202Ex2Sp05Sols

# 202Ex2Sp05Sols - EE-202 Ex 2 EE-202 Exam II March 8 2005...

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Unformatted text preview: EE-202, Ex 2, EE-202 Exam II March 8, 2005 Name: __________________________________ Student ID: _________________ CIRCLE YOUR DIVISION Division: 201-1 (morning) 201-2 (afternoon)INSTRUCTIONSThere are 11 multiple choice worth 5 points each; there is 1 workout problem worth 40 points. This is a closed book, closed notes exam. No scrap paper or calculators are permitted. A transform table and properties table are attached at the back of the exam. All students are expected to abide by the usual ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. As a reminder, at the very minimum, cheating will result in a zero on the exam and possibly an F in the course. EE-202, Ex 2, 1.For the circuit shown below, the Quality factor Q and BW are: (1) Q=0.5, BW=1 rad/sec (2) Q=2, BW=o.25 rad/sec (3) Q=1, BW=1 rad/sec (4) Q=1, BW=0.5 rad/sec (5) Q=0.5, BW=0.5 rad/sec (6) None of the above Since Req seen by the C and L in parallel = [(2Ω//2Ω) + 1Ω] // 2Ω= 1 Ω, BW = 1/ReqC = ½!p=12 * 2=0.5"Q=!pBW=12.The output voltage of a circuit has the following transfer function, v(0+) and v(∞) are: Vo=s+1s(s2+3s+1)(1)v(0+)=0, v(∞)=1 (2) v(0+)=1, v(∞)=0 (3) v(0+)=∞, v(∞)=0 (4) v(0+)=1, v(∞)=1 (5) v(0+)=0, v(∞)=∞(6) v(0+)=0, v(∞)=0 V+( )=lims!"sVs( )=s+1s2+3s+1=V!( )=lims"sVs( )=s+1s2+3s+1=11Ω2F 2H Vi2Ω2Ω2ΩEE-202, Ex 2, 3.For the pole-zero diagram shown below assuming that H(0)=2, H(s) is (1) H(s)=!8s2+24s+32s2+16(2) H(s)=!8s2!24s+32s2+16(3) H(s)=!8s2!24s+32s2!16(4) H(s)=!8s2!24s!32s2!16(5) H(s)=s2+16!8s2!24s+32(6) H(s)=8s2!24s!32s2+16By inspection H s( )=Ks2+3s!4s2+16. K=!8, thus, H s( )=!8s2+24s+32s2+16××-4 4j -4j 1 σjωEE-202, Ex 2, 4. Circuit shown below has a transfer function H(s) and is ??...
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202Ex2Sp05Sols - EE-202 Ex 2 EE-202 Exam II March 8 2005...

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