Name: Jiggy Lawrence Bathan Date: October 7,2021

BSME-IV

Transform Functions

2. Using the Laplace Transform pairs of Table 2.1 and the Laplace Transform theorems
of Table 2.2, derive the Laplace Transforms for the following time functions:
a. 𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 𝑢(𝑡)
• First, we will use Laplace Transform theorem.
• Let 𝑒 −𝑎𝑡 =s=s+a
• Then get the Laplace Transform of sin 𝜔𝑡.
𝑤
𝐿(𝑠𝑖𝑛𝑤𝑡) =
𝑠2 + 𝑤 2
• Substitute s by s+a.
`
𝑤
𝐿(𝑒 −𝑎𝑡 𝑠𝑖𝑛𝑤𝑡 𝑢(𝑡)) =
(𝑠 + 𝑎)2 + 𝑤 2

b. 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 𝑢(𝑡)

• Use Laplace Transform Theorem.
• Let 𝑒 −𝑎𝑡 =s=s+a
• Then get the Laplace Transform of cos 𝜔𝑡.
𝑠
𝐿(𝑐𝑜𝑠𝑤𝑡) = 2
𝑠 + 𝑤2
• Substitute s by s+a

𝑠+𝑎
𝐿(𝑒 −𝑎𝑡 𝑐𝑜𝑠𝑤𝑡 𝑢(𝑡)) =
(𝑠 + 𝑎)2 + 𝑤 2

c. 𝑡 3 𝑢(𝑡)
• From the definition of Laplace Transform, we can say that
𝑛!
𝐿(𝑡 𝑛 ) =
𝑠 𝑛+1
• Substitute n by 3 and we can get
3!
𝐿(𝑡 3 𝑢(𝑡)) =
𝑠 3+1

6
𝐿(𝑡 3 𝑢(𝑡)) =
𝑠4
3. Repeat problem 18 in chapter 1 using Laplace Transforms. Assume that the forcing
functions are zero prior to t=0.
• By seeing the equation, the equivalent loop equation is
𝑑𝑖 1
𝑅𝑖 + 𝐿 + ∫ 𝑖(𝑡)𝑑𝑡 + 𝑣(𝑡)
𝑑𝑡 𝐶
• Therefore, our differential equation by substituting the values we get

𝑑𝑖
(2)𝑖 + (1) + (25) ∫ 𝑖(𝑡)𝑑𝑡 = 𝑣(𝑡)
𝑑𝑡

𝑑𝑖
+ 2𝑖 + 25 ∫ 𝑖(𝑡)𝑑𝑡 = 𝑣(𝑡)
𝑑𝑡

• Applying Laplace Transform with zero initial conditions.

𝐿(𝑖 ′ ) + 2𝐿(𝑖) + 25𝐿 (∫ 𝑖(𝑡)𝑑𝑡) ∫ = 𝐿(𝑣(𝑡))

𝐼(𝑠) 1
𝑠𝐼(𝑠) + 21(𝑠) + 25 =
𝑠 𝑠
• Simplify the equation and then isolate I(s) by multiplying (s) to both
sides of the above equation.

𝑠 2 𝐼(𝑠) + 2𝑠𝐼(𝑠) = 1

• Factor out I(s) in the equation.

𝑠 2 𝐼(𝑠) + 2𝑠𝐼(𝑠) = 1
• Rearranging the equation to apply ILT,
1
𝐼(𝑠) =
𝑠2 + 2𝑠 + 1 + 24
1
𝐼(𝑠) = 2
(𝑠 + 1)2 + (√24)

• It is almost like a Laplace Transform of sin(at) with the application of

√24
the frequency shift theorem, we can multiply to the right side of the
√24
𝑎
equation to satisfy the Laplace Transform of sin(at) which is ;𝑎 =
𝑠2 +𝑠2
√24.

1 √24
𝐼(𝑠) = ( 2)
√24 (𝑠 + 1)2 + (√24)

• Apply Inverse Laplace Transform to the equation.

1 √24
𝐿−1 (𝐼(𝑠)) = 𝐿−1 ((𝑠+1)2 )
√24 +(√24𝑠)

• Therefore, the current 𝑖(𝑡) of the system network is.

 1
𝑖(𝑡) = 𝑒 −𝑡 𝑠𝑖𝑛(√24𝑡)
√24

4. Repeat problem 19 in chapter 1, using Laplace Transforms. Use the following initial
conditions for each part as follows: (a) x(0)=4, x’(0)=-4; (b) x(0)=4, x’(0)=1; (c) x(0)=2,
x’(0)=3, where x’(0)=dx/dt(0). Assume that the forcing functions are zero prior to t=0.
𝑑𝑥
𝑎. + 7𝑥 = 5𝑐𝑜𝑠2𝑡 ; 𝑥(0) = 4 𝑎𝑛𝑑 𝑥(0) = −4
𝑑𝑡
• Apply Laplace transform:
L(𝑥 ′ ) + 7𝐿(𝑥) = 5𝐿(𝑐𝑜𝑠2𝑡)
𝑠
𝑠𝑥(𝑠) − 4 + 7[𝑥(𝑠)] = 5[ 2 ]
𝑠 + 22
5𝑠
(𝑠 + 7)𝑥(𝑠) = 2 −4
𝑠 +4
5𝑠 − 4𝑠 2 − 16
(𝑠 + 7)𝑥(𝑠) =
𝑠2 + 4

5𝑠 − 4𝑠 2 − 16 5𝑠 − 4𝑠 2 − 16
𝑥(𝑠) = 𝑜𝑟 𝑥(𝑠) =
(𝑠 2 + 4)(𝑠 + 7) (𝑠 + 2)2 (𝑠 + 7)

𝑑2 𝑥 6𝑑𝑥
b. 𝑑𝑡 2 + 𝑑𝑡
+ 8𝑥 = 5sin (3t); x(0)=0; x’(0)=1
• Apply Laplace Transform

L(x ′′ ) + 6𝐿(𝑥 ′ ) + 8𝐿(𝑥) = 5𝐿(sin (3𝑡)

3 𝑠
𝑠 2 𝐹(𝑠) − 𝑠𝑓(0) − 𝑓′(0) + 6𝑠𝐹(𝑠) − 6𝑓(0) + 8𝐹(𝑠) = 5 ( ) ; L sin(at) =
𝑠 2 +9 𝑠 2 +𝑎2
 15
𝑠 2 𝐹(𝑠) − 0 − 1 + 6𝑠𝐹(𝑠) + 8𝐹(𝑠) =
𝑠2 +9
15
F(s)[𝑠 2 + 6𝑠 + 8] = 𝑠2 +9 +1

𝑠 2 + 15
F(s) =
(𝑠 2 + 6𝑠 + 8)(𝑠 2 + 9)

𝑑2 𝑥 𝑑𝑥
c. 𝑑𝑡 2 + 8 𝑑𝑡 + 25𝑥 = 10 𝑢(𝑡); 𝑥(0) = 2, 𝑥 ′ (0) = 3

• Take Laplace Transform of the equation

𝑠 2 𝑥(𝑠) − 𝑠𝑥(0) − 𝑥 ′ (0) + 8(𝑠𝑥(𝑠) − 𝑥 ′ (0)) + 25𝑥(𝑠) = 10𝑢(𝑠)

𝑠 2 𝑥(𝑠) − 𝑠𝑥(0) − 𝑥 ′ (0) + 8𝑠𝑥(𝑠) − 8𝑥 ′ (0) + 25𝑥(𝑠) = 10𝑢(𝑠)

• Substitute the initial conditions.

𝑠 2 𝑥(𝑠) − 𝑠(2) − 3 + 8𝑠𝑥(𝑠) − 8(3) + 25𝑥(𝑠) = 10𝑢(𝑠)

𝑠 2 𝑥(𝑠) − 2𝑠 − 3 + 8𝑠𝑥(𝑠) − 24 + 25𝑥(𝑠) = 10𝑢(𝑠)

• Isolate the terms with x(s)

𝑠 2 𝑥(𝑠) + 8𝑠𝑥(𝑠) + 25𝑥(𝑠) = 10𝑢(𝑠) + 2𝑠 + 27

• We can now factor x(s).

(𝑠 2 + 8𝑠 + 25)𝑥(𝑠) = 10𝑢(𝑠) + 2𝑠 + 27

• Simplify the equation to get x(s).

10𝑢(𝑠) + 2𝑠 + 27
𝑥(𝑠) =
𝑠 2 + 8𝑠 + 25

7. A system is described in the following differential equation:

𝑑3 𝑦 𝑑2 𝑦 𝑑𝑦 𝑑3 𝑥 𝑑2 𝑥 𝑑𝑥
3
+ 3 2
+ 5 + 𝑦 = 3
+ 4 2
+6 + 8𝑥
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
Find the expression for the transfer function of the system, Y(s)/X(s).
• Get the Laplace Transform.
𝑠 3 𝑌(𝑠) + 𝑠 2 𝑌(𝑠) + 5𝑠𝑌(𝑠) + 𝑌(𝑠) = 𝑠 3 𝑋(𝑠) + 4𝑠 2 𝑥(𝑠) + 6𝑠𝑋(𝑠) + 8𝑋(𝑠)

• Assuming zero initial conditions, the simplified Laplace Transform of

the equation is

(𝑠 3 + 3𝑠 2 + 5𝑠 + 1)𝑌(𝑠) = (𝑠 3 + 4𝑠 2 + 6𝑠 + 8)𝑋(𝑠)

• Therefore, the Transfer function is

 𝑌(𝑠) 𝑠 3 + 4𝑠 2 + 6𝑠 + 8
=
𝑋(𝑠) 𝑠 3 + 3𝑠 2 + 5𝑠 + 1

8. For each of the following transfer functions, write the corresponding differential
equation.
𝑋(𝑠) 7
a. =
𝐹(𝑠) 𝑠2 +5𝑠+10
• By simplifying using cross multiplication, the equation will be

(𝑠 2 + 5𝑠 + 10)𝑋(𝑠) = 7𝐹(𝑠)
• By taking the inverse Laplace Transform, we will get

𝑑2 𝑥 𝑑𝑥
2
+5 + 10𝑥 = 7𝑓
𝑑𝑡 𝑑𝑡
𝑋(𝑠) 15
b. 𝐹(𝑠)
= (𝑠+10)(𝑠+11)
• By simplifying using cross multiplication, the equation will be
(𝑠 + 10)(𝑠 + 11)𝑋(𝑠) = 15𝐹(𝑠)

• Expand the left side of the equation.

(𝑠 2 + 21𝑠 + 110)𝑋(𝑠) = 15𝐹(𝑠)

• Take the inverse Laplace Transform of the equation.

𝑑2 𝑥 𝑑𝑥
2
+ 21 + 110𝑥 = 15𝑓
𝑑𝑡 𝑑𝑡

𝑋(𝑠) 𝑠+3
c. =
𝐹(𝑠) 𝑠3 +11𝑠2 +12𝑠+18
• Cross multiply the equation.
(𝑠 3 + 11𝑠 2 + 12𝑠 + 18)𝑋(𝑠) = (𝑠 + 3)𝐹(𝑠)

• Take the Laplace transform of the equation.

𝑑3 𝑥 𝑑2 𝑥 𝑑𝑥 𝑑𝑓𝑡
3
+ 11 2
+ 12 + 18𝑥 = + 3𝑓
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
9. Write the differential equation for the system shown.

𝑠 5 + 2𝑠 4 + 4𝑠 3 + 𝑠 2 + 4
𝑠 6 + 7𝑠 5 + 3𝑠 4 + 2𝑠 3 + 𝑠 2 + 5
R(s) C(s)

• From the above given, we can say that output over input is C(s)/R(s).
Therefore
𝐶(𝑠) 𝑠 5 + 2𝑠 4 + 4𝑠 3 + 𝑠 2 + 4
= 6
𝑅(𝑠) 𝑠 + 7𝑠 5 + 3𝑠 4 + 2𝑠 3 + 𝑠 2 + 5
• By cross multiplication, we can achieve this equation.

(𝑠 6 + 7𝑠 5 + 3𝑠 4 + 2𝑠 3 + 𝑠 2 + 5)𝐶(𝑠) = (𝑠 5 + 2𝑠 4 + 4𝑠 3 + 𝑠 2 + 4)𝑅(𝑠)

• Taking the inverse Laplace Transform of the equation gives us

𝑑6 𝑐 𝑑5 𝑐 𝑑4 𝑐 𝑑3 𝑐 𝑑2 𝑐 𝑑5 𝑟 𝑑4 𝑟 𝑑3 𝑐 𝑑2 𝑐
+ 7 + 3 + 2 + + 5𝑐 = + 2 + 4 + + 4𝑟
𝑑𝑡 6 𝑑𝑡 5 𝑑𝑡 4 𝑑𝑡 3 𝑑𝑡 2 𝑑𝑡 5 𝑑𝑡 4 𝑑𝑡 3 𝑑𝑡 2

10. Write the differential equation that is mathematically equivalent to the block diagram
shown. Assume that r(t)=3𝑡 3 .

𝑠 4 + 3𝑠 3 + 2𝑠 2 + 𝑠 + 1
𝑠 5 + 4𝑠 4 + 3𝑠 3 + 2𝑠 2 + 3𝑠 + 2
R(s) C(s)

• The transfer function of the equation will be the output over input
which is C(s)/R(s).
𝐶(𝑠) 𝑠 4 + 3𝑠 3 + 2𝑠 2 + 𝑠 + 1
= 5
𝑅(𝑠) 𝑠 + 4𝑠 4 + 3𝑠 3 + 2𝑠 2 + 3𝑠 + 2

• By cross multiplication, we can have the equation below.

(𝑠 5 + 4𝑠 4 + 3𝑠 3 + 2𝑠 2 + 3𝑠 + 2)𝐶(𝑠) = (𝑠 4 + 3𝑠 3 + 2𝑠 2 + 𝑠 + 1) 𝑅(𝑠)

• Take differential for the side of the equation

Assuming: 𝑟(𝑡) = 3𝑡 3

𝑟′(𝑡) = 9𝑡 2 ; 𝑟′′(𝑡) = 18𝑡; 𝑟′′′(𝑡) = 18

𝑑4 𝑟
=0
𝑑𝑡 4
𝑑5 𝑐 𝑑4 𝑐 𝑑3 𝑐 𝑑2 𝑐 𝑑𝑐
+ 4 + 3 + 2 + 3 = 0 + 3(18) + 2(18𝑡) + 9𝑡 2 + 3𝑡 3
𝑑𝑡 5 𝑑𝑡 4 𝑑𝑡 3 𝑑𝑡 3 𝑑𝑡
 𝑑5 𝑐 𝑑4 𝑐 𝑑3 𝑐 𝑑2 𝑐 𝑑𝑐
5
+ 4 4
+ 3 3
+ 2 3
+3 = 54 + 36𝑡 + 9𝑡 2 + 3𝑡 3
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡

11. A system is described by the following differential equation:

𝑑2 𝑥 𝑑𝑥
2
+2 + 3𝑥 = 1
𝑑𝑡 𝑑𝑡
With the initial conditions x(0)=1, x’(0)=-1 Show a block diagram of the system, giving its
transfer function and all pertinent inputs and outputs.

• Take the Laplace transform of the differential equation

𝑠 2 𝑥(𝑠) − 𝑠𝑥(0) − Ẋ(0) + 2[𝑠𝑥(𝑠) − Ẋ(0)] + 3𝑥(𝑠) = 𝑅(𝑠)

𝑠 2 𝑥(𝑠) − 𝑠𝑥(0) − Ẋ(0) + 2𝑠𝑥(𝑠) − 2𝑥(0) + 3𝑥(𝑠) = 𝑅(𝑠)

• Substitute its initial condition which is x (0) =1 and Ẋ(0) = −1

𝑠 2 𝑥(𝑠) − 𝑠 + 1 + 2𝑠𝑥(𝑠) − 2 + 3𝑥(𝑠) = 𝑅(𝑠)

• Isolate the terms with x(s).

𝑠 2 𝑥(𝑠) + 2𝑠𝑥(𝑠) + 3𝑥(𝑠) = 𝑅(𝑠) + 𝑠 + 1

• Factor out the left side of the equation

(𝑠 2 + 2𝑠 + 3)𝑥(𝑠) = 𝑅(𝑠) + 𝑠 + 1

• From the equation above, we can solve for x(s)

𝑅(𝑠) + 𝑠 + 1
𝑥(𝑠) =
𝑠 2 + 2𝑠 + 3
𝑅(𝑠) 𝑠+1
𝑥(𝑠) = + 2
𝑠2 + 2𝑠 + 3 𝑠 + 2𝑠 + 3

1
Block diagram show below where R(s)=𝑠

𝑅(𝑠) + 1 𝑥(𝑠)
2
𝑠 + 2𝑠 + 3

𝑠+1