Fourier Series
16
Assessment Problems
AP 16.1 av = 1 T 2 T
2T /3 0
Vm dt +
1 T
T 2T /3
Vm 3
T
7 dt = Vm = 7 V 9 Vm cos k0 t dt 3
ak = = bk = =
2T /3 0
Vm cos k0 t dt + sin 4k 3 =
2T /3
4Vm 3k0 T 2 T
2T /3 0
6 4k sin k 3
T 2T /3
Vm sin k

Active Filter Circuits
15
Assessment Problems
AP 15.1 H(s) = -(R2 /R1 )s s + (1/R1 C) R1 = 1 , . C = 1 F
1 = 1 rad/s; R1 C R2 = 1, R1 .
. R2 = R1 = 1 -s s+1
Hprototype (s) =
AP 15.2 H(s) =
-20,000 -(1/R1 C) = s + (1/R2 C) s + 5000 C = 5 F
1

14
Introduction to Frequency-Selective Circuits
Assessment Problems
AP 14.1 fc = 8 kHz, c = 1 ; RC c = 2fc = 16 krad/s R = 10 k;
. C =
1 1 = = 1.99 nF c R (16 103 )(104 )
AP 14.2 [a] c = 2fc = 2(2000) = 4 krad/s L= 5000 R = = 0.40 H c 4000 4000

11 Balanced Three-Phase Circuits
Assessment Problems
AP 11.1 Make a sketch:
We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC . VAB = VAN + VN

10 Sinusoidal Steady State Power
Calculations
Assessment Problems
AP 10.1 [a] V = 100/- 45 V, I = 20/15 A Therefore 1 P = (100)(20) cos[-45 - (15)] = 500 W, 2 Q = 1000 sin -60 = -866.03 VAR, [b] V = 100/- 45 , I = 20/165 BA AB
AB
BA
P = 1000 cos(-

7 Response of First-Order RL and RC
Circuits
Assessment Problems
AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 resistor from the circuit.
First combine the 30 an

The Operational Amplifier
5
Assessment Problems
AP 5.1 [a] This is an inverting amplifier, so vo = (-Rf /Ri )vs = (-80/16)vs , vs ( V) 0.4 2.0 so vo = -5vs
3.5 -0.6 -1.6 -2.4
vo ( V) -2.0 -10.0 -15.0 3.0 8.0 10.0 Two of the vs values, 3.5 V and -

4 Techniques of Circuit Analysis
Assessment Problems
AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:
The two node voltage equations are v1 v1 - v2 v1 + + = 0 -15 + 60 15 5 v2 v2 - v1 5+ + = 0 2 5 Place these equ

Simple Resistive Circuits
3
Assessment Problems
AP 3.1
Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 resistor and the 10 r