ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
WEEK 7 | 24 FEBRUARY 2024
β’
The natural logarithm of a complex number can be obtained when it is expressed in exponential form, where
π§ = ππ ππ :
π₯π§(π) = π₯π§(π) + π(π½ + ππ
π) ; π ∈ β€
or
π = ππ(π½+ππ
π) ; π ∈ β€
π
π½ = πππ§−π ( ) ; π = √ππ + ππ
π
β’
If π = π, then π is the principal value.
TWO TYPES OF LOGARITHMS
1.) Common (Briggsian) Logarithm
β’
Notation: π₯π¨π π (π) ; π ∈ β€
β’
Base: ππ, i.e., π₯π¨π ππ (π)
2.) Natural Logarithm
β’
Notation: π₯π§(π)
β’
Base: π = 2.71828 … , i.e., π₯π¨π π (π) = π₯π§(π)
PROPERTIES OF LOGARITHM
1.) log π (π π₯ ) = π₯ ; π ∈ β€
5.) ln(π₯ π ) = π ln(π₯) ; π ∈ β€
2.) log π (π π₯ ) = ln(π π₯ ) = π₯
6.) log π (π₯π¦) = log π (π₯) + log π (π¦)
3.) π ln(π₯) = π₯
7.) log π ( ) = log π (π₯) − log π (π¦)
π₯
π¦
4.) 10log(π₯) = π₯
Examples:
β
π = π₯π§(π∠ππ°)
ln(π§) = ln(π) + π(2ππ) ; π ∈ β€
π
)
π
)
Eu(6∠30°) = 6π π(30°)(180° = 6π π( 6
π
π
)
ln (6π π( 6 ) = ln(6) + π ( + 2ππ) ; π ∈ β€
6
π
∴ π₯π§(π∠ππ°) = π. πππ + π ( + ππ
π) ; π ∈ β€
π
β
Take the natural logarithm of π = (π + ππ)π .
π₯π§(π) = π₯π§[(π + ππ)π ]
4 ln(1 + π2) = 4 ln(2.24π π(1.11+2ππ) ) = 4[ln(2.24) + π(1.11 + 2ππ)] = 3.22 + π(4.44 + 8ππ)
∴ π₯π§(π) = π. ππ + π(π. ππ + ππ
π) ; π ∈ β€
β
Evaluate π₯π¨π π−π (π + π√π)
πΏππ‘ π = log1−π (1 + π√3)
β΅ π logπ (π₯) = π₯
∴ (1 − π)π = (1 − π)log1−π (1+π√3) ⇒ (1 − π)π = 1 + π√3
ln[(1 − π)π ] = ln(1 + π√3)
π ln(1 − π) = ln(2∠60°)
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
π
π
ln(2) + π ( + 2ππ)
3
π=
=
= −0.7951 + π(1.2304 + 2ππ) ; π ∈ β€
π
π
π(− +2ππ)
ln (1.41π 4
) ln(1.41) + π (− 4 + 2ππ)
ln (2π π( 3 +2ππ) )
∴ π΅ = π₯π¨π π−π (π + π√π) = π. ππππ∠πππ. ππππ°
β
Determine the general value of (π + ππ)(π+ππ)
πΏππ‘ π = (3 + π4)(3+π4)
β΅ π₯π§(ππ ) = π π₯π§(π)
∴ ln(π) = ln[(3 + π4)(3+π4) ]
ln(π ) = (3 + π4) ln(3 + π4) = (3 + π4) ln(5π π(0.93+2ππ) ) = (5∠53.13°)[ln(5) + π(0.93 + 2ππ)]
ln(π) = (5∠53.13°)(1.86∠30.02°) = 9.3∠83.15° = 1.11 + π9.23
π ln(π) = π (1.11+π9.23)
π = π 1.11 π π9.23 = 3.03π π9.23+2ππ = 3.03∠528.84° + (360°π)
∴ π΅ = (π + ππ)(π+ππ) = π. ππ∠πππ. ππ° + (πππ°π) ; π ∈ β€
EULER’S THEOREM
ππ¨π¬ π½ + π π¬π’π§ π½ = (
β’
πππ½ + π−ππ½
πππ½ − π−ππ½
) + π(
)
π
ππ
Trigonometric Functions of Complex Numbers
π ππ + π −ππ
cos π =
2
sin π =
π ππ − π −ππ
π2
π ππ − π −ππ
tan π = −π ( ππ
)
π + π −ππ
β’
arccos π = −π ln (π ± √π 2 − 1)
arctan π = −π ln (√
β’
csc π =
1
π2
= ππ
sin π π − π −ππ
sec π =
1
2
=
cos π π ππ + π −ππ
Inverse Trigonometric Functions of Complex Numbers
arcsin π = −π ln (ππ ± √1 − π 2 )
β’
π ππ + π −ππ
cot π = π ( ππ
)
π − π −ππ
1 + ππ
)
1 − ππ
arccot π = −π ln (√
arccsc π = −π ln (
π+π
)
π−π
π ± √π 2 − 1
)
π
arcsec π = −π ln (
1 ± √1 − π 2
)
π
Hyperbolic Functions of Complex Numbers
π π + π −π
π π − π −π
sinh π =
π π − π −π
2
cosh π =
π π + π −π
2
csch π =
1
2
=
sinh π π π − π −π
tanh π =
π π − π −π
π π + π −π
sech π =
1
2
=
cosh π π π + π −π
coth π =
Inverse Hyperbolic Functions of Complex Numbers
arcsinh π = ln (π + √π 2 + 1)
arccoth π =
1
π+1
ln (
)
2
π−1
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
arccosh π = ln (π ± √π 2 − 1)
arctanh π =
β’
arccsch π = ln (
1
1+π
ln (
)
2
1−π
arcsech π = ln (
1 ± √1 − π 2
)
π
Hyperbolic Function Identities
cosh2 π − sinh2 π = 1
sinh(π ± π½) = sinh π cosh π½ ± cosh π sinh π½
sech2 π + tanh2 π = 1
cosh(π ± π½) = cosh π cosh π½ ± sinh π sinh π½
coth2 π − csch2 π = 1
β’
1 ± √1 + π 2
)
π
tanh(π ± π½) =
tanh π ± tanh π½
1 ± tanh π tanh π½
Relations between Hyperbolic and Trigonometric Functions
sin ππ = π sinh π
sinh ππ = π sin π
cos ππ = cosh π
cosh ππ = cos π
tan ππ = π tanh π
tanh ππ = π tan π
Examples:
β
Evaluate using Euler’s Theorem: ππ¨π¬(π. πππ + ππ. πππ)
β΅ cos(π ± π½) = cos π cos π½ β sin π sin π½ ; π = 0.573 ; π½ = π0.783
∴ cos(0.573 + π0.783) = cos(0.573) cos(π0.783) − sin(0.573) sin(π0.783)
radians → degrees
180°
0.573 (
) = 32.83°
π
β΅ cos ππ = cosh π ; sin ππ = π sinh π
∴ cos(π0.783) = cosh(0.783) ; sin(π0.783) = π sinh(0.783)
cos(0.573) cos(π0.783) − sin(0.573) sin(π0.783) ⇒ cos(32.83°) cosh(0.783) − sin(32.83°) π sinh(0.783)
= (0.840)(1.323) − π(0.542)(0.865) = 1.111 − π0.469
∴ ππ¨π¬(π. πππ + ππ. πππ) = π. πππ − ππ. πππ
β
Evaluate ππ«ππ¬π’π§(π + ππ).
β΅ arcsin π = −π ln (ππ ± √1 − π 2 ) ; π = 3 + π4
∴ arcsin(3 + π4) = −π ln [π(3 + π4) ± √1 − (3 + π4)2 ]
1
arcsin(3 + π4) = −π ln [(−4 + π3) ± √1 − (−7 + π24)] = −π ln[(−4 + π3) ± √8 − π24] = −π ln [(−4 + π3) ± (25.3∠ − 71.6°)2 ]
arcsin(3 + π4) = −π ln[(−4 + π3) ± (5.03∠ − 35.8°)]
Positive Root
Negative Root
arcsin(3 + π4) = −π ln[(−4 + π3) + (5.03∠ − 35.8°)]
arcsin(3 + π4) = −π ln[(−4 + π3) − (5.03∠ − 35.8°)]
arcsin(3 + π4) = −π ln(0.098∠35.9°)
arcsin(3 + π4) = −π ln(10.03∠143.67°)
arcsin(3 + π4) = −π ln(0.098π
π0.63 )
arcsin(3 + π4) = −π ln(10.03π π2.51 )
arcsin(3 + π4) = −π[ln(0.098) + π0.63]
arcsin(3 + π4) = −π[ln(10.03) + π2.51]
arcsin(3 + π4) = −π(−2.32 + π0.63)
arcsin(3 + π4) = −π(2.31 + π2.51)
arcsin(3 + π4) = 0.63 + π2.32
arcsin(3 + π4) = 2.51 − π2.31
∴ ππ«ππ¬π’π§(π + ππ) = π. πππ∠ππ. π°
∴ ππ«ππ¬π’π§(π + ππ) = π. πππ∠ − ππ. πππ°
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
β
π
πππ§π‘ (π )
π
β΅ tanh ππ = π tan π ; π =
π
3
π
π
∴ tanh (π ) = π tan ( )
3
3
π
tanh (π ) = π0.0183 = 0.0193∠90°
3
π
∴ πππ§π‘ (π ) = π. ππππ∠ππ°
π
β
Evaluate ππ«ππ¬π’π§π‘(π. π∠ππ°)
β΅ arcsinh π = ln (π ± √π 2 + 1) ; π = 0.4∠30°
∴ arcsinh(0.4∠30°) = ln [(0.4∠30°) + √(0.4∠30°)2 + 1]
arcsinh(0.4∠30°) = ln[(0.4∠30°) + √1.089∠7.311°] = ln[(0.4∠30°) + (1.044∠3.66°)]
arcsinh(0.4∠30°) = ln(1.414∠10.872°) = ln(1.414π π0.19 ) = ln(1.414) + π0.19 = 0.346 + π0.19
∴ ππ«ππ¬π’π§π‘(π. π∠ππ°) = π. πππ∠ππ. ππ°
β
Evaluate π¬π’π§π‘(π. πππ − ππ. πππ)
β΅ sinh(π ± π½) = sinh π cosh π½ ± cosh π sinh π½ ; π = 0.346 ; π½ = π0.548
∴ sinh(0.346 − π0.548) = sinh(0.346) cosh(π0.548) − cosh(0.346) sinh(π0.548)
β΅ cosh ππΌ = cos πΌ ; ∴ cosh(π0.548) = cos(0.548)
β΅ sinh ππΌ = π sin πΌ ; ∴ sinh(π0.548) = π sin(0.548)
sinh(0.346 − π0.548) = sinh(0.346) cos(0.548) − cosh(0.346) π sin(0.548)
radians → degrees
180°
0.548 (
) = 31.4° ; Argument of Hyperbolic Function should be in π«πππ’ππ§π¬
π
sinh(0.346) cos(0.548) − cosh(0.346) π sin(0.548) ⇒ sinh(0.346) cos(31.4°) − cosh(0.346) π sin(31.4°)
sinh(0.346 − π0.548) = 0.301 − π0.553
∴ π¬π’π§π‘(π. πππ − ππ. πππ) = π. ππππ∠ − ππ. ππππ°
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
WEEK 8 | 2 MARCH 2024
β’
Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules
or equations. The best way to convert differential equations into algebraic equations is the use of Laplace
transformation.
β’
Laplace transform is the integral transform of the given derivative function with real variable π to convert into a
complex function with variable π.
β’
The Laplace transform of π(π), that is denoted by π{π(π)} or π(π) is defined by the Laplace transform formula:
∞
π{π(π)} = π(π) = ∫ π(π)π−ππ π
π
−∞
β
π(π) = new function in the complex frequency domain
β
π(π) = function in the domain
β
π = complex frequency domain
β
π = time domain
β
π−ππ = kernel of transformation
TABLE OF TRANSFORMATIONS
π(π)
π{π(π)} = π(π)
1
1
π
π‘
1
π 2
π‘π
π!
π (π+1)
π βππ‘
1
(π ± π)
π‘ π π βππ‘
π!
(π ± π)(π+1)
sin(ππ‘)
cos(ππ‘)
π
(π 2 + π 2 )
π
(π 2 + π 2 )
π βππ‘ sin(ππ‘)
π
[(π ± π)2 + π 2 ]
π βππ‘ cos(ππ‘)
(π ± π)
[(π ± π)2 + π 2 ]
sinh(ππ‘)
π
(π 2 − π 2 )
cosh(ππ‘)
π
(π 2 − π 2 )
π βππ‘ sinh(ππ‘)
π
[(π ± π)2 − π 2 ]
π βππ‘ cosh(ππ‘)
(π ± π)
[(π ± π)2 − π 2 ]
π‘ sin(ππ‘ )
π‘ cos(ππ‘)
2ππ
(π 2 + π 2 )2
(π 2 − π 2 )
(π 2 + π 2 )2
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
√π
3
2 √π
√π‘
1
√π⁄π
√π‘
THEOREM 1: LINEARITY PROPERTY
β’
If π(π) and π(π) are any functions whose Laplace transforms exist and “π” and “π” are any constants, then:
π{ππ(π) ± ππ(π)} = π{ππ(π)} ± π{ππ(π)} = ππ{π(π)} ± ππ{π(π)}
Examples:
β
π{ππ + π}
1
1
2 3
β{2π‘ + 3} = β{2π‘} + β{3} = 2β{π‘} + 3β{1} = 2 ( 2 ) + 3 ( ) = 2 +
π
π
π
π
∴ π{ππ + π} =
β
π{ππ π−ππ }
β΅ π‘ π π βππ‘ =
π!
; π = 3 ; π = −4
(π ± π)(π+1)
∴ π{ππ π−ππ } =
β
ππ + πππ
ππ
3!
π
=
(π + 4)(3+1)
(π + π)π
π{πππ πππ − π ππ¨π¬(ππ)}
π!
; π =7; π =3
(π ± π)(π+1)
π
β΅ cos(ππ‘) = 2
; π=6
(π + π 2 )
β΅ π‘ π π βππ‘ =
∴ β{2π‘ 7 π 3π‘ − 4 cos(6π‘)} = β{π‘ 7 π 3π‘ } − 4β{cos(6π‘)}
β{2π‘ 7 π 3π‘ − 4 cos(6π‘)} =
7!
π
10080
4π
− 4[ 2
]=
− 2
(7+1)
2
8
(π + (6) )
(π − 3)
(π − 3)
π + 36
∴ π{πππ πππ − π ππ¨π¬(ππ)} =
β
π
πππππ(ππ + ππ) − ππ(π − π)π
(π − π)π (ππ + ππ)
π
π {π−ππ π¬π’π§(ππ) + π−ππ π¬π’π§π‘(π)}
β΅ β{π βππ‘ sin(ππ‘)} =
π
1
; π=− ; π=2
2
2
[(π ± π) + π ]
2
β΅ β{π βππ‘ sinh(ππ‘)} =
1
1
π
1
; π=− ; π=1
[(π ± π)2 − π 2 ]
2
1
π
π
∴ π {π−ππ π¬π’π§(ππ) + π−ππ π¬π’π§π‘(π)} =
2
1
∴ β {π −2π‘ sin(2π‘) + π −2π‘ sinh(π‘)} = β {π −2π‘ sin(2π‘)} + β {π −2π‘ sinh(π‘)} =
π
ππ + π +
ππ
π
+
1 2
(π + ) + 4
2
+
1
1 2
(π − ) − 1
2
π
π
π
(π + ) (π − )
π
π
THEOREM 2: FIRST SHIFTING PROPERTY
β’
If π{π(π)} = π(π) wen π > π, then:
π{πππ π(π)} = π(π − π) ; π ∈ β
β’
The substitution of (π − π) for π in the transform corresponds to the multiplication of the original function by
πππ .
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
Examples:
β
π{π−ππ ππ¨π¬(ππ)}
β{π(π‘)} = πΉ(π ) ⇒ β{cos(5π‘)} =
π
(π 2 + 25)
β{π −ππ‘ π(π‘)} = πΉ(π − π) ; π −ππ‘ = π −3π‘ = (π + 3)
(π + 3)
(π + 3)
=
(π + 3)2 + 25 π 2 + 6π + 9 + 25
∴ β{π −3π‘ cos(5π‘)} =
π+π
ππ + ππ + ππ
∴ π{π−ππ ππ¨π¬(ππ)} =
β
π{ππ πππ }
β{π‘ 3 } =
β
3!
π 3+1
6
π 4
∴ π{ππ πππ } =
π
(π − π)π
β{sin(π‘)} =
2
π 2 + 1
π{π−ππ π¬π’π§(π)}
β{π −2π‘ sin(π‘)} =
2
2
=
(π + 2)2 + 1 π 2 + 2π + 4 + 1
π
∴ π{π−ππ π¬π’π§(π)} =
β
=
ππ + ππ + π
π{ππ ππ¨π¬π‘(π)}
π
π 2 − 1
(π − 1)
π −1
π −1
∴ β{π π‘ cosh(π‘)} =
=
=
(π − 1)2 − 1 [(π − 1) − 1][(π − 1) + 1] (π − 2)(π )
β{cosh(π‘)} =
∴ π{ππ ππ¨π¬π‘(π)} =
β
π−π
ππ − ππ
π{π − ππ¨π¬(ππ)}
β{π‘ − cos(5π‘)} = β{π‘} − β{cos(5π‘)}
β{π‘ − cos(5π‘)} =
(π 2 + 25) − π 3
1
π
−
=
π 2 π 2 + 25
π 2 (π 2 + 25)
∴ π{π − ππ¨π¬(ππ)} =
β
−ππ + ππ + ππ
ππ + ππππ
π{ππ − ππ + π}
β{π‘ 3 − 8π‘ + 1} =
3!
8 1
6
8 1
− + = − +
π (3+1) π 2 π π 4 π 2 π
∴ π{ππ − ππ + π} =
β
π − πππ + ππ
ππ
π{[π¬π’π§(π) − ππ¨π¬(π)]π }
[sin(π‘) − cos(π‘)]2 = sin2 (π‘) − 2 sin(π‘) cos(π‘) + cos 2(π‘) = 1 − sin(2π‘)
∴ β{1 − sin(2π‘)} =
1
2
π 2 + 4 − 2π
− 2
=
π π +4
π (π 2 + 4)
∴ π{π − π¬π’π§(ππ)} =
β
ππ − ππ + π
ππ + ππ
π{πππ π−π }
3!
24
β{4π‘ 3 } = 4 ( (3+1) ) = 4
π
π
∴ π{πππ π−π } =
ππ
(π + π)π
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
β
π{π π¬π’π§π‘(ππ) − πππ−ππ }
β{4 sinh(3π‘) − 18π −5π‘ } = β{4 sinh(3π‘)} − β{18π −5π‘ }
3
1
β{4 sinh(3π‘) − 18π −5π‘ } = 4 ( 2
) − 18 (
)
π −9
π +5
π{π π¬π’π§π‘(ππ) − πππ−ππ } =
β
ππ
ππ
−
ππ − π π + π
π{ππ¨π¬ π (ππ)}
1
[cos(4π‘) + 1]
2
1
1
β{cos 2 (2π‘)} = β { [cos(4π‘) + 1]} = β{cos(4π‘) + 1}
2
2
cos 2 (2π‘) =
1
1
π
1
π
1
2π 2 + 2π 2 + 32 4π 2 + 32
π 2 + 8
β { [cos(4π‘) + 1]} = [ 2
+ ]= 2
+
=
= 3
= 3
3
2
2 π + 16 π
2π + 32 2π
4π + 64π
4π + 64 π + 16
∴ π{ππ¨π¬ π (ππ)} =
ππ + π
ππ + ππ
THEOREM 3: CHANGE OF SCALE PROPERTY
β’
If π{π(π)} = π(π) wen π > π, then:
π{π(ππ)} =
π
π
π( ) ; π ∈ β
π π
Example:
β
π{ππ¨π¬(ππ)}
β{cos(π‘)} =
π
π 2 + 1
(π ⁄3)
1
1
π
1
9π
π
∴ β{cos(3π‘)} = [
]= ( 2
)= ( 2
)= π
2
3 (π ⁄ ) + 1
9 π
9 π +9
π +π
+1
3
9
β
π¬π’π§(π)
Given that π {
π
π
π¬π’π§(ππ)
π
π
} = πππ§−π ( ), find π {
}
sin(π‘)
1
β{
} = tan−1 ( )
π‘
π
πβ {
sin(2π‘)
1
1
} = π ( ) tan−1 [π ]
ππ‘
2
⁄2
π¬π’π§(ππ)
π
∴ π{
} = πππ§−π ( )
π
π
β
ππ −π+π
If π{π(π)} = (ππ+π)π(π−π), find π{π(ππ)}
2
2
2
[π − 2π + 4⁄4]
(π ⁄2) − (π ⁄2) + 1
1
1 π ⁄4 − π ⁄2 + 1
4
π 2 − 2π + 4
π{π(ππ)} = [
]
=
[
]
=
[
]
=
π −2 4
(4π 2 + 8π + 4)(2π − 4)
2 [2(π ⁄ ) + 1]2 [(π ⁄ ) + 1]
2 (π + 1)2 (π − 2)
[(π 2 + 2π + 1) (
)]
2
2
2
2
β{π(2π‘)} =
π 2 − 2π + 4
8π 3 + 16π 2 + 8π − 16π 2 − 32π − 16
∴ π{π(ππ)} =
ππ − ππ + π
πππ − πππ − ππ
ELEN2213
ENGINEERING MATHEMATICS FOR E.E. | NOTES
THEOREM 4: MULTIPLICATION BY POWER OF π
β’
If π{π(π)} = π(π) wen π > π, then:
π{ππ π(π)} = (−π)π
ππ π
[π (π)] ; π ∈ β€
ππ¬ π
Examples:
β
π{π π¬π’π§(π)}
β΅ π(π‘) = sin(π‘) ⇒ πΉ(π ) =
∴ β{π‘ sin(π‘)} = (−1)1
ππ
ππ + πππ + π
π{ππ π¬π’π§(ππ)}
β΅ β{sin(3π‘)} =
= β{π‘ 2 sin(3π‘)} = (−1)2
=[
3
π 2 + 9
π2
3
π (0)(π 2 + 9) − (3)(2π )
π
−6π
[
]
=
[
]=
[ 2
]
2
2
2
2
(π + 9)
ππ π + 9
ππ
ππ (π + 9)2
(−6)(π 2 + 9)2 − (−6π )(2)(π 2 + 9)(2π )
−6(π 2 + 9) + (6π )(2)(2π ) −6π 2 − 54 + 24π 2
]=
=
2
4
(π + 9)
(π 2 + 9)3
(π 2 + 9)3
∴ π{ππ π¬π’π§(ππ)} =
β
; π=1
π
1
−2π
[ 2
] = −[ 2
]
(π + 1)2
ππ π + 1
∴ π{π π¬π’π§(π)} =
β
1
π 2 + 1
ππ π¬π’π§π‘(ππ)
π{
π
ππππ − ππ
(ππ + π)π
}
β{
π‘ 2 sinh(3π‘)
1
1
π2
} = β{π‘ 2 sinh(3π‘)} = (−1)2 2 [β{sinh(3π‘)}]
3
3
3
ππ
β΅ β{sinh(3π‘)} =
∴
3
π 2 − 9
1
d2
3
1 π (0)(π 2 − 9) − (3)(2π )
1 π
−6π
(−1)2 2 [ 2
]=
[
]=
[ 2
]
2
2
(π − 9)
3
dπ π − 9
3 ππ
3 ππ (π − 9)2
1 (−6)(π 2 − 9)2 − (−6π )(2)(π 2 − 9)(2π )
1 (−6)(π 2 − 9) + (6π )(2)(2π )
1 −6π 2 + 54 + 24π 2
= [
]= [
]= [
]
2
4
2
3
(π − 9)
(π − 9)
(π 2 − 9)3
3
3
3
ππ π¬π’π§π‘(ππ)
ππππ + ππ πππ + ππ
∴ π{
}=
=
π
π(ππ − π)π (ππ − π)π
