---
title: "Five-element Yagi design for 100 MHz"
subject: "Propagation and Antenna"
chapter: 3
key: "semester_6/propagation_and_antenna:yagi-design-100mhz"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/yagi-design-100mhz
site: BCT Engineering
---

# Five-element Yagi design for 100 MHz

Propagation and Antenna · Chapter 3: Antenna Classification

- Asked in: 2076 Bhadra, 7 marks; 2072 Asoj, 6 marks (asked 2 times)
- Video: [Examples and Designing of Yagi Uda Antenna in Antenna and Wave Propagation by Engineering Funda](https://www.youtube.com/watch?v=-V-JTevhogU) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/yagi-design-100mhz
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-3

## Question

With neat diagram design the 5 element Yagi-Uda antenna for receiving 100 MHz radio signal showing design steps. Consider effective dipole length equal to 0.48λc and spacing between elements is 0.15λc.

## Short answer

Wavelength is 3 m. Using reflector 5% longer and each director 5% shorter than the driven element, lengths are: reflector 1.512 m, driven 1.44 m, directors 1.368 m, 1.300 m, 1.235 m; spacing 0.45 m, boom 1.80 m.

## Answer

A 5-element Yagi-Uda antenna has **1 reflector + 1 driven dipole + 3 directors**. Design rules used (standard empirical rules):
- Driven element (DE) = 0.48λc (given effective length).
- Reflector ≈ 5 % longer than DE; each director ≈ 5 % shorter than the element before it.
- Spacing between adjacent elements = 0.15λc (given).

#### Step 1: Wavelength

```text
λc = c / f = (3 × 10⁸) / (100 × 10⁶) = 3 m
```

#### Step 2: Driven element

```text
L_DE = 0.48 λc = 0.48 × 3 = 1.44 m
```

#### Step 3: Reflector

```text
L_R = 1.05 × L_DE = 1.05 × 1.44 = 1.512 m  (≈ 0.504λ)
```

#### Step 4: Directors

```text
L_D1 = 0.95 × 1.44   = 1.368 m  (0.456λ)
L_D2 = 0.95 × 1.368  = 1.300 m  (0.433λ)
L_D3 = 0.95 × 1.2996 = 1.235 m  (0.412λ)
```

#### Step 5: Spacing and boom length

```text
S = 0.15 λc = 0.15 × 3 = 0.45 m
Boom (R to D3) = 4 gaps × 0.45 = 1.80 m
```

#### Design summary

| Element | Length (m) | Position from R (m) |
|---|---|---|
| Reflector | 1.512 | 0 |
| Driven (folded dipole) | 1.440 | 0.45 |
| Director 1 | 1.368 | 0.90 |
| Director 2 | 1.300 | 1.35 |
| Director 3 | 1.235 | 1.80 |

```text
  R      DE      D1      D2      D3
  |      |       |       |       |
 1.512  1.44   1.368   1.300   1.235 m
  |     -o-      |       |       |    ====> to
  |   feed       |       |       |   transmitter
  |<0.45>|<0.45->|<0.45->|<0.45->|
  <------------ boom 1.80 m ----->
```

Practical notes: elements are aluminium tubes of diameter about 0.005λ–0.01λ (≈ 1.5–3 cm here), mounted on a metal boom at their centres (voltage nulls). The driven element is usually a folded dipole fed with 300 Ω line or a balun to 75 Ω coax. The array points towards the directors, i.e. D3 faces the FM transmitter.

**Answer: λc = 3 m; reflector 1.512 m, driven 1.44 m, directors 1.368 m, 1.300 m, 1.235 m; spacing 0.45 m; boom length 1.80 m.**

## Check yourself

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

1. The driven element length is
   A. 1.512 m
   B. 1.368 m
   C. 0.48 m
   D. 1.44 m
   Answer: D. 0.48 x 3 m = 1.44 m. (Key point: Driven element)

2. The reflector, 5% longer than the driven element, is
   A. 1.512 m
   B. 1.368 m
   C. 1.44 m
   D. 1.30 m
   Answer: A. 1.05 x 1.44 = 1.512 m. (Key point: Reflector)

3. With 0.15 lambda spacing, boom length for 5 elements is
   A. 0.45 m
   B. 1.80 m
   C. 2.25 m
   D. 1.35 m
   Answer: B. Spacing is 0.45 m and there are 4 gaps: 4 x 0.45 = 1.80 m. (Key point: Spacing and boom)

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