※ 引述《balalenwo (BAGO)》之銘言：
: 是個小問題
: 作業上的
: 我也不知道這樣條件夠不夠解
: 一個放大器 gain=20dB , f(3dB)=1M Hz
: 若此放大器phase margin= 60度
: 則non-dominant pole為?
fp1 = 1MHz
ft = A*fp1 = 10MHz
速解1:(假設pole1造成的Phase shift是90deg)
PM = 180 - 90(pole1) - arctan(ft/fp2) = 60(deg)
ft/fp2 = tan(30deg) = 1/sqrt(3)
fp2 = ft*sqrt(3) = 17.3MHz
速解2:(考慮pole1造成的phase shift)
PM = 180 - arctan(ft/fp1) - arctan(ft/fp2) = 60(deg)
arctan(ft/fp2) = 120 - arctan(10) = 35.7(deg)
ft/fp2 = tan 35.7deg = 0.72
fp2 = ft/0.72 = 13.9MHz
精確解應該為
A(s) = 10/(1+s/p1)(1+s/p2)
p1 = 2*pi*1e6
A(jw) = 10/(1+jw/p1)(1+jw/p2)
ft: |A(jw)| = 1
10 = |(1+jw/p1)(1+jw/p2)| = sqrt(1+(w/p1)^2)sqrt(1+(w/p2)^2)
PM = 180 - arctan(w/p1) - arctan(w/p2) = 60
arctan(w/p1) + arctan(w/p2) = 120
1: w/p1 = 8
sqrt(1+(w/p2)^2) = 10/8.062 = 1.24
w/p2 = 0.54
PM = 180 - arctan(8) - arctan(0.54) = 68.8 deg
2: w/p1 = 7.5
w/p2 = 0.75
PM = 180 - arctan(7.5) - arctan(0.75) = 60.725deg
p2 = w/0.747
w = p1*7.5
p2 = p1*7.5/0.747 ~ 10MHz #
ft = 7.5MHz #
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