1. MINISTRY OF SCIENTIFIC EDUCATION AND HIGHER RESEARCHES
NORTHERN TECHNICAL UNIVERSITY
ENGINEERING TECHNICAL COLLEGE / MOSUL
DEPARTMENT OF COMPUTER TECHNOLOGY
1
ENGINEERING ANALYSIS LECTURE
DEPARTMENT OF COMPUTER
TECHNOLOGY –THIRD CLASS
2018 -2019
ARJUWAN MOHAMMED ABDULJAWAD
ALJAWADI
LECTURER
2. 2
- ENGINEERING ANALYSIS
One of important transforms used in linear- system analysis. It is
named in honor of the great French mathematician, Pierre Simon De
Laplace (1749-1827).
3. 3
- Purpose of Laplace Transform
• To convert from one type of operation to another
operations of different types in more simple form.
• A well-known technique for solving differential
equations.
5. - THE TIME DOMAIN SIGNAL IS CONTINUOUS , EXTENDS
TO:
1. POSITIVE AND NEGATIVE INFINITY.
2. PERIODIC OR APERIODIC SIGNAL
5
- Laplace Transform Definition:
The Laplace transform F(s) of a time function F (t) is given by the
integral:
6. 6
This definition is called the bilateral, or two-sided, Laplace transform—hence, the
subscript b. Notice that the bilateral Laplace transform integral becomes the Fourier
transform integral if is replaced by (𝑗𝑤 ) . The Laplace transform variable is
complex 𝑠 = 𝛼 + 𝑗𝑤, we can rewrite (1) as:
𝐹 𝑠 =
−∞
∞
𝑓(𝑡)𝑒−(𝜎+𝑗𝜔)𝑡
𝑑𝑡
=
−∞
∞
𝑓(𝑡)𝑒−𝜎𝑡
𝑒−𝑗𝑤
𝑑𝑡
7. 7
𝑓 𝑡 𝑖𝑠 𝑧𝑒𝑟𝑜 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 < 0 Thus the first integral in above equation
is zero.
The resulting transform, called the unilateral, or single-sided Laplace
transform, is given by:
8. 8
THE LAPLACE-TRANSFORM VARIABLE S IS COMPLEX, AND WE
DENOTE ITS REAL PART AS ∝ AND ITS IMAGINARY PART AS
𝑗𝑤 THAT IS: 𝑠 = ∝ +𝑗𝑤
The S-plane
9. 9
Some Elementary Functions F(t) and their Laplace Transform
F(t) F(s)
U(t) 1
s
t 1/s2
tn n!/ sn+1
e−at
1
s + a
eat
1
s − a
sin wt w/s2+w2
cos at s/s2+a2
sin hat w/s2- w2
cos hat s/s2- w2
10. 10
- Laplace Transform of some important functions
1. Laplace Transform of a unit –step function 𝑓(𝑡) = 1.
12. 12
3. Laplace Transform of 𝑓(𝑡) = 𝑡𝑛
Where n= (1,2,3,4,…………………………….)
f s = 0
∞
e−st tn .dt
udv = uv − v . du
U=tn , du = (𝑛 𝑡𝑛−1) , dv= e−st , v=-
e−st
s
f s = − e−st.tn/s |- 0
∞
𝑛
𝑒−𝑠𝑡
𝑠
.tn-1.dt
13. 13
The first term limits will be form (0 to ∞) by substituting it yields zero while the second
term by substitution we get:
f s = 0
∞ −ne−st
s
n tn-1.dt
u=ntn-1 , du=n(n-1)tn-2 , dv=
e−st
s
, v= - e−st/s2
f s = ne−st
tn-1 /s2 | - 0
∞
−n(n − 1)tn-2/s2 .e−st
. dt
f s = 0
∞
n(n − 1)e−st tn-2 /s2 .dt
Then after n times integration we have :
f s =
0
∞
n!
sn
tn−ne−st. dt
f s =
0
∞
n!
sn
e−st. dt
=
n!
sn+1 e−st
| =
n!
sn+1
19. 19
3. Multiplication by exponential:
𝑓1 𝑡 = 𝑓 𝑡 𝑒−𝑎𝑡
ℒ 𝑓 𝑡 𝑒−𝑎𝑡
=
0
∞
𝑓 𝑡 𝑒−𝑎𝑡
𝑒−𝑠𝑡
0
∞
𝑓 𝑡 𝑒− 𝑠+𝑎 𝑡
. 𝑑𝑡 Then 𝑓(𝑠 + 𝑎)
The transform ℒ 𝑓 𝑡 𝑒−𝑎𝑡
is thus the same as ℒ 𝑓 𝑡
with everywhere in the result replaced by (𝑠 + 𝑎).
25. 25
- Example: Use Laplace transform in solving for the current in an electric circuit. Consider the RL-circuit in the
following figure, where 𝑉 is constant. The loop equation for this circuit is given by:
𝐿.
𝑑𝑖(𝑡)
𝑑𝑡
+ 𝑅𝑖 𝑡 = 𝑉𝑢 𝑡 𝑓𝑜𝑟 𝑡 > 0
26. 26
Since the switch is closed at 𝑡 = 0 . The Laplace transform of this equation yields:
𝐿 𝑠𝐼 𝑠 − 𝑖 0 + 𝑅𝐼 𝑠 =
𝑉
𝑠
The initial current is zero 0 = 0 , 𝑖(𝑡) is zero for negative time since the switch is open for 𝑡 < 0 and the
current in an inductance cannot change instantaneously.
𝐼 𝑠 =
𝑉
𝑠 𝐿𝑠 + 𝑅
=
𝑉
𝐿
𝑠 𝑠 +
𝑅
𝐿
𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 𝐿
Solve with partial fraction:
𝐼 𝑠 =
𝑉
𝐿
𝑠 𝑠 +
𝑅
𝐿
=
𝑎
𝑠
+
𝑏
𝑠 +
𝑅
𝐿
=
𝑎𝑠 + 𝑎
𝑅
𝐿
+ 𝑏𝑠
𝑠 𝑠 + 𝑅
𝐿
30. 30
Since 𝜏 is the variable of integration and can be replaced with , the integral on the right side is
𝑓(𝑠) , hence the Laplace transform of the shifted time function is given by:
ℒ 𝑓 𝑡 − 𝑡0 𝑢 𝑡 − 𝑡0 = 𝑒−𝑡0𝑠𝑓(𝑠)
𝑡0 ≥ 0 And ℒ 𝑓 𝑡 = 𝑓(𝑠). This relationship called the real translation, properly applies for a
function of the type in the figure:
It is necessary that the function to be zero in time less than 𝑡0 , the amount of the shift.
33. 33
∴ ℒ[
0
𝑡
𝑓 𝜏 . 𝑑𝜏] =
1
𝑠 0
∞
𝑓 𝑡 𝑒−𝑠𝑡
. 𝑑𝑡
∴
𝑓(𝑠)
𝑠
- Example:
Find the Laplace transform for 𝑓 𝑡 = 0
∞
𝑠𝑖𝑛𝑡. 𝑑𝑡
By integration the sine the result will be cosine and have the Laplace
𝑠
𝑠2+1
By using the Laplace transform properties, the result will be ℒ
1
𝑠2+1
Then the Laplace transform for integration the sine is
1
𝑠2+1
𝑠
Which leads the result to
𝑠
𝑠2+1
34. 34
8.Initial Value Theorem
The initial value 𝑓(0) of the function 𝑓(𝑡) whose L.T is 𝑓(𝑠) is:
𝑓 0 = lim
𝑡→0
𝑓(𝑡) = lim
𝑠→∞
𝑠𝑓(𝑠)
- Example: For the function 𝑓 𝑡 = 3𝑒−2𝑡, prove the initial – value theorem:
ℒ3𝑒−2𝑡 =
3
𝑠2 + 2
𝑓 0 = lim
𝑡→0
3𝑒−2𝑡 = lim
𝑠→∞
𝑠
3
𝑠2 + 2
3 = 3
According to Lobital rule in limits that states when there is
∞
∞
then the limit is taking the
derivative of the function of both the nominators and denominators which is in this case
3
1
35. 35
9. Final value theorem:
The final value of the function 𝑓(𝑡) whose L.T is𝑓 ∞ =
lim
𝑡→∞
𝑓(𝑡) = lim
𝑠→0
𝑠𝑓(𝑠)
- Example: For the function 𝑓 𝑡 = 3𝑒−2𝑡
, prove the final –
value theorem:
𝑓 ∞ = lim
𝑡→∞
3𝑒−2𝑡 = lim
𝑠→0
𝑠
3
𝑠2 + 2
0 = 0