N02
page 1 of 13
Passive Components
Resistor
units = Ohms 
1 Volt = 1 Ohm x 1 Amp
R1
series
R2
R  R1  R 2
R1
parallel
R
R 1R 2
R1  R 2
1
1
1
1




R R1 R 2 R 3
i
V2  V1  V  i R
V2
complex impedance
nominal wattage
1/8 inch body ~ 1/8 W
1/4 inch body ~ 1/4 W
1/2 inch body ~ 1/2 W
ZR = R
V1
ZR
R2
N02
page 2 of 13
Capacitor
+
units = Farads F
series
C
1 Ohm x 1 Farad = 1 second
C1C 2
C1  C 2
1
1
1
1




C C1 C 2 C 3
C1
C1
parallel
C  C1  C2
C2
i
1
V2  V1  V   i dt
C
complex impedance
using
j  1
V2
ZC 
and
1
j 2 f C
V1
ZC
V  Vo e j 2  f t  Vo cos2 f t   j Vo sin 2 f t 
C2
N02
page 3 of 13
Inductor
units = Henry H
i
V2  V1  V  L
di
dt
complex impedance
using
j  1
V2
V1
ZL  j 2 f L
ZL
V  Vo e j 2  f t  Vo cos2 f t   j Vo sin 2 f t 
and
Transformer
(coils have resistance and inductance)
Mechanical relays
SPDT
SPST
NC
C
C
NO
NO
COIL
COIL
Stepping motor
D
A
e
B
C
C
D
B
A
N02
Automotive relay
page 4 of 13
N02
page 5 of 13
Impedance Loading
prevent current leakage out of sensors and measurement circuits
VIN
prevent current leakage into measuring devices
R1
VOUT
R2
VIN
VIN
i
R1  R 2
R1
VOUT  i R 2
VOUT
i
no leakage
VOUT
R2

VIN
R1  R 2
R2
+15V
test both circuits with voltmeter
VA = +5V
no current
leakage
2K
VB
VA
1K
OK
VOUT
1K
VOUT = ½ VB
OK
connect together
+15V
2K
1K
VOUT = 2.5 V ?????
VA
VB
+15V
2K
VA
1K
VOUT
1K
1K
1K
2K
N02
VB  VA  15V
667
 3.75V
2000  667
+15V
page 6 of 13
2K
VA
VOUT = ½ VB = 1.875 V
667 
need an impedance buffer between circuit components
+15V
2K
VA
+
1K
-
VB
1K
VOUT
1K
N02
page 7 of 13
Basic Circuits
Voltage divider
R1
VIN
gain  G 
VOUT
VOUT
R2

VIN
R1  R 2
R2
First order low pass filter
cutoff frequency
1
fC 
2 R C
[Hz]
VIN
R
VOUT
C
log(G)
0
-1
log(f)
phase
fC
10 fC
0
log(f)
-90deg
fC
N02
ZR  R
i
ZC 
VIN
ZR  ZC
V
G  OUT
VIN
for DC
1
j 2 f C
page 8 of 13
j  1
VIN
ZR
VOUT  i Z C
VOUT
i
 1 


j 2 f C 
ZC



ZR  ZC 
1 
 R 

j 2 f C 

f=0
1
 

0
G   
1 R 

R  
0

for very high frequencies
f=∞
no leakage
ZC
G=1
passes DC
1
 
0

G   
1  R0

R  


G=0
removes high frequencies
N02
page 9 of 13
use MATLAB to plot frequency response
for R = 1.6 K, C = 0.1 F, f C 
1
= 994.7 Hz
2 R C
lf = (0 : 0.1 : 6);
f = 10 .^ lf;
R = 1.6e3;
C = 0.1e-6;
ZR = R;
ZC = 1 ./ (j*2*pi*f*C);
G = ZC ./ (ZR+ZC);
lG = log10( abs(G) );
phase = angle(G) * 180/pi;
subplot( 2,
plot( lf,
subplot( 2,
plot( lf,
1, 1 )
lG )
1, 2 )
phase )
0
-1
-2
-3
-4
0
1
2
3
4
5
6
0
1
2
3
4
5
6
0
-20
-40
-60
-80
-100
N02
page 10 of 13
Filters
First order high pass filter
C
VIN
cutoff frequency
1
fC 
2 R C
VOUT
[Hz]
R
G
VOUT
ZR
R 


VIN
ZR  ZC 
1 
 R 

j 2 f C 

for DC
f=0
G
R 

R
R
f=∞
G
1

R  
0

for very high frequencies
G=0
R 
1

R  


removes DC

R
R0
G=1
passes high frequencies
0
-1
-2
-3
0
1
2
3
4
5
6
0
1
2
3
4
5
6
100
80
60
40
20
0
N02
page 11 of 13
Higher order Butterworth Filters
simple Butterworth
f 
log G    N log  
 fc 
actual Butterworth
G=1
for f < fC
for f > fC
G
(unity gain in pass band)
N = order of filter = 1, 2, 3, 4, etc.
1
1  f / f c 
2N
N = order of filter = 1, 2, 3, 4, etc.
dB = 20 log ( G )
low pass, fc = 100 Hz, unity gain in pass band
20
0
-20
-40
Gain [dB]
-60
-80
First order N=1
Second order N=2
Fourth order N=4
Sixth order N=6
-100
-120
-140
-160
-180
0
0.5
1
1.5
2
2.5
log (frequency) [Hz]
3
3.5
4
N02
page 12 of 13
low pass, fc = 100 Hz, unity gain in pass band
10
First order N=1
Second order N=2
Fourth order N=4
Sixth order N=6
Gain [dB]
5
0
-5
-10
1.75
1.8
1.85
1.9
1.95
2
2.05
log (frequency) [Hz]
2.1
2.15
2.2
% gain_butterworth.m - gain for Butterworth filters
% HJSIII, 14.03.28
clear
% cutoff and orders
fc = 100;
N1 = 1;
N2 = 2;
N4 = 4;
N6 = 6;
% frequency spectrum
lf = ( 0 : 0.01 : 4 )';
f = 10 .^ lf;
% gain only - no phase
G1 = sqrt( 1 ./ ( 1 + ( (f/fc).^(2*N1) ) ) );
dB1 = 20 * log10( G1 );
G2 = sqrt( 1 ./ ( 1 + ((f/fc).^(2*N2) ) ) );
dB2 = 20 * log10( G2 );
G4 = sqrt( 1 ./ ( 1 + ((f/fc).^(2*N4) ) ) );
dB4 = 20 * log10( G4 );
G6 = sqrt( 1 ./ ( 1 + ((f/fc).^(2*N6) ) ) );
dB6 = 20 * log10( G6 );
% plot
figure( 1 )
clf
plot( lf,dB1,'r', lf,dB2,'g', lf,dB4,'b', lf,dB6,'k' )
xlabel( 'log (frequency) [Hz]' )
ylabel( 'Gain [dB]' )
axis( [ 0 4 -180 20 ] )
title( 'low pass, fc = 100 Hz, unity gain in pass band' )
legend( 'First order N=1', 'Second order N=2', 'Fourth order N=4', 'Sixth order N=6' )
% bottom of gain_butterworth
N02
page 13 of 13
Second order Butterworth?
NO!
VIN
R
R
VOUT
a) Impedance loading
b) -6Bd at fc
C
C
Need an impedance buffer between circuit components
(still -6dB at fc)
VIN
R
+
R
VOUT
C
C