# Microelectronic Circuit Design

• Notes
• 78

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CHAPTER 16 16.1 A v s ( ) = 50 s 2 s + 2 ( ) s + 30 ( ) | A mid = 50 | F L s ( ) = s 2 s + 2 ( ) s + 30 ( ) | Poles: - 2,-30 | Zeros: 0,0 Yes, s = 30 | A v s ( ) 50 s s + 30 ( ) | ω L 30 rad s | f L = ω L 2 π 30 2 π = 4.77 Hz f L = 1 2 π 30 2 + 2 2 2 0 ( ) 2 2 0 ( ) 2 = 4.79 Hz A v j ω ( ) = 50 ω 2 ω 2 + 2 2 ω 2 + 30 2 | MATLAB : f L = − 4.80 Hz 16.2 A v s ( ) = 400 s 2 2 s 2 + 1400 s + 100,000 = 200 s 2 s + 619 ( ) s + 80.8 ( ) | A mid = 200 | F L s ( ) = s 2 s + 619 ( ) s + 80.8 ( ) Poles: - 619,-80.8 rad s | Zeros: 0, 0 | Yes, a 5:1 split is sufficient | s = 619 A v s ( ) 200 s s + 619 ( ) | ω L 619 rad s | f L 619 2 π = 98.5 Hz f L 1 2 π 80.8 2 + 619 2 2 0 ( ) 2 2 0 ( ) 2 = 99.4 Hz A v j ω ( ) = 200 ω 2 ω 2 + 80.8 2 ω 2 + 619 2 | MATLAB : 100 H z 16.3 A v s ( ) = − 150 s s + 15 ( ) s + 12 ( ) s + 20 ( ) | A mid = − 150 | F L s ( ) = s s + 15 ( ) s + 12 ( ) s + 20 ( ) Poles: -12, - 20 rad s | Zeros: 0, -15 rad s | N o, the poles and zeros are closely spaced. f L 1 2 π 12 2 + 20 2 2 0 ( ) 2 2 15 ( ) 2 = 1.54 Hz A v j ω ( ) = 150 ω ω 2 + 15 2 ω 2 + 12 2 ω 2 + 20 2 | MATLAB : f L = 2.72 H z | ω L = 17.1 rad s Note that ω L =16.1 rad/s does not satisfy the assumption used to obtain Eq. (16.15), and the estimate using Eq. (16.15) is rather poor. 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-1

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