The value of L − 1 { G ( s ) } .

Question

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Answer 1

Question

Chapter 14, Problem 33E

(a)

To determine

The value of $L−1{G(s)}$ .

(a)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3s(s2+2)2(s+2)$

Calculation:

The function is simplified as,

$G(s)=3s(s2+2)2(s+2)=12s(s+4)2(s+2)$ (1)

Apply the partial fraction on the above expression.

$12s(s+4)2(s+2)=A(s+4)2+B(s+4)+C(s+2)$ (2)

The equation is written as,

$12s=A(s+2)+B(s+4)(s+2)+C(s+4)2$ (3)

Substitute $−2$ for $s$ in equation (3).

$−24=A(0)+B(0)+C(−2+4)2−24=4CC=−6$

Substitute $−4$ for $s$ in equation (3).

$−48=A(−4+2)+B(0)+C(0)−48=−2AA=24$

The simplification of equation (3) is written as,

$12s=(B+C)s2+(A+6B+8C)s+(2A+8B+16C)$ (4)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−6$ for $C$ in the above expression.

$B−6=0B=6$

Substitute $24$ for $A$ , $6$ for $B$ and $−6$ for $C$ in equation (2).

$12s(s+4)2(s+2)=24(s+4)2+6(s+4)+−6(s+2)=24(s+4)2+6(s+4)−6(s+2)$

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $12s(s+4)2(s+2)$ in equation (1).

$G(s)=24(s+4)2+6(s+4)−6(s+2)$

The Laplace transform of $e−atu(t)$ is written as,

$L[e−atu(t)]=1s+a$

The Laplace transform of $te−atu(t)$ is written as,

$L[te−atu(t)]=1(s+a)2$

The properties for Laplace transform are written as,

$L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}$

$L−1{kF(s)}=kL−1{F(s)}$

The inverse Laplace of the given function is written as,

$g(t)=L−1{G(s)}$ (5)

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $G(s)$ in equation (5).

$g(t)=L−1{24(s+4)2+6(s+4)−6(s+2)}=24L−1{1(s+4)2}+6L−1{1(s+4)}−6L−1{1(s+2)}=24te−4tu(t)+6e−4tu(t)−6e−2tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

(b)

To determine

The value of $L−1{G(s)}$ .

(b)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3−3s(2s2+24s+70)(s+5)$

Calculation:

The function is simplified as,

$G(s)=3−3s(2s2+24s+70)(s+5)=3−3s2(s2+12s+35)(s+5)=3−1.5s(s+5)(s+7)(s+5)=3−1.5s(s+7)(s+5)2$ (6)

Apply the partial fraction on the above expression.

$1.5s(s+7)(s+5)2=A(s+5)2+B(s+5)+C(s+7)$ (7)

The equation is written as,

$1.5s=A(s+7)+B(s+7)(s+5)+C(s+5)2$ (8)

Substitute $−7$ for $s$ in equation (8).

$−10.5=A(0)+B(0)+C(−7+5)2−10.5=4CC=−2.625$

Substitute $−5$ for $s$ in equation (8).

$−7.5=A(−5+7)+B(0)+C(0)−7.5=2AA=−3.75$

The simplification of equation (8) is written as,

$1.5s=(B+C)s2+(A+12B+10C)s+(7A+35B+25C)$ (9)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−2.625$ for $C$ in the above expression.

$B−2.625=0B=2.625$

Substitute $−3.75$ for $A$ , $2.625$ for $B$ and $−2.625$ for $C$ in equation (7).

$1.5s(s+7)(s+5)2=−3.75(s+5)2+2.625(s+5)+−2.625(s+7)=−3.75(s+5)2+2.625(s+5)−2.625(s+7)$

Substitute $−3.75(s+5)2+2.625(s+5)−2.625(s+7)$ for $1.5s(s+7)(s+5)2$ in equation (6).

$G(s)=3−[−3.75(s+5)2+2.625(s+5)−2.625(s+7)]=3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$

The Laplace transform of $δ(t)$ is written as,

$L[δ(t)]=1$

Substitute $3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$ for $G(s)$ in equation (5).

$g(t)=L−1{3+3.75(s+5)2−2.625(s+5)+2.625(s+7)}=3L−1{1}+3.75L−1{1(s+5)2}−2.625L−1{1(s+5)}+2.625L−1{1(s+7)}=3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

(c)

To determine

The value of $L−1{G(s)}$ .

(c)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=2−1(s+100)+1s2+100$

Calculation:

The function is simplified as,

$G(s)=2−1(s+100)+1s2+100=2−1(s+100)+1s2+102$

The Laplace transform of $cosωtu(t)$ is written as,

$L[cosωtu(t)]=1s2+ω2$

Substitute $2−1(s+100)+1s2+102$ for $G(s)$ in equation (5).

$g(t)=L−1{2−1(s+100)+1s2+102}=2L−1{1}−L−1{1(s+100)}+L−1{1s2+102}=2δ(t)−e−100tu(t)+cos(10t)u(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

(d)

To determine

The value of $L−1{G(s)}$ .

(d)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $tu(2t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=L[tu(2t)]$

Calculation:

Substitute $L[tu(2t)]$ for $G(s)$ in equation (5).

$g(t)=L−1{L[tu(2t)]}=tu(2t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $tu(2t)_$ .

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5. Three single-phase transformers rated at 250 kVA, 7200 V/600 V, 60 Hz, are connected in wye-delta on a 12470 V, 3-phase line. If the load is 450 kVA, calculate the following currents: (1) In the incoming and outgoing transmission lines (2) In the primary and secondary windings (3) If this transformer is used to raise the voltage of a 3-phase, 600 V line to 7.2 kV. (a) How must they be connected? (b) Calculate the line currents for a 600 kVA load. (c) Calculate the corresponding primary and secondary currents.

please answer handwritten not chatgbt

Answer 2

Question

Chapter 14, Problem 33E

(a)

To determine

The value of $L−1{G(s)}$ .

(a)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3s(s2+2)2(s+2)$

Calculation:

The function is simplified as,

$G(s)=3s(s2+2)2(s+2)=12s(s+4)2(s+2)$ (1)

Apply the partial fraction on the above expression.

$12s(s+4)2(s+2)=A(s+4)2+B(s+4)+C(s+2)$ (2)

The equation is written as,

$12s=A(s+2)+B(s+4)(s+2)+C(s+4)2$ (3)

Substitute $−2$ for $s$ in equation (3).

$−24=A(0)+B(0)+C(−2+4)2−24=4CC=−6$

Substitute $−4$ for $s$ in equation (3).

$−48=A(−4+2)+B(0)+C(0)−48=−2AA=24$

The simplification of equation (3) is written as,

$12s=(B+C)s2+(A+6B+8C)s+(2A+8B+16C)$ (4)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−6$ for $C$ in the above expression.

$B−6=0B=6$

Substitute $24$ for $A$ , $6$ for $B$ and $−6$ for $C$ in equation (2).

$12s(s+4)2(s+2)=24(s+4)2+6(s+4)+−6(s+2)=24(s+4)2+6(s+4)−6(s+2)$

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $12s(s+4)2(s+2)$ in equation (1).

$G(s)=24(s+4)2+6(s+4)−6(s+2)$

The Laplace transform of $e−atu(t)$ is written as,

$L[e−atu(t)]=1s+a$

The Laplace transform of $te−atu(t)$ is written as,

$L[te−atu(t)]=1(s+a)2$

The properties for Laplace transform are written as,

$L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}$

$L−1{kF(s)}=kL−1{F(s)}$

The inverse Laplace of the given function is written as,

$g(t)=L−1{G(s)}$ (5)

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $G(s)$ in equation (5).

$g(t)=L−1{24(s+4)2+6(s+4)−6(s+2)}=24L−1{1(s+4)2}+6L−1{1(s+4)}−6L−1{1(s+2)}=24te−4tu(t)+6e−4tu(t)−6e−2tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

(b)

To determine

The value of $L−1{G(s)}$ .

(b)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3−3s(2s2+24s+70)(s+5)$

Calculation:

The function is simplified as,

$G(s)=3−3s(2s2+24s+70)(s+5)=3−3s2(s2+12s+35)(s+5)=3−1.5s(s+5)(s+7)(s+5)=3−1.5s(s+7)(s+5)2$ (6)

Apply the partial fraction on the above expression.

$1.5s(s+7)(s+5)2=A(s+5)2+B(s+5)+C(s+7)$ (7)

The equation is written as,

$1.5s=A(s+7)+B(s+7)(s+5)+C(s+5)2$ (8)

Substitute $−7$ for $s$ in equation (8).

$−10.5=A(0)+B(0)+C(−7+5)2−10.5=4CC=−2.625$

Substitute $−5$ for $s$ in equation (8).

$−7.5=A(−5+7)+B(0)+C(0)−7.5=2AA=−3.75$

The simplification of equation (8) is written as,

$1.5s=(B+C)s2+(A+12B+10C)s+(7A+35B+25C)$ (9)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−2.625$ for $C$ in the above expression.

$B−2.625=0B=2.625$

Substitute $−3.75$ for $A$ , $2.625$ for $B$ and $−2.625$ for $C$ in equation (7).

$1.5s(s+7)(s+5)2=−3.75(s+5)2+2.625(s+5)+−2.625(s+7)=−3.75(s+5)2+2.625(s+5)−2.625(s+7)$

Substitute $−3.75(s+5)2+2.625(s+5)−2.625(s+7)$ for $1.5s(s+7)(s+5)2$ in equation (6).

$G(s)=3−[−3.75(s+5)2+2.625(s+5)−2.625(s+7)]=3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$

The Laplace transform of $δ(t)$ is written as,

$L[δ(t)]=1$

Substitute $3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$ for $G(s)$ in equation (5).

$g(t)=L−1{3+3.75(s+5)2−2.625(s+5)+2.625(s+7)}=3L−1{1}+3.75L−1{1(s+5)2}−2.625L−1{1(s+5)}+2.625L−1{1(s+7)}=3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

(c)

To determine

The value of $L−1{G(s)}$ .

(c)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=2−1(s+100)+1s2+100$

Calculation:

The function is simplified as,

$G(s)=2−1(s+100)+1s2+100=2−1(s+100)+1s2+102$

The Laplace transform of $cosωtu(t)$ is written as,

$L[cosωtu(t)]=1s2+ω2$

Substitute $2−1(s+100)+1s2+102$ for $G(s)$ in equation (5).

$g(t)=L−1{2−1(s+100)+1s2+102}=2L−1{1}−L−1{1(s+100)}+L−1{1s2+102}=2δ(t)−e−100tu(t)+cos(10t)u(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

(d)

To determine

The value of $L−1{G(s)}$ .

(d)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $tu(2t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=L[tu(2t)]$

Calculation:

Substitute $L[tu(2t)]$ for $G(s)$ in equation (5).

$g(t)=L−1{L[tu(2t)]}=tu(2t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $tu(2t)_$ .

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Answer 3

Question

Chapter 14, Problem 33E

(a)

To determine

The value of $L−1{G(s)}$ .

(a)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3s(s2+2)2(s+2)$

Calculation:

The function is simplified as,

$G(s)=3s(s2+2)2(s+2)=12s(s+4)2(s+2)$ (1)

Apply the partial fraction on the above expression.

$12s(s+4)2(s+2)=A(s+4)2+B(s+4)+C(s+2)$ (2)

The equation is written as,

$12s=A(s+2)+B(s+4)(s+2)+C(s+4)2$ (3)

Substitute $−2$ for $s$ in equation (3).

$−24=A(0)+B(0)+C(−2+4)2−24=4CC=−6$

Substitute $−4$ for $s$ in equation (3).

$−48=A(−4+2)+B(0)+C(0)−48=−2AA=24$

The simplification of equation (3) is written as,

$12s=(B+C)s2+(A+6B+8C)s+(2A+8B+16C)$ (4)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−6$ for $C$ in the above expression.

$B−6=0B=6$

Substitute $24$ for $A$ , $6$ for $B$ and $−6$ for $C$ in equation (2).

$12s(s+4)2(s+2)=24(s+4)2+6(s+4)+−6(s+2)=24(s+4)2+6(s+4)−6(s+2)$

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $12s(s+4)2(s+2)$ in equation (1).

$G(s)=24(s+4)2+6(s+4)−6(s+2)$

The Laplace transform of $e−atu(t)$ is written as,

$L[e−atu(t)]=1s+a$

The Laplace transform of $te−atu(t)$ is written as,

$L[te−atu(t)]=1(s+a)2$

The properties for Laplace transform are written as,

$L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}$

$L−1{kF(s)}=kL−1{F(s)}$

The inverse Laplace of the given function is written as,

$g(t)=L−1{G(s)}$ (5)

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $G(s)$ in equation (5).

$g(t)=L−1{24(s+4)2+6(s+4)−6(s+2)}=24L−1{1(s+4)2}+6L−1{1(s+4)}−6L−1{1(s+2)}=24te−4tu(t)+6e−4tu(t)−6e−2tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

(b)

To determine

The value of $L−1{G(s)}$ .

(b)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3−3s(2s2+24s+70)(s+5)$

Calculation:

The function is simplified as,

$G(s)=3−3s(2s2+24s+70)(s+5)=3−3s2(s2+12s+35)(s+5)=3−1.5s(s+5)(s+7)(s+5)=3−1.5s(s+7)(s+5)2$ (6)

Apply the partial fraction on the above expression.

$1.5s(s+7)(s+5)2=A(s+5)2+B(s+5)+C(s+7)$ (7)

The equation is written as,

$1.5s=A(s+7)+B(s+7)(s+5)+C(s+5)2$ (8)

Substitute $−7$ for $s$ in equation (8).

$−10.5=A(0)+B(0)+C(−7+5)2−10.5=4CC=−2.625$

Substitute $−5$ for $s$ in equation (8).

$−7.5=A(−5+7)+B(0)+C(0)−7.5=2AA=−3.75$

The simplification of equation (8) is written as,

$1.5s=(B+C)s2+(A+12B+10C)s+(7A+35B+25C)$ (9)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−2.625$ for $C$ in the above expression.

$B−2.625=0B=2.625$

Substitute $−3.75$ for $A$ , $2.625$ for $B$ and $−2.625$ for $C$ in equation (7).

$1.5s(s+7)(s+5)2=−3.75(s+5)2+2.625(s+5)+−2.625(s+7)=−3.75(s+5)2+2.625(s+5)−2.625(s+7)$

Substitute $−3.75(s+5)2+2.625(s+5)−2.625(s+7)$ for $1.5s(s+7)(s+5)2$ in equation (6).

$G(s)=3−[−3.75(s+5)2+2.625(s+5)−2.625(s+7)]=3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$

The Laplace transform of $δ(t)$ is written as,

$L[δ(t)]=1$

Substitute $3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$ for $G(s)$ in equation (5).

$g(t)=L−1{3+3.75(s+5)2−2.625(s+5)+2.625(s+7)}=3L−1{1}+3.75L−1{1(s+5)2}−2.625L−1{1(s+5)}+2.625L−1{1(s+7)}=3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

(c)

To determine

The value of $L−1{G(s)}$ .

(c)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=2−1(s+100)+1s2+100$

Calculation:

The function is simplified as,

$G(s)=2−1(s+100)+1s2+100=2−1(s+100)+1s2+102$

The Laplace transform of $cosωtu(t)$ is written as,

$L[cosωtu(t)]=1s2+ω2$

Substitute $2−1(s+100)+1s2+102$ for $G(s)$ in equation (5).

$g(t)=L−1{2−1(s+100)+1s2+102}=2L−1{1}−L−1{1(s+100)}+L−1{1s2+102}=2δ(t)−e−100tu(t)+cos(10t)u(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

(d)

To determine

The value of $L−1{G(s)}$ .

(d)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $tu(2t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=L[tu(2t)]$

Calculation:

Substitute $L[tu(2t)]$ for $G(s)$ in equation (5).

$g(t)=L−1{L[tu(2t)]}=tu(2t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $tu(2t)_$ .

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Answer 4

Question

Chapter 14, Problem 33E

(a)

To determine

The value of $L−1{G(s)}$ .

(a)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3s(s2+2)2(s+2)$

Calculation:

The function is simplified as,

$G(s)=3s(s2+2)2(s+2)=12s(s+4)2(s+2)$ (1)

Apply the partial fraction on the above expression.

$12s(s+4)2(s+2)=A(s+4)2+B(s+4)+C(s+2)$ (2)

The equation is written as,

$12s=A(s+2)+B(s+4)(s+2)+C(s+4)2$ (3)

Substitute $−2$ for $s$ in equation (3).

$−24=A(0)+B(0)+C(−2+4)2−24=4CC=−6$

Substitute $−4$ for $s$ in equation (3).

$−48=A(−4+2)+B(0)+C(0)−48=−2AA=24$

The simplification of equation (3) is written as,

$12s=(B+C)s2+(A+6B+8C)s+(2A+8B+16C)$ (4)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−6$ for $C$ in the above expression.

$B−6=0B=6$

Substitute $24$ for $A$ , $6$ for $B$ and $−6$ for $C$ in equation (2).

$12s(s+4)2(s+2)=24(s+4)2+6(s+4)+−6(s+2)=24(s+4)2+6(s+4)−6(s+2)$

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $12s(s+4)2(s+2)$ in equation (1).

$G(s)=24(s+4)2+6(s+4)−6(s+2)$

The Laplace transform of $e−atu(t)$ is written as,

$L[e−atu(t)]=1s+a$

The Laplace transform of $te−atu(t)$ is written as,

$L[te−atu(t)]=1(s+a)2$

The properties for Laplace transform are written as,

$L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}$

$L−1{kF(s)}=kL−1{F(s)}$

The inverse Laplace of the given function is written as,

$g(t)=L−1{G(s)}$ (5)

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $G(s)$ in equation (5).

$g(t)=L−1{24(s+4)2+6(s+4)−6(s+2)}=24L−1{1(s+4)2}+6L−1{1(s+4)}−6L−1{1(s+2)}=24te−4tu(t)+6e−4tu(t)−6e−2tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

(b)

To determine

The value of $L−1{G(s)}$ .

(b)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3−3s(2s2+24s+70)(s+5)$

Calculation:

The function is simplified as,

$G(s)=3−3s(2s2+24s+70)(s+5)=3−3s2(s2+12s+35)(s+5)=3−1.5s(s+5)(s+7)(s+5)=3−1.5s(s+7)(s+5)2$ (6)

Apply the partial fraction on the above expression.

$1.5s(s+7)(s+5)2=A(s+5)2+B(s+5)+C(s+7)$ (7)

The equation is written as,

$1.5s=A(s+7)+B(s+7)(s+5)+C(s+5)2$ (8)

Substitute $−7$ for $s$ in equation (8).

$−10.5=A(0)+B(0)+C(−7+5)2−10.5=4CC=−2.625$

Substitute $−5$ for $s$ in equation (8).

$−7.5=A(−5+7)+B(0)+C(0)−7.5=2AA=−3.75$

The simplification of equation (8) is written as,

$1.5s=(B+C)s2+(A+12B+10C)s+(7A+35B+25C)$ (9)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−2.625$ for $C$ in the above expression.

$B−2.625=0B=2.625$

Substitute $−3.75$ for $A$ , $2.625$ for $B$ and $−2.625$ for $C$ in equation (7).

$1.5s(s+7)(s+5)2=−3.75(s+5)2+2.625(s+5)+−2.625(s+7)=−3.75(s+5)2+2.625(s+5)−2.625(s+7)$

Substitute $−3.75(s+5)2+2.625(s+5)−2.625(s+7)$ for $1.5s(s+7)(s+5)2$ in equation (6).

$G(s)=3−[−3.75(s+5)2+2.625(s+5)−2.625(s+7)]=3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$

The Laplace transform of $δ(t)$ is written as,

$L[δ(t)]=1$

Substitute $3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$ for $G(s)$ in equation (5).

$g(t)=L−1{3+3.75(s+5)2−2.625(s+5)+2.625(s+7)}=3L−1{1}+3.75L−1{1(s+5)2}−2.625L−1{1(s+5)}+2.625L−1{1(s+7)}=3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

(c)

To determine

The value of $L−1{G(s)}$ .

(c)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=2−1(s+100)+1s2+100$

Calculation:

The function is simplified as,

$G(s)=2−1(s+100)+1s2+100=2−1(s+100)+1s2+102$

The Laplace transform of $cosωtu(t)$ is written as,

$L[cosωtu(t)]=1s2+ω2$

Substitute $2−1(s+100)+1s2+102$ for $G(s)$ in equation (5).

$g(t)=L−1{2−1(s+100)+1s2+102}=2L−1{1}−L−1{1(s+100)}+L−1{1s2+102}=2δ(t)−e−100tu(t)+cos(10t)u(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

(d)

To determine

The value of $L−1{G(s)}$ .

(d)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $tu(2t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=L[tu(2t)]$

Calculation:

Substitute $L[tu(2t)]$ for $G(s)$ in equation (5).

$g(t)=L−1{L[tu(2t)]}=tu(2t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $tu(2t)_$ .

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Answer 5

Chapter 14, Problem 33E

(a)

To determine

The value of $L−1{G(s)}$ .

(a)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3s(s2+2)2(s+2)$

Calculation:

The function is simplified as,

$G(s)=3s(s2+2)2(s+2)=12s(s+4)2(s+2)$ (1)

Apply the partial fraction on the above expression.

$12s(s+4)2(s+2)=A(s+4)2+B(s+4)+C(s+2)$ (2)

The equation is written as,

$12s=A(s+2)+B(s+4)(s+2)+C(s+4)2$ (3)

Substitute $−2$ for $s$ in equation (3).

$−24=A(0)+B(0)+C(−2+4)2−24=4CC=−6$

Substitute $−4$ for $s$ in equation (3).

$−48=A(−4+2)+B(0)+C(0)−48=−2AA=24$

The simplification of equation (3) is written as,

$12s=(B+C)s2+(A+6B+8C)s+(2A+8B+16C)$ (4)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−6$ for $C$ in the above expression.

$B−6=0B=6$

Substitute $24$ for $A$ , $6$ for $B$ and $−6$ for $C$ in equation (2).

$12s(s+4)2(s+2)=24(s+4)2+6(s+4)+−6(s+2)=24(s+4)2+6(s+4)−6(s+2)$

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $12s(s+4)2(s+2)$ in equation (1).

$G(s)=24(s+4)2+6(s+4)−6(s+2)$

The Laplace transform of $e−atu(t)$ is written as,

$L[e−atu(t)]=1s+a$

The Laplace transform of $te−atu(t)$ is written as,

$L[te−atu(t)]=1(s+a)2$

The properties for Laplace transform are written as,

$L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}$

$L−1{kF(s)}=kL−1{F(s)}$

The inverse Laplace of the given function is written as,

$g(t)=L−1{G(s)}$ (5)

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $G(s)$ in equation (5).

$g(t)=L−1{24(s+4)2+6(s+4)−6(s+2)}=24L−1{1(s+4)2}+6L−1{1(s+4)}−6L−1{1(s+2)}=24te−4tu(t)+6e−4tu(t)−6e−2tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

(b)

To determine

The value of $L−1{G(s)}$ .

(b)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3−3s(2s2+24s+70)(s+5)$

Calculation:

The function is simplified as,

$G(s)=3−3s(2s2+24s+70)(s+5)=3−3s2(s2+12s+35)(s+5)=3−1.5s(s+5)(s+7)(s+5)=3−1.5s(s+7)(s+5)2$ (6)

Apply the partial fraction on the above expression.

$1.5s(s+7)(s+5)2=A(s+5)2+B(s+5)+C(s+7)$ (7)

The equation is written as,

$1.5s=A(s+7)+B(s+7)(s+5)+C(s+5)2$ (8)

Substitute $−7$ for $s$ in equation (8).

$−10.5=A(0)+B(0)+C(−7+5)2−10.5=4CC=−2.625$

Substitute $−5$ for $s$ in equation (8).

$−7.5=A(−5+7)+B(0)+C(0)−7.5=2AA=−3.75$

The simplification of equation (8) is written as,

$1.5s=(B+C)s2+(A+12B+10C)s+(7A+35B+25C)$ (9)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−2.625$ for $C$ in the above expression.

$B−2.625=0B=2.625$

Substitute $−3.75$ for $A$ , $2.625$ for $B$ and $−2.625$ for $C$ in equation (7).

$1.5s(s+7)(s+5)2=−3.75(s+5)2+2.625(s+5)+−2.625(s+7)=−3.75(s+5)2+2.625(s+5)−2.625(s+7)$

Substitute $−3.75(s+5)2+2.625(s+5)−2.625(s+7)$ for $1.5s(s+7)(s+5)2$ in equation (6).

$G(s)=3−[−3.75(s+5)2+2.625(s+5)−2.625(s+7)]=3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$

The Laplace transform of $δ(t)$ is written as,

$L[δ(t)]=1$

Substitute $3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$ for $G(s)$ in equation (5).

$g(t)=L−1{3+3.75(s+5)2−2.625(s+5)+2.625(s+7)}=3L−1{1}+3.75L−1{1(s+5)2}−2.625L−1{1(s+5)}+2.625L−1{1(s+7)}=3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

(c)

To determine

The value of $L−1{G(s)}$ .

(c)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=2−1(s+100)+1s2+100$

Calculation:

The function is simplified as,

$G(s)=2−1(s+100)+1s2+100=2−1(s+100)+1s2+102$

The Laplace transform of $cosωtu(t)$ is written as,

$L[cosωtu(t)]=1s2+ω2$

Substitute $2−1(s+100)+1s2+102$ for $G(s)$ in equation (5).

$g(t)=L−1{2−1(s+100)+1s2+102}=2L−1{1}−L−1{1(s+100)}+L−1{1s2+102}=2δ(t)−e−100tu(t)+cos(10t)u(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

(d)

To determine

The value of $L−1{G(s)}$ .

(d)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $tu(2t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=L[tu(2t)]$

Calculation:

Substitute $L[tu(2t)]$ for $G(s)$ in equation (5).

$g(t)=L−1{L[tu(2t)]}=tu(2t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $tu(2t)_$ .

Answer 6

Chapter 14, Problem 33E

(a)

To determine

The value of $L−1{G(s)}$ .

(a)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3s(s2+2)2(s+2)$

Calculation:

The function is simplified as,

$G(s)=3s(s2+2)2(s+2)=12s(s+4)2(s+2)$ (1)

Apply the partial fraction on the above expression.

$12s(s+4)2(s+2)=A(s+4)2+B(s+4)+C(s+2)$ (2)

The equation is written as,

$12s=A(s+2)+B(s+4)(s+2)+C(s+4)2$ (3)

Substitute $−2$ for $s$ in equation (3).

$−24=A(0)+B(0)+C(−2+4)2−24=4CC=−6$

Substitute $−4$ for $s$ in equation (3).

$−48=A(−4+2)+B(0)+C(0)−48=−2AA=24$

The simplification of equation (3) is written as,

$12s=(B+C)s2+(A+6B+8C)s+(2A+8B+16C)$ (4)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−6$ for $C$ in the above expression.

$B−6=0B=6$

Substitute $24$ for $A$ , $6$ for $B$ and $−6$ for $C$ in equation (2).

$12s(s+4)2(s+2)=24(s+4)2+6(s+4)+−6(s+2)=24(s+4)2+6(s+4)−6(s+2)$

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $12s(s+4)2(s+2)$ in equation (1).

$G(s)=24(s+4)2+6(s+4)−6(s+2)$

The Laplace transform of $e−atu(t)$ is written as,

$L[e−atu(t)]=1s+a$

The Laplace transform of $te−atu(t)$ is written as,

$L[te−atu(t)]=1(s+a)2$

The properties for Laplace transform are written as,

$L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}$

$L−1{kF(s)}=kL−1{F(s)}$

The inverse Laplace of the given function is written as,

$g(t)=L−1{G(s)}$ (5)

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $G(s)$ in equation (5).

$g(t)=L−1{24(s+4)2+6(s+4)−6(s+2)}=24L−1{1(s+4)2}+6L−1{1(s+4)}−6L−1{1(s+2)}=24te−4tu(t)+6e−4tu(t)−6e−2tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Answer 7

Chapter 14, Problem 33E

(a)

To determine

The value of $L−1{G(s)}$ .

Answer 8

(a)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data:

The function is given as,

$G(s)=3s(s2+2)2(s+2)$

Calculation:

The function is simplified as,

$G(s)=3s(s2+2)2(s+2)=12s(s+4)2(s+2)$ (1)

Apply the partial fraction on the above expression.

$12s(s+4)2(s+2)=A(s+4)2+B(s+4)+C(s+2)$ (2)

The equation is written as,

$12s=A(s+2)+B(s+4)(s+2)+C(s+4)2$ (3)

Substitute $−2$ for $s$ in equation (3).

$−24=A(0)+B(0)+C(−2+4)2−24=4CC=−6$

Substitute $−4$ for $s$ in equation (3).

$−48=A(−4+2)+B(0)+C(0)−48=−2AA=24$

The simplification of equation (3) is written as,

$12s=(B+C)s2+(A+6B+8C)s+(2A+8B+16C)$ (4)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−6$ for $C$ in the above expression.

$B−6=0B=6$

Substitute $24$ for $A$ , $6$ for $B$ and $−6$ for $C$ in equation (2).

$12s(s+4)2(s+2)=24(s+4)2+6(s+4)+−6(s+2)=24(s+4)2+6(s+4)−6(s+2)$

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $12s(s+4)2(s+2)$ in equation (1).

$G(s)=24(s+4)2+6(s+4)−6(s+2)$

The Laplace transform of $e−atu(t)$ is written as,

$L[e−atu(t)]=1s+a$

The Laplace transform of $te−atu(t)$ is written as,

$L[te−atu(t)]=1(s+a)2$

The properties for Laplace transform are written as,

$L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}$

$L−1{kF(s)}=kL−1{F(s)}$

The inverse Laplace of the given function is written as,

$g(t)=L−1{G(s)}$ (5)

Substitute $24(s+4)2+6(s+4)−6(s+2)$ for $G(s)$ in equation (5).

$g(t)=L−1{24(s+4)2+6(s+4)−6(s+2)}=24L−1{1(s+4)2}+6L−1{1(s+4)}−6L−1{1(s+2)}=24te−4tu(t)+6e−4tu(t)−6e−2tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $24te−4tu(t)+6e−4tu(t)−6e−2tu(t)_$ .

Answer 13

(b)

To determine

The value of $L−1{G(s)}$ .

(b)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=3−3s(2s2+24s+70)(s+5)$

Calculation:

The function is simplified as,

$G(s)=3−3s(2s2+24s+70)(s+5)=3−3s2(s2+12s+35)(s+5)=3−1.5s(s+5)(s+7)(s+5)=3−1.5s(s+7)(s+5)2$ (6)

Apply the partial fraction on the above expression.

$1.5s(s+7)(s+5)2=A(s+5)2+B(s+5)+C(s+7)$ (7)

The equation is written as,

$1.5s=A(s+7)+B(s+7)(s+5)+C(s+5)2$ (8)

Substitute $−7$ for $s$ in equation (8).

$−10.5=A(0)+B(0)+C(−7+5)2−10.5=4CC=−2.625$

Substitute $−5$ for $s$ in equation (8).

$−7.5=A(−5+7)+B(0)+C(0)−7.5=2AA=−3.75$

The simplification of equation (8) is written as,

$1.5s=(B+C)s2+(A+12B+10C)s+(7A+35B+25C)$ (9)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−2.625$ for $C$ in the above expression.

$B−2.625=0B=2.625$

Substitute $−3.75$ for $A$ , $2.625$ for $B$ and $−2.625$ for $C$ in equation (7).

$1.5s(s+7)(s+5)2=−3.75(s+5)2+2.625(s+5)+−2.625(s+7)=−3.75(s+5)2+2.625(s+5)−2.625(s+7)$

Substitute $−3.75(s+5)2+2.625(s+5)−2.625(s+7)$ for $1.5s(s+7)(s+5)2$ in equation (6).

$G(s)=3−[−3.75(s+5)2+2.625(s+5)−2.625(s+7)]=3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$

The Laplace transform of $δ(t)$ is written as,

$L[δ(t)]=1$

Substitute $3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$ for $G(s)$ in equation (5).

$g(t)=L−1{3+3.75(s+5)2−2.625(s+5)+2.625(s+7)}=3L−1{1}+3.75L−1{1(s+5)2}−2.625L−1{1(s+5)}+2.625L−1{1(s+7)}=3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Answer 14

(b)

To determine

The value of $L−1{G(s)}$ .

Answer 15

(b)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given data:

The function is given as,

$G(s)=3−3s(2s2+24s+70)(s+5)$

Calculation:

The function is simplified as,

$G(s)=3−3s(2s2+24s+70)(s+5)=3−3s2(s2+12s+35)(s+5)=3−1.5s(s+5)(s+7)(s+5)=3−1.5s(s+7)(s+5)2$ (6)

Apply the partial fraction on the above expression.

$1.5s(s+7)(s+5)2=A(s+5)2+B(s+5)+C(s+7)$ (7)

The equation is written as,

$1.5s=A(s+7)+B(s+7)(s+5)+C(s+5)2$ (8)

Substitute $−7$ for $s$ in equation (8).

$−10.5=A(0)+B(0)+C(−7+5)2−10.5=4CC=−2.625$

Substitute $−5$ for $s$ in equation (8).

$−7.5=A(−5+7)+B(0)+C(0)−7.5=2AA=−3.75$

The simplification of equation (8) is written as,

$1.5s=(B+C)s2+(A+12B+10C)s+(7A+35B+25C)$ (9)

Equate the coefficient of $s2$ of equation (4).

$B+C=0$

Substitute $−2.625$ for $C$ in the above expression.

$B−2.625=0B=2.625$

Substitute $−3.75$ for $A$ , $2.625$ for $B$ and $−2.625$ for $C$ in equation (7).

$1.5s(s+7)(s+5)2=−3.75(s+5)2+2.625(s+5)+−2.625(s+7)=−3.75(s+5)2+2.625(s+5)−2.625(s+7)$

Substitute $−3.75(s+5)2+2.625(s+5)−2.625(s+7)$ for $1.5s(s+7)(s+5)2$ in equation (6).

$G(s)=3−[−3.75(s+5)2+2.625(s+5)−2.625(s+7)]=3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$

The Laplace transform of $δ(t)$ is written as,

$L[δ(t)]=1$

Substitute $3+3.75(s+5)2−2.625(s+5)+2.625(s+7)$ for $G(s)$ in equation (5).

$g(t)=L−1{3+3.75(s+5)2−2.625(s+5)+2.625(s+7)}=3L−1{1}+3.75L−1{1(s+5)2}−2.625L−1{1(s+5)}+2.625L−1{1(s+7)}=3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $3δ(t)+3.75te−5tu(t)−2.625e−5tu(t)+2.625e−7tu(t)_$ .

Answer 20

(c)

To determine

The value of $L−1{G(s)}$ .

(c)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=2−1(s+100)+1s2+100$

Calculation:

The function is simplified as,

$G(s)=2−1(s+100)+1s2+100=2−1(s+100)+1s2+102$

The Laplace transform of $cosωtu(t)$ is written as,

$L[cosωtu(t)]=1s2+ω2$

Substitute $2−1(s+100)+1s2+102$ for $G(s)$ in equation (5).

$g(t)=L−1{2−1(s+100)+1s2+102}=2L−1{1}−L−1{1(s+100)}+L−1{1s2+102}=2δ(t)−e−100tu(t)+cos(10t)u(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Answer 21

(c)

To determine

The value of $L−1{G(s)}$ .

Answer 22

(c)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given data:

The function is given as,

$G(s)=2−1(s+100)+1s2+100$

Calculation:

The function is simplified as,

$G(s)=2−1(s+100)+1s2+100=2−1(s+100)+1s2+102$

The Laplace transform of $cosωtu(t)$ is written as,

$L[cosωtu(t)]=1s2+ω2$

Substitute $2−1(s+100)+1s2+102$ for $G(s)$ in equation (5).

$g(t)=L−1{2−1(s+100)+1s2+102}=2L−1{1}−L−1{1(s+100)}+L−1{1s2+102}=2δ(t)−e−100tu(t)+cos(10t)u(t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $2δ(t)−e−100tu(t)+cos(10t)u(t)_$ .

Answer 27

(d)

To determine

The value of $L−1{G(s)}$ .

(d)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $tu(2t)_$ .

Explanation of Solution

Given data:

The function is given as,

$G(s)=L[tu(2t)]$

Calculation:

Substitute $L[tu(2t)]$ for $G(s)$ in equation (5).

$g(t)=L−1{L[tu(2t)]}=tu(2t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $tu(2t)_$ .

Answer 28

(d)

To determine

The value of $L−1{G(s)}$ .

Answer 29

(d)

Expert Solution

Answer to Problem 33E

The value of $L−1{G(s)}$ is $tu(2t)_$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given data:

The function is given as,

$G(s)=L[tu(2t)]$

Calculation:

Substitute $L[tu(2t)]$ for $G(s)$ in equation (5).

$g(t)=L−1{L[tu(2t)]}=tu(2t)$

Conclusion:

Therefore, the value of $L−1{G(s)}$ is $tu(2t)_$ .

Answer 34

Ch. 14.1 - Identify all the complex frequencies present in...Ch. 14.1 - Use real constants A, B, C, , and so forth, to...Ch. 14.2 - Let f (t) = 6e2t [u(t + 3) u(t 2)]. Find the (a)...Ch. 14.3 - Prob. 4P Ch. 14.3 - Prob. 5P Ch. 14.4 - Prob. 6P Ch. 14.4 - Prob. 7P Ch. 14.4 - Prob. 8P Ch. 14.4 - Prob. 9P Ch. 14.5 - Prob. 10P

The value of L − 1 { G ( s ) } .

Videos

The value of $L−1{G(s)}$ .

Answer to Problem 33E

Explanation of Solution

The value of $L−1{G(s)}$ .

Answer to Problem 33E

Explanation of Solution

The value of $L−1{G(s)}$ .

Answer to Problem 33E

Explanation of Solution

The value of $L−1{G(s)}$ .

Answer to Problem 33E

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

The value of L − 1 { G ( s ) } .

Videos

The value of L−1{G(s)}<m:math><m:mrow><m:msup><m:mi>L</m:mi><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow></m:msup><m:mrow><m:mo>{</m:mo><m:mrow><m:mi>G</m:mi><m:mrow><m:mo>(</m:mo><m:mi>s</m:mi><m:mo>)</m:mo></m:mrow></m:mrow><m:mo>}</m:mo></m:mrow></m:mrow></m:math>.

Answer to Problem 33E

Explanation of Solution

The value of L−1{G(s)}<m:math><m:mrow><m:msup><m:mi>L</m:mi><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow></m:msup><m:mrow><m:mo>{</m:mo><m:mrow><m:mi>G</m:mi><m:mrow><m:mo>(</m:mo><m:mi>s</m:mi><m:mo>)</m:mo></m:mrow></m:mrow><m:mo>}</m:mo></m:mrow></m:mrow></m:math>.

Answer to Problem 33E

Explanation of Solution

The value of L−1{G(s)}<m:math><m:mrow><m:msup><m:mi>L</m:mi><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow></m:msup><m:mrow><m:mo>{</m:mo><m:mrow><m:mi>G</m:mi><m:mrow><m:mo>(</m:mo><m:mi>s</m:mi><m:mo>)</m:mo></m:mrow></m:mrow><m:mo>}</m:mo></m:mrow></m:mrow></m:math>.

Answer to Problem 33E

Explanation of Solution

The value of L−1{G(s)}<m:math><m:mrow><m:msup><m:mi>L</m:mi><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow></m:msup><m:mrow><m:mo>{</m:mo><m:mrow><m:mi>G</m:mi><m:mrow><m:mo>(</m:mo><m:mi>s</m:mi><m:mo>)</m:mo></m:mrow></m:mrow><m:mo>}</m:mo></m:mrow></m:mrow></m:math>.

Answer to Problem 33E

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

The value of $L−1{G(s)}$ .

The value of $L−1{G(s)}$ .

The value of $L−1{G(s)}$ .

The value of $L−1{G(s)}$ .