EngMathG
Worksheet1 (solution)
March, 2022
[1] Find all the (complex) roots of the following equations:
(a) 𝑥𝑥 2 + 4𝑥𝑥 + 5 = 0
Sol)
𝑥𝑥 2 + 4𝑥𝑥 + 5 = (𝑥𝑥 + 2)2 + 1 = (𝑥𝑥 + 2)2 + 12 = 0 ⇒ Roots: 𝑥𝑥 = −2 ± 𝑗𝑗1
Roots: 𝑥𝑥 = −2 ± √22 − 5 = −2 ± 𝑗𝑗1 (Using the quadratic formula)
(b) 𝑥𝑥 2 + 4𝑥𝑥 − 5 = 0
Sol)
𝑥𝑥 2 + 4𝑥𝑥 − 5 = (𝑥𝑥 + 2)2 − 9 = (𝑥𝑥 + 2)2 − 32 = 0 ⇒ Roots: 𝑥𝑥 = −2 ± 3 = 1, −5
Roots: 𝑥𝑥 = −2 ± √22 + 5 = −2 ± 3 = 1, −5 (Using the quadratic formula)
(c) 𝑥𝑥 2 − 4𝑥𝑥 + 5 = 0
Sol)
𝑥𝑥 2 − 4𝑥𝑥 + 5 = (𝑥𝑥 − 2)2 + 1 = (𝑥𝑥 − 2)2 + 12 = 0 ⇒ Roots: 𝑥𝑥 = 2 ± 𝑗𝑗1
Roots: 𝑥𝑥 = 2 ± �(−2)2 − 5 = 2 ± 𝑗𝑗1 (Using the quadratic formula)
(d) 𝑥𝑥 2 − 4𝑥𝑥 − 5 = 0
Sol)
𝑥𝑥 2 − 4𝑥𝑥 − 5 = (𝑥𝑥 − 2)2 − 9 = (𝑥𝑥 − 2)2 − 32 = 0 ⇒ Roots: 𝑥𝑥 = 2 ± 3 = 5, −1
Roots: 𝑥𝑥 = 2 ± �(−2)2 + 5 = 2 ± 3 = 5, −1 (Using the quadratic formula)
(e) 𝑥𝑥 2 + 4𝑥𝑥 + 4 = 0
Sol)
𝑥𝑥 2 + 4𝑥𝑥 + 4 = (𝑥𝑥 + 2)2 + 0 = 0 ⇒ Roots: 𝑥𝑥 = −2 ± 0 = −2, −2 (double roots)
Roots: 𝑥𝑥 = −2 ± √22 − 4 = −2 ± 0 = −2, −2 (double roots) (Using the quadratic formula)
(f) 𝑥𝑥 2 + 3𝑥𝑥 + 3 = 0
Sol)
2
3 2 3
3 2
3
√3
√3
2
𝑥𝑥 + 3𝑥𝑥 + 3 = �𝑥𝑥 + � + = �𝑥𝑥 + � + � � = 0 ⇒ Roots: 𝑥𝑥 = − ± 𝑗𝑗
2
4
2
2
2
2
Roots: 𝑥𝑥 =
−3±√32 −12
2
3
= − ± 𝑗𝑗
2
√3
(Using the quadratic formula)
2
(g) 𝑥𝑥 2 + 3𝑥𝑥 − 3 = 0
Sol)
2
3 2 21
3 2
3 √21
√21
2
𝑥𝑥 + 3𝑥𝑥 − 3 = �𝑥𝑥 + � −
= �𝑥𝑥 + � − �
� = 0 ⇒ Roots: 𝑥𝑥 = − ±
2
4
2
2
2
2
Roots: 𝑥𝑥 =
−3±�(3)2 +12
2
3
=− ±
2
√21
(Using the quadratic formula)
2
(h) 𝑥𝑥 2 + 3𝑥𝑥 + 2 = 0
Sol)
3 2 1
3 2
1 2
3 1
𝑥𝑥 + 3𝑥𝑥 + 2 = �𝑥𝑥 + � − = �𝑥𝑥 + � − � � = 0 ⇒ Roots: 𝑥𝑥 = − ± = −1, −2
2
4
2
2
2 2
2
Roots: 𝑥𝑥 =
−3±√32 −8
2
3
1
= − ± = −1, −2 (Using the quadratic formula)
2
2
[2] Given four complex numbers 𝑧𝑧1 = 2 + 𝑗𝑗 , 𝑧𝑧2 = −2 + 𝑗𝑗, 𝑧𝑧3 = 1 + 2𝑗𝑗, 𝑧𝑧4 = −4 − 2𝑗𝑗 , evaluate
the following quantities using the rectangular-form operations, and represent them in the rectangular
form.
(a) 𝑧𝑧2 ∙ 𝑧𝑧3
Sol)
𝑧𝑧2 ∙ 𝑧𝑧3 = (−2 + 𝑗𝑗) ∙ (1 + 2𝑗𝑗) = (−2) ∙ 1 + (−2) ∙ 2𝑗𝑗 + 𝑗𝑗 ∙ 1 + 𝑗𝑗 ∙ 2𝑗𝑗 = −2 − 4𝑗𝑗 + 𝑗𝑗 + 2𝑗𝑗 2
= −4 − 3𝑗𝑗
(b) 𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧1
Sol)
𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧1 = (1 + 2𝑗𝑗) ∙ (−4 − 2𝑗𝑗) ∙ (2 + 𝑗𝑗) = {(1 + 2𝑗𝑗) ∙ (−4 − 2𝑗𝑗)} ∙ (2 + 𝑗𝑗)
= {1 ∙ (−4) + 1 ∙ (−2𝑗𝑗) + 2𝑗𝑗 ∙ (−4) + 2𝑗𝑗 ∙ (−2𝑗𝑗)} ∙ (2 + 𝑗𝑗) = {−4 − 2𝑗𝑗 − 8𝑗𝑗 − 4𝑗𝑗 2 } ∙ (2 + 𝑗𝑗)
= (−10𝑗𝑗) ∙ (2 + 𝑗𝑗) = −20𝑗𝑗 − 10𝑗𝑗 2 = 10 − 20𝑗𝑗
(c) 𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧�1 ∙ 𝑧𝑧2
Sol)
(2 + 𝚥𝚥) ∙ (−2 + 𝑗𝑗)
𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧�1 ∙ 𝑧𝑧2 = (1 + 2𝑗𝑗) ∙ (−4 − 2𝑗𝑗) ∙ ���������
= (1 + 2𝑗𝑗) ∙ (−4 − 2𝑗𝑗) ∙ (2 − 𝑗𝑗) ∙ (−2 + 𝑗𝑗) = (−2 + 𝑗𝑗) ∙ (1 + 2𝑗𝑗) ∙ {−2 ∙ (2 + 𝑗𝑗) ∙ (2 − 𝑗𝑗)}
= (−4 − 3𝑗𝑗) ∙ {−2 ∙ (22 + 1)} = 40 + 30𝑗𝑗
(d) 𝑧𝑧1 /𝑧𝑧2
Sol)
𝑧𝑧1
2 + 𝑗𝑗
(2 + 𝑗𝑗) ∙ (−2 − 𝑗𝑗)
−(2 + 𝑗𝑗)2
−(4 + 4𝑗𝑗 − 1) −3 − 4𝑗𝑗
=
=
=
=
=
2
2
𝑧𝑧2 −2 + 𝑗𝑗 (−2 + 𝑗𝑗) ∙ (−2 − 𝑗𝑗) (−2) + 1
5
5
(e) 𝑧𝑧4 /𝑧𝑧3
Sol)
𝑧𝑧4 −4 − 2𝑗𝑗 (−4 − 2𝑗𝑗) ∙ (1 − 2𝑗𝑗) (−4 − 2𝑗𝑗) ∙ (1 − 2𝑗𝑗) −8 + 6𝑗𝑗
=
=
=
=
𝑧𝑧3
1 + 2𝑗𝑗
(1 + 2𝑗𝑗) ∙ (1 − 2𝑗𝑗)
12 + 22
5
(f) (𝑧𝑧1 ∙ 𝑧𝑧3 )/(𝑧𝑧2 ∙ 𝑧𝑧4 )
Sol)
(2 + 𝑗𝑗) ∙ (1 + 2𝑗𝑗)
5𝑗𝑗 1
𝑧𝑧1 ∙ 𝑧𝑧3
=
=
= 𝑗𝑗
𝑧𝑧2 ∙ 𝑧𝑧4 (−2 + 𝑗𝑗) ∙ (−4 − 2𝑗𝑗) 10 2
(g) (𝑧𝑧1 ∙ 𝑧𝑧2 )/(𝑧𝑧�3 ∙ 𝑧𝑧4 )
Sol)
(2 + 𝑗𝑗) ∙ (−2 + 𝑗𝑗)
𝑧𝑧1 ∙ 𝑧𝑧2
−5
−5 ∙ (−8 − 6𝑗𝑗)
−5 ∙ (−8 − 6𝑗𝑗)
=
=
=
=
�����������
(−8)2 + 62
𝑧𝑧�3 ∙ 𝑧𝑧4 (1 + 2𝚥𝚥) ∙ (−4 − 2𝑗𝑗) −8 + 6𝑗𝑗 (−8 + 6𝑗𝑗) ∙ (−8 − 6𝑗𝑗)
=
40 + 30𝑗𝑗 4 + 3𝑗𝑗
=
100
10
[3] Given four complex numbers 𝑧𝑧1 = 2 + 𝑗𝑗, 𝑧𝑧2 = −2 + 𝑗𝑗, 𝑧𝑧3 = 1 + 2𝑗𝑗, 𝑧𝑧4 = −4 − 2𝑗𝑗, represent
the four complex numbers in the Euler’s form. For the Euler’s form, no a numerical value, but an
𝑦𝑦
expression of tan−1 � �-form may be allowable.
𝑥𝑥
Sol)
−1 �1�
𝑧𝑧1 = 2 + 𝑗𝑗 = �22 + 12 𝑒𝑒 𝑗𝑗 tan
𝑧𝑧2 = −2 + 𝑗𝑗 = �(−2)2 + 12 𝑒𝑒
2
−1 �1�
= √5𝑒𝑒 𝑗𝑗 tan
1
𝑗𝑗 tan−1 � �
−2
2
−1 �1��
= √5𝑒𝑒 𝑗𝑗�π−tan
2
2
−1
−1
𝑧𝑧3 = 1 + 2𝑗𝑗 = �12 + 22 𝑒𝑒 𝑗𝑗 tan �1� = √5𝑒𝑒 𝑗𝑗 tan (2)
−1 −2
−1 1
𝑧𝑧4 = −4 − 2𝑗𝑗 = �(−4)2 + (−2)2 𝑒𝑒 𝑗𝑗 tan �−4� = 2√5𝑒𝑒 𝑗𝑗�−π+tan �2��
[4] Given four complex numbers 𝑧𝑧1 = 2 + 𝑗𝑗 , 𝑧𝑧2 = −2 + 𝑗𝑗, 𝑧𝑧3 = 1 + 2𝑗𝑗, 𝑧𝑧4 = −4 − 2𝑗𝑗 , evaluate
the following quantities using the Euler-form operations, and represent them in the Euler’s form.
(a) 𝑧𝑧2 ∙ 𝑧𝑧3
Sol)
For this problem, we will use the results in Problem [3].
−1 �1��
𝑧𝑧2 ∙ 𝑧𝑧3 = √5𝑒𝑒 𝑗𝑗�π−tan
2
−1 �1� +tan−1 (2)�
= 5𝑒𝑒 𝑗𝑗�π−tan
2
−1 (2)
∙ √5𝑒𝑒 𝑗𝑗 tan
−1 �1��+𝑗𝑗 tan−1 (2)
= 5𝑒𝑒 𝑗𝑗�π−tan
2
Note: −tan−1 𝑥𝑥 = tan−1 (−𝑥𝑥). If 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ∈ 𝑅𝑅, then
𝑠𝑠1 − 𝑠𝑠3 + 𝑠𝑠5 − 𝑠𝑠7 + ⋯
�
tan−1 𝑥𝑥1 + tan−1 𝑥𝑥2 + ⋯ + tan−1 𝑥𝑥𝑛𝑛 = tan−1 �
1 − 𝑠𝑠2 + 𝑠𝑠4 − 𝑠𝑠6 + ⋯
where 𝑠𝑠𝑘𝑘 = 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑥𝑥₁, 𝑥𝑥₂, … 𝑥𝑥𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
For example, tan−1 𝑥𝑥 + tan−1 𝑦𝑦 = tan−1 �
𝑥𝑥+𝑦𝑦
1−𝑥𝑥𝑥𝑥
� , tan−1 𝑥𝑥 + tan−1 𝑦𝑦 + tan−1 𝑧𝑧 = tan−1 �
1
3
(𝑥𝑥+𝑦𝑦+𝑧𝑧)−𝑥𝑥𝑥𝑥𝑥𝑥
1−(𝑥𝑥𝑥𝑥+𝑥𝑥𝑥𝑥+𝑦𝑦𝑦𝑦)
− +2
1
1
3
−tan−1 � � + tan−1 (2) = tan−1 �− � + tan−1 (2) = tan−1 � 2 1 � = tan−1 � 2 � = tan−1 � �
2
2
4
1−(−1)
1−(− ∙2)
�
2
−1 �3��
4
𝑧𝑧2 ∙ 𝑧𝑧3 = 5𝑒𝑒 𝑗𝑗�π+tan
−1 �3��
4
= 5𝑒𝑒 𝑗𝑗�−π+tan
3
�∵ π + tan−1 � � > π, −2π adjustment�
4
(Supplementary)
−1 �𝑦𝑦�
𝑥𝑥
Note: Conversion from the Euler’s form to the rectangular form 𝑒𝑒 𝑗𝑗 tan
−1 �3��
4
5𝑒𝑒 𝑗𝑗�−π+tan
or
−1 �3��
4
5𝑒𝑒 𝑗𝑗�−π+tan
−1 �3��
4
= 5 ∙ (−1) ∙ 𝑒𝑒 𝑗𝑗�tan
−1 �−3��
−4
= 5𝑒𝑒 𝑗𝑗�tan
=5∙
= −5 ∙
1
√42 +32
1
�(−4)2 +(−3)2
↔
1
�𝑥𝑥 2 +𝑦𝑦 2
(𝑥𝑥 + 𝑗𝑗𝑗𝑗)
(4 + 𝑗𝑗3) = −4 − 3𝑗𝑗 (cf. Prob. [2a])
(−4 − 𝑗𝑗3) = −4 − 3𝑗𝑗 (cf. Prob. [2a])
(b) 𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧1
Sol)
For this problem, we will use the results in Problem [3].
𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧1
−1 (2)
= √5𝑒𝑒 𝑗𝑗 tan
−1 �1��
∙ 2√5𝑒𝑒 𝑗𝑗�−π+tan
2
−1 �1�
∙ √5𝑒𝑒 𝑗𝑗 tan
2
−1 �1�+tan−1 �1�+tan−1 (2)�
= 10√5𝑒𝑒 𝑗𝑗�−π+tan
Note: −tan−1 𝑥𝑥 = tan−1 (−𝑥𝑥). If 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ∈ 𝑅𝑅, then
2
2
𝑠𝑠1 − 𝑠𝑠3 + 𝑠𝑠5 − 𝑠𝑠7 + ⋯
�
tan−1 𝑥𝑥1 + tan−1 𝑥𝑥2 + ⋯ + tan−1 𝑥𝑥𝑛𝑛 = tan−1 �
1 − 𝑠𝑠2 + 𝑠𝑠4 − 𝑠𝑠6 + ⋯
where 𝑠𝑠𝑘𝑘 = 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑥𝑥₁, 𝑥𝑥₂, … 𝑥𝑥𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
For example, tan−1 𝑥𝑥 + tan−1 𝑦𝑦 + tan−1 𝑧𝑧 = tan−1 �
(𝑥𝑥+𝑦𝑦+𝑧𝑧)−𝑥𝑥𝑥𝑥𝑥𝑥
1−(𝑥𝑥𝑥𝑥+𝑥𝑥𝑥𝑥+𝑦𝑦𝑦𝑦)
1 1
�
11
1
2
4
� + +2�− ∙ ∙2
3−
1
1
−π + tan−1 � � + tan−1 � � + tan−1 (2) = −π + tan−1 � 2 121 1 2 12 � = −π + tan−1 � 29�
2
2
1−
1−� ∙ + ∙2+ ∙2�
22 2
5
= −π + tan−1 � 25 � = −π + tan−1 [−2] + 𝜋𝜋 = −tan−1 [2]
−
𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧1
4
−1 (2)
= 10√5𝑒𝑒 −𝑗𝑗 tan
−1 (−2)
= 10√5𝑒𝑒 𝑗𝑗 tan
(Supplementary)
−1 �𝑦𝑦�
𝑥𝑥
Note: Conversion from the Euler’s form to the rectangular form 𝑒𝑒 𝑗𝑗 tan
−1 (2)
10√5𝑒𝑒 −𝑗𝑗 tan
(cf. Prob. [2b])
−1 �−2��
1
= 10√5𝑒𝑒 𝑗𝑗 tan (−2) = 10√5 ∙ 𝑒𝑒𝑗𝑗�tan
−1
= 10√5 ∙
1
�(1)2 +(−2)2
↔
1
�𝑥𝑥 2 +𝑦𝑦 2
(𝑥𝑥 + 𝑗𝑗𝑗𝑗)
(1 − 𝑗𝑗2) = 10 − 20𝑗𝑗
(c) 𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧�1 ∙ 𝑧𝑧2
Sol)
For this problem, we will use the results in Problem [3].
𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧�1 ∙ 𝑧𝑧2
= √5𝑒𝑒 𝑗𝑗 tan
−1 (2)
∙ 2√5𝑒𝑒
1
1
1
1
𝑗𝑗�−π+tan−1 � ��
−𝑗𝑗 tan−1 � �
𝑗𝑗�π−tan−1 � ��
𝑗𝑗�tan−1 (2)−tan−1 � ��
2 ∙ √5𝑒𝑒
2 ∙ √5𝑒𝑒
2 = 50𝑒𝑒
2
Note: −tan−1 𝑥𝑥 = tan−1 (−𝑥𝑥). If 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ∈ 𝑅𝑅, then
𝑠𝑠1 − 𝑠𝑠3 + 𝑠𝑠5 − 𝑠𝑠7 + ⋯
�
tan−1 𝑥𝑥1 + tan−1 𝑥𝑥2 + ⋯ + tan−1 𝑥𝑥𝑛𝑛 = tan−1 �
1 − 𝑠𝑠2 + 𝑠𝑠4 − 𝑠𝑠6 + ⋯
where 𝑠𝑠𝑘𝑘 = 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑥𝑥₁, 𝑥𝑥₂, … 𝑥𝑥𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
For example, tan−1 𝑥𝑥 + tan−1 𝑦𝑦 + tan−1 𝑧𝑧 = tan−1 �
1
1
(𝑥𝑥+𝑦𝑦+𝑧𝑧)−𝑥𝑥𝑥𝑥𝑥𝑥
1−(𝑥𝑥𝑥𝑥+𝑥𝑥𝑥𝑥+𝑦𝑦𝑦𝑦)
tan−1 (2) − tan−1 � � = tan−1 (2) + tan−1 �− � = tan−1 �
2
𝑧𝑧3 ∙ 𝑧𝑧4 ∙ 𝑧𝑧�1 ∙ 𝑧𝑧2
2
3
2
�
1−(−1)
3
� = tan−1 � �
4
−1 �3�
= 50𝑒𝑒 𝑗𝑗 tan
4
(Supplementary)
−1 �𝑦𝑦�
𝑥𝑥
Note: Conversion from the Euler’s form to the rectangular form 𝑒𝑒 𝑗𝑗 tan
−1 �3�
4
50𝑒𝑒 𝑗𝑗 tan
= 50 ∙
1
�(4)2 +(3)2
(4 + 𝑗𝑗3) = 40 + 30𝑗𝑗 (cf. Prob. [2c])
1
(𝑥𝑥 + 𝑗𝑗𝑗𝑗)
1
(𝑥𝑥 + 𝑗𝑗𝑗𝑗)
↔
�𝑥𝑥 2 +𝑦𝑦 2
↔
�𝑥𝑥 2 +𝑦𝑦 2
(d) 𝑧𝑧1 /𝑧𝑧2
Sol)
For this problem, we will use the results in Problem [3].
𝑧𝑧1 /𝑧𝑧2
=
√5𝑒𝑒
√5𝑒𝑒
1
𝑗𝑗 tan−1 �2�
1
𝑗𝑗�π−tan−1 �2��
−1 �1�
= 𝑒𝑒 𝑗𝑗 tan
−1 �1��
2 𝑒𝑒 −𝑗𝑗�π−tan
−1 �1��
= 𝑒𝑒 𝑗𝑗�− π+2tan
2
2
Note: −tan−1 𝑥𝑥 = tan−1 (−𝑥𝑥). If 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ∈ 𝑅𝑅, then
𝑠𝑠1 − 𝑠𝑠3 + 𝑠𝑠5 − 𝑠𝑠7 + ⋯
�
tan−1 𝑥𝑥1 + tan−1 𝑥𝑥2 + ⋯ + tan−1 𝑥𝑥𝑛𝑛 = tan−1 �
1 − 𝑠𝑠2 + 𝑠𝑠4 − 𝑠𝑠6 + ⋯
where 𝑠𝑠𝑘𝑘 = 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑥𝑥₁, 𝑥𝑥₂, … 𝑥𝑥𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
For example, tan−1 𝑥𝑥 + tan−1 𝑦𝑦 + tan−1 𝑧𝑧 = tan−1 �
1
1
1
2 tan−1 � � = tan−1 � � + tan−1 � � = tan−1 �
2
𝑧𝑧1 /𝑧𝑧2
2
−1 �4��
= 𝑒𝑒 𝑗𝑗�− π+tan
3
2
1
1
4
(𝑥𝑥+𝑦𝑦+𝑧𝑧)−𝑥𝑥𝑥𝑥𝑥𝑥
1−(𝑥𝑥𝑥𝑥+𝑥𝑥𝑥𝑥+𝑦𝑦𝑦𝑦)
4
� = tan−1 � �
1−� �
�
3
−1 �−4��
= 𝑒𝑒 𝑗𝑗�tan
−3
(Supplementary)
−1 �𝑦𝑦�
𝑥𝑥
Note: Conversion from the Euler’s form to the rectangular form 𝑒𝑒 𝑗𝑗 tan
−1 �−4�
−3
𝑒𝑒 𝑗𝑗 tan
(e) 𝑧𝑧4 /𝑧𝑧3
Sol)
=
1
�(−3)2 +(−4)2
3
4
(−3 − 𝑗𝑗4) = − − 𝑗𝑗 (cf. Prob. [2d])
5
5
For this problem, we will use the results in Problem [3].
𝑧𝑧4 /𝑧𝑧3
=
−1 �1��
2√5𝑒𝑒 𝑗𝑗�−π+tan
√5𝑒𝑒
𝑗𝑗 tan−1 (2)
2
−1 �1��
= 2𝑒𝑒 𝑗𝑗�− π+tan
2
−1 (2)
𝑒𝑒 −𝑗𝑗 tan
Note: −tan−1 𝑥𝑥 = tan−1 (−𝑥𝑥). If 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ∈ 𝑅𝑅, then
−1 �1�+tan−1 (−2)�
= 2𝑒𝑒 𝑗𝑗�−π+tan
2
𝑠𝑠1 − 𝑠𝑠3 + 𝑠𝑠5 − 𝑠𝑠7 + ⋯
�
tan−1 𝑥𝑥1 + tan−1 𝑥𝑥2 + ⋯ + tan−1 𝑥𝑥𝑛𝑛 = tan−1 �
1 − 𝑠𝑠2 + 𝑠𝑠4 − 𝑠𝑠6 + ⋯
where 𝑠𝑠𝑘𝑘 = 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑥𝑥₁, 𝑥𝑥₂, … 𝑥𝑥𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
For example, tan−1 𝑥𝑥 + tan−1 𝑦𝑦 + tan−1 𝑧𝑧 = tan−1 �
1
tan−1 � � + tan−1 (−2) = tan−1 �
2
𝑧𝑧4 /𝑧𝑧3
−1 �−3��
4
= 2𝑒𝑒 𝑗𝑗�−π+tan
−1 �−3��
= 2𝑒𝑒 𝑗𝑗�+π+tan
4
3
−2
(𝑥𝑥+𝑦𝑦+𝑧𝑧)−𝑥𝑥𝑥𝑥𝑥𝑥
1−(𝑥𝑥𝑥𝑥+𝑥𝑥𝑥𝑥+𝑦𝑦𝑦𝑦)
−3
� = tan−1 � 4 �
1−(−1)
�
−3
(−π + tan−1 � � < −π이므로, +2π 보정)
4
−1 � 3 ��
= 2𝑒𝑒 𝑗𝑗�tan
−4
(추가 설명)!!!
−1 �−3��
2𝑒𝑒 𝑗𝑗�π+tan
=2∙
4
1
−1 �−3�
= 2𝑒𝑒 𝑗𝑗π 𝑒𝑒 𝑗𝑗 tan
�42 + (−3)2
4
(−4 + 𝑗𝑗3) = 2 ∙
= 2 ∙ (−1) ∙
1
1
�42 + (−3)2
�42 + (−3)2
(4 − 𝑗𝑗3)
−1 � 3 �
�(−4)2 + (3)2 𝑒𝑒 𝑗𝑗 tan
−4
−1 � 3 �
= 2𝑒𝑒 𝑗𝑗 tan
−4
정리하면, ( 𝑒𝑒 ±𝑗𝑗π = −1 )이므로, 위상을 구할 때, (±)π 를 제거하는 대신에, 더해지는
tan^-1(a/b)에서 a와 b의 부호를 모두 바꾸어 주면 [a->(-a), b->(-b)] 동일한 결과를 얻
을 수 있습니다.
(Supplementary)
−1 �𝑦𝑦�
𝑥𝑥
Note: Conversion from the Euler’s form to the rectangular form 𝑒𝑒 𝑗𝑗 tan
−1
3
2𝑒𝑒𝑗𝑗 tan �−4� =
2
�(−4)2 +(3)2
(f) (𝑧𝑧1 ∙ 𝑧𝑧3 )/(𝑧𝑧2 ∙ 𝑧𝑧4 )
8
6
(−4 + 𝑗𝑗3) = − + 𝑗𝑗 (cf. Prob. [2e])
5
5
↔
1
�𝑥𝑥 2 +𝑦𝑦 2
(𝑥𝑥 + 𝑗𝑗𝑗𝑗)
Sol)
For this problem, we will use the results in Problem [3].
(𝑧𝑧1 ∙ 𝑧𝑧3 )/(𝑧𝑧2 ∙ 𝑧𝑧4 )
=
√5𝑒𝑒
√5𝑒𝑒
1
𝑗𝑗 tan−1 �2�
1
𝑗𝑗�π−tan−1 �2��
−1 (2)
∙ √5𝑒𝑒 𝑗𝑗 tan
∙ 2√5𝑒𝑒
1 𝑗𝑗�tan−1�1�+tan−1(2)� −𝑗𝑗{0} 1 𝑗𝑗�tan−1�1�+tan−1(2)�
2
2
=
𝑒𝑒
𝑒𝑒
= 𝑒𝑒
1
2
2
𝑗𝑗�−π+tan−1 �2��
Note: −tan−1 𝑥𝑥 = tan−1 (−𝑥𝑥). If 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ∈ 𝑅𝑅, then
𝑠𝑠1 − 𝑠𝑠3 + 𝑠𝑠5 − 𝑠𝑠7 + ⋯
�
tan−1 𝑥𝑥1 + tan−1 𝑥𝑥2 + ⋯ + tan−1 𝑥𝑥𝑛𝑛 = tan−1 �
1 − 𝑠𝑠2 + 𝑠𝑠4 − 𝑠𝑠6 + ⋯
where 𝑠𝑠𝑘𝑘 = 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑥𝑥₁, 𝑥𝑥₂, … 𝑥𝑥𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
For example, tan−1 𝑥𝑥 + tan−1 𝑦𝑦 + tan−1 𝑧𝑧 = tan−1 �
1
tan−1 � � + tan−1 (2) = tan−1 �
2
3
2
(𝑥𝑥+𝑦𝑦+𝑧𝑧)−𝑥𝑥𝑥𝑥𝑥𝑥
1−(𝑥𝑥𝑥𝑥+𝑥𝑥𝑥𝑥+𝑦𝑦𝑦𝑦)
3
π
� = tan−1 � 02 � = 2
1−(1)
�
(𝑧𝑧1 ∙ 𝑧𝑧3 )/(𝑧𝑧2 ∙ 𝑧𝑧4 )
1 π
= 𝑒𝑒 𝑗𝑗� 2 �
2
(Supplementary)
−1 �𝑦𝑦�
𝑥𝑥
Note: Conversion from the Euler’s form to the rectangular form 𝑒𝑒 𝑗𝑗 tan
1
2
π
1
𝑒𝑒 𝑗𝑗� 2 � = 𝑗𝑗 (cf. Prob. [2f])
↔
1
�𝑥𝑥 2 +𝑦𝑦 2
(𝑥𝑥 + 𝑗𝑗𝑗𝑗)
2
(g) (𝑧𝑧1 ∙ 𝑧𝑧2 )/(𝑧𝑧�3 ∙ 𝑧𝑧4 )
Sol)
For this problem, we will use the results in Problem [3].
(𝑧𝑧1 ∙ 𝑧𝑧2 )/(𝑧𝑧̅3 ∙ 𝑧𝑧4 )
=
√5𝑒𝑒
1
1
𝑗𝑗 tan−1 � �
𝑗𝑗�π−tan−1 � ��
2 ∙ √5𝑒𝑒
2
√5𝑒𝑒 −𝑗𝑗 tan
−1 (2)
∙ 2√5𝑒𝑒
1
𝑗𝑗�−π+tan−1 � ��
2
=
1 𝑗𝑗π −𝑗𝑗�−π+tan−1 �1�−tan−1 (2)� 1 𝑗𝑗�− tan−1 �1�+tan−1 (2)�
2
2
𝑒𝑒 𝑒𝑒
= 𝑒𝑒
2
2
Note: −tan−1 𝑥𝑥 = tan−1 (−𝑥𝑥). If 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ∈ 𝑅𝑅, then
𝑠𝑠1 − 𝑠𝑠3 + 𝑠𝑠5 − 𝑠𝑠7 + ⋯
�
tan−1 𝑥𝑥1 + tan−1 𝑥𝑥2 + ⋯ + tan−1 𝑥𝑥𝑛𝑛 = tan−1 �
1 − 𝑠𝑠2 + 𝑠𝑠4 − 𝑠𝑠6 + ⋯
where 𝑠𝑠𝑘𝑘 = 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑥𝑥₁, 𝑥𝑥₂, … 𝑥𝑥𝑛𝑛 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑘𝑘 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
For example, tan−1 𝑥𝑥 + tan−1 𝑦𝑦 + tan−1 𝑧𝑧 = tan−1 �
(𝑥𝑥+𝑦𝑦+𝑧𝑧)−𝑥𝑥𝑥𝑥𝑥𝑥
1−(𝑥𝑥𝑥𝑥+𝑥𝑥𝑥𝑥+𝑦𝑦𝑦𝑦)
�
3
1
1
3
− tan−1 � � + tan−1 (2) = tan−1 �− � + tan−1 (2) = tan−1 � 2 � = tan−1 � �
2
2
4
1−(−1)
(𝑧𝑧1 ∙ 𝑧𝑧2 )/(𝑧𝑧̅3 ∙ 𝑧𝑧4 )
1
−1 3
= 𝑒𝑒 𝑗𝑗 tan �4�
2
(Supplementary)
−1 �𝑦𝑦�
𝑥𝑥
Note: Conversion from the Euler’s form to the rectangular form 𝑒𝑒 𝑗𝑗 tan
1
2
−1 �3�
4
𝑒𝑒 𝑗𝑗 tan
=
1
1
2 �(4)2 +(3)2
(4 + 𝑗𝑗3) =
4
10
+
3
10
𝑗𝑗 (cf. Prob. [2g])
↔
1
�𝑥𝑥 2 +𝑦𝑦 2
(𝑥𝑥 + 𝑗𝑗𝑗𝑗)
[5] Find all the (complex) roots of the following equations:
(a) 𝑥𝑥 8 − 1 = 0
Sol)
𝑥𝑥 8 = 1
→ 𝑥𝑥 8 = 1 → 𝑥𝑥 8 = 𝑒𝑒 𝑗𝑗2𝜋𝜋𝜋𝜋 , 𝑘𝑘 = 0,1,2, … ,7
2𝜋𝜋𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 𝑗𝑗 8 , 𝑘𝑘 = 0,1,2, … ,7
𝜋𝜋
2𝜋𝜋
3𝜋𝜋
4𝜋𝜋
5𝜋𝜋
6𝜋𝜋
7𝜋𝜋
→ 𝑥𝑥 = 1, 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4
3𝜋𝜋
2𝜋𝜋
𝜋𝜋
𝜋𝜋
2𝜋𝜋
3𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 −𝑗𝑗𝑗𝑗 , 𝑒𝑒 −𝑗𝑗 4 , 𝑒𝑒 −𝑗𝑗 4 , 𝑒𝑒 −𝑗𝑗 4 , 1, 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4 , 𝑒𝑒 𝑗𝑗 4
(b) (𝑗𝑗𝑗𝑗)8 + 𝑗𝑗 = 0
Sol)
(𝑗𝑗𝑗𝑗)8 = −𝑗𝑗
8
𝜋𝜋
𝜋𝜋
→ (𝑗𝑗𝑗𝑗)8 = −𝑗𝑗 → �𝑒𝑒 𝑗𝑗 2 𝑥𝑥� = 𝑒𝑒 𝑗𝑗�− 2 +2𝜋𝜋𝜋𝜋� , 𝑘𝑘 = 0,1,2, … ,7
𝜋𝜋
𝑗𝑗 2
→ 𝑒𝑒 𝑥𝑥 = 𝑒𝑒
→ 𝑥𝑥 = 𝑒𝑒
𝑗𝑗
𝜋𝜋
�− 2 +2𝜋𝜋𝜋𝜋�
8
, 𝑘𝑘 = 0,1,2, … ,7
𝜋𝜋
�− +2𝜋𝜋𝜋𝜋� 𝜋𝜋
𝑗𝑗� 2
−2�
8
9𝜋𝜋
5𝜋𝜋
𝜋𝜋𝜋𝜋 9𝜋𝜋
= 𝑒𝑒 𝑗𝑗� 4 −16 � , 𝑘𝑘 = 0,1,2, … ,7
𝜋𝜋
3𝜋𝜋
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
19𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16
9𝜋𝜋
5𝜋𝜋
𝜋𝜋
3𝜋𝜋
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
13𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16
13𝜋𝜋
9𝜋𝜋
5𝜋𝜋
𝜋𝜋
3𝜋𝜋
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16
Another Sol) – 학생 질문에 따라 𝑗𝑗 8 = 1을 반영하여 풀이 (결과는 동일함.)
(𝑗𝑗𝑗𝑗)8 = −𝑗𝑗
𝜋𝜋
→ (𝑗𝑗𝑗𝑗)8 = −𝑗𝑗 → 𝑗𝑗 8 𝑥𝑥 8 = −𝑗𝑗 → 𝑥𝑥 8 = −𝑗𝑗 → 𝑥𝑥 8 = 𝑒𝑒 𝑗𝑗�− 2 +2𝜋𝜋𝜋𝜋� , 𝑘𝑘 = 0,1,2, … ,7
→ 𝑥𝑥 = 𝑒𝑒
𝑗𝑗
𝜋𝜋
�− 2 +2𝜋𝜋𝜋𝜋�
8
𝜋𝜋
3𝜋𝜋
𝜋𝜋𝜋𝜋
𝜋𝜋
= 𝑒𝑒 𝑗𝑗� 4 −16� , 𝑘𝑘 = 0,1,2, … ,7
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
19𝜋𝜋
23𝜋𝜋
27𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 −𝑗𝑗16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16
𝜋𝜋
3𝜋𝜋
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
13𝜋𝜋
9𝜋𝜋
5𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 −𝑗𝑗16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16
13𝜋𝜋
9𝜋𝜋
5𝜋𝜋
𝜋𝜋
3𝜋𝜋
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16
(c) 𝑥𝑥 −8 + 𝑗𝑗 = 0
Sol)
𝑥𝑥 −8 + 𝑗𝑗 = 0
𝜋𝜋
→ 𝑥𝑥 −8 = −𝑗𝑗 → 𝑥𝑥 −8 = 𝑒𝑒 𝑗𝑗�− 2 +2𝜋𝜋𝜋𝜋� , 𝑘𝑘 = 0,1,2, … ,7
→ 𝑥𝑥 = 𝑒𝑒
𝑗𝑗
𝜋𝜋
�− 2 +2𝜋𝜋𝜋𝜋�
−8
𝜋𝜋
3𝜋𝜋
𝜋𝜋
𝜋𝜋𝜋𝜋
= 𝑒𝑒 𝑗𝑗�16− 4 � , 𝑘𝑘 = 0,1,2, … ,7
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
19𝜋𝜋
23𝜋𝜋
27𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 𝑗𝑗16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16
13𝜋𝜋
9𝜋𝜋
5𝜋𝜋
𝜋𝜋
3𝜋𝜋
7𝜋𝜋
11𝜋𝜋
15𝜋𝜋
→ 𝑥𝑥 = 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗 16 , 𝑒𝑒 𝑗𝑗16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16 , 𝑒𝑒 −𝑗𝑗 16
(d) 𝑥𝑥 8 − 64 = 0
Sol)
𝑥𝑥 8 − 64 = 0
→ 𝑥𝑥 8 = 64 → 𝑥𝑥 8 = 64𝑒𝑒 𝑗𝑗(2𝜋𝜋𝜋𝜋) = 26 𝑒𝑒 𝑗𝑗(2𝜋𝜋𝜋𝜋) , 𝑘𝑘 = 0,1,2, … ,7
6
2𝜋𝜋𝜋𝜋
→ 𝑥𝑥 = 28 𝑒𝑒 𝑗𝑗� 8 � , 𝑘𝑘 = 0,1,2, … ,7
3
3
𝜋𝜋
3
2𝜋𝜋
3
3𝜋𝜋
→ 𝑥𝑥 = 24 , 24 𝑒𝑒 𝑗𝑗 4 , 24 𝑒𝑒 𝑗𝑗 4 , 24 𝑒𝑒 𝑗𝑗 4 ,
3
3
3𝜋𝜋
3
2𝜋𝜋
3
𝜋𝜋
3
3
𝜋𝜋
3
2𝜋𝜋
3
3𝜋𝜋
→ 𝑥𝑥 = 24 𝑒𝑒 −𝑗𝑗𝑗𝑗 , 24 𝑒𝑒 −𝑗𝑗 4 , 24 𝑒𝑒 −𝑗𝑗 4 , 24 𝑒𝑒 −𝑗𝑗 4 , 24 , 24 𝑒𝑒 𝑗𝑗 4 , 24 𝑒𝑒 𝑗𝑗 4 , 24 𝑒𝑒 𝑗𝑗 4
(e) (2𝑥𝑥)8 + 64 = 0
Sol)
(2𝑥𝑥)8 + 64 = 0
→ (2𝑥𝑥)8 = −64 → (2𝑥𝑥)8 = 26 𝑒𝑒 𝑗𝑗(𝜋𝜋+2𝜋𝜋𝜋𝜋) , 𝑘𝑘 = 0,1,2, … ,7
6
→ 2𝑥𝑥 = 28 𝑒𝑒 𝑗𝑗�
→ 𝑥𝑥 = 2
1
�−4�
1
𝜋𝜋+2𝜋𝜋𝜋𝜋
�
8
𝜋𝜋
𝑗𝑗 8
, 𝑘𝑘 = 0,1,2, … ,7
1
3𝜋𝜋
1
5𝜋𝜋
1
7𝜋𝜋
1
9𝜋𝜋
1
11𝜋𝜋
1
13𝜋𝜋
1
15𝜋𝜋
𝑒𝑒 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8
7𝜋𝜋
1
5𝜋𝜋
1
3𝜋𝜋
1
𝜋𝜋
1
𝜋𝜋
1
3𝜋𝜋
1
5𝜋𝜋
1
7𝜋𝜋
→ 𝑥𝑥 = 2�−4� 𝑒𝑒 −𝑗𝑗 8 , 2�−4� 𝑒𝑒 −𝑗𝑗 8 , 2�−4� 𝑒𝑒 −𝑗𝑗 8 , 2�−4� 𝑒𝑒 −𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8 , 2�−4� 𝑒𝑒 𝑗𝑗 8
(f) 𝑥𝑥 4 − 16 = 0
Sol)
𝑥𝑥 4 − 16 = 0
→ 𝑥𝑥 4 = 16 → 𝑥𝑥 4 = 24 𝑒𝑒 𝑗𝑗(2𝜋𝜋𝜋𝜋) , 𝑘𝑘 = 0,1,2,3
2𝜋𝜋𝜋𝜋
→ 𝑥𝑥 = 2𝑒𝑒 𝑗𝑗� 4 � , 𝑘𝑘 = 0,1,2,3
𝜋𝜋
3𝜋𝜋
→ 𝑥𝑥 = 2,2𝑒𝑒 𝑗𝑗 2 , 2𝑒𝑒 𝑗𝑗𝑗𝑗 , 2𝑒𝑒 𝑗𝑗 2
𝜋𝜋
𝜋𝜋
→ 𝑥𝑥 = 2𝑒𝑒 −𝑗𝑗𝑗𝑗 , 2𝑒𝑒 −𝑗𝑗 2 , 2,2𝑒𝑒 𝑗𝑗 2
𝑥𝑥 4
(g) � � + 16 = 0
2
Sol)
𝑥𝑥 4
� � + 16 = 0
2
𝑥𝑥 4
𝑥𝑥 4
→ � � = −16 → � � = 24 𝑒𝑒 𝑗𝑗(𝜋𝜋+2𝜋𝜋𝜋𝜋) , 𝑘𝑘 = 0,1,2,3
→
𝑥𝑥
2
2
= 2𝑒𝑒
𝑗𝑗
(𝜋𝜋+2𝜋𝜋𝜋𝜋)
4
𝜋𝜋
𝑗𝑗 4
→ 𝑥𝑥 = 4𝑒𝑒 , 4𝑒𝑒
3𝜋𝜋
2
, 𝑘𝑘 = 0,1,2,3
3𝜋𝜋
𝑗𝑗 4
5𝜋𝜋
7𝜋𝜋
, 4𝑒𝑒 𝑗𝑗 4 , 4𝑒𝑒 𝑗𝑗 4
𝜋𝜋
𝜋𝜋
3𝜋𝜋
→ 𝑥𝑥 = 4𝑒𝑒 −𝑗𝑗 4 , 4𝑒𝑒 −𝑗𝑗 4 , 4𝑒𝑒 𝑗𝑗 4 , 4𝑒𝑒 𝑗𝑗 4
[6] Find the 𝑛𝑛-th power of the complex number 𝑧𝑧 = 𝑒𝑒 𝑗𝑗(𝜋𝜋/2) .
(a) 𝑛𝑛 = 2
Sol)
𝑧𝑧 2 = 𝑒𝑒 𝑗𝑗(2𝜋𝜋/2) = 𝑒𝑒 −𝑗𝑗𝑗𝑗
(b) 𝑛𝑛 = 3
Sol)
𝑧𝑧 3 = 𝑒𝑒 𝑗𝑗(3𝜋𝜋/2) = 𝑒𝑒 −𝑗𝑗(𝜋𝜋/2)
(c) 𝑛𝑛 = 4
Sol)
𝑧𝑧 4 = 𝑒𝑒 𝑗𝑗(4𝜋𝜋/2) = 1
(d) 𝑛𝑛 = 5
Sol)
𝑧𝑧 5 = 𝑒𝑒 𝑗𝑗(5𝜋𝜋/2) = 𝑒𝑒 𝑗𝑗(𝜋𝜋/2) = 𝑧𝑧
(e) 𝑛𝑛 = 6
Sol)
𝑧𝑧 6 = 𝑒𝑒 𝑗𝑗(6𝜋𝜋/2) = 𝑒𝑒 𝑗𝑗(2𝜋𝜋/2) = 𝑒𝑒 −𝑗𝑗𝑗𝑗 = 𝑧𝑧 2
[7] Find the 𝑛𝑛-th power of the complex number 𝑧𝑧 = 2𝑒𝑒 𝑗𝑗(𝜋𝜋/2) .
(a) 𝑛𝑛 = 2
Sol)
𝑧𝑧 2 = 4𝑒𝑒 𝑗𝑗(2𝜋𝜋/2) = 4𝑒𝑒 −𝑗𝑗𝑗𝑗
(b) 𝑛𝑛 = 3
Sol)
𝑧𝑧 3 = 8𝑒𝑒 𝑗𝑗(3𝜋𝜋/2) = 8𝑒𝑒 −𝑗𝑗(𝜋𝜋/2)
(c) 𝑛𝑛 = 4
Sol)
𝑧𝑧 4 = 16𝑒𝑒 𝑗𝑗(4𝜋𝜋/2) = 16
(d) 𝑛𝑛 = 5
Sol)
𝑧𝑧 5 = 32𝑒𝑒 𝑗𝑗(5𝜋𝜋/2) = 32𝑒𝑒 𝑗𝑗(𝜋𝜋/2) = 16(= 24 )𝑧𝑧
(e) 𝑛𝑛 = 6
Sol)
𝑧𝑧 6 = 64𝑒𝑒 𝑗𝑗(6𝜋𝜋/2) = 64𝑒𝑒 −𝑗𝑗𝑗𝑗 = 16(= 24 )𝑧𝑧 2
[8] Find 𝑥𝑥, 𝑦𝑦 when (2 + 𝑗𝑗)2 + (1 + 𝑗𝑗)(𝑥𝑥 − 𝑗𝑗𝑗𝑗) = 5 + 𝑗𝑗8, 𝑥𝑥, 𝑦𝑦 ∈ 𝑅𝑅.
Sol)
(2 + 𝑗𝑗)2 + (1 + 𝑗𝑗)(𝑥𝑥 − 𝑗𝑗𝑗𝑗) = 4 + 𝑗𝑗4 − 1 + 𝑥𝑥 + 𝑦𝑦 + 𝑗𝑗(𝑥𝑥 − 𝑦𝑦) = 3 + 𝑥𝑥 + 𝑦𝑦 + 𝑗𝑗(4 + 𝑥𝑥 − 𝑦𝑦)
We get
3 + 𝑥𝑥 + 𝑦𝑦 = 5, 4 + 𝑥𝑥 − 𝑦𝑦 = 8 ⟹ 𝑥𝑥 = 3, 𝑦𝑦 = −1
[9] Find real numbers 𝑎𝑎, 𝑏𝑏 when �12 − 8𝑗𝑗 = 𝑎𝑎 + 𝑗𝑗𝑗𝑗.
Sol)
�12 − 8𝑗𝑗 = 𝑎𝑎 + 𝑗𝑗𝑗𝑗 ⟹ 12 − 8𝑗𝑗 = 𝑎𝑎2 − 𝑏𝑏 2 + 𝑗𝑗2𝑎𝑎𝑎𝑎 ⟹ 𝑎𝑎2 − 𝑏𝑏 2 = 12, 𝑎𝑎𝑎𝑎 = −4
4 2
⟹ 𝑎𝑎2 − �− �
𝑎𝑎
= 12 ⟹ 𝑎𝑎4 − 12𝑎𝑎2 − 16 = 0 ⟹ (𝑎𝑎2 − 16)(𝑎𝑎2 + 1) = 0
⟹ 𝑎𝑎2 = 16 (∵ (𝑎𝑎2 + 1) ≠ 0)
⟹ 𝑎𝑎 = ±4, 𝑏𝑏 = ∓1
Therefore, 𝑎𝑎 = +4, 𝑏𝑏 = −1 or 𝑎𝑎 = −4, 𝑏𝑏 = +1
That is, �12 − 8𝑗𝑗 = 4 − 𝑗𝑗 or �12 − 8𝑗𝑗 = −4 + 𝑗𝑗
[10] Find the complex number 𝑧𝑧 which satisfies 𝑧𝑧 2 + 2𝑧𝑧 − 11 + 𝑗𝑗8 = 0.
Sol)
𝑧𝑧 = −1 ± �1 − (−11 + 𝑗𝑗8) = −1 ± �12 − 𝑗𝑗8
Using the result of Problem [9],
(i) 𝑎𝑎 = +4, 𝑏𝑏 = −1 (i.e., �12 − 8𝑗𝑗 = 4 − 𝑗𝑗)
𝑧𝑧 = −1 ± �12 − 𝑗𝑗8 = −1 ± (4 − 𝑗𝑗) = 3 − 𝑗𝑗, −5 + 𝑗𝑗
(ii) 𝑎𝑎 = −4, 𝑏𝑏 = +1 (i.e., �12 − 8𝑗𝑗 = −4 + 𝑗𝑗)
𝑧𝑧 = −1 ± �12 − 𝑗𝑗8 = −1 ± (−4 + 𝑗𝑗) = −5 + 𝑗𝑗, 3 − 𝑗𝑗
Therefore, 𝑧𝑧 = 3 − 𝑗𝑗 or 𝑧𝑧 = −5 + 𝑗𝑗
[11] Simplify the following quantities.
(a)
Sol)
cos 3𝜃𝜃+𝑗𝑗 sin 3𝜃𝜃
sin 𝜃𝜃−𝑗𝑗 cos 𝜃𝜃
From De Moivre’s Formula
cos 3𝜃𝜃 + 𝑗𝑗 sin 3𝜃𝜃 = (cos 𝜃𝜃 + 𝑗𝑗 sin 𝜃𝜃)3
sin 𝜃𝜃 − 𝑗𝑗 cos 𝜃𝜃 = −𝑗𝑗(cos 𝜃𝜃 + 𝑗𝑗 sin 𝜃𝜃)
cos 3𝜃𝜃+𝑗𝑗 sin 3𝜃𝜃
sin 𝜃𝜃−𝑗𝑗 cos 𝜃𝜃
= 𝑗𝑗(cos 𝜃𝜃 + 𝑗𝑗 sin 𝜃𝜃)2 = 𝑗𝑗(cos 2𝜃𝜃 + 𝑗𝑗 sin 2𝜃𝜃) = − sin 2𝜃𝜃 + 𝑗𝑗 cos 2𝜃𝜃
2
3
4
𝜋𝜋
𝜋𝜋 2
2𝜋𝜋
+ 𝑗𝑗 sin
2𝜋𝜋
�
𝜋𝜋 3
3𝜋𝜋
+ 𝑗𝑗 sin
3𝜋𝜋
�
(b) �1 + 𝑗𝑗√3�, �1 + 𝑗𝑗√3� , �1 + 𝑗𝑗√3� , �1 + 𝑗𝑗√3�
Sol)
From De Moivre’s Formula
2
1
�1 + 𝑗𝑗√3� = 22 � + 𝑗𝑗
3
2
1
2
√3
�
2
= 4 �− + 𝑗𝑗
1
2
3
√3
� = −2 + 𝑗𝑗2√3
2
3
�1 + 𝑗𝑗√3� = 23 � + 𝑗𝑗
√3
�
2
4
√3
�
2
2
= 4 �cos + 𝑗𝑗 sin � = 4 �cos
𝜋𝜋
= 8 �cos + 𝑗𝑗 sin � = 8 �cos
= 8(−1 − 𝑗𝑗0) = −8
1
�1 + 𝑗𝑗√3� = 24 � + 𝑗𝑗
2
1
= 16 �− − 𝑗𝑗
2
4
3
3
3
𝜋𝜋
𝜋𝜋 4
3
3
= 16 �cos + 𝑗𝑗 sin � = 16 �cos
3
√3
� = −8 − 𝑗𝑗8√3
2
3
4𝜋𝜋
3
3
3
+ 𝑗𝑗 sin
4𝜋𝜋
3
�
[12] Show that cos 3𝜃𝜃 = cos3 𝜃𝜃 − 3 cos 𝜃𝜃 sin2 𝜃𝜃.
Sol)
From De Moivre’s Formula
cos 3𝜃𝜃 + 𝑗𝑗 sin 3𝜃𝜃 = (cos 𝜃𝜃 + 𝑗𝑗 sin 𝜃𝜃)3
= cos3 𝜃𝜃 + 3 cos2 𝜃𝜃 (𝑗𝑗 sin 𝜃𝜃) + 3 cos 𝜃𝜃 (𝑗𝑗 sin 𝜃𝜃)2 + (𝑗𝑗 sin 𝜃𝜃)3
= cos 3 𝜃𝜃 + 𝑗𝑗3 cos2 𝜃𝜃 sin 𝜃𝜃 − 3 cos 𝜃𝜃 sin2 𝜃𝜃 − 𝑗𝑗 sin3 𝜃𝜃
Comparing the real parts at both sides, we get
cos 3𝜃𝜃 = cos3 𝜃𝜃 − 3 cos 𝜃𝜃 sin2 𝜃𝜃
[13] Given a transfer function 𝐻𝐻(𝑠𝑠) =
1
, answer the following questions:
2𝜋𝜋
2𝜋𝜋
𝑗𝑗
−𝑗𝑗
�𝑠𝑠−𝑒𝑒 3 ��𝑠𝑠−𝑒𝑒 3 �
(a) Convert the transfer function to a rational function (e.g., 𝐻𝐻(𝑠𝑠) =
𝑁𝑁(𝑠𝑠)
𝐷𝐷(𝑠𝑠)
, with 𝑁𝑁(𝑠𝑠) and 𝐷𝐷(𝑠𝑠)
are the enumerated numerator polynomial and enumerated denominator polynomial,
respectively.)
Sol)
1
𝐻𝐻(𝑠𝑠) = (𝑠𝑠2
+𝑠𝑠+1)
(b) Find the frequency response expression 𝐻𝐻(𝑗𝑗𝑗𝑗) as a function of 𝜔𝜔, in the rectangular form.
Sol)
𝐻𝐻(𝑗𝑗𝑗𝑗) = 𝐻𝐻(𝑠𝑠)]𝑠𝑠=𝑗𝑗𝑗𝑗 = (𝑠𝑠2
=
1
(1−𝜔𝜔2 )+𝑗𝑗𝑗𝑗
�1−𝜔𝜔2 �−𝑗𝑗𝑗𝑗
= (1−𝜔𝜔2 )2
+𝜔𝜔2
1
�
+𝑠𝑠+1) 𝑠𝑠=𝑗𝑗𝑗𝑗
=
1−𝜔𝜔2
= (−𝜔𝜔2
1−𝜔𝜔2 +𝜔𝜔4
− 𝑗𝑗
1
+𝑗𝑗𝑗𝑗+1)
𝜔𝜔
1−𝜔𝜔2 +𝜔𝜔4
(c) Find the frequency response expression 𝐻𝐻(𝑗𝑗𝑗𝑗) as a function of 𝜔𝜔, in the Euler’s form.
Sol)
𝐻𝐻(𝑗𝑗𝑗𝑗) =
1−𝜔𝜔2
1−𝜔𝜔2 +𝜔𝜔4
− 𝑗𝑗
𝜔𝜔
1−𝜔𝜔2 +𝜔𝜔4
=
1
1−𝜔𝜔2 +𝜔𝜔4
�(1 − 𝜔𝜔2 ) − 𝑗𝑗𝑗𝑗�
Note: 1 − 𝜔𝜔2 + 𝜔𝜔4 = (1 − 𝜔𝜔2 )2 + 𝜔𝜔2 > 0
𝐻𝐻(𝑗𝑗𝑗𝑗) =
𝐻𝐻(𝑗𝑗𝑗𝑗) =
1
1−𝜔𝜔2 +𝜔𝜔
1
2)
− 𝑗𝑗𝑗𝑗� =
4 �(1 − 𝜔𝜔
�(1−𝜔𝜔2 +𝜔𝜔4 )
𝑒𝑒
−𝑗𝑗 tan−1 �
𝜔𝜔
�
1−𝜔𝜔2
If |𝜔𝜔| ≤ 1, then (1 − 𝜔𝜔2 ) ≥ 0
𝐻𝐻(𝑗𝑗𝑗𝑗) =
𝑒𝑒
4
�(1−𝜔𝜔2 +𝜔𝜔 )
2)
(1
If 𝜔𝜔 > 1, then
𝐻𝐻(𝑗𝑗𝑗𝑗) =
1
−𝑗𝑗 tan−1 �
1−𝜔𝜔2 +𝜔𝜔
𝜔𝜔
�
1−𝜔𝜔2
− 𝜔𝜔
< 0 & 𝜔𝜔 > 0
1
𝑒𝑒
1
𝑒𝑒
4
�(1−𝜔𝜔2 +𝜔𝜔 )
−𝑗𝑗�tan−1 �
𝜔𝜔
�+𝜋𝜋�
1−𝜔𝜔2
−𝑗𝑗�tan−1 �
𝜔𝜔
�−𝜋𝜋�
1−𝜔𝜔2
If 𝜔𝜔 < −1, then (1 − 𝜔𝜔2 ) < 0 & 𝜔𝜔 < 0
𝐻𝐻(𝑗𝑗𝑗𝑗) =
�(1−𝜔𝜔2 +𝜔𝜔4 )
1
�(1 − 𝜔𝜔 2 + 𝜔𝜔 4 )𝑒𝑒
4 ∙
𝑗𝑗 tan−1 �
−𝜔𝜔
�
1−𝜔𝜔2
[14] Given a transfer function 𝐻𝐻(𝑠𝑠) =
𝑠𝑠
, answer the following questions:
2𝜋𝜋
2𝜋𝜋
𝑗𝑗
−𝑗𝑗
�𝑠𝑠−𝑒𝑒 3 ��𝑠𝑠−𝑒𝑒 3 �
(a) Convert the transfer function to a rational function (e.g., 𝐻𝐻(𝑠𝑠) =
𝑁𝑁(𝑠𝑠)
𝐷𝐷(𝑠𝑠)
, with 𝑁𝑁(𝑠𝑠) and 𝐷𝐷(𝑠𝑠)
are the enumerated numerator polynomial and enumerated denominator polynomial,
respectively.)
Sol)
𝑠𝑠
𝐻𝐻(𝑠𝑠) = (𝑠𝑠2
+𝑠𝑠+1)
(b) Find the frequency response expression 𝐻𝐻(𝑗𝑗𝑗𝑗) as a function of 𝜔𝜔, in the rectangular form.
Sol)
𝐻𝐻(𝑗𝑗𝑗𝑗) = 𝐻𝐻(𝑠𝑠)]𝑠𝑠=𝑗𝑗𝑗𝑗 = (𝑠𝑠2
=
𝑗𝑗𝑗𝑗
(1−𝜔𝜔2 )+𝑗𝑗𝑗𝑗
=
𝑠𝑠
�
+𝑠𝑠+1) 𝑠𝑠=𝑗𝑗𝑗𝑗
𝑗𝑗𝑗𝑗��1−𝜔𝜔2 �−𝑗𝑗𝑗𝑗�
(1−𝜔𝜔2 )2 +𝜔𝜔2
=
= (−𝜔𝜔2
𝑗𝑗𝑗𝑗
+𝑗𝑗𝑗𝑗+1)
𝜔𝜔2 +𝑗𝑗𝑗𝑗�1−𝜔𝜔2 �
1−𝜔𝜔2 +𝜔𝜔4
=
𝜔𝜔2
1−𝜔𝜔2 +𝜔𝜔
4 + 𝑗𝑗
𝜔𝜔−𝜔𝜔3
1−𝜔𝜔2 +𝜔𝜔4
(c) Find the frequency response expression 𝐻𝐻(𝑗𝑗𝑗𝑗) as a function of 𝜔𝜔, in the Euler’s form.
Sol)
𝐻𝐻(𝑗𝑗𝑗𝑗) =
𝜔𝜔2
1−𝜔𝜔2 +𝜔𝜔
4 + 𝑗𝑗
𝜔𝜔−𝜔𝜔3
1−𝜔𝜔2 +𝜔𝜔4
=
1
1−𝜔𝜔2 +𝜔𝜔4
�𝜔𝜔2 + 𝑗𝑗(𝜔𝜔 − 𝜔𝜔3 )�
Note: 1 − 𝜔𝜔2 + 𝜔𝜔4 = (1 − 𝜔𝜔2 )2 + 𝜔𝜔2 > 0
𝐻𝐻(𝑗𝑗𝑗𝑗) =
𝐻𝐻(𝑗𝑗𝑗𝑗) =
1
1−𝜔𝜔2 +𝜔𝜔4
|𝜔𝜔|
2
�𝜔𝜔 + 𝑗𝑗(𝜔𝜔 − 𝜔𝜔
�(1−𝜔𝜔2 +𝜔𝜔4 )
𝑒𝑒
3 )�
�𝜔𝜔−𝜔𝜔3 �
𝑗𝑗 tan−1 �
�
𝜔𝜔2
Note: 𝜔𝜔: −∞~ + ∞ → 𝜔𝜔2 ≥ 0
=
1
1−𝜔𝜔2 +𝜔𝜔4
3
−1 ��𝜔𝜔−𝜔𝜔 ��
2
𝜔𝜔
𝑗𝑗 tan
∙ �𝜔𝜔 2 (1 − 𝜔𝜔 2 + 𝜔𝜔 4 )𝑒𝑒
[15] Find the phasor of the following sinusoids:
(a) 𝑥𝑥(𝑡𝑡) = 10 cos(20𝜋𝜋𝜋𝜋 + π/4)
Sol)
π
𝐗𝐗 = 10𝑒𝑒 𝑗𝑗 4 (𝜔𝜔 = 20𝜋𝜋 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
(b) 𝑥𝑥(𝑡𝑡) = 10 cos(20𝑡𝑡 + π/4)
Sol)
π
𝐗𝐗 = 10𝑒𝑒 𝑗𝑗 4 (𝜔𝜔 = 20 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
(c) 𝑥𝑥(𝑡𝑡) = 5 sin(20𝑡𝑡 + π/4)
Sol)
𝑥𝑥(𝑡𝑡) = 5 𝑠𝑠𝑠𝑠𝑠𝑠(20𝑡𝑡 + π/4) = 5 𝑐𝑐𝑐𝑐𝑐𝑐(20𝑡𝑡 + 𝜋𝜋/4 − 𝜋𝜋/2) = 5 cos(20𝑡𝑡 − π/4)
π
𝐗𝐗 = 5𝑒𝑒 −𝑗𝑗 4 (𝜔𝜔 = 20 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
(d) 𝑥𝑥(𝑡𝑡) = −5 cos(20𝑡𝑡 + π/4)
Sol)
𝑥𝑥(𝑡𝑡) = −5 𝑐𝑐𝑐𝑐𝑐𝑐(20𝑡𝑡 + 𝜋𝜋/4) = 5 𝑐𝑐𝑐𝑐𝑐𝑐(20𝑡𝑡 + 𝜋𝜋/4 − 𝜋𝜋) = 5 cos(20𝑡𝑡 − 3π/4)
3π
𝐗𝐗 = 5𝑒𝑒 −𝑗𝑗 4 (𝜔𝜔 = 20 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
(e) 𝑥𝑥(𝑡𝑡) = −5 sin(50𝑡𝑡 + π/4)
Sol)
𝑥𝑥(𝑡𝑡) = −5 𝑠𝑠𝑠𝑠𝑠𝑠(50𝑡𝑡 + 𝜋𝜋/4) = 5 𝑐𝑐𝑐𝑐𝑐𝑐(50𝑡𝑡 + 𝜋𝜋/4 + 𝜋𝜋/2) = 5 cos(50𝑡𝑡 + 3π/4)
3π
𝐗𝐗 = 5𝑒𝑒 𝑗𝑗 4 (𝜔𝜔 = 50 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
[16] Find the corresponding sinusoids of the following phasors:
(a) 𝐗𝐗 = 10𝑒𝑒 𝑗𝑗45° (𝜔𝜔 = 10 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
Sol)
𝑥𝑥(𝑡𝑡) = 10 cos(10𝑡𝑡 + 45°)
(b) 𝐗𝐗 = 10𝑒𝑒 𝑗𝑗45° (𝜔𝜔 = 100 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
Sol)
𝑥𝑥(𝑡𝑡) = 10 cos(100𝑡𝑡 + 45°)
(c) 𝐗𝐗 = 10𝑒𝑒 𝑗𝑗45° − 10 (𝜔𝜔 = 100 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
Sol)
𝐗𝐗 = 10𝑒𝑒 𝑗𝑗45° − 10 = �−10 + 5√2� + 𝑗𝑗5√2 = 10�2 − √2𝑒𝑒 𝑗𝑗112.5°
𝑥𝑥(𝑡𝑡) = 10�2 − √2 cos(100𝑡𝑡 + 112.5°)
(d) 𝐗𝐗 = 8 + 𝑗𝑗6 (𝜔𝜔 = 100 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
Sol)
−1 3
𝐗𝐗 = 8 + 𝑗𝑗6 = 10𝑒𝑒 𝑗𝑗 tan �4� = 10𝑒𝑒 𝑗𝑗36.87°
𝑥𝑥(𝑡𝑡) = 10 cos(100𝑡𝑡 + 36.87°)
π
𝜋𝜋
[17] Given 𝑥𝑥1 (𝑡𝑡) = 10 cos �𝑡𝑡 + � and 𝑥𝑥2 (𝑡𝑡) = 10 𝑐𝑐𝑐𝑐𝑐𝑐 �2𝑡𝑡 + �, answer the following questions:
4
4
(a) Find the corresponding phasors of the sinusoids 𝑥𝑥1 (𝑡𝑡) and 𝑥𝑥2 (𝑡𝑡).
Sol)
π
π
𝐗𝐗 𝟏𝟏 = 10𝑒𝑒 𝑗𝑗 4 (𝜔𝜔 = 1 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠), 𝐗𝐗 𝟐𝟐 = 10𝑒𝑒 𝑗𝑗 4 (𝜔𝜔 = 2 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
(b) Find the phasors of 𝑧𝑧1 (𝑡𝑡) =
Sol)
Note: 𝒫𝒫 �
𝑑𝑑 𝑛𝑛 𝑥𝑥(𝑡𝑡)
𝑑𝑑𝑡𝑡 𝑛𝑛
𝑑𝑑 2 𝑥𝑥1 (𝑡𝑡)
� = (𝑗𝑗𝑗𝑗)𝑛𝑛 𝐗𝐗
𝑑𝑑 2 𝑥𝑥1 (𝑡𝑡)
𝑑𝑑𝑡𝑡 2
+3
𝑑𝑑𝑥𝑥1 (𝑡𝑡)
𝑑𝑑𝑑𝑑
+ 4𝑥𝑥1 (𝑡𝑡) and 𝑧𝑧2 (𝑡𝑡) =
𝑑𝑑 2 𝑥𝑥2 (𝑡𝑡)
𝑑𝑑𝑡𝑡 2
+3
𝑑𝑑𝑥𝑥2 (𝑡𝑡)
𝑑𝑑𝑑𝑑
+ 4𝑥𝑥2 (𝑡𝑡).
𝑑𝑑𝑥𝑥 (𝑡𝑡)
𝑧𝑧1 (𝑡𝑡) =
+ 3 1 + 4𝑥𝑥1 (𝑡𝑡) (𝜔𝜔 = 1 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
𝑑𝑑𝑡𝑡 2
𝑑𝑑𝑑𝑑
→ 𝐙𝐙𝟏𝟏 = (𝑗𝑗𝑗𝑗)2 𝐗𝐗 𝟏𝟏 + 3(𝑗𝑗𝑗𝑗)𝐗𝐗 𝟏𝟏 + 4𝐗𝐗 𝟏𝟏 = ((𝑗𝑗𝑗𝑗)2 + 3(𝑗𝑗𝑗𝑗) + 4)𝐗𝐗 𝟏𝟏 = (−𝜔𝜔2 + 3𝜔𝜔𝜔𝜔 + 4)𝐗𝐗 𝟏𝟏
π
π
−1 (1)�
→ 𝐙𝐙𝟏𝟏 = (−12 + 3 ∙ 1𝑗𝑗 + 4) ∙ 10𝑒𝑒 𝑗𝑗 4 = (3 + 3𝑗𝑗) ∙ 10𝑒𝑒 𝑗𝑗 4 = 3√2𝑒𝑒 𝑗𝑗�tan
π
→ 𝐙𝐙𝟏𝟏 = 30√2𝑒𝑒 𝑗𝑗 2 (𝜔𝜔 = 1 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
𝑑𝑑 2 𝑥𝑥 (𝑡𝑡)
π
∙ 10𝑒𝑒 𝑗𝑗 4
𝑑𝑑𝑥𝑥 (𝑡𝑡)
2
𝑧𝑧2 (𝑡𝑡) =
+ 3 2 + 4𝑥𝑥2 (𝑡𝑡) (𝜔𝜔 = 2 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
𝑑𝑑𝑡𝑡 2
𝑑𝑑𝑑𝑑
→ 𝐙𝐙𝟐𝟐 = (𝑗𝑗𝑗𝑗)2 𝐗𝐗 𝟐𝟐 + 3(𝑗𝑗𝑗𝑗)𝐗𝐗 𝟐𝟐 + 4𝐗𝐗 𝟐𝟐 = ((𝑗𝑗𝑗𝑗)2 + 3(𝑗𝑗𝑗𝑗) + 4)𝐗𝐗 𝟐𝟐 = (−𝜔𝜔2 + 3𝜔𝜔𝜔𝜔 + 4)𝐗𝐗 𝟐𝟐
π
π
3π
→ 𝐙𝐙𝟐𝟐 = (−22 + 3 ∙ 2𝑗𝑗 + 4) ∙ 10𝑒𝑒 𝑗𝑗 4 = (6𝑗𝑗) ∙ 10𝑒𝑒 𝑗𝑗 4 = 60𝑒𝑒 𝑗𝑗 4 (𝜔𝜔 = 2 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠)
(c) Find the corresponding sinusoids of 𝑧𝑧1 (𝑡𝑡) and 𝑧𝑧2 (𝑡𝑡).
Sol)
π
𝐙𝐙𝟏𝟏 = 30√2𝑒𝑒 𝑗𝑗 2
→ 𝑧𝑧1 (𝑡𝑡) = 30√2 cos(𝑡𝑡 + π/2)
3π
𝐙𝐙𝟐𝟐 = 60𝑒𝑒 𝑗𝑗 4
→ 𝑧𝑧2 (𝑡𝑡) = 60 cos(2𝑡𝑡 + 3π/4)
[18] Find the impedances 𝑍𝑍𝑎𝑎𝑎𝑎 of the following circuit parts in terms of 𝜔𝜔, 𝑅𝑅, 𝐿𝐿, and 𝐶𝐶:
(a)
(b)
(c)
(d)
(e)
Sol)
Impedances for 𝑅𝑅, 𝐿𝐿, and 𝐶𝐶: 𝑍𝑍𝑅𝑅 = 𝑅𝑅, 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝑗𝑗𝑗𝑗, 𝑍𝑍𝐶𝐶 = 1/𝑗𝑗𝑗𝑗𝑗𝑗
(a)
Sol)
𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑅𝑅 ||𝑍𝑍𝐶𝐶 =
𝑍𝑍𝑅𝑅 𝑍𝑍𝐶𝐶
𝑍𝑍𝑅𝑅 +𝑍𝑍𝐶𝐶
=
1
𝑅𝑅∙𝑗𝑗𝑗𝑗𝑗𝑗
1
𝑅𝑅+𝑗𝑗𝑗𝑗𝑗𝑗
=
𝑅𝑅
1+𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗
(b)
Sol)
𝑍𝑍 𝑍𝑍
𝑅𝑅∙𝑗𝑗𝑗𝑗𝑗𝑗
𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗
𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑅𝑅 ||𝑍𝑍𝐿𝐿 = 𝑅𝑅 𝐿𝐿 =
=
𝑅𝑅+𝑗𝑗𝑗𝑗𝑗𝑗
𝑍𝑍𝑅𝑅 +𝑍𝑍𝐿𝐿
(c)
Sol)
𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑅𝑅 + 𝑍𝑍𝐶𝐶 = 𝑅𝑅 +
1
𝑗𝑗𝑗𝑗𝑗𝑗
(d)
Sol)
𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑅𝑅 + 𝑍𝑍𝐿𝐿 = 𝑅𝑅 + 𝑗𝑗𝑗𝑗𝑗𝑗
=
𝑅𝑅+𝑗𝑗𝑗𝑗𝑗𝑗
1+𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗
𝑗𝑗𝑗𝑗𝑗𝑗
(e)
Sol)
𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑅𝑅 + 𝑍𝑍𝐿𝐿 + 𝑍𝑍𝐶𝐶 = 𝑅𝑅 + 𝑗𝑗𝑗𝑗𝑗𝑗 +
1
𝑗𝑗𝑗𝑗𝑗𝑗
= 𝑅𝑅 + 𝑗𝑗 �𝜔𝜔𝜔𝜔 −
1
𝜔𝜔𝜔𝜔
�
[19] Answer the following questions for the circuit below:
(a) Find the impedance 𝑍𝑍𝑎𝑎𝑎𝑎 as a function of 𝜔𝜔.
Sol)
𝑍𝑍𝑎𝑎𝑎𝑎 = 104 + 𝑗𝑗 �10−3 𝜔𝜔 −
1
10−9 𝜔𝜔
�
(b) Find the value of 𝜔𝜔 such that 𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑅𝑅 = 10𝑘𝑘Ω.
Sol)
1
𝑍𝑍𝑎𝑎𝑎𝑎 = 104 + 𝑗𝑗 �10−3 𝜔𝜔 − −9 � = 104
10 𝜔𝜔
1
1
→ 10−3 𝜔𝜔 − −9 = 0 → 10−3 𝜔𝜔 = −9
→ 𝜔𝜔2 = 1012
10 𝜔𝜔
10 𝜔𝜔
→ 𝜔𝜔 = 106 rads/s
(c) Given 𝑣𝑣𝑖𝑖 (𝑡𝑡) = 10 cos(106 𝑡𝑡 + π/4), find 𝑣𝑣𝐿𝐿 (𝑡𝑡) using the phasor approach.
Sol)
𝜔𝜔 = 106 rads/s
𝑍𝑍𝑎𝑎𝑎𝑎 = 104 + 𝑗𝑗 �10−3 ∙ 106 −
π
𝐕𝐕𝐢𝐢 = 10𝑒𝑒 𝑗𝑗 4
𝐈𝐈 =
𝐕𝐕𝐢𝐢
𝑍𝑍𝑎𝑎𝑎𝑎
=
π
𝑗𝑗
10𝑒𝑒 4
104
1
10−9 ∙106
π
= 10−3 𝑒𝑒 𝑗𝑗 4
π
3π
𝐕𝐕𝐋𝐋 = 𝐈𝐈𝑍𝑍𝐿𝐿 = 10−3 𝑒𝑒 𝑗𝑗 4 ∙ 𝑗𝑗103 = 𝑒𝑒 𝑗𝑗 4
→ 𝑣𝑣𝐿𝐿 (𝑡𝑡) = cos(106 𝑡𝑡 + 3π/4)
� = 104
[20] A 2nd-order moving average filter is described by the following difference equation.
𝑦𝑦[𝑛𝑛] = 𝑥𝑥[𝑛𝑛] + 𝑥𝑥[𝑛𝑛 − 1] + 𝑥𝑥[𝑛𝑛 − 2]
(a) Find the transfer function 𝐻𝐻(𝑧𝑧).
Sol)
By inspection,
𝐻𝐻(𝑧𝑧) = 1 + 𝑧𝑧 −1 + 𝑧𝑧 −2 or 𝐻𝐻(𝑧𝑧) =
𝑧𝑧 2 +𝑧𝑧+1
(Supplementary)
Using Z-transform
𝑧𝑧 2
𝑦𝑦[𝑛𝑛] = 𝑥𝑥 [𝑛𝑛] + 𝑥𝑥 [𝑛𝑛 − 1] + 𝑥𝑥 [𝑛𝑛 − 2] ↔ 𝑌𝑌(𝑧𝑧) = 𝑋𝑋(𝑧𝑧) + 𝑧𝑧 −1 𝑋𝑋(𝑧𝑧) + 𝑧𝑧 −2 𝑋𝑋(𝑧𝑧) = (1 + 𝑧𝑧 −1 + 𝑧𝑧 −2 )𝑋𝑋(𝑧𝑧)
Transfer function 𝐻𝐻(𝑧𝑧)
𝐻𝐻(𝑧𝑧) ≡
𝑌𝑌(𝑧𝑧)
𝑋𝑋(𝑧𝑧)
= 1 + 𝑧𝑧 −1 + 𝑧𝑧 −2 or 𝐻𝐻(𝑧𝑧) =
𝑧𝑧 2 +𝑧𝑧 1 +1
(b) Find the poles and zeros of 𝐻𝐻(𝑧𝑧)
Sol)
𝑧𝑧 2
𝑧𝑧 2 +𝑧𝑧 1 +1
From 𝐻𝐻(𝑧𝑧) =
𝑧𝑧 2
Poles: 𝑝𝑝1 = 0, 𝑝𝑝2 = 0
1
Zeros: 𝑧𝑧1 = − + 𝑗𝑗
2
√3
,
2
(Supplementary) 𝐻𝐻(𝑧𝑧) =
1
𝑧𝑧2 = − − 𝑗𝑗
𝑁𝑁(𝑧𝑧)
𝐷𝐷(𝑧𝑧)
2
√3
2
Poles: Roots of 𝐷𝐷(𝑧𝑧) = 𝑧𝑧 2 → 𝑝𝑝1 = 0, 𝑝𝑝2 = 0
1
Zeros: Roots of 𝑁𝑁(𝑧𝑧) = 𝑧𝑧 2 + 𝑧𝑧 + 1 = 0 → 𝑧𝑧1 = − + 𝑗𝑗
(c) Plot the corresponding pole-zero diagram.
Sol)
2
√3
,
2
1
𝑧𝑧2 = − − 𝑗𝑗
2
√3
2
(d) What is the analog frequency 𝑓𝑓 that this filter can remove, when the sampling rate 𝑓𝑓𝑠𝑠 =
24,000 Hz.
Sol)
Note that a signal can be removed completely at a zero
1
𝑧𝑧1 = − + 𝑗𝑗
2𝜋𝜋
𝑒𝑒 𝑗𝑗 3 = 𝑒𝑒
𝑓𝑓
2
𝑓𝑓
2𝜋𝜋
√3
= 𝑒𝑒 𝑗𝑗 3
2
𝑗𝑗2𝜋𝜋� 𝑓𝑓𝑟𝑟 �
𝑠𝑠
𝑓𝑓𝑟𝑟 = 𝑠𝑠 = 8,000 Hz
3
[21] The difference equation of a digital filter is given by
1
1
3
1
1
𝑦𝑦[𝑛𝑛] = 𝑦𝑦[𝑛𝑛 − 1] + 𝑦𝑦[𝑛𝑛 − 2] − 𝑦𝑦[𝑛𝑛 − 3] + 𝑥𝑥[𝑛𝑛] − 𝑥𝑥[𝑛𝑛 − 1] − 𝑥𝑥[𝑛𝑛 − 2]
4
4
8
2
2
(a) Find the transfer function 𝐻𝐻(𝑧𝑧).
Sol)
1
1
3
1
1
𝑦𝑦[𝑛𝑛] − 𝑦𝑦[𝑛𝑛 − 1] − 𝑦𝑦[𝑛𝑛 − 2] + 𝑦𝑦[𝑛𝑛 − 3] = 𝑥𝑥[𝑛𝑛] − 𝑥𝑥[𝑛𝑛 − 1] − 𝑥𝑥[𝑛𝑛 − 2]
4
𝐻𝐻(𝑧𝑧) =
4
1
1
1−2𝑧𝑧 −1 −2𝑧𝑧 −2
1
1
3
1−4𝑧𝑧 −1 −4𝑧𝑧 −2 +8𝑧𝑧 −3
1
1
𝑧𝑧 3 −2𝑧𝑧 2 −2𝑧𝑧
𝐻𝐻(𝑧𝑧) = 3 1 2 1
8
2
3
𝑧𝑧 −4𝑧𝑧 −4𝑧𝑧+8
(b) Find the poles and zeros of 𝐻𝐻(𝑧𝑧).
Sol)
1
1
𝑧𝑧 3 − 𝑧𝑧 2 − 𝑧𝑧
From 𝐻𝐻(𝑧𝑧) = 3 1 22 1 2 3
Poles:
𝑧𝑧 −4𝑧𝑧 −4𝑧𝑧+8
3
1
1
𝑝𝑝1 = − , 𝑝𝑝2 = + 𝑗𝑗 ,
4
2
2
Zeros: 𝑧𝑧1 = 0, 𝑧𝑧2 = 1, 𝑧𝑧3 = −
1
𝑝𝑝3 = − 𝑗𝑗
1
2
(c) Plot the pole-zero diagram of 𝐻𝐻(𝑧𝑧).
Sol)
2
1
2
2
[22] The poles and zeros of a digital filter are as follows:
𝜋𝜋
𝜋𝜋
𝜋𝜋
𝜋𝜋
1
1
1
1
Poles: 𝑝𝑝1 = 𝑒𝑒 𝑗𝑗 4 , 𝑝𝑝2 = 𝑒𝑒 −𝑗𝑗 4 , 𝑝𝑝3 = 𝑒𝑒 𝑗𝑗 2 , 𝑝𝑝4 = 𝑒𝑒 −𝑗𝑗 2
√2
2
√2
Zeros: 𝑧𝑧1 = 0, 𝑧𝑧2 = 0, 𝑧𝑧3 = 1, 𝑧𝑧4 = −1
2
(a) Plot the pole-zero diagram of the digital filter.
Sol)
(b) Find the transfer function 𝐻𝐻(𝑧𝑧).
Sol)
From pole-zero to transfer function
𝐻𝐻(𝑧𝑧) =
𝐻𝐻(𝑧𝑧) =
(𝑧𝑧−𝑧𝑧1 )(𝑧𝑧−𝑧𝑧2 )(𝑧𝑧−𝑧𝑧3 )(𝑧𝑧−𝑧𝑧4 )
(𝑧𝑧−𝑝𝑝1 )(𝑧𝑧−𝑝𝑝2 )(𝑧𝑧−𝑝𝑝3 )(𝑧𝑧−𝑝𝑝4 )
(𝑧𝑧−0)(𝑧𝑧−0)(𝑧𝑧−(+1))(𝑧𝑧−(−1))
1 1
1 1
1
1
�𝑧𝑧−�2−𝑗𝑗2���𝑧𝑧−�2+𝑗𝑗 2���𝑧𝑧−�+𝑗𝑗 2���𝑧𝑧−�−𝑗𝑗2��
𝐻𝐻(𝑧𝑧) = 4
𝐻𝐻(𝑧𝑧) =
𝑧𝑧 4 −𝑧𝑧 2
𝑧𝑧 −𝑧𝑧 3 +34𝑧𝑧 2 −14𝑧𝑧+18
1−𝑧𝑧 −2
1−𝑧𝑧 −1 +34𝑧𝑧 −2 −14𝑧𝑧 −3 +18𝑧𝑧 −4
(c) Find the difference equation [The current output should be expressed as a function of previous
outputs and current and previous inputs].
Sol)
By inspection from 𝐻𝐻(𝑧𝑧) =
1−𝑧𝑧 −2
1−𝑧𝑧 −1 +34𝑧𝑧 −2 −14𝑧𝑧 −3 +18𝑧𝑧 −4
1
1
4
8
𝑦𝑦[𝑛𝑛] − 𝑦𝑦[𝑛𝑛 − 1] + 34𝑦𝑦[𝑛𝑛 − 2] − 𝑦𝑦[𝑛𝑛 − 3] + 𝑦𝑦[𝑛𝑛 − 4] = 𝑥𝑥[𝑛𝑛] − 𝑥𝑥[𝑛𝑛 − 2]
Difference equation for the current output
𝑦𝑦[𝑛𝑛] = 𝑦𝑦[𝑛𝑛 − 1] − 34𝑦𝑦[𝑛𝑛 − 2] + 14𝑦𝑦[𝑛𝑛 − 3] − 18𝑦𝑦[𝑛𝑛 − 4] + 𝑥𝑥[𝑛𝑛] − 𝑥𝑥[𝑛𝑛 − 2]
[23] You are asked to design a digital filter that will remove the two frequencies 𝑓𝑓𝑟𝑟1 = 5,500 Hz and
𝑓𝑓𝑟𝑟2 = 11,000 Hz in the signal, whose spectrum is shown below. Note that the sampling frequency
𝑓𝑓𝑠𝑠 = 55,000 Hz.
(a) Find the zeros and poles by setting the corresponding poles to 𝑝𝑝i = 0.99𝑧𝑧𝑖𝑖 .
Sol)
Zeros: 𝑧𝑧1,2 = 𝑒𝑒
𝑓𝑓
±𝑗𝑗 2𝜋𝜋� 𝑓𝑓𝑟𝑟1 �
Poles: 𝑝𝑝1,2 = 0.99𝑒𝑒
𝑠𝑠
𝜋𝜋
= 𝑒𝑒 ±𝑗𝑗 5 , 𝑧𝑧3,4 = 𝑒𝑒
𝑓𝑓
±𝑗𝑗 2𝜋𝜋� 𝑓𝑓𝑟𝑟1 �
𝑠𝑠
𝜋𝜋
𝑓𝑓
±𝑗𝑗 2𝜋𝜋� 𝑓𝑓𝑟𝑟2 �
𝑠𝑠
= 0.99𝑒𝑒 ±𝑗𝑗 5 , 𝑝𝑝3,4 = 0.99𝑒𝑒
(b) Plot the pole-zero diagram.
Sol)
(c) Obtain the transfer function 𝐻𝐻(𝑧𝑧).
Sol)
From pole-zero to transfer function
(𝑧𝑧−𝑧𝑧1 )(𝑧𝑧−𝑧𝑧2 )(𝑧𝑧−𝑧𝑧3 )(𝑧𝑧−𝑧𝑧4 )
𝐻𝐻(𝑧𝑧) =
(𝑧𝑧−𝑝𝑝1 )(𝑧𝑧−𝑝𝑝2 )(𝑧𝑧−𝑝𝑝3 )(𝑧𝑧−𝑝𝑝4 )
𝐻𝐻(𝑧𝑧) =
𝜋𝜋
𝑗𝑗
�𝑧𝑧−𝑒𝑒 5 ��𝑧𝑧−𝑒𝑒
𝜋𝜋
𝑗𝑗
�𝑧𝑧−0.99𝑒𝑒 5 ��𝑧𝑧−0.99𝑒𝑒
𝐻𝐻(𝑧𝑧) = 4
𝜋𝜋
2𝜋𝜋
2𝜋𝜋
−𝑗𝑗
𝑗𝑗
−𝑗𝑗
5 ��𝑧𝑧−𝑒𝑒 5 ��𝑧𝑧−𝑒𝑒
5 �
𝜋𝜋
2𝜋𝜋
2𝜋𝜋
−𝑗𝑗
𝑗𝑗
−𝑗𝑗
5 ��𝑧𝑧−0.99𝑒𝑒 5 ��𝑧𝑧−0.99𝑒𝑒
5 �
𝑧𝑧 4 −√5𝑧𝑧 3 +3𝑧𝑧 2 −√5𝑧𝑧+1
𝑧𝑧 −√5(0.99)𝑧𝑧 3 +3(0.99)2 𝑧𝑧 2 −√5(0.99)3 𝑧𝑧+(0.99)4
2𝜋𝜋
= 𝑒𝑒 ±𝑗𝑗 5
𝑓𝑓
±𝑗𝑗 2𝜋𝜋� 𝑓𝑓𝑟𝑟2 �
𝑠𝑠
2𝜋𝜋
= 0.99𝑒𝑒 ±𝑗𝑗 5
𝐻𝐻(𝑧𝑧) =
1−√5𝑧𝑧 −1 +3𝑧𝑧 −2 −√5𝑧𝑧 −3 +𝑧𝑧 −4
1−√5(0.99)𝑧𝑧 −1 +3(0.99)2 𝑧𝑧 −2 −√5(0.99)3 𝑧𝑧 −3 +(0.99)4 𝑧𝑧 −4
(d) Obtain the difference equation [The current output should be expressed as a function of
previous outputs and current and previous inputs].
Sol)
From 𝐻𝐻(𝑧𝑧) =
1−√5𝑧𝑧 −1 +3𝑧𝑧 −2 −√5𝑧𝑧 −3 +𝑧𝑧 −4
1−√5(0.99)𝑧𝑧 −1 +3(0.99)2 𝑧𝑧 −2 −√5(0.99)3 𝑧𝑧 −3 +(0.99)4 𝑧𝑧 −4
2
[𝑛𝑛
𝑦𝑦[𝑛𝑛] − √5(0.99)𝑦𝑦
− 1] + 3(0.99) 𝑦𝑦[𝑛𝑛 − 2] − √5(0.99)3 𝑦𝑦[𝑛𝑛 − 3] + (0.99)4 𝑦𝑦[𝑛𝑛 − 4]
= 𝑥𝑥[𝑛𝑛] − √5𝑥𝑥[𝑛𝑛 − 1] + 3𝑥𝑥[𝑛𝑛 − 2] − √5𝑥𝑥[𝑛𝑛 − 3] + 𝑥𝑥[𝑛𝑛 − 4]
Difference equation for the current output
𝑦𝑦[𝑛𝑛] = √5(0.99)𝑦𝑦[𝑛𝑛 − 1] − 3(0.99)2 𝑦𝑦[𝑛𝑛 − 2] + √5(0.99)3 𝑦𝑦[𝑛𝑛 − 3] − (0.99)4 𝑦𝑦[𝑛𝑛 − 4]
+𝑥𝑥[𝑛𝑛] − √5𝑥𝑥[𝑛𝑛 − 1] + 3𝑥𝑥[𝑛𝑛 − 2] − √5𝑥𝑥[𝑛𝑛 − 3] + 𝑥𝑥[𝑛𝑛 − 4]
Hint: Fill the table.
-----------------------------------------------------------------------------------------------------------------------------------Sampling frequency
Target frequency
Zeros
Poles (with 𝑟𝑟 = 0.99)
-----------------------------------------------------------------------------------------------------------------------------------𝑓𝑓𝑠𝑠 = 55,000 Hz
𝑓𝑓𝑟𝑟1 = 5,500 Hz
𝑧𝑧1 =
𝑝𝑝1 =
𝑧𝑧2 =
𝑝𝑝2 =
𝑓𝑓𝑟𝑟2 = 11,000 Hz
𝑧𝑧3 =
𝑝𝑝3 =
𝑧𝑧4 =
𝑝𝑝4 =