{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EEE_Exam Question 3

EEE_Exam Question 3 - Givm HUQkZL(1w LCRSRLHmLm RL R5RL R5...

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

View Full Document Right Arrow Icon
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Givm HUQkZL (1w) LCRSRLHmLm +RL)+R5RL R5 =5 k9 C=5nF RL=50kQ L=2 mH ja)[2mx50k) [jmfpmx 511 x skxsok}+ jazm(sk+50k)+[skx50k) le'GaJ —w2{25x1073)+}110w+[250x106) H050}: H00): (3) Agra—>0,H[ja))—>0 Andas (0—)00, H(jm}—>0 So, this is a Second order Band pass filter. (b) 100 j .3 0) Hum) : 25x10 7 +j 1115fI w 250><HTS 2.5x 1.0a 2.5xlO’E . 1400004.: H :— (W) in): +144000w+101l . a) _ _ _ git—”m Comparing thls With standard form: HUM) : —0Jl +j%m+r:0§ Resonant frequency a)” = #10” (q, =316.23 hadfs {0) fl : 44000 Q 6’“ =22000 2Q 1 _ 22000 29 60., 1 22000 E—3lé.23x103 1 —=0.07 29 Damping ratio 4 =$ g: 0.7 {d} E = 44000 Q .'.BW = 44 kradfs {6) Q Halfpowcr frequencies; 5912 :17 + 212 =:22000+‘ 220002 +[31623x103r q; 21220004415994 all =—22000+316994 5-)] =295 kradfs 612 =22000+316994 a)z = 339 kradfs ...
View Full Document

{[ snackBarMessage ]}