--- title: "Yagi driven element and 760 MHz design" subject: "Propagation and Antenna" chapter: 3 key: "semester_6/propagation_and_antenna:yagi-driven-element-design-760" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/yagi-driven-element-design-760 site: BCT Engineering --- # Yagi driven element and 760 MHz design Propagation and Antenna · Chapter 3: Antenna Classification - Asked in: 2080 Chaitra, 3+6 marks - 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-driven-element-design-760 - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-3 ## Question What is the main role of driven element of the Yagi-Uda antenna? Design five-element Yagi-Uda antenna with operating frequency 760 MHz and length of driven element 0.48λc. ## Short answer Driven element is the only fed element; it excites parasitic elements through mutual coupling, determines resonant frequency and input impedance, and its position sets the beam phase. Design: λ=39.47 cm; R=19.89 cm, DE=18.95 cm, D1=18.00, D2=17.10, D3=16.25 cm; S_R=9.87 cm, S_D=5.92 cm, boom=27.63 cm. ## Answer #### Role of the driven element The **driven element** is the only element of a Yagi-Uda antenna connected to the feed line; usually a resonant λ/2 dipole or folded dipole. - It receives RF power from the transmitter (or delivers received power to the receiver) — the single feed point of the array. - Its radiated field induces currents in the reflector and directors by mutual coupling; without it the parasitic elements would carry no current. - It fixes the **operating (resonant) frequency** and the **input impedance**; a folded dipole (≈ 292 Ω alone) compensates for the impedance drop caused by the close parasitic elements and matches the feeder. - Its position between reflector and directors sets the phase relationships that produce the forward beam. #### Design of 5-element Yagi at 760 MHz Design rules used (standard empirical Yagi rules): ```text λc = c / fc L_DE = 0.48 λc (driven, given) L_R = 1.05 L_DE (≈ 0.5λc) (reflector 5 % longer) L_D1 = 0.95 L_DE, L_Dk+1 = 0.95 L_Dk (5 % shorter each) S_R = 0.25 λc (reflector to driven) S_D = 0.15 λc (driven to D1 and between directors) ``` **Step 1: Wavelength** ```text λc = 3×10⁸ / 760×10⁶ = 0.3947 m = 39.47 cm ``` **Step 2: Element lengths** ```text L_DE = 0.48 × 39.47 = 18.95 cm L_R = 1.05 × 18.95 = 19.89 cm L_D1 = 0.95 × 18.95 = 18.00 cm L_D2 = 0.95 × 18.00 = 17.10 cm L_D3 = 0.95 × 17.10 = 16.25 cm ``` **Step 3: Spacings and boom** ```text S_R = 0.25 × 39.47 = 9.87 cm S_D = 0.15 × 39.47 = 5.92 cm Boom = 9.87 + 3 × 5.92 = 27.63 cm ``` | Element | Length (cm) | Spacing from previous (cm) | |---|---|---| | Reflector | 19.89 | – | | Driven | 18.95 | 9.87 | | Director 1 | 18.00 | 5.92 | | Director 2 | 17.10 | 5.92 | | Director 3 | 16.25 | 5.92 | ```text R DE D1 D2 D3 | | | | | 19.89 18.95 18.00 17.10 16.25 cm | (o) | | | ===> beam |<9.87>|<5.92>|<5.92>|<5.92>| <------ boom 27.63 cm ------> ``` **Answer: λc = 39.47 cm; R = 19.89 cm, DE = 18.95 cm, D1 = 18.00 cm, D2 = 17.10 cm, D3 = 16.25 cm; S_R = 9.87 cm, S_D = 5.92 cm; boom ≈ 27.6 cm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement describes the driven element of a Yagi-Uda antenna? A. It is the longest, unconnected parasitic element B. It is the only element connected to the feed line C. It is the shortest element and guides the beam D. All elements are connected to the feed Answer: B. Only the driven element is fed; the reflector and directors are parasitic. (Key point: Role of driven element) 2. Why are the parasitic elements dependent on the driven element? A. They have their own transmitters B. They are connected by the boom as a feed line C. The driven element's radiated field induces currents in them by mutual coupling D. They resonate by absorbing ground waves Answer: C. Without the driven element, the parasitic elements would carry no current. (Key point: Parasitic excitation) 3. Why is a folded dipole (~292 Ω) often used as the driven element? A. It makes the pattern omnidirectional B. It raises the reflector's length C. It removes the need for a reflector D. It compensates for the impedance drop caused by close parasitic elements and matches the feeder Answer: D. Parasitic elements lower the driven impedance; the folded dipole's high impedance restores a good match. (Key point: Folded dipole as driven element) 4. For 760 MHz, what are lambda_c and the driven element length 0.48 lambda_c? A. 39.47 cm and 18.95 cm B. 35.71 cm and 17.14 cm C. 19.74 cm and 9.47 cm D. 39.47 cm and 19.89 cm Answer: A. lambda = 3x10^8/760x10^6 = 39.47 cm; 0.48 x 39.47 = 18.95 cm. 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