{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

329fall10hw12sol - ECE 329 Fall 2010 Homework 12 Solution...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

Unformatted text preview: ECE 329 Fall 2010 Homework 12 - Solution Due: Dec. 8, 2010 1. a) (i) z L = Z L /Z = 2 . Locating the point z L on the Smith Chart, we can nd the corresponding Γ L = 0 . 333 . (ii) (iii) Rotating along the constant | Γ | circle towards the generator (clockwisely) by a distance l = 0 . 2 λ , we nd Γ( l ) = 0 . 333 ∠ θ , where θ =- . 2 λ . 25 λ × 180 o =- 144 o , and z ( l ) = 0 . 54- j . 24 . Then, Z ( l ) = Z · z ( l ) = 27 .- j 12 . 0 (Ω) . b) V ( l ) = Z ( l ) Z ( l ) + Z g V g = 27 .- j 12 . 27 .- j 12 . 0 + 50 × 10 = 3 . 66- j . 99 = 3 . 79 ∠- 15 . 1 o ( V ) . c) V ( l ) = V + e jβl + Γ L e- jβl = V + ( e j . 4 π + 0 . 333 e- j . 4 π ) , ∴ V + = 1 . 54- j 4 . 77 = 5 . 01 ∠- 72 . 1 o ( V ) . d) V (0) = V + (1 + Γ L ) = V + × (1 + 0 . 333) = 2 . 05- j 6 . 36 = 6 . 68 ∠- 72 . 1 o ( V ) . e) I (0) = V (0) Z L = 0 . 0205- j . 0636 = 0 . 0668 ∠- 72 . 1 o ( A ) . 1 ECE 329 Fall 2010 2. a) VSWR = | V | max | V | min = 8 . 4 ( V ) 2 . 1 ( V ) = 4 . b) Adjacent voltage minimums are separated by a distance of . 5 λ , and d . 5 λ = 5+ . 092 λ . 5 λ , therefore d min = 0 . 092 λ. c) Draw the constant | Γ | circle crossing the point A : z = VSWR (corresponding to voltage maxi- mum) on the real Γ axis. Then the other crossing point B (opposite to the point A ) corresponds...
View Full Document

{[ snackBarMessage ]}