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Electronic InstrumentationExperiment 2
* Part A: Intro to Transfer Functions and AC Sweeps* Part B: Phasors, Transfer Functions and Filters* Part C: Using Transfer Functions and RLC Circuits* Part D: Equivalent Impedance and DC Sweeps
Part A
Introduction to Transfer Functions and Phasors
Complex Polar Coordinates Complex Impedance (Z) AC Sweeps
Transfer Functions
The transfer function describes the behavior of a circuit at Vout for all possible Vin.
in
out
V
VH
Simple Example
321
32*
RRR
RRVV inout
kkk
kkVV inout 321
32*
6
5
in
out
V
VH
VktVtVthen
VktVtVif
out
in
10)2
2sin(5)(
12)2
2sin(6)(
More Complicated Example
What is H now?
H now depends upon the input frequency ( = 2f) because the capacitor and inductor make the voltages change with the change in current.
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
We will define Phasors
A phasor is a function of the amplitude and phase of a sinusoidal signal
Phasors allow us to manipulate sinusoids in terms of amplitude and phase changes.
Phasors are based on complex polar coordinates.
Using phasors and complex numbers we will be able to find transfer functions for circuits.
),( AfV
Review of Polar Coordinates
point P is at( rpcosp , rpsinp )
22
1tan
PPP
P
PP
yxr
x
y
Review of Complex Numbers
zp is a single number represented by two numbers
zp has a “real” part (xp) and an “imaginary” part (yp)
jj
jj
j
1
1
1
Complex Polar Coordinates
z = x+jy where x is A cos and y is A sin t cycles once around the origin once for each
cycle of the sinusoidal wave (=2f)
Now we can define Phasors
.),(
sincos,
)sin()cos(
,)cos()(
droppedisitsotermeachtocommonist
jAAVsimplyor
tjAtAV
letthentAtVif
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.
The “V=IR” of Phasors
The influence of each component is given by Z, its complex impedance
Once we have Z, we can use phasors to analyze circuits in much the same way that we analyze resistive circuits – except we will be using the complex polar representation.
ZIV
Magnitude and Phase
Vofphasex
yV
VofmagnitudeAyxV
jyxAjAV
1
22
tan
sincos
Phasors have a magnitude and a phase derived from polar coordinates rules.
Influence of Resistor on Circuit
Resistor modifies the amplitude of the signal by R
Resistor has no effect on the phase
RIV RR
)sin(*)(
)sin()(
tARtVthen
tAtIif
R
R
Influence of Inductor on Circuit
Inductor modifies the amplitude of the signal by L
Inductor shifts the phase by +/2
dt
dILV L
L
)2
sin(*)(
)cos(*)(
)sin()(
tALtVor
tALtVthen
tAtIif
L
L
L
Note: cos=sin(+/2)
Influence of Capacitor on Circuit
Capacitor modifies the amplitude of the signal by 1/C
Capacitor shifts the phase by -/2
dtIC
V CC
1
)2
sin(*1
)2
sin(*1
)(
)cos(*1
)cos(*1
)(
)sin()(
tAC
tAC
tVor
tAC
tAC
tVthen
tAtIif
C
C
C
Understanding the influence of Phase
x
yV 1tan
180
)(0
tan
00:
9020
tan
00:
9020
tan
00:
00
tan
00:
1
1
1
1
orx
Vthen
xandyifreal
yVthen
yandxifj
yVthen
yandxifj
xVthen
xandyifreal
Complex Impedance
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/jC • change the amplitude by 1/C
• shift the phase -90 (1/j=-j)
• Inductors: ZL=jL • change the amplitude by L
• shift the phase +90 (j)
ZIV
AC SweepsAC Source sweeps from 1Hz to 10K Hz
Time
200ms 250ms 300ms 350ms 400msV(R1:2) V(C1:1)
-1.0V
0V
1.0V
Time
20ms 25ms 30ms 35ms 40msV(R1:2) V(C1:1)
-1.0V
0V
1.0V
Time
2.0ms 2.5ms 3.0ms 3.5ms 4.0msV(R1:2) V(C1:1)
-1.0V
0V
1.0V
Transient at 10 Hz Transient at 100 Hz Transient at 1k Hz
Notes on Logarithmic Scales
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
Phasors Complex Transfer Functions Filters
Definition of a Phasor
sincos
,)cos()(
jAAV
letthentAtVif
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.
Phasor References
http://ccrma-www.stanford.edu/~jos/filters/Phasor_Notation.html
http://www.ligo.caltech.edu/~vsanni/ph3/ExpACCircuits/ACCircuits.pdf
http://ptolemy.eecs.berkeley.edu/eecs20/berkeley/phasors/demo/phasors.html
Magnitude and Phase
Vofphasex
yV
VofmagnitudeAyxV
jyxAjAV
1
22
tan
sincos
Phasors have a magnitude and a phase derived from polar coordinates rules.
Euler’s Formula
sincos je j
2132
13
)(
2
1
2
1
2
13
,
sincos
21
2
1
andr
rrtherefore
er
r
er
er
z
zzthen
rejrrjyxzif
jj
j
j
214214
)(2121214
,
2121
andrrrtherefore
errererzzzand jjj
Manipulating Phasors (1)
2132
13
)(
2
1
2
1)(
2
)(1
2
13
)(
,
)sin()cos(
21
2
1
2
1
XandA
AXtherefore
eA
A
e
e
e
e
A
A
eA
eA
V
VX
AetjtAV
jj
tj
tj
tj
tj
tj
Note t is eliminated by the ratio• This gives the phase change between
signal 1 and signal 2
Manipulating Phasors (2)
22
22
21
2
2
1
3
1
yx
yx
V
VX
333
222111
jyxV
jyxVjyxV
2
21
1
11213 tantan
x
y
x
yVVX
Complex Transfer Functions
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
)(
)()(
jV
jVjH
in
out
Complex Impedance
Z defines the influence of a component on the amplitude and phase of a circuit• Resistors: ZR = R
• Capacitors: ZC=1/jC
• Inductors: ZL=jL
We can use the rules for resistors to analyze circuits with capacitors and inductors if we use phasors and complex impedance.
ZIV
Simple Example
1
1)(
1
1
)(
)(
)()(
RCjjH
Cj
Cj
CjR
CjjH
ZZ
Z
IZZ
IZ
jV
jVjH
CR
C
CR
C
in
out
CjZ
RZ
C
R
1
Simple Example (continued)
1
1)(
RCjjH
222
22
)(1
1
)(1
01
1
01)(
RCRCRCj
jjH
)(tan1
tan1
0tan)(
)1()01()(
111 RCRC
jH
RCjjjH
2)(1
1)(
RCjH
)(tan)( 1 RCjH
High and Low Pass FiltersHigh Pass Filter
H = 0 at 0
H = 1 at
at c
Frequency
1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(C1:2) / V(R1:2)
0
0.5
1.0
Frequency
1.0Hz 100Hz 10KHz 1.0MHz 100MHzV1(R1) / V(C1:2)
0
0.5
1.0
Low Pass Filter
H = 1 at 0
H = 0 at
at c
c=2fc
fc
fc
c=2fc
Corner Frequency The corner frequency of an RC or RL circuit
tells us where it transitions from low to high or visa versa.
We define it as the place where
For RC circuits:
For RL circuits:
2
1)( cjH
L
Rc
RCc
1
Corner Frequency of our example
212 RC 2
2
1 RC
RCjjH
1
1)(
2
1)( jH
2
1
)(1
1
2
1
)(1
1)(
22
RCRCjH
RCc
1
H(j), c, and filters We can use the transfer function, H(j), and the
corner frequency, c, to easily determine the characteristics of a filter.
If we consider the behavior of the transfer function as 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:
2
ccf
Taking limits
012
2
012
2)(bbb
aaajH
At low frequencies, (ie. =10-3), lowest power of dominates
At high frequencies (ie. =10+3), highest power of dominates
0
00
03
16
2
00
31
62
101010
101010)(
b
a
bbb
aaajH
2
20
03
16
2
00
31
62
101010
101010)(
b
a
bbb
aaajH
Taking limits -- Example
523
159)(
2
2
jH
At low frequencies, (lowest power)
At high frequencies, (highest power)
35
15)( jH LO
33
9)(
2
2
jH HI
Our example at low frequencies
)(01
0tan)(
110)(
1 axisxonjH
asjH
LOW
LOW
RCjjH
1
1)(
101
1)(
jH LOW
Our example at high frequencies
220
0tan
1
0tan)(
011
)(
11
RCjH
RCjasjH
HIGH
HIGH
RCjjH
1
1)(
RCjjH HIGH 1
)(
Our example is a low pass filter
Frequency
1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(C1:2) / V(R1:2)
0
0.5
1.0
RCf c
c
2
1
2
01 HIGHLOW HH
What about the phase?
Our example has a phase shift
0
1
HIGH
LOW
H
H
Frequency
1.0Hz 10KHz 100MHzVP(C1:2)
-100d
-50d
0dV(R1:2) / V(V1:+)
0
0.5
1.0
SEL>>
90)(
0)(
jH
jH
HIGH
LOW
Part C
Using Transfer Functions Capacitor Impedance Proof More Filters Transfer Functions of RLC Circuits
Using H to find Vout
in
out
in
out
jin
jout
tjjin
tjjout
in
out
eA
eA
eeA
eeA
V
VjH
)(
inout
inout
jin
jHjjout
jin
jout
eAejHeA
eAjHeA
)()(
)(
inout AjHA )( inout jH )(
Simple Example (with numbers)
157.0)2(1
1)(
2
2
jH 41.11
2tan0)( 1
jH
)4
2cos(2)(
11
tkVtV
kRFC
in
)41.1785.02cos(2*157.0)( tkVtVout
12
1
1112
1
1
1)(
jkkjRCjjH
)625.02cos(314.0)( tkVtVout
Capacitor Impedance Proof
)(1
)()()(
)(
)()cos()(
Re)(
)()(
)sin()cos()(
)cos()()(
)(
)()(
)(
tICj
tVtVCjdt
tdVCtI
tVjtAjdt
jVd
dt
tdV
jVjeAjdt
dAe
dt
jVd
AetjAtAjV
tAtVanddt
tdVCtI
CCCC
C
CCC
Ctj
tjC
tjC
CC
C
CjZC
1Prove:
Band Filters
Frequency
1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(L1:1) / V(V1:+)
0
0.5
1.0
Frequency
1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(R1:1)/ V(R1:2)
0
0.5
1.0
f0
f0
Band Pass Filter
H = 0 at 0
H = 0 at
at 0=2f0
Band Reject Filter
H = 1 at 0
H = 1 at
at 0 =2f0
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).
The equation for the resonant frequency of an RLC circuit is:
LC
10
Another Example
1
11
1
)(22
LCjRCj
CjLjR
CjjH
CjZ
LjZRZ
C
LR
1
RCjLCjH
)1(
1)(
2
At Very Low Frequencies
0)(
10)(
11
1)(
jH
jH
jH
LOW
LOW
LOW
At Very High Frequencies
orjH
jH
LCjH
HIGH
HIGH
HIGH
)(
01
)(
1)(
2
At the Resonant Frequency
LC
10
RCjLCjH
)1(
1)(
2
LC
RCjRC
LCjLC
LC
jH
)11(
1
1)
11(
1)( 20
2)(
)(
)(
0
0
0
jH
RC
LCjH
RC
LCjjH
if L=1mH, C=0.1uF and R=100
krad/secfkHz
LCf
2
10
radiansH2
Frequency
1.0Hz 100Hz 10KHz 1.0MHzV(C1:1)/V(L1:1)
0
0.4
0.8
1.2p(V(C1:1)/V(L1:1))
-200d
-100d
0d
SEL>>
Our example is a low pass filterPhase
= 0 at 0
= - at
Magnitude
= 1 at 0
= 0 at
fkHz
Actual circuit resonance is only at the theoretical resonant frequency, f0, when there is no resistance.
Part D
Equivalent Impedance Transfer Functions of More
Complex Circuits
Equivalent Impedance
Even though this filter has parallel components, we can still handle it.
We can combine complex impedances like resistors to find the equivalent impedance of the components combined.
Equivalent Impedance
LC
Lj
LCj
Lj
CjLj
CjLj
ZZ
ZZZ
CL
CLCL 222 111
1
Determine H
LC
LjZCL 21
LjLCR
LjjH
)1()(
2
LC
LCbymultiply
LCLj
R
LCLj
jH2
2
2
2
1
1
1
1)(
CL
CL
ZR
ZjH
)(
At Very Low Frequencies
2)(
00)(
)(
jH
jHR
LjjH
LOW
LOW
LOW
At Very High Frequencies
2)(
01
)(
)(2
jH
jH
RC
j
LRC
LjjH
HIGH
HIGH
HIGH
At the Resonant Frequency
11
)1
1(
1
)( 20
LLC
jLCLC
R
LLC
j
jH
0)(1)( 00 jHjH
LC
10
Frequency
1.0Hz 10KHz 100MHzVP(R1:1)
-100d
0d
100d
SEL>>
V1(R1) / V(V1:+)0
0.5
1.0
Our example is a band pass filter
Phase
= 90 at 0
= 0 at
= - at
Magnitude
= 0 at 0
H=1 at
= 0 at
f
