Transcript Experiment 2 - Rensselaer Polytechnic Institute

Electronic Instrumentation
Experiment 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
In Class Solution
P1
Question 1: From diagram A: Which of the following
statements is true given the direction of the current flow.
c.) P2 > P1
Current flow from high pressure to low pressure
therefore P2 must be greater than P1.
Question 2: Draw the circuit equivalent of diagram A,
label current flow and voltage (+ and – on the voltage
source).
Question 3: Draw an AC signal with the following
parameters
Vp-p=6V
Vave=0V
Frequency=2KHz
Label the axis, label the amplitude and period
I
Pump
I
Constriction
P2
Diagram A
Vsource
R1
In Class Solution
4.0V
Amp
3V
0V
Period 0.5ms
-4.0V
0s
0.5ms
1.0ms
1.5ms
V(V1:+)
Time
2.0ms
2.5ms
3.0ms
Circuit Analysis (Combination Method)
Part A

Introduction to Transfer
Functions and Phasors
 Complex Polar Coordinates
 Complex Impedance (Z)
 AC Sweeps
Transfer Functions
Vout
H
Vin

The transfer function describes the
behavior of a circuit at Vout for all
possible Vin.
Simple Example
Vout
R 2  R3
 Vin *
R1  R 2  R3
Vout
2k  3k
 Vin *
1k  2k  3k
Vout 5
H

Vin 6
if Vin (t )  6V sin(2kt 

2
)  12V
then Vout (t )  5V sin(2kt 

2
)  10V
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.
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

V  f ( A,  )

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.
Review of Polar Coordinates
point P is at
( rpcosqp , rpsinqp )
 yP 
q P  tan  
 xP 
1
rP  xP2  y P2
Review of 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)
Complex Polar Coordinates

z = x+jy where x is A
cos and y is A sin
 wt cycles once around
the origin once for each
cycle of the sinusoidal
wave (w=2f)
Now we can define Phasors
if V (t )  A cos(w t   ) , then let

V  A cos(w t   )  jA sin(w t   )

or sim ply, V  A cos  jA sin 
(wt is com m onto each term, so it is dropped.)



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
 
V  IZ


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.
Magnitude and Phase

V  A cos  j A sin   x  jy


2
2
V  x  y  A m agnitudeof V

1  y 
V  tan    
 x


phaseof V
Phasors have a magnitude and a phase
derived from polar coordinates rules.
Influence of Resistor on Circuit
VR  I R R
if I R (t )  A sin(wt )
then VR (t )  R * A sin(wt )


Resistor modifies the amplitude of the
signal by R
Resistor has no effect on the phase
Influence of Inductor on Circuit
dI L
VL  L
dt
Note:
cosq=sin(q+/2)
if I L (t )  A sin(wt )
then VL (t )  w L * A cos(wt )

or VL (t )  w L * A sin(wt  )
2


Inductor modifies the amplitude of the
signal by wL
Inductor shifts the phase by +/2
Influence of Capacitor on Circuit
1
VC   I C dt
C
if I C (t )  A sin(wt )
1
1
then VC (t ) 
* A cos(wt ) 
* A cos(wt   )
wC
wC
1

1

or VC (t ) 
* A sin(wt    ) 
* A sin(wt  )
wC
2
wC
2


Capacitor modifies the amplitude of the
signal by 1/wC
Capacitor shifts the phase by -/2
Understanding the influence of Phase
 real : if y  0 and x  0

 0 
then V  t an1     0
x 
 j:
if x  0 and y  0
 j:

1  y 
V  tan  
x



1  y 
then V  t an     90
 0  2
if x  0 and y  0



1  y 
then V  t an      90
2
 0 
 real : if y  0 and x  0

 0 
then V  t an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)
AC Sweeps
AC Source
sweeps from
1Hz to 10K Hz
1.0V
1.0V
1.0V
0V
0V
0V
-1.0V
200ms
250ms
300ms
V(R1:2)
V(C1:1)
Time
350ms
Transient at 10 Hz
400ms
-1.0V
20ms
25ms
30ms
V(R1:2)
V(C1:1)
Time
35ms
Transient at 100 Hz
40ms
-1.0V
2.0ms
2.5ms
3.0ms
V(R1:2)
V(C1:1)
Time
3.5ms
Transient at 1k Hz
4.0ms
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
if V (t )  A cos(w t   ) , then let

V  A cos  jA sin 



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://ccrmawww.stanford.edu/~jos/filters/Phasor_Notat
ion.html
http://www.ligo.caltech.edu/~vsanni/ph3/Ex
pACCircuits/ACCircuits.pdf
http://ptolemy.eecs.berkeley.edu/eecs20/ber
keley/phasors/demo/phasors.html
Phasor Applet
Adding Phasors & Other Applets
Magnitude and Phase

V  A cos  j A sin   x  jy


2
2
V  x  y  A m agnitudeof V

1  y 
V  tan    
 x


phaseof V
Phasors have a magnitude and a phase
derived from polar coordinates rules.
Euler’s Formula
e
jq
 cosq  j sin q
if z  x  jy  r cosq  jr sin q  re jq
z1 r 1e jq1 r1 j (q1 q 2 )
then z3 

 e
jq 2
z 2 r2 e
r2
r1
therefore, r3 
r2
and z4  z1  z2 r1e
and q 3  q1  q 2
jq1
therefore, r4  r1  r2
 r2e
jq 2
 r1  r2e
j (q1 q 2 )
and q 4  q1  q 2
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
X3   


e
V2 A2 e j (w t 2 ) A2 e jw t e2 A2


A1
therefore, X 3 
and X 3  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
X3   
2
2
V2
x2  y2



1  y1 
1  y2 
X 3  V1  V2  tan    tan  
 x1 
 x2 
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
 
Complex Impedance V  I Z

Z defines the influence of a component
on the amplitude and phase of a circuit
• Resistors: ZR = R
• Capacitors: ZC=1/jwC
• Inductors: ZL=jwL

We can use the rules for resistors to
analyze circuits with capacitors and
inductors if we use phasors and complex
impedance.
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  0 2
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)
In Class Problems
R4
R1
7 ohms
Vin
2Vdc
Vout
7 ohms
R2
12 ohms
R3
6 ohms
R5
5 ohms
Question 1: What is the equation for Rtotal?
(Combining R1, R2, R3, R4, and R5?)
Question 2: What is the value for Rtotal?
Question 3: What is the transfer function for the above circuit?
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
1
1
H ( jw ) 

2
2
1  (w RC)
2  1  wRC
2
1
2

w
2
RC
1
1

2
1  (w RC)
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
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
Taking limits -- Example
9w 2  15w
H ( jw ) 
2
3w  2w  5

At low frequencies, (lowest power)
15w
H LO ( jw ) 
 3w
5

At high frequencies, (highest power)
9w 2
H HI ( jw ) 
3
2
3w
Our example at low frequencies
1
H ( jw ) 
1  jw RC
H LOW
1
( 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 ) asw   
 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
Part C

Using Transfer Functions
 Capacitor Impedance Proof
 More Filters
 Transfer Functions of RLC Circuits
Using H to find Vout

Vout Aout e jout e jwt Aout e jout
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
jin
jH ( jw )
Aine
jin
Aout  H ( jw ) Ain out  H ( jw )  in
Simple Example (with numbers)
C  1 F
R  1k

Vin (t )  2V cos(2k t  )
4

H ( jw ) 
H ( jw ) 
1
1
1


jw RC  1 j 2k 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)
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   )  jA sin(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
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).
 The equation for the resonant frequency of an
RLC circuit is:
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 ) 
(1  w 2 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
1
H ( jw ) 
(1  w 2 LC )  jw RC
H ( jw 0 ) 
w0 
1
LC
1
2
 1 
(1  
 LC ) 
 LC 
LC
H ( jw 0 )   j
RC
LC
H ( jw 0 ) 
RC
H ( j w 0 )  

2
 1 
j
 RC
 LC 

f0 
1
2 LC
1
 RC 
(1  1)  j

 LC 
if L=1mH, C=0.1uF and R=100
w0100k rad/sec f016k Hz

H   radians
|H0|1
2
Our example is a low pass filter
0d
Phase
90
 = 0 at w0
-100d
 = -180 at w 
SEL>>
-200d
p(V(C1:1)/V(L1:1))
1.2
1
Magnitude
0.8
H = 1 at w0
0.4
H = 0 at w 
0
1.0Hz
100Hz
V(C1:1)/V(L1:1)
10KHz
1.0MHz
Frequency
Actual circuit resonance is only at the theoretical
resonant frequency, f0, when there is no resistance.
f016k Hz
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
jw L 
Z CL 
1
Z L  ZC
jw L
jw L
jw C

 2 2

2
1
Z L  ZC
j
w
LC

1
1

w
LC
jw L 
jw C
Determine H
Z CL
jw L

2
1  w LC
jw L
2
1

w
LC
H ( jw ) 
jw L
R
1  w 2 LC
Z CL
H ( jw ) 
R  Z CL
1  w 2 LC
m ultiplyby
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 Resonant 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 
In Class Problems
R4
R1
7 ohms
Vin
2Vdc
Vout
7 ohms
R2
12 ohms
R3
6 ohms
R5
5 ohms
Question 1: What is the equation for Rtotal?
(Combining R1, R2, R3, R4, and R5?)
[(R5+R4)*(R2*R3/R2+R3)]/ [(R5+R4)+(R2*R3/R2+R3)] + R1
Question 2: What is the value for Rtotal? 10 ohms
Question 3: What is the transfer function for the above circuit?
(next slides)
Find voltage at this point, then use
voltage divider (only use this in series)
R4
R1
7 ohms
7 ohms
Vin
R2
12 ohms
2Vdc
R1
R3
6 ohms
VB
Vin  2V
R5
5 ohms
R2345  3 
R2345
1 ohm
VB  Vin 
R2345  R1
V1
2Vdc
Vout
R2345
3 ohms
VB  0.6V
R5
Vout  VB
R5  R4
R1  7
R5  5
R4  7
VB=0.6V
R4
R1
7 ohms
Vin
7 ohms
R2
12 ohms
2Vdc
Vout
R3
6 ohms
R5
5 ohms
Now use the voltage
divider to find Vout
VB  0 .6V
R5
Vo ut   VB
R5  R4
Vo ut  0 .25V
We’re not done though…we are looking for the transfer function
H=Vout/Vin so remember V  2V
in
V o ut
Vin
 0 .12 5