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Unformatted text preview: ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo 33. RA RB (a) 1 1, 2 2, 3 KCL at Vf: 1 , 2 4 1 , 1 1 (1) KCL at Vp: Substitute Equation (2) into Equation (1) and solve for Vp: 234 234 124 123 224 221 123 234 124 123 224 KCL at Vn: 1 (2) 221 Ve Vp – Vn, Vo a f Ve Ve cVi dVo – KVo By letting: a(f) = the open-loop gain of the amplifier, 1 K Page 1 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo Assuming a(f)b is very large over the frequency of operation, 1 1 1 1 1 Plugging in the generalized impedance terms gives the ideal transfer function with impedance terms: 12 34 1 3 2 3 11 1 11 4 1 0 (b) 1.8 1.2 2 1.8 1.6 1.4 magnitude response 1.2 1 0.8 0.6 0.4 0.2 0 1 1212 1 1 21 22 1 1212 X: 1.115 Y: 1.324 0 0.5 1 1.5 2 2.5 3 Frequency r/s 3.5 4 4.5 5 Page 2 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo 0 -20 -40 phase response[degree] -60 -80 -100 -120 -140 -160 -180 0 0.5 1 1.5 2 2.5 3 Frequency r/s 3.5 4 4.5 5 >> n=1.8*[1]; >> d=[1 1.2 1]; >> w=0:0...
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