡 ⋅ 𝑒 𝑑𝑡 lim

→𝑠𝐹 𝑠 𝑓 0

lim→

𝑑𝑓𝑑𝑡 𝑒 𝑑𝑡 0 lim

→𝑠 𝐹 𝑠 𝑓 0 ⇒ 𝑓 0 lim

→𝑠 𝐹 𝑠

𝑓 ∞

lim→

𝑑𝑓𝑑𝑡 ⋅ 𝑒 𝑑𝑡

𝑑𝑓𝑑𝑡 ⋅ lim

→𝑒 𝑑𝑡 lim

→𝑠𝐹 𝑠 𝑓 0


EXAMPLE ON LAPLACE TRANSFORM (1)

•

•

– Using the initial and final value theorems

– But the final value theorem is not valid because

t

f(t)

3

2 6

𝑓 𝑡

1.5𝑡 for 0 𝑡 23 for 2 𝑡 60 for 6 𝑡 0 for 𝑡 0

𝑓 𝑡 1.5𝑡 S 𝑡 1.5 𝑡 2 S 𝑡 2 3 S 𝑡 6

∴ 𝐹 𝑠 𝔏 𝑓 𝑡1.5𝑠 1 𝑒

3𝑠 𝑒

For 𝐹 𝑠2

𝑠 5 , find 𝑓 0 and 𝑓 ∞ .

𝑓 0 lim→

𝑠 𝐹 𝑠 lim→

2𝑠𝑠 5 2 𝑓 ∞ lim

→𝑠 𝐹 𝑠 lim

→

2𝑠𝑠 5 0

lim→

𝑓 𝑡 lim→

2 𝑒∞


EXAMPLE ON LAPLACE TRANSFORM (2)

• What is the final value of the following system?

– Actually, cannot be defined due to sin t term.

• Find the Laplace transform for ?

𝑥 𝑥 𝑥 sin 𝑡 ; 𝑥 0 𝑥 0 0

⇒ 𝑠 𝑋 𝑠 𝑠𝑋 𝑠 𝑋1

𝑠 1 ⇒ x s1

𝑠 1 𝑠 𝑠 1

𝑥 ∞ lim→

𝑠𝑠 1 𝑠 𝑠 1 0

𝑥 ∞

From 𝑑𝐹 𝑠

𝑑𝑠 𝔏 𝑡 ⋅ 𝑓 𝑡

𝔏 𝑡 ⋅ sin 𝜔 𝑡𝑑

𝑑𝑠𝜔

𝑠 𝜔2𝜔𝑠

𝑠 𝜔


INVERSE LAPLACE TRANSFORM

• Used to recover the solution in time domain

– From the table– By partial fraction expansion– By inversion using contour integral

• Partial fraction expansion– After the partial fraction expansion, it requires to know some

simple formula of inverse Laplace transform such as

𝑓 𝑡 𝔏 𝐹 𝑠1

2𝜋𝑗 𝑒 𝐹 𝑠 𝑑𝑠

1𝜏𝑠 1 ,

𝑠𝑠 𝑏 𝜔 ,

𝑛 1 !𝑠 ,

𝑒𝜏 𝑠 2𝜁𝜏𝑠 1 , etc.


PARTIAL FRACTION EXPANSION

• Case I: All pi’s are distinct and real– By a root-finding technique, find all roots (time-consuming)– Find the coefficients for each fraction

• Comparison of the coefficients after multiplying the denominator• Replace some values for s and solve linear algebraic equation• Use of Heaviside expansion

– Multiply both side by a factor, (s+pi), and replace s with –pi.

– Inverse LT:

𝐹 𝑠𝑁 𝑠𝐷 𝑠

𝑁 𝑠𝑠 𝑝 ⋯ 𝑠 𝑝

𝛼𝑠 𝑝 ⋯

𝛼𝑠 𝑝

𝛼 𝑠 𝑝𝑁 𝑠𝐷 𝑠

𝑓 𝑡 𝛼 𝑒 𝛼 𝑒 ⋯ 𝛼 𝑒


• Case II: Some roots are repeated

– Each repeated factors have to be separated first.– Same methods as Case I can be applied.– Heaviside expansion for repeated factors

– Inverse LT


𝑁 𝑠𝑠 𝑝

𝑏 𝑠 ⋯ 𝑏𝑠 𝑝

𝛼𝑠 𝑝 ⋯

𝛼𝑠 𝑝

𝛼1𝑖!

𝑑𝑑𝑠

𝑁 𝑠𝐷 𝑠 𝑠 𝑝 𝑖 0, ⋯ , 𝑟 1

𝑓 𝑡 𝛼 𝑒 𝛼 𝑡𝑒 ⋯𝛼

𝑟 1 ! 𝑡 𝑒


• Case III: Some roots are complex

– Each repeated factors have to be separated first.– Then,

– Inverse LT


𝑐 𝑠 𝑐𝑠 𝑑 𝑠 𝑑

𝛼 𝑠 𝑏 𝛽 𝜔𝑠 𝑏 𝜔

𝛼 𝑠 𝑏 𝛽 𝜔𝑠 𝑏 𝜔 𝛼

𝑠 𝑏𝑠 𝑏 𝜔 𝛽

𝜔𝑠 𝑏 𝜔

𝑓 𝑡 𝛼 𝑒 cos 𝜔 𝑡 𝛽 𝑒 sin 𝜔 𝑡

where 𝑏 𝑑 /2, 𝜔 𝑑 𝑑 /4

𝛼 𝑐 , 𝛽 𝑐 𝛼 𝑏 /𝜔


EXAMPLES ON INVERSE LAPLACE TRANSFORM

•

– Multiply each factor and insert the zero value

𝐹 𝑠𝑠 5

𝑠 𝑠 1 𝑠 2 𝑠 3𝐴𝑠

𝐵𝑠 1

𝐶𝑠 2

𝐷𝑠 3 distinct

𝑠 5𝑠 1 𝑠 2 𝑠 3 𝐴 𝑠

𝐵𝑠 1 𝑠

𝐶𝑠 2 𝑠

𝐷𝑠 3 ⇒ 𝐴 5/6

𝑠 5𝑠 𝑠 2 𝑠 3

𝐴 𝑠 1𝑠 𝐵

𝐶 𝑠 1𝑠 2

𝐷 𝑠 1𝑠 3 ⇒ 𝐵 2

𝑠 5𝑠 𝑠 1 𝑠 3

𝐴 𝑠 2𝑠

𝐵 𝑠 2𝑠 1 𝐶

𝐷 𝑠 2𝑠 3 ⇒ 𝐶 3/2

𝑠 5𝑠 𝑠 1 𝑠 2

𝐴 𝑠 3𝑠

𝐵 𝑠 3𝑠 1

𝐶 𝑠 3𝑠 2 𝐷 ⇒ 𝐷 1/3

∴ 𝑓 𝑡 𝔏 𝐹 𝑠56 2𝑒

32 𝑒

13 𝑒


•

– Use of Heaviside expansion

𝐹 𝑠1

𝑠 1 𝑠 2𝐴𝑠 𝐵𝑠 𝐶

𝑠 1𝐷

𝑠 2 repeated

𝛼1𝑖!

𝑑𝑑𝑠

𝑁 𝑠𝐷 𝑠 𝑠 𝑝 𝑖 0, ⋯ , 𝑟 1

1 𝐴𝑠 𝐵𝑠 𝐶 𝑠 2 𝐷 𝑠 1𝐴 𝐷 𝑠 2𝐴 𝐵 3𝐷 𝑠 2𝐵 𝐶 3𝐷 𝑠 2𝐶 𝐷

∴ 𝐴 𝐷, 2𝐴 𝐵 3𝐷 0, 2𝐵 𝐶 3𝐷 0, 2𝐶 𝐷 1

⇒ 𝐴 1, 𝐵 1, 𝐶 1, 𝐷 1

𝑠 𝑠 1𝑠 1

𝛼𝑠 1

𝛼𝑠 1

𝛼𝑠 1

𝑖 1 : 𝛼11!

𝑑𝑑𝑠 𝑠 𝑠 1 1

𝑖 2 : 𝛼12!

𝑑𝑑𝑠 𝑠 𝑠 1 1

𝑖 0 : 𝛼 𝑠 𝑠 1 1

∴ 𝑓 𝑡 𝔏 𝐹 𝑠 𝑒 𝑡𝑒12 𝑡 𝑒 𝑒


• 𝐹 𝑠𝑠 1

𝑠 𝑠 4𝑠 5𝐴 𝑠 2 𝐵

𝑠 2 1𝐶𝑠 𝐷

𝑠 complex

𝑠 1 𝐴 𝑠 2 𝑠 𝐵𝑠 𝐶𝑠 𝐷 𝑠 4𝑠 5𝐴 𝐶 𝑠 2𝐴 𝐵 4𝐶 𝐷 𝑠 5𝐶 4𝐷 𝑠 5𝐷

∴ 𝐴 𝐶, 2𝐴 𝐵 4𝐶 𝐷 0, 5𝐶 4𝐷 1, 5𝐷 1

⇒ 𝐴 1/25, 𝐵 7/25, 𝐶 1/25, 𝐷 1/5

𝐴 𝑠 2 𝐵𝑠 2 1

125

𝑠 2𝑠 2 1

725

𝐵𝑠 2 1

𝐶𝑠 𝐷𝑠

125

1𝑠

15

1𝑠

∴ 𝑓 𝑡 𝔏 𝐹 𝑠1

25 𝑒 cos 𝑡7

25 𝑒 sin 𝑡1

2515 𝑡


• 𝐹 𝑠1 𝑒

4𝑠 1 3𝑠 1𝐴

4𝑠 1𝐵

3𝑠 1 1 𝑒 Time delay

𝐴 1/ 3𝑠 1/

4, 𝐵 1/ 4𝑠 1/

3

∴ 𝑓 𝑡 𝔏 𝐹 𝑠 𝔏4

4𝑠 13

3𝑠 1 𝔏4𝑒4𝑠 1

3𝑒3𝑠 1

𝑒 / 𝑒 / 𝑒 / 𝑒 / S 𝑡 2


SOLVING ODE BY LAPLACE TRANSFORM

• Procedure1. Given linear ODE with initial condition,2. Take Laplace transform and solve for output3. Inverse Laplace transform

• Example: Solve for 5𝑑𝑦𝑑𝑡 4𝑦 2; 𝑦 0 1

𝔏 5𝑑𝑦𝑑𝑡 𝔏 4𝑦 𝔏 2 ⇒ 5 𝑠𝑌 𝑠 𝑦 0 4𝑌 𝑠

2𝑠

5𝑠 4 𝑌 𝑠2𝑠 5 ⇒ 𝑌 𝑠

5𝑠 2𝑠 5𝑠 4

∴ 𝑦 𝑡 𝔏 𝑌 𝑠 𝔏0.5𝑠

2.55𝑠 4 0.5 0.5𝑒 .


TRANSFER FUNCTION (1)

• Definition– An algebraic expression for the dynamic relation between the

input and output of the process model

• How to find transfer function1. Find the equilibrium point2. If the system is nonlinear, then linearize around equil. point3. Introduce deviation variables4. Take Laplace transform and solve for output5. Do the Inverse Laplace transform and recover the original

variables from deviation variables

TransferFunction, G(s)5

𝑑𝑦𝑑𝑡 4𝑦 𝑢; 𝑦 0 1

Let 𝑦 𝑦 1 and 𝑢 𝑢 4

5𝑠 4 𝑌 𝑠 𝑈 𝑠 ⇒ 𝑌 𝑠𝑈 𝑠

15𝑠 4

0.251.25𝑠 1 𝐺 𝑠

𝑈 𝑠 𝑌 𝑠



• Benefits– Once TF is known, the output response to various given inputs

can be obtained easily.

– Interconnected system can be analyzed easily. • By block diagram algebra

– Easy to analyze the qualitative behavior of a process, such as stability, speed of response, oscillation, etc.

• By inspecting “Poles” and “Zeros”• Poles: all s’s satisfying D(s)=0• Zeros: all s’s satisfying N(s)=0

𝑦 𝑡 𝔏 𝑌 𝑠 𝔏 𝐺 𝑠 𝑈 𝑠 𝔏 𝐺 𝑠 𝔏 𝑈 𝑠

G1 G2

G3

+-

X Y 𝑌 𝑠𝑋 𝑠

𝐺1 𝑠 𝐺2 𝑠1 𝐺1 𝑠 𝐺2 𝑠 𝐺3 𝑠



• Steady-state Gain: The ratio between ultimate changes in input and output

– For a unit step change in input, the gain is the change in output– Gain may not be definable: for example, integrating processes

and processes with sustaining oscillation in output– From the final value theorem, unit step change in input with

zero initial condition gives

– The transfer function itself is an impulse response of the system

Gain 𝐾ΔouputΔinput

𝑦 ∞ 𝑦 0𝑢 ∞ 𝑢 0

𝐾𝑦 ∞

1 lim→

𝑠 𝑌 𝑠 lim→

𝑠 𝐺 𝑠1𝑠 lim

→𝐺 𝑠

𝑌 𝑠 𝐺 𝑠 𝑈 𝑠 𝐺 𝑠 𝔏 𝛿 𝔱 𝐺 𝑠


EXAMPLE

• Horizontal cylindrical storage tank (Ex4.7)

– Equilibrium point:(if , can be any value in .)

– Linearization:

L q

qi

Rh

wi

𝑑𝑚𝑑𝑡 𝜌

𝑑𝑉𝑑𝑡 𝜌𝑞 𝜌𝑞

𝑉 ℎ 𝐿𝑤 ℎ 𝑑ℎ ⇒𝑑𝑉𝑑𝑡 𝐿𝑤 ℎ

𝑑ℎ𝑑𝑡

𝑤 𝐿𝑑ℎ𝑑𝑡 𝑞 𝑞 ⇒

𝑑ℎ𝑑𝑡

12𝐿 𝐷 ℎ ℎ

𝑞 𝑞 Nonlinear ODE

𝑤 ℎ /2 𝑅 𝑅 ℎ 2𝑅 ℎ ℎ

0 𝑞 𝑞 / 2𝐿 𝐷 ℎ ℎ𝑞 , 𝑞, ℎ𝑞 𝑞 ℎ 0 ℎ 𝐷

𝑑ℎ𝑑𝑡 𝑓 ℎ, 𝑞 , 𝑞

𝜕𝑓𝜕ℎ , ̄ , ̄

ℎ ℎ𝜕𝑓𝜕𝑞 , ̄ , ̄

𝑞 𝑞𝜕𝑓𝜕𝑞 , ̄ , ̄

𝑞 𝑞


𝑠𝐻 𝑠 𝑘𝑄 𝑠 𝑘𝑄 𝑠

𝜕𝑓𝜕ℎ , ̄ , ̄

𝑞 𝑞𝜕

𝜕ℎ1

2𝐿 𝐷 ℎ ℎ0 ∵ 𝑞 𝑞

𝜕𝑓𝜕𝑞 , ̄ , ̄

1

2𝐿 𝐷 ℎ ℎ,

𝜕𝑓𝜕𝑞 , ̄ , ̄

1

2𝐿 𝐷 ℎ ℎ

Let this term be k

• Transfer function between (integrating)

• Transfer function between (integrating)

• If is near 0 or D, k becomes very large and is around D/2, kbecomes minimum.

The model could be quite different depending on the operating condition used for the linearization.

The best suitable range for the linearization in this case is around D/2. (less change in gain)

Linearized model would be valid in very narrow range near 0.

𝐻 𝑠 and 𝑄 𝑠 : 𝑘𝑠

𝐻 𝑠 and 𝑄 𝑠 : 𝑘𝑠

ℎ ℎ


PROPERTIES OF TRANSFER FUNCTION

• Additive property

• Multiplicative property

• Physical realizability – In a transfer function, the order of numerator(m) is greater

than that of denominator(n): called “physically unrealizable”– The order of derivative for the input is higher than that of

output. (requires future input values for current output)

G1(s)

G2(s)

X1(s)

X2(s)

Y(s)

++

Y1(s)

Y2(s)

𝑌 𝑠 𝑌 𝑠 𝑌 𝑠𝐺 𝑠 𝑋 𝑠 𝐺 𝑠 𝑋 𝑠

G1(s) G2(s)X1(s) X2(s) X3(s)

𝑋 𝑠 𝐺 𝑠 𝑋 𝑠𝐺 𝑠 𝐺 𝑠 𝑋 𝑠


EXAMPLES ON TWO TANK SYSTEM

• Two tanks in series (Ex3.7)– No reaction

– Initial condition: c1(0)= c2(0)=1 kg mol/m3 (Use deviation var.)– Parameters: V1/q=2 min., V2/q=1.5 min.– Transfer functions

𝑉𝑑𝑐𝑑𝑡 𝑞𝑐 𝑞𝑐

𝑉𝑑𝑐𝑑𝑡 𝑞𝑐 𝑞𝑐

𝐶 𝑠𝐶 𝑠

1𝑉 /𝑞 𝑠 1

𝐶 𝑠𝐶 𝑠

1𝑉 /𝑞 𝑠 1

𝐶 𝑠𝐶 𝑠

𝐶 𝑠𝐶 𝑠

𝐶 𝑠𝐶 𝑠

1𝑉 /𝑞 𝑠 1 𝑉 /𝑞 𝑠 1

Ci

C1

C2

V1

V2

q


• Pulse input

• Equivalent impulse input

• Pulse response vs. Impulse response

t

5

0.25

𝑐

𝐶 𝑠5𝑠 1 𝑒 .

𝐶 𝑠 𝔏 5 0.25 𝛿 𝑡 1.25

𝐶 𝑠1

2𝑠 1 𝐶 𝑠5

𝑠 2𝑠 1 1 𝑒 .

5𝑠

102𝑠 1 1 𝑒 .

⇒ 𝑐 𝑡 5 1 𝑒 /

5 1 𝑒 . / S 𝑡 0.25

𝐶 𝑠1

2𝑠 1 𝐶 𝑠1.25

2𝑠 1

⇒ 𝑐 0.625𝑒 /


𝐶 𝑠1

2𝑠 1 1.5𝑠 1 𝐶 𝑠

1.252𝑠 1 1.5𝑠 1

52𝑠 1

3.751.5𝑠 1

⇒ 𝑐 2.5𝑒 / 2.5𝑒 / .

𝐶 𝑠1

2𝑠 1 1.5𝑠 1 𝐶 𝑠5

𝑠 2𝑠 1 1.5𝑠 1 1 𝑒 .

5𝑠

402𝑠 1

22.51.5𝑠 1 1 𝑒 .

⇒ 𝑐 𝑡 5 20𝑒 / 15𝑒 / .

5 20𝑒 . / 15𝑒 . / . S 𝑡 0.25

LAPLACE TRANSFORM AND TRANSFER FUNCTION · 2019-10-22 · CHBE320 Process Dynamics and Control Korea University 5-1 CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor

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