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Laplace Transform of
Periodic Functions
- Ayushi Bhagat
(19102A0014)
Aditi Shahasane (19102A0054)
Laplace Transform
The Laplace Transform is an operator that uses a definite
integral to change a function of one variable denoted by t to
another function denoted by s.
If f(t) is a function of t satisfying certain conditions, then the
definite integral is denoted as L[f(t)] , it is defined by
L[f(t)] = 0
∞
𝑒−𝑠𝑡. f(t)dt
Periodic Function
A function f(t) is said to be periodic if there exists constant
period T (T>0) such that f(t + T) = f(t) for all values of t.
f(t + 2T) = f(t + T + T) = f(t + T) = f(t)
In general, f(t + nT) = f(t) for all t, where n is an integer and T is
The period of the function.
Thus Formula for Laplace Transform of Periodic function is:
L[f(t)] =
𝟏
𝟏−𝒆−𝒂𝒔 𝟎
𝒂
𝒆−𝒂𝒕f(t)dt
Proof of Theorem
If f(t) is a periodic function with period T, then f(t + T) = f(t)
L[ f(t) ] = 0
∞
𝑒−𝑠𝑡
f(t)dt
= 0
𝑇
𝑒−𝑠𝑡 f(t)dt + 𝑇
2𝑇
𝑒−𝑠𝑡 f(t)dt + 2𝑇
3𝑇
𝑒−𝑠𝑡 f(t)dt + …
= 𝐼1 + 𝐼2 + 𝐼3 + …
Now Substitute t = T+ u in 𝐼2 … (i)
∴ 𝐼2 = 𝑇
2𝑇
𝑒−𝑠𝑡
f(t)dt
= 0
𝑇
𝑒−𝑠(𝑇+𝑢) 𝑓(𝑇 + 𝑢) du
= 𝑒−𝑠𝑇
0
𝑇
𝑒−𝑠𝑢
𝑓 𝑢 𝑑𝑢 𝑓 𝑇 + 𝑢 = 𝑓 𝑢
= 𝑒−𝑠𝑇 𝐼1
Diff (i) w.r.t,
1= 0+
𝑑𝑢
𝑑𝑡
dt= du
When t=T then u=0
When t=2T then
u=T
Similarly ,put t=2T+u …(ii)
∴ 𝐼3 = 2𝑇
3𝑇
𝑒−𝑠𝑡
𝑓 𝑡 𝑑𝑡
= 0
𝑇
𝑒−𝑠(2𝑇+𝑢)
𝑓 2𝑇 + 𝑢 𝑑𝑢
= 𝑒−2𝑠𝑇
0
𝑇
𝑒−𝑠𝑢
𝑓 𝑢 𝑑𝑢 [𝑓 2𝑇 + 𝑢 = 𝑓(𝑢)]
= 𝑒−2𝑠𝑇
𝐼1
On continuing the same way, we get
𝐼4 = 𝑒−3𝑠𝑇
𝐼1
𝐼5 = 𝑒−4𝑠𝑇
𝐼1
𝐼6 = 𝑒−6𝑠𝑇
𝐼1 & so on…
∴ L[f(t)] = 𝐼1 + 𝐼2 + 𝐼3+ 𝐼4 + … .
∴ L[f(t)] = 𝐼1 + 𝑒−𝑠𝑇
𝐼1+ 𝑒−2𝑠𝑇
𝐼1 + 𝑒−3𝑠𝑇
𝐼1 + …
But it is geometric progression (infinite series)
Here first term → 𝐼1 & common ratio → 𝑒−𝑠𝑇
[ r < 1]
The sum of infinite GP is
𝑎
1−𝑟
→
𝐼1
1− 𝑒−𝑠𝑇 =
1
1− 𝑒−𝑠𝑇 0
𝑇
𝑒−𝑠𝑡
f(t)dt
Diff (ii) w.r.t,
1= 0+
𝑑𝑢
𝑑𝑡
dt=du
When t=2T then
u=0
When t= 3T then
u=T
L[f(t)] =
1
1−𝑒−𝑠𝑇 0
𝑇
𝑒−𝑠𝑡
𝑓(𝑡 𝑑𝑡)
Find the Laplace transform of f(t)= | sin p t|, t >=0.
Sol: We first note that f [t +
П
𝑝
] = |sin p [t +
П
𝑝
]|
= |sin (pt+П)|= |sin pt|
‫؞‬ f(t) is a periodic function with period
П
𝑝
Now, L f(t) =
1
1− 𝑒−П𝑠/𝑝 0
П/𝑝
𝑒−𝑠𝑡
|sin pt|dt
=
1
1− 𝑒−П𝑠/𝑝 0
П/𝑝
𝑒−𝑠𝑡
sin pt dt [∵ sin pt > 0, for 0≤ t ≤
П
𝑝
]
=
1
1− 𝑒−П𝑠/𝑝 [
𝑒−𝑠𝑡
𝑠2+𝑝2 ( −s sin pt − pcos pt)] П/𝑝
0
Examples
𝑒𝑎𝑥
sin 𝑏𝑥 𝑑𝑥 =
𝑒𝑎𝑥
𝑎2+ 𝑏2 [ a sin 𝑏𝑥 - b cos 𝑏𝑥 ] + c
=
1
1− 𝑒−П𝑠/𝑝 .
1
𝑠2+𝑝2 [𝑒−𝑠П/𝑝
0 + 𝑝 − (0 − 𝑝) ]
=
1
𝑠2+𝑝2 .
1
1− 𝑒−П𝑠/𝑝 .p. (1+ 𝑒−П𝑠/𝑝)
=
𝑝
𝑠2+𝑝2 (
𝑒П𝑠/2𝑝+𝑒−П𝑠/2𝑝
𝑒П𝑠/2𝑝 − 𝑒−П𝑠/2𝑝 )
=
𝑝
𝑠2+𝑝2 cot h (
П𝑠
2𝑝
)
The function f(t)= |sin p t |is known as full –sine wave rectifier and its
graph
shown below.
Find laplace transform of f(t) = 1, for 0 ≤ t < a and f(t) = -1, a < t
< 2a and f(t) is periodic with period 2a.
Sol. Since f(t) is periodic with period 2a,
L[f(t)] =
1
1 − 𝑒−2𝑎𝑠 0
2𝑎
𝑒−𝑠𝑡
𝑓 𝑡 𝑑𝑡
=
1
1 − 𝑒−2𝑎𝑠 [ 0
𝑎
𝑒−𝑠𝑡
(1)𝑑𝑡 + 𝑎
2𝑎
𝑒−𝑠𝑡
−1 𝑑𝑡 ]
=
1
1 − 𝑒−2𝑎𝑠 [{-
𝑒−𝑠𝑡
𝑠
}𝑎
0
+ {
𝑒−𝑠𝑡
𝑠
}2𝑎
𝑎
]
=
1
𝑠
.
1
1 − 𝑒−2𝑎𝑠 . (1 − 𝑒−𝑎𝑠)
2
=
1
𝑠
.
1 − 𝑒−𝑎𝑠
1 + 𝑒−𝑎𝑠
=
1
𝑠
.
𝑒𝑎𝑠/2 − 𝑒−𝑎𝑠/2
𝑒𝑎𝑠/2+ 𝑒−𝑎𝑠/2
=
1
𝑠
tanh{
𝑎𝑠
2
}
The function is known as “square-wave” function and its graph is
shown below-
tanh 𝑥 =
𝑒𝑥 − 𝑒−𝑥
𝑒𝑥 + 𝑒−𝑥
Applications of Laplace transform
• Using laplace transform, we can change periodic function
of one variable denoted by t into another variable denoted
by s.
• The laplace transform is used to solve differential
equations and is extensively used in mechanical and
electrical engineering.
• The laplace transform reduces a linear differential
equation to an algebraic equation, which can be solved by
the formal rules of algebra.
 Applied Mathematics – III By G.V.
Kumbhojkar
 Higher Engineering Mathematics by
B. S. Grewal
 https://youtu.be/1cuB7fm0M4Q
 Class notes
EM3 mini project Laplace Transform

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EM3 mini project Laplace Transform

  • 1. Laplace Transform of Periodic Functions - Ayushi Bhagat (19102A0014) Aditi Shahasane (19102A0054)
  • 2. Laplace Transform The Laplace Transform is an operator that uses a definite integral to change a function of one variable denoted by t to another function denoted by s. If f(t) is a function of t satisfying certain conditions, then the definite integral is denoted as L[f(t)] , it is defined by L[f(t)] = 0 ∞ 𝑒−𝑠𝑡. f(t)dt
  • 3. Periodic Function A function f(t) is said to be periodic if there exists constant period T (T>0) such that f(t + T) = f(t) for all values of t. f(t + 2T) = f(t + T + T) = f(t + T) = f(t) In general, f(t + nT) = f(t) for all t, where n is an integer and T is The period of the function. Thus Formula for Laplace Transform of Periodic function is: L[f(t)] = 𝟏 𝟏−𝒆−𝒂𝒔 𝟎 𝒂 𝒆−𝒂𝒕f(t)dt
  • 4. Proof of Theorem If f(t) is a periodic function with period T, then f(t + T) = f(t) L[ f(t) ] = 0 ∞ 𝑒−𝑠𝑡 f(t)dt = 0 𝑇 𝑒−𝑠𝑡 f(t)dt + 𝑇 2𝑇 𝑒−𝑠𝑡 f(t)dt + 2𝑇 3𝑇 𝑒−𝑠𝑡 f(t)dt + … = 𝐼1 + 𝐼2 + 𝐼3 + … Now Substitute t = T+ u in 𝐼2 … (i) ∴ 𝐼2 = 𝑇 2𝑇 𝑒−𝑠𝑡 f(t)dt = 0 𝑇 𝑒−𝑠(𝑇+𝑢) 𝑓(𝑇 + 𝑢) du = 𝑒−𝑠𝑇 0 𝑇 𝑒−𝑠𝑢 𝑓 𝑢 𝑑𝑢 𝑓 𝑇 + 𝑢 = 𝑓 𝑢 = 𝑒−𝑠𝑇 𝐼1 Diff (i) w.r.t, 1= 0+ 𝑑𝑢 𝑑𝑡 dt= du When t=T then u=0 When t=2T then u=T
  • 5. Similarly ,put t=2T+u …(ii) ∴ 𝐼3 = 2𝑇 3𝑇 𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 = 0 𝑇 𝑒−𝑠(2𝑇+𝑢) 𝑓 2𝑇 + 𝑢 𝑑𝑢 = 𝑒−2𝑠𝑇 0 𝑇 𝑒−𝑠𝑢 𝑓 𝑢 𝑑𝑢 [𝑓 2𝑇 + 𝑢 = 𝑓(𝑢)] = 𝑒−2𝑠𝑇 𝐼1 On continuing the same way, we get 𝐼4 = 𝑒−3𝑠𝑇 𝐼1 𝐼5 = 𝑒−4𝑠𝑇 𝐼1 𝐼6 = 𝑒−6𝑠𝑇 𝐼1 & so on… ∴ L[f(t)] = 𝐼1 + 𝐼2 + 𝐼3+ 𝐼4 + … . ∴ L[f(t)] = 𝐼1 + 𝑒−𝑠𝑇 𝐼1+ 𝑒−2𝑠𝑇 𝐼1 + 𝑒−3𝑠𝑇 𝐼1 + … But it is geometric progression (infinite series) Here first term → 𝐼1 & common ratio → 𝑒−𝑠𝑇 [ r < 1] The sum of infinite GP is 𝑎 1−𝑟 → 𝐼1 1− 𝑒−𝑠𝑇 = 1 1− 𝑒−𝑠𝑇 0 𝑇 𝑒−𝑠𝑡 f(t)dt Diff (ii) w.r.t, 1= 0+ 𝑑𝑢 𝑑𝑡 dt=du When t=2T then u=0 When t= 3T then u=T L[f(t)] = 1 1−𝑒−𝑠𝑇 0 𝑇 𝑒−𝑠𝑡 𝑓(𝑡 𝑑𝑡)
  • 6. Find the Laplace transform of f(t)= | sin p t|, t >=0. Sol: We first note that f [t + П 𝑝 ] = |sin p [t + П 𝑝 ]| = |sin (pt+П)|= |sin pt| ‫؞‬ f(t) is a periodic function with period П 𝑝 Now, L f(t) = 1 1− 𝑒−П𝑠/𝑝 0 П/𝑝 𝑒−𝑠𝑡 |sin pt|dt = 1 1− 𝑒−П𝑠/𝑝 0 П/𝑝 𝑒−𝑠𝑡 sin pt dt [∵ sin pt > 0, for 0≤ t ≤ П 𝑝 ] = 1 1− 𝑒−П𝑠/𝑝 [ 𝑒−𝑠𝑡 𝑠2+𝑝2 ( −s sin pt − pcos pt)] П/𝑝 0 Examples 𝑒𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 = 𝑒𝑎𝑥 𝑎2+ 𝑏2 [ a sin 𝑏𝑥 - b cos 𝑏𝑥 ] + c
  • 7. = 1 1− 𝑒−П𝑠/𝑝 . 1 𝑠2+𝑝2 [𝑒−𝑠П/𝑝 0 + 𝑝 − (0 − 𝑝) ] = 1 𝑠2+𝑝2 . 1 1− 𝑒−П𝑠/𝑝 .p. (1+ 𝑒−П𝑠/𝑝) = 𝑝 𝑠2+𝑝2 ( 𝑒П𝑠/2𝑝+𝑒−П𝑠/2𝑝 𝑒П𝑠/2𝑝 − 𝑒−П𝑠/2𝑝 ) = 𝑝 𝑠2+𝑝2 cot h ( П𝑠 2𝑝 ) The function f(t)= |sin p t |is known as full –sine wave rectifier and its graph shown below.
  • 8. Find laplace transform of f(t) = 1, for 0 ≤ t < a and f(t) = -1, a < t < 2a and f(t) is periodic with period 2a. Sol. Since f(t) is periodic with period 2a, L[f(t)] = 1 1 − 𝑒−2𝑎𝑠 0 2𝑎 𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 = 1 1 − 𝑒−2𝑎𝑠 [ 0 𝑎 𝑒−𝑠𝑡 (1)𝑑𝑡 + 𝑎 2𝑎 𝑒−𝑠𝑡 −1 𝑑𝑡 ] = 1 1 − 𝑒−2𝑎𝑠 [{- 𝑒−𝑠𝑡 𝑠 }𝑎 0 + { 𝑒−𝑠𝑡 𝑠 }2𝑎 𝑎 ] = 1 𝑠 . 1 1 − 𝑒−2𝑎𝑠 . (1 − 𝑒−𝑎𝑠) 2
  • 9. = 1 𝑠 . 1 − 𝑒−𝑎𝑠 1 + 𝑒−𝑎𝑠 = 1 𝑠 . 𝑒𝑎𝑠/2 − 𝑒−𝑎𝑠/2 𝑒𝑎𝑠/2+ 𝑒−𝑎𝑠/2 = 1 𝑠 tanh{ 𝑎𝑠 2 } The function is known as “square-wave” function and its graph is shown below- tanh 𝑥 = 𝑒𝑥 − 𝑒−𝑥 𝑒𝑥 + 𝑒−𝑥
  • 10. Applications of Laplace transform • Using laplace transform, we can change periodic function of one variable denoted by t into another variable denoted by s. • The laplace transform is used to solve differential equations and is extensively used in mechanical and electrical engineering. • The laplace transform reduces a linear differential equation to an algebraic equation, which can be solved by the formal rules of algebra.
  • 11.  Applied Mathematics – III By G.V. Kumbhojkar  Higher Engineering Mathematics by B. S. Grewal  https://youtu.be/1cuB7fm0M4Q  Class notes