Download Experiment 5 - Plasma Dynamics Lab

Electronic Instrumentation
Experiment 5
* AC Steady State Theory
* Part A: RC and RL Circuits
* Part B: RLC Circuits
AC Steady State Theory

Transfer Functions
 Phasors
 Complex Impedance
 Complex Transfer Functions
Transfer Functions
Vout
H
Vin

The transfer function describes the
behavior of a circuit at Vout for all
possible Vin.
Simple Example
Vout
Vout
R 2  R3
 Vin *
R1  R 2  R3
2 K  3K
 Vin *
1K  2 K  3K
Vout 5
H

Vin 6

if Vin (t )  6V sin( 2 Kt  )  12V
2

then Vout (t )  5V sin( 2 Kt  )  10V
2
More Complicated Example
What is H now?

H now depends upon the input frequency
(w = 2f) because the capacitor and
inductor make the voltages change with
the change in current.
Influence of Resistor on Circuit
VR  I R R

if I R (t )  A sin( wt )  A cos(wt  )
2

then VR (t )  R * A sin( wt )  R * A cos(wt  )
2


Resistor modifies the amplitude of the
signal by R
Resistor has no effect on the phase
Influence of Capacitor on Circuit
1
VC   I C dt
C

if I C (t )  A sin( wt )  A cos(wt  )
2
1
1

then VC (t ) 
* A cos(wt ) 
* A sin( wt  )
wC
wC
2
or VC (t ) 


1

1
* A sin( wt  ) 
* A cos(wt   )
wC
2 wC
Capacitor modifies the amplitude of the
signal by 1/wC
Capacitor shifts the phase by -/2
Influence of Inductor on Circuit
dI L
VL  L
dt

if I L (t )  A sin( wt )  A cos(wt  )
2

then VL (t )  w L * A cos(wt )  w L * A sin( wt  )
2

or VL (t )  w L * A sin( wt  )  w L * A cos(wt )
2


Inductor modifies the amplitude of the
signal by wL
Inductor shifts the phase by +/2
How do we model H?


We want a way to combine the effect of
the components in terms of their
influence on the amplitude and the
phase.
We can only do this because the signals
are sinusoids
• cycle in time
• derivatives and integrals are just phase
shifts and amplitude changes
Phasors



 
V  IZ
Phasors allow us to manipulate
sinusoids in terms of amplitude and
phase changes
Phasors are based on complex polar
coordinates
The influence of each component is
given by Z, its complex impedance
Phasor References



http://ccrmawww.stanford.edu/~jos/filters/Phasor_Notat
ion.html
http://www.ligo.caltech.edu/~vsanni/ph3/Ex
pACCircuits/ACCircuits.pdf
http://robotics.eecs.berkeley.edu/~cssharp/P
hasors/Phasors.html
Phasor Applet
Complex Polar Coordinates
V (t )  A coswt   
Review of Polar Coordinates
point P is at
( rpcosqp , rpsinqp )
 yP 
q P  tan  
 xP 
1
rP  xP2  y P2
Introduction to Complex Numbers
j  1
j  j  1
1
j
j

zp is a single number represented by two numbers
 zp has a “real” part (xp) and an “imaginary” part (yp)
Now we can define Phasors
if V (t )  A cos(w t   ) , then let

V  A cos(w t   )  jA sin( w t   )



The real part is our signal.
The two parts allow us to determine the
influence of the phase and amplitude
changes mathematically.
After we manipulate the numbers, we
discard the imaginary part.
Magnitude and Phase

V  A cos(w t   )  j A sin( w t   )  x  jy


2
2
V  x  y  A magnitude of V

1  y 
V  tan    w t  
x


phase of V
Phasors have a magnitude and a phase
derived from polar coordinates rules.
Euler’s Formula
e
jq
 cosq  j sin q
z  x  jy  r cos q  jr sin q  re
jq
jq1
z1 r 1e
r1 j (q1 q 2 )
z3  
 e
jq 2
z 2 r2 e
r2
r1
therefore, r3 
r2
and q 3  q1  q 2
Phasor Applet
Manipulating Phasors (1)

j (w t  )
V  A cos(w t   )  j sin( w t   )  Ae

 V1 A1e j (w t 1 ) A1 e jwt e j1 A1 j (1 2 )
V3   

 e
j (w t  2 )
2
jw t
V2 A2e
A2 e e
A2
 A1

therefore, V3 
and V3  1   2
A2

Note wt is eliminated by the ratio
• This gives the phase change between
signal 1 and signal 2
Manipulating Phasors (2)


V1  x1  jy1 V2  x2  jy2

V3  x3  jy3

2
2
x1  y1
 V1
V3   
2
2
V2
x2  y2



1  y1 
1  y2 
V3  V1  V2  tan    tan  
 x1 
 x2 
Understanding the influence of Phase
 real : if y  0 and x  0

 0 
then V  tan 1     0
x 
 j:
if x  0 and y  0
 j:

1  y 
V  tan  
x



1  y 
then V  tan     90
 0  2
if x  0 and y  0



1  y 
then V  tan      90
2
 0 
 real : if y  0 and x  0

 0 
then V  tan 1      (or   )
x 
 180
 
Complex Impedance V  I Z

Z defines the influence of a component
on the amplitude and phase of a circuit
• Resistors:
ZR = R
• change the amplitude by R
• Capacitors: ZC=1/jwC
• change the amplitude by 1/wC
• shift the phase -90 (1/j=-j)
• Inductors: ZL=jwL
• change the amplitude by wL
• shift the phase +90 (j)
Capacitor Impedance Proof
Prove:
ZC 
1
jw C
dV (t )
I C (t )  C C
and VC (t )  A cos(w t   )
dt

VC ( jw )  A cos(w t   )  jAsin( w t   )  Ae j (w t  )


dVC ( jw ) dAe j (w t  )
j ( w t  )

 Ajwe
 jw VC ( jw )
dt
dt

 dVC ( jw ) 
dVC (t )
 Re 
  jwA cos(w t   )  jw VC (t )
dt
dt


I C (t )  C
dVC (t )
1
 Cjw VC (t ) VC (t ) 
I C (t )
dt
jw C
Complex Transfer Functions


Vout ( jw )
H ( jw )  
Vin ( jw )



If we use phasors, we can define H for
all circuits in this way.
If we use complex impedances, we can
combine all components the way we
combine resistors.
H and V are now functions of j and w
Simple Example
ZR  R
ZC 
1
jw C



Vout ( jw )
ZC I
ZC

H ( jw )  

Vin ( jw ) Z R  Z C I Z R  Z C 
1

H ( jw ) 
jw C
jw C

1
jw C
R
jw C

H ( jw ) 
1
jw RC  1
Simple Example (continued)

H ( jw ) 
H ( jw ) 
1  j0
1  jw RC
1
jw RC  1

12  02
1  (w RC )
2
2

1
1  (w RC ) 2
H ( jw )  (1  j 0)  (1  jw RC )
0
1  w RC 
1
H ( jw )  tan    tan 
   tan (w RC )
1
 1 
1
H ( jw ) 
1
1  (w RC )
2
H ( jw )   tan 1 (w RC )
Using H to find Vout

jout jwt
jout
Vout Aoute e
Aoute
H ( jw )   

jin jwt
jin
Aine e
Vin
Aine
Aout e
Aout e
j out
j out
 H ( jw ) Aine
 H ( jw ) e
Aout  H ( jw ) Ain
jin
jH ( jw )
Aine
jin
out  H ( jw )  in
Simple Example (with numbers)
C  1 F
R  1K

Vin (t )  2V cos( 2 K t  )
4

H ( jw ) 
H ( jw ) 
1
1
1


jw RC  1 j 2 K 1K1  1 2 j  1
12
1  (2 ) 2
 0.157
 2 
H ( jw )  0  tan 
  1.41
 1 
1
Vout (t )  0.157 * 2V cos(2K t  0.785  1.41)
Vout (t )  0.314V cos(2K t  0.625)
Adding Phasors & Other Applets
Part A -- RC and RL Circuits

High and Low Pass Filters
 H and Filters
High and Low Pass Filters
1.0
High Pass Filter
wc=2fc
H = 0 at w0
0.5
H = 1 at w 
0
1.0Hz
100Hz
V1(R1) / V(C1:2)
10KHz
Frequency
1.0MHz
100MHz
H0.707at wc
fc
1.0
Low Pass Filter
wc=2fc
H = 1 at w0
0.5
H = 0 at w 
0
1.0Hz
100Hz
V(C1:2) / V(R1:2)
10KHz
Frequency
1.0MHz
fc
100MHz
H0.707at wc
Corner Frequency

The corner frequency of an RC or RL circuit
tells us where it transitions from low to high or
visa versa.
1
 We define it as the place where H ( jw c ) 
2
1
 For RC circuits: w c 
RC
R
 For RL circuits: w c 
L
Corner Frequency of our example
1
H ( jw ) 
1  jw RC
1
H ( jw ) 
2
H ( jw ) 
1
1  (w RC ) 2
2  1  wRC 
2

1
2
1
2

w
RC 2
1
1

1  (w RC ) 2 2
1
wc 
RC
H(jw), wc, and filters

We can use the transfer function, H(jw), and the
corner frequency, wc, to easily determine the
characteristics of a filter.
 If we consider the behavior of the transfer
function as w approaches 0 and infinity and
look for when H nears 0 and 1, we can identify
high and low pass filters.
 The corner frequency gives us the point where
the filter changes:
wc
fc 
2
Our example at low frequencies
1
H ( jw ) 
1  jw RC
1
H LOW ( jw ) 
1
1 0
H LOW ( jw ) asw  0  1  1
0
H LOW ( jw )  tan    0 (on  x axis)
1
1
Our example at high frequencies
1
H ( jw ) 
1  jw RC
1
H HIGH ( jw ) 
jw RC
1
1
H HIGH ( jw ) as w   
 0
jw RC



0
1  w RC 
H HIGH ( jw )  tan    tan 
  0  
2
2
1
 0 
1
Our example is a low pass filter
H LOW  1 H HIGH  0
wc
1
fc 

2 2 RC
1.0
What about the phase?
0.5
0
1.0Hz
100Hz
V(C1:2) / V(R1:2)
10KHz
Frequency
1.0MHz
100MHz
Our example has a phase shift
1.0
H LOW  1
H HIGH  0
0.5
SEL>>
0
V(R1:2) / V(V1:+)
0d
H LOW ( jw )  0
H HIGH ( jw )  90
-50d
-100d
1.0Hz
VP(C1:2)
10KHz
Frequency
100MHz
Taking limits
a2w 2  a1w  a0
H ( jw ) 
b2w 2  b1w  b0

At low frequencies, (ie. w=10-3), lowest
power of w dominates
a210 6  a110 3  a0100 a0
H ( jw ) 

6
3
0
b210  b110  b010
b0

At high frequencies (ie. w =10+3), highest
power of w dominates
a210 6  a110 3  a0100 a2
H ( jw ) 

6
3
0
b210  b110  b010
b2
Capture/PSpice Notes

Showing the real and imaginary part of the signal
• in Capture: PSpice->Markers->Advanced
• ->Real Part of Voltage
• ->Imaginary Part of Voltage
• in PSpice: Add Trace
• real part: R( )
• imaginary part: IMG( )

Showing the phase of the signal
• in Capture:
• PSpice->Markers->Advanced->Phase of Voltage
• in PSPice: Add Trace
• phase: P( )
Part B -- RLC Circuits

Band Filters
 Another example
 A more complex example
Band Filters
Band Pass Filter
1.0
H = 0 at w0
0.5
H = 0 at w 
0
1.0Hz
100Hz
V(R1:1)/ V(R1:2)
10KHz
f0
1.0MHz
100MHz
H1at w0=2f0
Frequency
1.0
Band Reject Filter
H = 1 at w0
0.5
H = 1 at w 
0
1.0Hz
V(L1:1) /
100Hz
V(V1:+)
10KHz
f0
Frequency
1.0MHz
100MHz
H0at w0 =2f0
Resonant Frequency

The resonant frequency of an RLC circuit tells
us where it reaches a maximum or minimum.
 This can define the center of the band (on a band
filter) or the location of the transition (on a high
or low pass filter).
 We are already familiar with the equation for the
resonant frequency of an RLC circuit:
w0 
1
LC
Another Example
Z R  R Z L  jw L
ZC 
1
jw C
1
H ( jw ) 
jw C
R  jw L 
1
jw C
1

jw RC  j 2w 2 LC  1
1
H ( jw ) 
2
(1  w LC )  jw RC
At Very Low Frequencies
1
H LOW ( jw )   1
1
H LOW ( jw ) w  0  1
H LOW ( jw )  0
At Very High Frequencies
H HIGH ( jw ) 
1
 w 2 LC
1
H HIGH ( jw ) w   
0

H HIGH ( jw )   or  
At the Resonant Frequency
H ( jw ) 
H ( jw 0 ) 
1
(1  w 2 LC )  jw RC
w0 
1
LC
1
2
 1 
(1  
 LC ) 
 LC 
LC
H ( jw 0 )   j
RC
LC
H ( jw 0 ) 
RC
H ( jw 0 )  

2
 1 
j
 RC
 LC 

f0 
1
2 LC
1
 RC 
(1  1)  j

 LC 
if L=10uH, C=1nF and R=1K 
w010Mrad/sec f01.6MHz

|H|0.1
H  
2
radians
Our example is a low pass filter
0d
Phase
90
 = 0 at w0
-100d
 = -180 at w 
SEL>>
-200d
P( V(R1:2))
1.0
Magnitude
H = 1 at w0
0.5
H = 0 at w 
0.1
0
1.0Hz
10KHz
V(R1:2) / V(V1:+)
Frequency
100MHz
f01.6MHz
A more complex example


Even though this filter has parallel components, we
can still handle it.
We can combine complex impedances like resistors
Determine H
jw L 
1
jw L
jw L
jw C
Z CL 
 2 2

2
1
j
w
LC

1
1

w
LC
jw L 
jw C
jw L
2
2
Z CL
1

w
LC
1

w
LC
H ( jw ) 

multiply by
R  Z CL R  jw L
1  w 2 LC
1  w 2 LC
jw L
H ( jw ) 
R(1  w 2 LC )  jw L
At Very Low Frequencies
jw L
H LOW ( jw ) 
R
H LOW ( jw ) w  0  0
H LOW ( jw ) 

2
At Very High Frequencies
jw L
j
H HIGH ( jw ) 

2
 w LRC w RC
H HIGH ( jw ) w   
H HIGH ( jw )  

2
1
0

At the Resonance Frequency
w0 
1
LC
 1 
j
L
 LC 
H ( jw 0) 
2
 1 
 1 
R(1  
 LC )  j 
L
 LC 
 LC 
H ( jw0 )  1
H ( jw0 )  0
 1
Our example is a band pass filter
Magnitude
1.0
H = 0 at w0
0.5
H=1 at w0
H = 0 at w 
0
V1(R1) / V(V1:+)
100d
Phase
 = 90 at w0
0d
SEL>>
-100d
1.0Hz
VP(R1:1)
 = 0 at w0
10KHz
Frequency
100MHz
f0
 = -90 at w 
Simple Filter Design

Step One: Pick a High Pass or Low Pass
Filter Configuration
• This is done by looking at the high or low
frequency limit of various simple circuits.
• For example, for an RC low pass filter, we
choose the following configuration:
R1
V1
V OFF =
V A MPL =
FREQ =
C1
0
Simple Filter Design
R1
Vin
Vout
V1
V OFF =
V A MPL =
FREQ =
C1
0

At very high frequencies, the capacitor becomes a
short circuit. (Impedance goes to zero.)
 At very low frequencies, the capacitor becomes an
open circuit. (Impedance goes to infinity.) The
output appears fully across the capacitor.
 Thus, if the output voltage is taken across the
capacitor, this is a low pass filter.
Simple Filter Design
Vin
R1
Vout
V1
V OFF =
V A MPL =
FREQ =
C1
0

Step Two: Pick values for R1 and C1.
• The impedance of C1 decreases from infinity to
zero as frequency increases from zero to
infinity.
• At the corner frequency of the circuit, the
magnitude of the impedance of C1 equals the
impedance of R1
Simple Filter Design
Vin
R1
Vout
V1
V OFF =
V A MPL =
FREQ =
C1
0

At the corner frequency,
Vout
H
Vin
1
1
1
jwC



1
1  jwRC 1  j
R
jwC
Simple Filter Design
Vin
R1
Vout
V1
V OFF =
V A MPL =
FREQ =
C1
0

The magnitude of H at the corner frequency
is 0.707.
Vout
1
1
H 


 0.707
Vin
1 j
2
Simple Filter Design
Vin
R1
Vout
V1
V OFF =
V A MPL =
FREQ =
C1
0


Thus, for frequencies less than the corner
frequency, Vout is comparable to Vin.
For frequencies greater than the corner
frequency, the output is much smaller than
the input.
1.0
Example
R1
Vin
R2
50
0.8
Vout
0.6
50
V
V1
VOFF = 0
VAMPL = 1
FREQ = 1k
V
C1
0.4
33uF
0.2
0
0
1.0Hz
3.0Hz
10Hz
30Hz
100Hz
300Hz
1.0KHz
3.0KHz
10KHz
V(R2:2)/ V(R1:2)
Frequency


For this example, we select components
from our boxes of parts to pass most
frequencies below 100Hz.
For R=50 and C=33uF, the corner frequency
is about 96Hz.