(a) Interpretation: The liquid level response, y ( t ) for the sudden change in u ( t ) from 0 to 1 m 3 at t = 0 is to be calculated. Concept introduction: For a function f ( t ) , the Laplace transform is given by, F ( s ) = L [ f ( t ) ] = ∫ 0 ∞ f ( f ) e − s t d t …… (1) Here, F ( s ) represents the Laplace transform, s is a variable which is complex and independent, f ( t ) is any function of time which is being transformed, and L is the operator which is defined by an integral. f ( t ) is calculated by taking inverse Laplace transform of the function F ( s ) . PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms. Laplace transform of higher order derivatives is given by: L ( d n f d t n ) = s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ( 1 ) ( 0 ) − ⋯ − s f ( n − 2 ) ( 0 ) − f ( n − 1 ) ( 0 ) …… (2)

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Answer 1

Question

Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, y(t) for the sudden change in u(t) from 0 to 1 m3 at t=0 is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt …… (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

L(dnfdtn)=snF(s)−sn−1f(0)−sn−2f(1)(0)−⋯−sf(n−2)(0)−f(n−1)(0) …… (2)

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, umax without the tank being overflowing is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt ........ (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

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Answer 2

Question

Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, y(t) for the sudden change in u(t) from 0 to 1 m3 at t=0 is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt …… (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

L(dnfdtn)=snF(s)−sn−1f(0)−sn−2f(1)(0)−⋯−sf(n−2)(0)−f(n−1)(0) …… (2)

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, umax without the tank being overflowing is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt ........ (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

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Answer 3

Question

Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, y(t) for the sudden change in u(t) from 0 to 1 m3 at t=0 is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt …… (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

L(dnfdtn)=snF(s)−sn−1f(0)−sn−2f(1)(0)−⋯−sf(n−2)(0)−f(n−1)(0) …… (2)

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, umax without the tank being overflowing is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt ........ (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

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Answer 4

Question

Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, y(t) for the sudden change in u(t) from 0 to 1 m3 at t=0 is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt …… (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

L(dnfdtn)=snF(s)−sn−1f(0)−sn−2f(1)(0)−⋯−sf(n−2)(0)−f(n−1)(0) …… (2)

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, umax without the tank being overflowing is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt ........ (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

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Answer 5

Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, y(t) for the sudden change in u(t) from 0 to 1 m3 at t=0 is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt …… (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

L(dnfdtn)=snF(s)−sn−1f(0)−sn−2f(1)(0)−⋯−sf(n−2)(0)−f(n−1)(0) …… (2)

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, umax without the tank being overflowing is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt ........ (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Answer 6

Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, y(t) for the sudden change in u(t) from 0 to 1 m3 at t=0 is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt …… (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

L(dnfdtn)=snF(s)−sn−1f(0)−sn−2f(1)(0)−⋯−sf(n−2)(0)−f(n−1)(0) …… (2)

Answer 7

Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, y(t) for the sudden change in u(t) from 0 to 1 m3 at t=0 is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt …… (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

L(dnfdtn)=snF(s)−sn−1f(0)−sn−2f(1)(0)−⋯−sf(n−2)(0)−f(n−1)(0) …… (2)

Answer 8

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Answer 9

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Answer 10

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, umax without the tank being overflowing is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt ........ (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Answer 11

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, umax without the tank being overflowing is to be calculated.

Concept introduction:

For a function f(t), the Laplace transform is given by,

F(s)=L[f(t)]=∫0∞f(f)e−stdt ........ (1)

Here, F(s) represents the Laplace transform, s is a variable which is complex and independent, f(t) is any function of time which is being transformed, and L is the operator which is defined by an integral.

f(t) is calculated by taking inverse Laplace transform of the function F(s).

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of y(t) can be calculated using Final Value theorem (FVM) as shown below:

limt→∞y(t)=lims→0[sY(s)] ........ (3)

This theorem is applicable only if lims→0[sY(s)] exists for all values of Re(s)≥0.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

Ch. 3 - Find the Laplace transforms of the following...Ch. 3 - Prob. 3.2E Ch. 3 - Prob. 3.3E Ch. 3 - Prob. 3.4E Ch. 3 - Prob. 3.5E Ch. 3 - Using PFE where required, find X(t) for a....Ch. 3 - Expand each of the following s-domain functions...Ch. 3 - For the integro-differential equation,...Ch. 3 - Prob. 3.9E Ch. 3 - Which solutions of the following equations will...

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