Because energy is conserved, insta sous and average power are conserved
If voltageand current are sinusoids then r active power is conserved, and
complex power.
Bandpass filter realized with a cascade connection of two first-order filters:
C1
R2
R1
Vi
R1
R2
C2
Vo
Vx
s j
Vi (s)
R1
Vx (s)
sC1Vx (s) 0
R1
Thus
Vx (s)
1
Vi (s)
sR1C1 1
so
Vo ( j )
Vx (s)
1
R2
sC2
V
Parallel RLC circuit
Case 1:
R
10
L
1
Z
Y
arg Z
z
10 10 3
C
ZdB
1 10 6
1
R
Y
C
1
LC
20 log Z
100 200 10 6
j
R
L
1
j
1
1
L
10 4
1
RC
10 3
1
10 5
30
20
ZdB
20
10
20
0
10
100
1 10
3
4
1 10
3
5
10 10
5
1
Bandpass filter realized with a cascade connection of two first-order filters:
C1
R2
R1
Vi
R1
R2
C2
Vo
Vx
s j
Vi (s)
R1
Vx (s)
sC1Vx (s) 0
R1
Thus
Vx (s)
1
Vi (s)
sR1C1 1
so
Vo ( j )
Vx (s)
1
R2
sC2
V
Calculate frequency response
Find the complete response
Calculate R that makes e circuit critically damped
4) compute frequency response. Sketch the graphs,mcalculate cut off frequency
Determine hi
Homework 1
Due Friday, January 23, 2015
1. For the filter circuit below, determine the frequency response function H( ) = Vout( )/Vin( ) and
calculate the cutoff frequency. Sketch the magnitude of H i
Final Exam Review
Spring 2015
ELCT 222
The Final Exam will be 9-11 AM on Wednesday, April 29, 2015, in the regular classroom.
The exam format will be closed book, but you may use one sheet of notes.
1
Second Order Butterworth Active Filter
C1
Introduction to Butterworth
Filters
vi
R
R
C2
vo
ELCT 222
Active Filter
Butterworth Low Pass
2
Second Order Butterworth Filter
1/j C1
Vo
Vi
R
Vx
R
1/j C2
Vo
A
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 = 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(
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 relationshi
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
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 re
ELCT 222
Exam 1 Review
1
Vi
Vo
Ra
Rf
Ra
Vo
=
H
Cf = 0
Vo j
=
Vi
3.3 k
Rf
1
Rf
Rf
Ra
Cf
Cf
1
j
1
1
j
Rf
Ra
33 nF
rad
s
Rf Cf
6.67
Rf Cf
1
6.67
vc 0 = 0
j
1
Cf =
V
Ra i
j
=
1.377 10 3
c
H
2
1
1
Ra
22 k
Voltage of the whole circuit is equal to the sum of ohms from the (capacitor plus
resistor) multiplied by the current (voltage around the resistor divided by
resistor)
Magnitude is a calculation like