---
title: "Smith chart, load 100+j150 on 75 Ω line"
subject: "Electromagnetism"
chapter: 5
key: "semester_3/electromagnetism:smith-chart-100-j150-load"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/smith-chart-100-j150-load
site: BCT Engineering
---

# Smith chart, load 100+j150 on 75 Ω line

Electromagnetism · Chapter 5: Transmission lines

- Asked in: 2068 Baisakh, 10 marks
- Video: [Smith Chart Example for Transmission line Parameters (VSWR, Reflection Coefficient, Input Impedance)](https://www.youtube.com/watch?v=X26nHbBRMjk) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/smith-chart-100-j150-load
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5

## Question

A load of 100 + j150 Ω is connected to a 75 Ω lossless line. Find using Smith Chart: (a) reflection coefficient (b) VSWR (c) the load admittance (d) Z_in at 0.4λ from the load (e) Z_in at generator if the line is 0.6λ long.

## Short answer

Normalize load to z_L = 1.333 + j2.0. On Smith chart, reflection coefficient circle gives |Γ| = 0.66 at 39.9°. VSWR = 4.88. Rotate 180° for y_L = 0.231 − j0.346 (Y_L = 3.08 − j4.62 mS). Move 0.4λ toward generator: Z_in = 22 + j47.6 Ω. For 0.6λ line (equivalent 0.1λ): Z_in = 133.6 − j165.7 Ω.

## Answer

Given: Z_L = 100 + j150 Ω, Z₀ = 75 Ω. Normalised load:

z_L = Z_L/Z₀ = 1.333 + j2.0

Plot z_L (r = 1.33 circle, x = +2 arc). With the compass centred on the chart centre, draw the circle through z_L.

#### (a) Reflection coefficient

Read |Γ| from the radial scale and the angle from the angle-of-reflection scale:

```text
Γ = (z_L − 1)/(z_L + 1) = (0.333 + j2)/(2.333 + j2)
  = 2.028∠80.5° / 3.073∠40.6°
  = 0.660 ∠39.9°
```

#### (b) VSWR

Where the |Γ| circle cuts the right side of the real axis:

```text
VSWR = (1 + 0.660)/(1 − 0.660) = 4.88 ≈ 4.9
```

#### (c) Load admittance

Draw the diameter through z_L and read the opposite point on the same circle (rotate 180° = λ/4):

```text
y_L = 1/z_L = 0.231 − j0.346
Y_L = y_L/Z₀ = (0.231 − j0.346)/75 = 3.08 − j4.62 mS
```

#### (d) Z_in at 0.4λ from the load

Move 0.4λ toward the generator (clockwise); the angle of Γ becomes 39.9° − 2×360°×0.4 = 39.9° − 288° = −248.1° ≡ 111.9°. Reading the chart at this point:

```text
z_in ≈ 0.29 + j0.64  →  Z_in = 75 × z_in ≈ 22 + j47.6 Ω   (52.4∠65.2°)
```

Check by formula with βl = 144°: Z_in = 21.97 + j47.61 Ω.

#### (e) Z_in at the generator, line 0.6λ long

0.6λ is more than a half wavelength; the chart repeats every 0.5λ, so it is the same as 0.1λ:

Angle of Γ = 39.9° − 72° = −32.1°.

```text
z_in ≈ 1.78 − j2.21  →  Z_in = 75 × z_in ≈ 133.6 − j165.7 Ω   (212.9∠−51.1° Ω)
```

**Answer: (a) 0.66∠39.9°; (b) 4.88; (c) y_L = 0.231 − j0.346 (Y_L = 3.08 − j4.62 mS); (d) 22 + j47.6 Ω; (e) 133.6 − j165.7 Ω.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the normalised load for Z_L = 100 + j150 on a 75 ohm line?
   A. 1.333 + j2.0
   B. 7500 + j11250
   C. 0.75 + j0.5
   D. 1.333 - j2.0
   Answer: A. Divide by Z0 = 75. (Key point: Normalised load)

2. Gamma is 0.66 angle 39.9 degrees. What is VSWR?
   A. (1 - 0.66)/(1 + 0.66) = 0.2
   B. 1/0.66 = 1.5
   C. 0.66
   D. (1 + 0.66)/(1 - 0.66) = 4.88
   Answer: D. VSWR is read on the right side of the real axis. (Key point: Reflection coefficient and VSWR)

3. How is y_L found on the Smith chart?
   A. Take the mirror image about the real axis
   B. Rotate by lambda/2
   C. Rotate z_L by 180 degrees (lambda/4): y_L = 0.231 - j0.346
   D. Add 1 to z_L
   Answer: C. A quarter-wave rotation inverts the normalised impedance. (Key point: Load admittance)

4. What is the angle of Gamma after moving 0.4 lambda toward the generator?
   A. 39.9 + 288 = 327.9 degrees
   B. 39.9 - 288 = -248.1, equivalent to 111.9 degrees
   C. 39.9 - 144 = -104.1 degrees
   D. 39.9 degrees (unchanged)
   Answer: B. The angle decreases by 2 beta l = 720 degrees x 0.4. Z_in = 22 + j47.6 ohm. (Key point: Z_in at 0.4 lambda)

5. Why can 0.6 lambda be treated as 0.1 lambda?
   A. The chart repeats every 0.5 lambda
   B. Impedance repeats every lambda only
   C. Lengths above 0.5 lambda are ignored
   D. The chart repeats every 0.25 lambda
   Answer: A. Rotation of 0.5 lambda is a full circle, so 0.6 lambda equals 0.1 lambda. (Key point: Line of 0.6 lambda)

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