{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

au10_312ps4_sol

# au10_312ps4_sol - ECE312 Autumn 2010 Problem Set 4...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE312 Autumn 2010: Problem Set 4 Solutions Problem 1 : Let Î· = radicalbig Î¼ /Ç« = 377 Î©. (a) Ëœ E = Ë† yje jÏ€z ; by comparing with Ëœ E = Ëœ E e âˆ’ j k Â· r , we can see that the the vector k = âˆ’ Ï€ Ë† z (i) The field propagates in the direction Ë† k = âˆ’ Ë† z (ii) The magnetic field Ëœ H = 1 Î· Ë† k Ã— Ëœ E = 1 Î· ( âˆ’ Ë† z ) Ã— Ë† yje jÏ€z = Ë† x 1 Î· je jÏ€z (iii) In free space k = Ï‰/c , therefore f = ck/ (2 Ï€ ) = 150 MHz. Also Î» = 2 Ï€/k = 2 m. (iv) Since this field has only a single component (i.e. Ë† y ), the polarization is linear . (b) Ëœ E = ( j Ë† y âˆ’ Ë† z ) e âˆ’ j 1000 x (i) Ë† k = Ë† x (ii) The magnetic field Ëœ H = 1 Î· Ë† k Ã— Ëœ E = 1 Î· Ë† x Ã— ( j Ë† y âˆ’ Ë† z ) e âˆ’ j 1000 x = 1 Î· (Ë† y + j Ë† z ) e âˆ’ j 1000 x (iii) k = 1000, so f = 1000 c/ (2 Ï€ ) = 47 . 75 GHz, and Î» = 2 Ï€/ 1000 = 6 . 28 mm. (iv) Here the field has two components (Ë† y and Ë† z ) of equal amplitude, 90 â—¦ out of phase, so the polarization is circular. To classify the handedness, look in the time domain: E ( t ) = âˆ’ Ë† y sin( Ï‰t âˆ’ 1000 x ) âˆ’ Ë† z cos( Ï‰t âˆ’ 1000 x ). Fix a point in space ( x = 0), we can see the field rotate from âˆ’ Ë† z to âˆ’ Ë† y as time goes from Ï‰t = 0 to Ï‰t = Ï€/ 2. By the right hand rule, curling with right-hand fingers in the direction of field rotation results in the right-hand thumb pointing in the ( âˆ’ Ë† z ) Ã— (...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

au10_312ps4_sol - ECE312 Autumn 2010 Problem Set 4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online