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EEE_Exam Question 8

# EEE_Exam Question 8 - Response Detail fro Given transfer...

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Unformatted text preview: Response Detail: fro Given transfer function is Hum] = 101? 1+ﬂ 1000 Then H(jm)=i j“? . ...... (1) 100 1+—Jw 1000 From the above transfer function, The corner frequencies are “1.1 =1 radfsec with slope 20ddeec and = 1000 rad/sec with slope 720dB/dec The magnitude at ml1=1radfsec is 1 M = 2010 — lame: 3(100] M Indian: : 2010g [0.01) M ' new: : ‘40 dB And the magnitude at a)“ =1000 mdfsec is 2010g[w]£] =1 +40 60 = x + 40 x = 20 '-"M1M-Gmd.eec :20 dB The magnitude plot is From {1) we have _. 1 '91 HOQ’FE 1+ij 1000 then the phase is (H ' =90°—tan'1 —‘° (”0) [1000] Then for (0:1, 1H[j(o]=90°7tan' 43050) =89,94° ZH[jm)=90°—0.573° AHUw) =89.42” For \$210051H[jm)=90°—tan_1{ 11109) = 900— 5.71“ 4H Um) : 84.290 —> a;( radfsec 1 1 1000 For (0:10, ZH[jm)=90“—tan'l[%] 100 1 000 J 1000 For a) =1000, AH (fro) = 90“ — tan'1[—) lH(j:o)=90°—45° zH(jm_)=45° For a) :10000, zH{ jco} : 90" — tan’1[— AH{jaJ)=90”—S4.29° zH(;w):5,71° Then the phase plot is AHUaJ) T 900 1000 10000 1000 100 1000 10000 a w(;adfsec] ...
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