hw1_sol - Then we have: (b) 1,0 12 1 10 2 th E mA k V G G g...

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ECE 342 Electronic Circuits Homework 1 Solutions 1. Given the circuit shown: (a) Find the effective resistance as seen by the capacitor terminals, R 12 (b) Determine the Thévenin voltage as seen from node 1 to ground, V . Th-1,0 (c) Determine the Thévenin voltage as seen from node 2 to ground, V . Th-2,0 . (a) Remove capacitor; open the current source; short the voltage source. Terminal 1 sees R 1 =10 k g to ground; terminal 2 sees two 10 k g resistors in parallel, equivalent to R eq =5 k g to ground. Those two resistances are in series, as seen by the capacitor terminals (the same current flowing through R 1 will flow through R eq The total resistance seen by the terminals 1 and 2 is then: 10k+5k=15 k g . as well). For calculating the Thevenin voltages, we leave all the independent sources in place.
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Unformatted text preview: Then we have: (b) 1,0 12 1 10 2 th E mA k V G G g (c) 2,0 10 12 6 10 10 th E V G 3. (P1.66) 4. (P1.67) 5. Sedra P1.76 1 1 1 1 1 ( ) 1/ 1 ( ) ( ) 1/ 1 i i s V s sC T s Low pass V s sC R sC R g g g G 11 6 1 1 1 1 3 15.9 16 2 2 10 10 dB frequency kHz kHz C R G g g g For T o (s) , the following equivalent circuit can be used: 3 2 2 3 2 3 2 2 2 3 ( ) ( ) 1 1/ o m m R s T s G R G R R High pass R R sC s C R R g G g G G g 9 3 2 2 3 1 1 3 53 2 2 100 10 30 10 dB frequency Hz C R R G g g g 3 1 ( ) ( ) ( ) 666.7 2 53 1 2 15.9 10 i o s T s T s T s s s g g G Bandwidth=16 kHz - 53 Hz g 16 kHz Magnitude= 20log | 666.7 | 56.5 dB G g...
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This note was uploaded on 11/06/2011 for the course ECE 342 taught by Professor Nareshshanbhag during the Spring '11 term at University of Illinois, Urbana Champaign.

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hw1_sol - Then we have: (b) 1,0 12 1 10 2 th E mA k V G G g...

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