ECE 202 - Linear Circuit Analyi II Final Exam Solution December 9, 2008 Solution Breaking F into partial fraction, F 2 9 9 + + 35 9 ft δt + [ + 35e 9t ]ut A 9 Hence 3 i the correct anwer. Solution 2 ft 3[ut ut ] 2[ut ut 2] 3ut 5ut + 2ut 2 F 3 5e + 2e 2 Hence i the correct anwer. Solution 3 Zin /8 I V Zin 2/ ia /3 ohm ia/6 Conider the impedance Z in looking right at the node with voltage V. We can then write, I i a 8 i a 6 + i a 3 V i a 3 0
I i a 3 + 3 V 3 + Z in V I 3 + Z in 2/ Z in 2 2 + 6 + 8 Hence 3 i the correct anwer. Solution Tranfer function i given by the tandard expreion, Hence i the correct anwer. Solution 5 Uing voltage diviion, H V out Z 2 V in Z Z 2 2/ 2 + 2 Z + / + H 2 + + 2 H V o V i + + 2 2 + 2 + + 2 ht te t ut Hence 5 i the correct anwer. Solution 6 v i t 2[ut ut ] V i 2 e V o HV i 2 e + 2 [ 2 e + 2 ] + v o t [2 2te t 2e t ]ut [2 2t e t 2e t ]ut 2
+ Hence i the correct anwer. Solution 7 The equivalent impedance of the LC combination at ω0.5 i zero. Thu voltage at the inverting terminal of the op-amp i v o. By virtual ground principle of ideal op-amp, v o t v t 2 in0.5t + 30 Hence 5 i the correct anwer. Solution 8 When the witch i cloed, v 5 t v 3 t v 2 t. Thu applying principle of conervation of charge, v 5 t 5 8 5 + 3 + 2 Hence i the correct anwer. Solution 9 Here common mitake i to forget converting the circuit from time domain to -domain and keeping the voltage ource a 2 intead of 2/. Analyzing the equivalent circuit in -domain, and applying voltage diviion, V out 2 + 2 + 3 6 + 2 + Hence i the correct anwer. Solution 0 Jut before t0 the circuit i in teady tate and the inductor act a a hort, while the capacitor i open. Thu initial current through the inductor/22 A. The circuit thu look a follow for t > 0 :- I 2 / Vc 3
Applying KVL around the loop, I V c V c + I + 2 0 V c + 2 + 2 0 V c 8 2 + 2 2 + v c t in2tut Hence 5 i the correct anwer. Solution Jut before t0, the capacitor i charged to an initial voltage of 2V. The equivalent -domain circuit for zero input repone i a hown below. Clearly, V c i given by, ohm / Vc 6 V c 6 / 2 + v c t 2e t ut Hence 6 i the correct anwer. Solution 2 The equivalent admittance between the two terminal i given by, 2 + 6 Y eq C + + C + 2 + + 6 C + 2 + + 6 Y eq j3 j3 C + 3 + 3j j3 C + j 6 Clearly, for reonance, Y eq j3 hould be real, which i poible if C/6. Hence 2 i the correct anwer.
Solution 3 Applying KVL to the -domain equivalent circuit, V I L R + L β + I L C Clearly, the circuit i reonant when, 0 V R + L + I L [ Y eq R + L + β + C β + C ] Hence 6 i the correct anwer. Solution Hence 2 i the correct anwer. Solution 5 ωrlc 2 β + β + or ω r LC K m 50 0.05 000 ω old 0.5 0.5 8 K f ω new ω old 00/0.5 200 C new C old K m K f 8 0 µf 000 200 5 ohm / Vi ohm ohm V 0/ + Vo The following KCL node equation can be written, V i V + 0 V + V 0 + Vo V 5 0 5
V o V Eliminating V from the above equation, we get, V o V i 0 2 + 22 + 2 Thi i clearly the tranfer function of a low pa filter with a DC gain0 of -0/2-5. Hence 2 i the correct anwer. Solution 6 Let input voltage be V and input current be I from the top. Here care ha to be taken to write the proper ign for drop due to mutual inductance. We can write, uing KVL and dot convention, V 5I 3I + 2I I 5I + 2I 0 6 Z in V I 5 + + 6/ Hence 5 i the correct anwer. Solution 7 Uing convolution algebra, Hence i the correct anwer. Solution 8 Again, uing convolution algebra, ft ht f t h t 3δt h t 3h t 3 t t 2 [ut ut 2] dt 0, t < 0 t 3, 0 t 2 8, t 2 Hence i the correct anwer. ft ht f t h t f t 2δt 2f t 2 t fτ dτ 0, t < 0 6t, 0 t < 2 2 6t, 2 t < 0, t 6
Solution 9 We know, h V I V2 0 h 22 I 2 V 2 I0 For finding h, we hort circuit the output ide and reflect the econdary impedance on the primary ide. The reflected impedance become a 2 Z 9 / 36/. Thu we get the following circuit, I V 36/ h V I 36/ + 36/ 36 2 + 36 Similarly, for finding h 22, we open circuit the input ide and reflect the primary impedance on the econdary ide. The reflected impedance become /a 2 Z p /9 /9. Thu we get the following circuit, /9 / I2 V2 h 22 I 2 V 2 9 + 9 2 + 36 Hence 2 i the correct anwer. Solution 20 Note that when we hort the input ide, the voltage appearing on the primary ide of the ideal tranformer become Ω I I. Alo, from the ideal tranformer relation, I /V 2 n. Thu, 7
t 2 V 2 I V0 n Hence 3 i the correct anwer. Solution 2 [T ] [T A ][T B ] 2 + 2 3 + + 2 3 + + 2 Hence 6 i the correct anwer. Solution 22 If we keep reflecting the load impedance Z L from the right to the left mot ide, we will get the reflected load impedance Z r a, Z r [N /N 2 2 N 3 /N 2 N 5 /N 6 2 ]Z L [2/3 2 2/2 2 6/2 2 ]Z L Z L Now, uing Maximum Power Tranfer theorem, Z L 2 Z L 3 Ω Hence 3 i the correct anwer. Wih you all the bet for your final grade 8