350sp13hw14sol

# 0125 watts 2 120 4 2 e making use of pr found in part

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: its matched load is given by Pr ⌘ Sinc A(✓, ) where Sinc = 2 4⇡ G ( ✓ , ˜ |Einc |2 2⌘o and A(✓, ) = ), we can write Pr = 4 (2⇡ )2 3 sin2 45 = 0.0125 Watts. 2 ⇥ 120⇡ 4⇡ 2 e) Making use of Pr found in part (d), we can calculate the open circuit voltage as | Vo | = q 8Pr Re{Zth } = p 8 ⇥ 0.0125 ⇥ 1 ⇡ 0.316 V. f) When the load is replaced by ZL = 1, the mismatch loss factor will be q= which implies that 4R 2 R L = 0.8 | ZL + Z2 | 2 PL = qPr = 0.01 Watts. 3. !&quot;! # a) Following from the below ﬁgure, &quot;% &quot;\$ &quot; ! &amp;\$ &quot;! &quot;! &quot;# &quot; \$% &amp; &quot;% # '&quot;! ()*+,-.#/0.#-./1)23#/)#/0.#2,40/#)-#/0.#5622.-/#7)625.#'7#'-#.86,9':.-/#,*;.&lt;'-5.&quot;# the equivalent impedance Zeq is given by Zeq = ZL + (Z2 ZC ) || ZC + (Z1 ZC ). Thus, we write 2 2 Z ( + ⇤= &quot; \$% &quot; # Z &quot; %= &quot; !1&gt; Z2&quot; ! Z2 ) &quot;\$ Zc&quot; ! &gt; Z1 2Re{Z2 } Zc . = = eq ⇤ Z2 + Z2 2Re{Z2 } Then, the average = &quot; &quot; is calculat...
View Full Document

## This note was uploaded on 02/03/2014 for the course ECE 350 taught by Professor Kudeki during the Spring '11 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online