Topics covered in this week:

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1 | P a g e Umm Al-Qura Universtiy, Makkah

Department of Electrical Engineering Electromagnetics (2) - (8022320)

Term 1: 2021/2022

Solution Homework 11 Dr. Waheed Ahmad Younis

Do not submit this homework. There will be a quiz from this homework on Wednesday, 08 Dec 2021.

Topics covered in this week:

• Smith chart:

o Stub matching

o Quarter wavelength impedance transformer

____________________________________________________________________________________

Q1. For each of the following, use Smith chart to find the normalized input impedance of a lossless line of length 𝑙 terminated in a normalized load impedance 𝑧

_𝐿

. Also find the reflection co-efficient and SWR:

a. 𝑙 = 0.25𝜆, 𝑧

_𝐿

= 1 + 𝑗0 b. 𝑙 = 0.5𝜆, 𝑧

_𝐿

= 1 + 𝑗 c. 𝑙 = 0.3𝜆, 𝑧

_𝐿

= 1 − 𝑗 d. 𝑙 = 1.2𝜆, 𝑧

_𝐿

= 0.5 − 𝑗0.5 e. 𝑙 = 0.1𝜆, 𝑧

_𝐿

= 𝑆ℎ𝑜𝑟𝑡 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 f. 𝑙 = 0.4𝜆, 𝑧

_𝐿

= 𝑗3

g. 𝑙 = 0.2𝜆, 𝑧

_𝐿

= 𝑂𝑝𝑒𝑛 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 Solution:

𝑍

_𝑖𝑛

= 𝑍(𝑙) = 𝑍

₀^𝑍^𝐿^+𝑗𝑍⁰^{tan 𝛽𝑙}

𝑍₀+𝑗𝑍_𝐿tan 𝛽𝑙

⇒ 𝑧

_𝑖𝑛

=

^𝑍^𝑖𝑛

𝑍₀

=

^𝑧^𝐿^{+𝑗 tan 𝛽𝑙}

1+𝑗𝑧_𝐿tan 𝛽𝑙

. Also, 𝑙 = 𝑎𝜆 = 𝑎

^2𝜋

𝛽

⇒ 𝛽𝑙 = 𝑎(2𝜋) Γ =

^𝑍^𝐿^−𝑍⁰

𝑍𝐿+𝑍0

=

𝑍𝐿 𝑍0−1 𝑍𝐿

𝑍0+1

=

^𝑧^𝐿⁻¹

𝑧𝐿+1

a. 𝑙 = 0.25𝜆, 𝑧

_𝐿

= 1 + 𝑗0. Impedance is matched with line. 1 + 𝑗0 is in the center of the Smith chart.

No change at any distance. Hence 𝑧

_𝑖𝑛

= 1 + 𝑗0

From Smith chart, Γ = 0 and SWR = 1 (SWR circle has zero radius).

𝑧

_𝑖𝑛

=

1+𝑗𝑧𝐿tan 𝛽𝑙

=

^{1+𝑗 tan 𝛽𝑙}

1+𝑗 tan 𝛽𝑙

= 1 + 𝑗0 Γ =

^𝑧^𝐿⁻¹

𝑧𝐿+1

=

¹⁻¹

1+1

= 0, 𝑆𝑊𝑅 =

^1+|Γ|

1−|Γ|

=

¹⁺⁰

1−0

= 1

b. 𝑙 = 0.5𝜆, 𝑧

_𝐿

= 1 + 𝑗. Since 𝑙 = 0.5𝜆, we will be at the same point on Smith chart after one complete revolution. No change of impedance. Hence 𝑧

_𝑖𝑛

= 1 + 𝑗.

𝑧

_𝑖𝑛

=

^𝑧^𝐿+𝑗 tan 0.5(2𝜋)

1+𝑗𝑧𝐿tan 0.5(2𝜋)

=

^𝑧^𝐿^+𝑗0

1+0

= 𝑧

_𝐿

= 1 + 𝑗 Γ =

^𝑧^𝐿⁻¹

𝑧𝐿+1

=

^1+𝑗−1

1+𝑗+1

=

^𝑗

2+𝑗

=

¹

√5

∠63.43°, 𝑆𝑊𝑅 =

^1+|Γ|

1−|Γ|

=

¹⁺

1

√5 1−¹

√5

= 2.618

(2)

2 | P a g e

c. 𝑙 = 0.3𝜆, 𝑧

_𝐿

= 1 − 𝑗. Impedance is marked as “A” (wtg = 0.338). For input impedance, wtg =

0.338+0.3 = 0.638 = (0.5)+0.138 ≡ 0.138. Hence the input impedance (“C”) 𝑧

_𝑖𝑛

= 0.76 + 𝑗0.84 𝑧

_𝑖𝑛

=

(1−𝑗)+𝑗 tan 0.3(2𝜋)

1+𝑗(1−𝑗) tan 0.3(2𝜋)

=

1−𝑗+𝑗(−3.078)

1+(𝑗+1)(−3.078)

=

^{1−𝑗4.078}

−2.078−𝑗3.078

= 1.132∠47.82°

= 0.76 + 𝑗0.84 Γ =

^𝑧^𝐿⁻¹

𝑧𝐿+1

=

^1−𝑗−1

1−𝑗+1

=

^−𝑗

2−𝑗

=

¹

√5

∠ − 63.43°, 𝑆𝑊𝑅 =

^1+|Γ|

1−|Γ|

=

¹⁺

1

√5 1−¹

√5

= 2.618

(3)

3 | P a g e

d. 𝑙 = 1.2𝜆, 𝑧

_𝐿

= 0.5 − 𝑗0.5. Impedance is marked as P0 (wtg = 0.412). For input impedance, wtg =

0.412+1.2 = 1.612 =3(0.5)+0.112 ≡ 0.112. Hence the input impedance (P1) 𝑧

_𝑖𝑛

= 0.59 + 𝑗0.65 𝑧

_𝑖𝑛

=

(0.5−𝑗0.5)+𝑗 tan 1.2(2𝜋)

1+𝑗(0.5−𝑗0.5) tan 1.2(2𝜋)

=

0.5−𝑗0.5+𝑗3.078

1+(𝑗0.5+0.5)(3.078)

=

^0.5+𝑗2.58

2.54+𝑗1.54

= 0.886∠47.8°

= 0.59 + 𝑗0.66 Γ =

^𝑧^𝐿⁻¹

𝑧_𝐿+1

=

^{0.5−𝑗0.5−1}

0.5−𝑗0.5+1

=

^{−0.5−𝑗0.5}

1.5−𝑗0.5

=

¹

√5

∠243.43°, 𝑆𝑊𝑅 =

^1+|Γ|

1−|Γ|

=

¹⁺

1

√5 1−¹

√5

= 2.618

e. 𝑙 = 0.1𝜆, 𝑧

_𝐿

= 𝑆ℎ𝑜𝑟𝑡 𝑐𝑖𝑟𝑐𝑢𝑖𝑡. Impedance is marked as “A” (wtg = 0). For input impedance, wtg = 0 +0.1 = 0.1. Hence the input impedance (“E”) 𝑧

_𝑖𝑛

= 0 + 𝑗0.73

𝑧

_𝑖𝑛

=

0+𝑗 tan 0.1(2𝜋)

1+𝑗(0) tan 0.1(2𝜋)

=

^1+𝑗0.73

1+𝑗0

= 𝑗0.73 Γ =

^𝑧^𝐿⁻¹

𝑧_𝐿+1

=

⁰⁻¹

0+1

= −1 = 1∠180°, 𝑆𝑊𝑅 =

^1+|Γ|

1−|Γ|

=

¹⁺¹

1−1

= ∞

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4 | P a g e

f. 𝑙 = 0.4𝜆, 𝑧

_𝐿

= 𝑗3. Impedance is marked as “A” (wtg = 0.199). For input impedance, wtg = 0.199

+0.4 = 0.599 ≡ 0.099. Hence the input impedance (“F”) 𝑧

_𝑖𝑛

= 0 + 𝑗0.72 𝑧

_𝑖𝑛

=

𝑗3+𝑗 tan 0.4(2𝜋)

1+𝑗(𝑗3) tan 0.4(2𝜋)

=

^{𝑗3−𝑗0.727}

1+2.18

= 𝑗0.715 Γ =

^𝑧^𝐿⁻¹

𝑧𝐿+1

=

^𝑗3−1

𝑗3+1

= 1∠36.87°, 𝑆𝑊𝑅 =

^1+|Γ|

1−|Γ|

=

¹⁺¹

1−1

= ∞

g. 𝑙 = 0.2𝜆, 𝑧

_𝐿

= 𝑂𝑝𝑒𝑛 𝑐𝑖𝑟𝑐𝑢𝑖𝑡. Impedance is marked as “A” (wtg = 0.25). For input impedance, wtg = 0.25 +0.2 = 0.45. Hence the input impedance (“G”) 𝑧

_𝑖𝑛

= 0 − 𝑗0.32

𝑧

_𝑖𝑛

=

¹⁺

1

𝑧𝐿𝑗 tan 0.2(2𝜋) 1

𝑧𝐿+𝑗 tan 0.2(2𝜋)

=

¹

𝑗3.078

= −𝑗0.325

Γ =

^𝑧^𝐿⁻¹

𝑧𝐿+1

=

¹⁻

1 𝑧𝐿 1+¹ 𝑧𝐿

= 1 = 1∠0°, 𝑆𝑊𝑅 =

^1+|Γ|

1−|Γ|

=

¹⁺¹

1−1

= ∞

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5 | P a g e

Q2.

An antenna operates at a wavelength of 2 m, its impedance is 𝑍_𝐿= (15 + 𝑗60) Ω. It is connected to a line with characteristic impedance 𝑍₀ = 75 Ω. Calculate (using Smith chart):

a. The length of the parallel, shorted stub and its location on the line to match the antenna to the line.

The line and stub have the same characteristic impedance.

b. Repeat (a) for open circuit stub.

Solution:

a. The normalized load impedance = 𝑧_𝐿

=

^𝑍_𝑍^𝐿

0=^15+𝑗60

75 = 0.2 + 𝑗0.8 (see “P0”).

The corresponding admittance is marked as “P1”, It is 𝑦_𝐿= 0.294 − 𝑗1.176. If we move from P1 to P2, the admittance will look to be 1 + 𝑗2.528. This distance is 0.141𝜆 from P1 to the short circuit point (left) + 0.196𝜆 from short circuit point to P2. So, location of stub would be 0.141𝜆 + 0.196𝜆 = 0.337𝜆 from the load (or 0.337x2 = 0.674m).

Now the shorted stub should contribute an admittance of −𝑗2.528. This admittance is marked as P3. The distance of this point from the point of infinite admittance or short-circuit (right) is 0.31𝜆 − 0.25𝜆 = 0.06𝜆.

Hence the length of the stub should be 0.06x2 = 0.12m.

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6 | P a g e Let us see the second solution now:

The normalized impedance and admittance are marked at P0 and P1 as before. If we move from P1 to P2 (towards generator), the admittance will look to be 1 − 𝑗2.528. This distance is 0.141𝜆 from P1 to the short circuit point (left) + 0.303𝜆 from short circuit point to P2. So, location of stub would be 0.141𝜆 + 0.303𝜆 = 0.444𝜆 (or 0.444x2 = 0.888m) from the load.

Now the shorted stub should contribute an admittance of 𝑗2.528. This admittance is marked as P3. The distance of this point from the point of infinite admittance or short-circuit (right) is 0.25𝜆 + 0.19𝜆 = 0.44𝜆.

Hence the length of the stub should be 0.44x2 = 0.88m.

b. Consider the first solution again. Now the stub has to be open circuit. So, it will be represented by zero admittance (left). The distance from this point is 0.25𝜆 + 0.06𝜆 = 0.31𝜆. Hence the length of the stub should be 0.31x2=0.62m.

Now consider the second solution. The length of stub should be 0.19𝜆 (or 0.19x2 = 0.38m).

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7 | P a g e

Q3

. Consider the same transmission line and load of Q2. Calculate the length and location of shorted series stub to match the load to the line.

Solution:

The normalized load impedance = 𝑧_𝐿

=

_𝑍^𝑍^𝐿

0=^15+𝑗60

75 = 0.2 + 𝑗0.8 (marked as “P0”). We will use the Smith chart as impedance chart. Moving from P0 to P1, the impedance will become 1 + 𝑗2.528. This distance is 0.196𝜆 − 0.109𝜆 = 0.087𝜆 or 0.087x2 = 0.174m. Hence the series stub has to be located at 0.174m from the load.

The stub should contribute an impedance of −𝑗2.528. This is marked as P2. Since this is a shorted stub, the distance of point P2 from the short circuit point (left) is 0.31𝜆 or 0.31x2 = 0.62m. Hence the length of stub should be 0.62m.

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8 | P a g e Second solution:

Moving from P0 to P1, the impedance will become 1 − 𝑗2.528. This distance is 0.303𝜆 − 0.109𝜆 = 0.194𝜆 or 0.194x2 = 0.388m. Hence the series stub has to be located at 0.338m from the load.

The stub should contribute an impedance of 𝑗2.528. This is marked as P2. Since this is a shorted stub, the distance of point P2 from the short circuit point (left) is 0.19𝜆 or 0.19x2 = 0.38m. Hence the length of stub should be 0.38m.