System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 2, Problem 2.8P
To determine

(a)

The inverse Laplace transform f(t) of the given function.

Expert Solution
Check Mark

Answer to Problem 2.8P

Inverse Laplace transform of the function F(s)=6s2+9 is f(t)=2sin3t.

Explanation of Solution

Given:

F(s)=6s2+9.

Concept Used:

Laplace transform of function f(t) is defined as

L(f(t))=0estf(t)dt

Where f(t) is piecewise continuo function defined for all t0.

L(tn)=n+1sn+1=nsn+1L(eatf(t))=F(s+a);L(f(t))=F(s)L(sinat)=L(eiateiat2i)=as2+a2L(cosat)=L(eiat+eiat2)=ss2+a2L(tnf(t))=(1)ndndsn(F(s));L(f(t))=F(s)

Inverse Laplace transform of given function F(s) is given by

f(t)=L1(F(s)).

Calculation:

Here, we have

F(s)=6s2+9=2(3)s2+32

Thus, the inverse Laplace transform f(t) of the given function is given as

f(t)=L1(F(s))f(t)=L1(6s2+9)f(t)=L1(2(3)s2+32)f(t)=2sin3t [L(sinat)=as2+a2].

To determine

(b)

The inverse Laplace transform f(t) of the given function.

Expert Solution
Check Mark

Answer to Problem 2.8P

Inverse Laplace transform of the function F(s)=5s2+4+4ss2+4 is f(t)=52sin(2t)+4cos(2t).

Explanation of Solution

Given:

F(s)=5s2+4+4ss2+4.

Concept Used:

Laplace transform of function f(t) is defined as

L(f(t))=0estf(t)dt

Where f(t) is piecewise continuo function defined for all t0.

L(tn)=n+1sn+1=nsn+1L(eatf(t))=F(s+a);L(f(t))=F(s)L(sinat)=L(eiateiat2i)=as2+a2L(cosat)=L(eiat+eiat2)=ss2+a2L(tnf(t))=(1)ndndsn(F(s));L(f(t))=F(s)

Inverse Laplace transform of given function F(s) is given by

f(t)=L1(F(s)).

Calculation:

Here, we have

F(s)=5s2+4+4ss2+4

Thus, the inverse Laplace transform f(t) of the given function is given as

f(t)=L1(5s2+4+4ss2+4)f(t)=L1(5s2+22)+L1(4ss2+22)f(t)=52L1(2s2+22)+4L1(ss2+22)f(t)=52sin(2t)+4cos(2t) [L(sinat)=as2+a2&L(cosat)=ss2+a2].

To determine

(c)

The inverse Laplace transform f(t) of the given function.

Expert Solution
Check Mark

Answer to Problem 2.8P

Inverse Laplace transform of the function F(s)=6s2+4s+13 is f(t)=2e2tsin(3t) .

Explanation of Solution

Given:

F(s)=6s2+4s+13.

Concept Used:

Laplace transform of function f(t) is defined as

L(f(t))=0estf(t)dt

Where f(t) is piecewise continuo function defined for all t0.

L(tn)=n+1sn+1=nsn+1L(eatf(t))=F(s+a);L(f(t))=F(s)L(sinat)=L(eiateiat2i)=as2+a2L(cosat)=L(eiat+eiat2)=ss2+a2L(tnf(t))=(1)ndndsn(F(s));L(f(t))=F(s)

Inverse Laplace transform of given function F(s) is given by

f(t)=L1(F(s)).

Calculation:

Here, we have

F(s)=6s2+4s+13

Thus, the inverse Laplace transform f(t) of the given function is given as

f(t)=L1(6s2+4s+13)f(t)=L1(6s2+2(2)s+4+9)f(t)=2L1(3(s+2)2+32)f(t)=2e2tsin(3t) [L(sinat)=as2+a2&L(eatf(t))=F(s+a)].

To determine

(d)

The inverse Laplace transform f(t) of the given function.

Expert Solution
Check Mark

Answer to Problem 2.8P

Inverse Laplace transform of the function F(s)=5s(s+3) is f(t)=53(1e3t) .

Explanation of Solution

Given:

F(s)=5s(s+3).

Concept Used:

Laplace transform of function f(t) is defined as

L(f(t))=0estf(t)dt

Where f(t) is piecewise continuo function defined for all t0.

L(tn)=n+1sn+1=nsn+1L(eatf(t))=F(s+a);L(f(t))=F(s)L(tnf(t))=(1)ndndsn(F(s));L(f(t))=F(s)

Inverse Laplace transform of given function F(s) is given by

f(t)=L1(F(s)).

Calculation:

Here, we have

F(s)=5s(s+3)

Thus, the inverse Laplace transform f(t) of the given function is given as

f(t)=L1(5s(s+3))f(t)=53L1(3s(s+3))f(t)=53L1(s+3ss(s+3))f(t)=53L1(1s1(s+3))f(t)=53L1(1s)53L1(1s+3)f(t)=53(1)53e3t [L(1)=1s&L(eatf(t))=F(s+a)]f(t)=53(1e3t) .

To determine

(e)

The inverse Laplace transform f(t) of the given function.

Expert Solution
Check Mark

Answer to Problem 2.8P

Inverse Laplace transform of the function F(s)=10(s+3)(s+7) is f(t)=52(e3te7t).

Explanation of Solution

Given:

F(s)=10(s+3)(s+7).

Concept Used:

Laplace transform of function f(t) is defined as

L(f(t))=0estf(t)dt

Where f(t) is piecewise continuo function defined for all t0.

L(tn)=n+1sn+1=nsn+1L(eatf(t))=F(s+a);L(f(t))=F(s)L(tnf(t))=(1)ndndsn(F(s));L(f(t))=F(s)

Inverse Laplace transform of given function F(s) is given by

f(t)=L1(F(s)).

Calculation:

Here, we have

F(s)=10(s+3)(s+7)

Thus, the inverse Laplace transform f(t) of the given function is given as

f(t)=L1(10(s+3)(s+7)) (1)

Now, we find the partial fraction of 10(s+3)(s+7) such that

10(s+3)(s+7)=A(s+3)+B(s+7) (2)

Here, A and B will be

A=lims3(s+3)10(s+3)(s+7)=104 (3)B=lims7(s+7)10(s+3)(s+7)=104 (4)

Using (3) and (4) in (1), we get

10(s+3)(s+7)=104(s+3)104(s+7) (5)

Using (5) in (1), we get

f(t)=L1(104(s+3)104(s+7))f(t)=L1(104(s+3))L1(104(s+7))f(t)=52L1(1s+3)52L1(1s+7)f(t)=52e3t52e7t [L(1)=1s&L(eatf(t))=F(s+a)]f(t)=52(e3te7t).

To determine

(f)

The inverse Laplace transform f(t) of the given function.

Expert Solution
Check Mark

Answer to Problem 2.8P

Inverse Laplace transform of the function F(s)=2s+8(s+3)(s+7) is f(t)=12e3t+32e7t.

Explanation of Solution

Given:

F(s)=2s+8(s+3)(s+7).

Concept Used:

Laplace transform of function f(t) is defined as

L(f(t))=0estf(t)dt

Where f(t) is piecewise continuo function defined for all t0.

L(tn)=n+1sn+1=nsn+1L(eatf(t))=F(s+a);L(f(t))=F(s)L(tnf(t))=(1)ndndsn(F(s));L(f(t))=F(s)

Inverse Laplace transform of given function F(s) is given by

f(t)=L1(F(s)).

Calculation:

Here, we have

F(s)=2s+8(s+3)(s+7)

Thus, the inverse Laplace transform f(t) of the given function is given as

f(t)=L1(2s+8(s+3)(s+7)) (1)

Now, we find the partial fraction of 2s+8(s+3)(s+7) such that

2s+8(s+3)(s+7)=A(s+3)+B(s+7) (2)

Here, A and B will be

A=lims3(s+3)2s+8(s+3)(s+7)=2(3)+84=12 (3)B=lims7(s+7)2s+8(s+3)(s+7)=2(7)+84=32 (4)

Using (3) and (4) in (1), we get

2s+8(s+3)(s+7)=12(s+3)+32(s+7) (5)

Using (5) in (1), we get

f(t)=L1(12(s+3)+32(s+7))f(t)=L1(12(s+3))+L1(32(s+7))f(t)=12L1(1s+3)+32L1(1s+7)f(t)=12e3t+32e7t [L(1)=1s&L(eatf(t))=F(s+a)].

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Chapter 2 Solutions

System Dynamics

Ch. 2 - Prob. 2.11PCh. 2 - Obtain the inverse Laplace transform xt for each...Ch. 2 - Solve the following problems: 5x=7tx0=3...Ch. 2 - Solve the following: 5x+7x=0x0=4 5x+7x=15x0=0...Ch. 2 - Solve the following problems: x+10x+21x=0x0=4x0=3...Ch. 2 - Solve the following problems: x+7x+10x=20x0=5x0=3...Ch. 2 - Solve the following problems: 3x+30x+63x=5x0=x0=0...Ch. 2 - Solve the following problems where x0=x0=0 ....Ch. 2 - Invert the following transforms: 6ss+5 4s+3s+8...Ch. 2 - Invert the following transforms: 3s+2s2s+10...Ch. 2 - Prob. 2.21PCh. 2 - Compare the LCD method with equation (2.4.4) for...Ch. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - (a) Prove that the second-order system whose...Ch. 2 - For each of the following models, compute the time...Ch. 2 - Prob. 2.27PCh. 2 - Prob. 2.28PCh. 2 - Prob. 2.29PCh. 2 - If applicable, compute , , n , and d for the...Ch. 2 - Prob. 2.31PCh. 2 - For each of the following equations, determine the...Ch. 2 - Prob. 2.33PCh. 2 - Obtain the transfer functions Xs/Fs and Ys/Fs for...Ch. 2 - a. Obtain the transfer functions Xs/Fs and Ys/Fs...Ch. 2 - Prob. 2.36PCh. 2 - Solve the following problems for xt . Compare the...Ch. 2 - Prob. 2.38PCh. 2 - Prob. 2.39PCh. 2 - Prob. 2.40PCh. 2 - Determine the general form of the solution of the...Ch. 2 - a. Use the Laplace transform to obtain the form of...Ch. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Obtain the inverse transform in the form xt=Asint+...Ch. 2 - Use the Laplace transform to solve the following...Ch. 2 - Express the oscillatory part of the solution of...Ch. 2 - Prob. 2.48PCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - 2.54 The Taylor series expansion for tan t...Ch. 2 - 2.55 Derive the initial value theorem: Ch. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Use MATLAB to obtain the inverse transform of the...Ch. 2 - Use MATLAB to obtain the inverse transform of the...Ch. 2 - Use MATLAB to solve for and plot the unit-step...Ch. 2 - Use MATLAB to solve for and plot the unit-impulse...Ch. 2 - Use MATLAB to solve for and plot the impulse...Ch. 2 - Use MATLAB to solve for and plot the response of...
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