20095ee110_1_Hw6_solutions

20095ee110_1_Hw6_solutions - Problem 3(b) Schematic:...

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Unformatted text preview: Problem 3(b) Schematic: Netlist: * source HW6 R_R1 VIN VOUT 15.915 C_C1 0 VOUT 1n V_V1 VIN 0 DC 0Vdc AC 1Vac Plot of ‐3dB @ 10MHz Problem 4(b): AC Analysis Schematic: Netlist: * source HW6 R_R1 VIN VOUT 83.3333333333 L_L1 VOUT N000051 2.652582384865 C_C1 0 N000051 2.652582384865u V_V1 VIN 0 DC 0Vdc AC 1Vac Magnitude Response: 0 @ 60Hz Phase Response: Problem 4(b): Transient Analysis Schematic: Netlist: * source HW6 R_R1 VIN VOUT 83.3333333333 L_L1 VOUT N000051 2.652582384865 C_C1 0 N000051 2.652582384865u V_V1 N00243 0 +SIN 0 0.1 1000 0 0 0 V_V2 VIN N00243 +SIN 0 2 60 0 0 0 Transient input and output Input = 2*cos(2*pi*60*t)+0.1*cos(2*pi*1000*t) Output =0.1*cos(2*pi*1000*t) (Component at 60Hz has been completely attenuated) ...
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This note was uploaded on 10/02/2010 for the course EE EE 110 taught by Professor Gupta during the Fall '09 term at UCLA.

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