9. Laplace transform of periodic functions

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7.Laplace transform of functions obtained with multiplication by

tⁿ

Definition 7.1

Leibniz's rule states that integration and differentiation are interchangeable i.e.

d dx∫

y1

y₂

f(x , y)dy=∫

y1

y₂

∂ f(x , y)

∂ x dy

with the condition that ^f⁽^{x , y}⁾ and ^{∂ f}⁽_{∂ x}^{x , y}⁾ are continuous within the interval ( ^x¹^{, x}²^¿ for the variable ^x and the corresponding interval (¿¿1^y, y₂)

¿

for the variable ^{y .} In general,

f(x , y)dy=¿ ∫

y1(x) y2(x)

∂ f(x , y)

∂ x dy+¿

d dx ∫

y1(x) y₂(x)

¿

f(x , y₂(x))d y₂

dx −f(x , y₁(x))d y₁ dx

In this problem ,the limits on ^t are independent of ^s . ∴ the last 2 terms are 0 .

tⁿf(t) } L¿

F(s)=L f(t)=∫

0

∞

e^−stf(t)dt

taking the derivative with respect to ^s on both the sides , we obtain

dF(s) ds = d

ds∫

0

∞

e^−stf(t)dt

dF(s) ds =∫

0

∞ ∂

∂ s e^−stf (t)dt ... Leibniz rule

dF(s) ds =−∫

0

∞

e^−stf (t)tdt=−L{tf (t) } D.U.

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If we consider the derivative once more , we obtain

d²F d s²=∫

0

∞

e^−stf (t)t²dt = ^L{t²^f⁽^t^)}

We can consider the higher order derivatives and obtain ,

L{tⁿf(t)}=(−1)ⁿdⁿF(s)

d sⁿ (12)

Example 7.1:

Determine ^L^{{t e}⁻⁴^t^}

From the above theorem , ^f⁽^t^)=e⁻⁴^t^;n=1 , ^F⁽^s)=_s+¹₄

d F(s)

ds = −1

(s+4)² therefore ^L^{^{t e}⁻⁴^t^}^=¿ ₍_s+⁻¹₄₎² Example 7.2:

Determine ^L{t²^{sin 2t}^}

In this case we need to consider the second derivative of ^L^{{sin 2t}^}.

d²

d s²

[

^(s²²⁺⁴⁾

]

^¿^ds^d

[

⁽^s⁻²²⁺⁴^s⁾²

]

⁼⁽^s⁻²²⁺⁴⁾²⁺⁽^s⁸²⁺^s⁴²⁾³

L{t²sin 2t}=¿ −2

(s²+4)²⁺

8s²

(s²+4)³ (13) Example 3:

Find ^L{t²^cos^at^}=⁽⁻¹⁾²_ds^d²²

[

^s²⁺^s^a²

]

D.U.

8s³

(a²+s²)³

−6s

(a²+s²)² or

L{t²cosat}= 8s³

(a²+s²)³⁻

6s

(a²+s²)² (14)

8. Laplace transform of functions which are divided by ^t

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L{f(t) t }=∫

s

∞

F(u)du (15)

Proof: ^F⁽^s) is ^L{f⁽^t^)}

Using the definition of Laplace transform, we obtain

∫0

∞

f(t)e^−tu [¿dt]du

∫

s

∞

F(u)du=∫

s

∞

¿

due to the convergence of the function we can change the order of integration to obtain

¿¿

f(t)e^−tudu

∫0

∞

∫s

∞

¿dt¿ = ∫

0

∞

−f(t)e^−st dt

−t = ^L^{^F_t^(t⁾ }

since the value of the integral at the upper limit is ⁰ . Example 8.1:

Evaluate ^L

{

^∫⁰^x ^sin^t ^t^dt

}

Comparing with the above formula ,

f(t)=sint and ^L

{

^sint ^t

}

⁼^∫^∞_s _u²¹₊₁^du=^¿ ^π²⁻^tan⁻¹^s=tan⁻¹¹^s

We next apply the previous theorem

L

{

^∫⁰^x ^f⁽^t⁾^dt

}

⁼^F^(s)^s

f(t)=sint

t and therefore

L

{

^∫⁰^x ^sin^t ^t^dt

}

⁼¹^s ^tan⁻¹¹^s (16)

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9. Laplace transform of periodic functions

We consider a periodic function ^f^(t) with a period ^T . Thus

f(t+T)=f(t)

substituting the expression for LT we have,

∫0

∞

f(t)e^−stdt=∫

0 T

f(t)e^−stdt +

f(t)e^−stdt+¿∫

2T 3T

f(t)e^−stdt+..

∫T 2T

¿

in the second integral, we substitute,

t−T=u ;t=u+T , dt=du; and the limits change to ⁽⁰^{, T}⁾ . Thus the second integral is

∫0 T

f(u+T)e^−s^(u^+T)du=e^−sT∫

0 T

f(u)e^−sudu as ^f^(u+T^)=f^(u)

for the next integral we substitute ^t^−2T⁼^u; and the integral reduces to

∫0 T

f(u+2T)e^−s⁽^u^+2T⁾du=e⁻²^sT∫

0 T

f(u)e^−sudu

similarly by substituting in the following integrals we obtain an infinite geometric series of the form

1+e (¿¿−sT+e⁻²^sT+e⁻³^sT+…)∫

0 T

f(u)e^−sudu=[∫

0 T

f(u)e^−sudu] 1 1−e^−sT

¿

Thus Laplace transform of a periodic function ^f^(t⁾ with period ^T is

L{f(t)}= 1 1−e^−sT∫

0 T

f (u)e^−sudu (17)

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Example 9.1:

Determine the Laplace transform of the periodic function ^f ⁽^t⁾ which has a period of 4. ^f⁽^t⁾⁼²^for⁰^{≤ t ≤}²^;f⁽^t⁾⁼⁻²^for²^{≤ t ≤}⁴

We evaluate ∫

0 4

f(t)e^−stdt=∫

0 2

2e^−stdt+∫

2 4

−2e^−stdt

= ²⁽^1−e_s ⁻²^s⁾⁺²_s⁽^e⁻⁴^s^−e⁻²^s⁾

Hence the Laplace transform of the given square wave is

2(1−2e⁻²^s+4e⁻⁴^s) 1−e⁻⁴^s

Example 9.2:

Determine the Laplace transform of a saw tooth wave with period 2

f(t)=t for0≤ t ≤2 ;

Fig 3 saw tooth wave with period 2

∫

0 2

t e^−stdt=2e^−2s

−s +∫

0 2 e^−st

s dt=2e⁻²^s

−s −[e⁻²^s−1 s² ]

Thus the Laplace transform of a sawtooth wave of period 2 is given by

2e^−2s

−s −[e⁻²^s−1 s² ] 1−e⁻²^s

Example 9.3 :

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Find the Laplace transform of the half wave rectified wave

f(t)=Vsin ωt for ⁰^{≤ ωt ≤ π}

f(t)=0 for ^{π ≤ ωt ≤2}^π

Fig 4 Output of half wave rectifier

∫

0 π/ω

e^{−t(s−iω)}−e^−t^(s+iω) 2i

Vsinωt e^−stdt+0=V[¿¿dt]

f(t)e^−stdt=∫

0 π ω

¿

∫0 T

¿

V

2i

[

^es−iω^−t^(s−iω)−e^−t^(s^+iω) s+iω

]

π/ω

0

=V

2i

[

¹⁻^es−iω^{−π(s−iω)/}^ω−1−e^−π^(s^+iω)/^ω s+iω

]

V

2i

[

²⁻^s−iω^e^πs/^ω⁻^2−e^s+iω^−πs/^ω

]

^¿₂^V_i

[

⁴^iω+^s

⁽

^e^−πs^ω ^−e^s²^πs⁺^ω

⁾

^ω^−iω(e² ^−πs^ω ^+e^πs^ω⁾

]

Thus the Laplace transform of the rectified half wave sine wave is

V

2i

[

⁴^iω+s

⁽

^e^−πs^ω ⁻^s^e²^πs⁺^ω

⁾

^ω^−iω(e² ^−πs^ω ^+e^πs^ω⁾

]

1−e⁻²^sπ/^ω

10. More Laplace transforms Example10.1:

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Determine ^L^{^sin√^t^}

We consider the sine series ^sin√^t=√^t−⁽√^t⁾³

3! +(√^t⁾⁵

5! −…

L{sin√^t}=L

{

^√^t⁻⁽^√3!^t⁾³+(√t)⁵

5! −…

}

L { √^t^}=Γ⁽³₂^{) /s}

3

2 ; _L

{

⁽^√3^t!⁾³

}

⁼ ^Γ⁽

5 2)

3! s⁵^/2 ; _L

{

⁽^√5^t!⁾⁵

}

⁼^Γ⁽

7 2) 5! s⁷^/2

^Γ

(

¹2

)

⁼^√^π where ^Γ represents the gamma function which is defined by the following integral.

^Γ⁽ⁿ⁺¹⁾⁼∫

0

∞

xⁿ⁻¹e^−xdx and ^Γ⁽ⁿ⁺¹⁾⁼^{n Γ}⁽ⁿ⁾ Substituting these values in eqn (1) ,

L { ^sin^√^t^}=₂^√_s^π3/2−√^π

s^5/2.1 2.3

2. 1 3!+√^π

s^7/2.1 2.3

2.5 2. 1

5!−…

^L { ^sin^√^t^}=¿ ₂^√_s^π^3/2

[

¹⁻⁴¹^s⁺³²¹^s²^−…

]

⁼ ²^√^s^π³^/2

[

¹⁻⁴¹^s⁺⁽⁴^s)¹²^2!^−…

]

^L { ^sin^√^t^}=¿ ₂^√_s^π3/2e

−1

4s (18)

11. Evaluation of integrals using Laplace transforms Example11.1:

Evaluate ∫

0

∞ sintx

x dx using Laplace transform.

L

{

^sint ^t

}

⁼^∫^∞_s _u²¹₊₁^du=¿ ^π²^−tan⁻¹^s=^tan⁻¹¹^s We can change the variable ^tx=^y and the result is unaltered.

s=0 , ∫

0

∞ sintx

x dx = ^π₂^−¿ ^tan⁻¹⁰^=¿ ^π₂

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L∫

0

∞ sintx

x dx=¿ π

2 (19)

12. Laplace Transforms of Integrals of Functions 1. Determine the Laplace transform of ∫

t

∞ e^−p p dp

Let ^f⁽^t⁾⁼∫

t

∞ e⁻^p

p dp

Taking the derivative on both sides and considering the derivative of an integral of a function is the function, we obtain , ^f^'⁽^t^)=¿ ^e_t^−t ;

t f^'(t)=e^−t

L{^{t f}^'^(t)}⁼ ¹

s+1 ; using ^L^{^tⁿ^g^(t⁾^}⁼⁽⁻¹⁾ⁿ_{d s}^dⁿnG(s) ^G^(s^)=¿ ^L^{g^(t^{) }}

Thus ^L^{^{t f}^'⁽^t⁾^}⁼^−d_ds ^L^{^f^'^(t⁾^}⁼^−d_ds ^[^sF^(s^)−f⁽⁰⁾^] = ⁻_ds^d^[^sF⁽^s)] as ^f⁽⁰⁾ is not a function of ^s .

−d

ds [sF(s)] = _s+1¹

integrating both sides with respect to ^s , ^−sF⁽^s)=ln_s+1¹ or

F(s)=1

sln(s+1) L∫

t

∞ e⁻^p

p dp=¿ 1

sln(s+1) (20)

Exercises:

1. Determine the Laplace transform of the following functions.

a) ^sinh²⁴^t

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b) ^sinh^⁡at^/^t c) ⁽^e^t⁾⁽^t⁺¹⁾ d) ^cos²⁽^2t⁾

e) ^e^t^(t⁺¹⁾²

2. Determine the Laplace transform of the following functions.

a) ^f⁽^t⁾^=¿ ( ^t−3^¿^u^(t⁻³⁾ ^t>⁰ b) ^f⁽^t⁾⁼⁰ for ^0<^{t ≤}³

f(t)=1, 3≤t ≤5

f(t)=0,t ≥5

3. Determine the Laplace transform of a step function.

f(t)=a for ^{t ≥}²^{; f}⁽^t⁾⁼⁰ for ^{0<t ≤ a} 4. Find Laplace transform of

(a) ^f^(t⁾⁼^e^3t_t⁻¹ (b) ^f⁽^t⁾⁼²^{t e}^−t^sin²^t

5. Evaluate the following integrals using Laplace transforms a) ∫

0

∞ e⁻²^tsin 2t

t dt b) ∫

0

∞

t e⁻²^tcost dt 6. Draw the output of a full wave rectifier and determine its Laplace transform given that ^f⁽^t⁾⁼^|^sin^ωt^| , ^a is the period of the output.

7. Verify the initial value theorem for the following functions a) ^t²^−cost b) ^t^{sin 2}^t

8. Verify the final value theorem for the following functions a) ^e^−t⁺⁴ b) ^e^−t^{cos 2}^t

9. Evaluate the Laplace transform of ^cos³^t .

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10. Draw a triangular wave and determine the Laplace transform of the function representing this wave. Treat the time period of the wave as 2a and its amplitude as 1.

11. Using ^L { ^sin^√^t^}=¿ ₂^√_s^π3/2e

−1

4s and ^L^{^f^'^(t⁾^}⁼^sF⁽^s⁾^−f⁽⁰⁾ , determine ^L^{^{2 cos}^√^t

√^t }.