Download ENGR-43_Scope_Phase-Angle_Tutorial

Engineering 43
Oscilloscope
Phase-Angle
Measurement
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Oscope Summarized
 An Oscope does ONE thing:
Draws a PLOT
of VOLTAGE vs
TIME
 And That’s IT!
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Amplitude Measurements

These are Easy
1. Check the
VOLTS/DIV setting
on the Scope
•
FILL screen
vertically
5.1 Div
High
2. Count VERTICAL
Deflection Divisions
•
i.e; Count Squares
3. Multiply DIVs times
VOLTS/DIV
Engineering-43: Engineering Circuit Analysis
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0.5V
V pp  5.1DIV 
 2.55V
DIV
VM  V pp 2  1.28V
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Vertical (V) Scale for Digital Scope
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Phase Angle, 
 The Equation for a Phase-SHIFTED
Sinusoidal Electrical-Potential Signal
vX t   VXM cos(t   )
 Where
• VXM  The AMPLITUDE (Max, or Peak
Value) of the Sinusoid in Volts
•   The PHASE Angle in DEGREES
– MAGNITUDE <180°
– SIGN can be POSITIVE or NEGATIVE
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Scope Phase-Angle


The Scope Trace Tells us
NOTHING about the MAGNITUDE
and SIGN of the Phase Angle
•
It Doesn’t Even give a Starting Point
•
All we get is TWO v(t) Traces
The Steps to Get to 
1. Define (pick) a BASELINE Signal
2. Get ±  from shifted-Signal LEAD or LAG
3. Get -Magnitude from TIME-SHIFT, td
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
1. Define the BaseLine Signal
 For ANY Steady-State AC Signal
(SS-AC) We, as Ckt Analysts, get to
PICK ONE Node-Voltage exOR BranchCurrent as having a ZERO Phase Angle
• i.e., We can SET the point where  = 0°
• Analogous to Selecting a GND
 Since the Scope ONLY measures
Potential we can Pick any Node
VOLTAGE as the BaseLine Signal
which has ZERO Phase
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
1. Define the BaseLine Signal
 The BaseLine Signal is USUALLY (not
Always) the +Side of the Supply
vS t   VSM cos(t  0)  VS  VSM 0
rads 

e.g. , 5V  cos 377
t   5V0  5V
sec 

 On the Scope The BaseLine Signal
is typically
• The “A” or CH1 Trace
• The Trigger Source
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
2. Determine the Sign of 
 Looking at the Traces we can
OBSERVE whether the Unknown, or “X”
Signal LEADS or LAGS the BaseLine
• See Next Slide
 The Question Then becomes: Does
• LEAD Imply POSITIVE-?
– Then Lag implies NEGATIVE-
• LAG Imply POSITIVE-?
– Then Lead implies NEGATIVE-
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
This is the BASELINE Signal
The X-Signal LAGS the BASELINE;
its PEAK occurs LATER in Time
vS(ωt)
Engineering-43: Engineering Circuit Analysis
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vX(ωt±||)
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
2. Lead or Lab = +/− by MATLAB
Vx LAGS by 53°
Vx LEADS by 53°
10
10
Vs(t)
Vx(t)
6
6
4
4
2
0
-2
-4
2
0
-2
-4
-6
-6
-8
-8
-10
0
1
2
3
4
time (mS)
5
6
7
8
vX t   VXM cos(t  53)
 LEADING →
POSITIVE 
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Vs(t)
Vx(t)
8
Electrical Potenial (V)
Electrical Potenial (V)
8
-10
0
1
2
3
4
time (mS)
5
6
7
vX t   VXM cos(t  53)
 LAGGING →
NEGATIVE 
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
8
3. -Magnitude
 Notice from the Scope Trace that ONE
Sinusoidal CYCLE-TIME-PERIOD, T,
corresponds to 360°: T↔ 360°
 Further Notice from the Dual-Trace
Display that the X-Signal will Lead or
Lag the BaseLine by the TIME-Shift, td
 Now Realize that td will be some
FRACTION of a Period; Thus
• Find td by SEC/DIV, Multiply by 360°/T
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
td = 1.6DIV
vX Lagging
VXpp = 4.6DIV
T = 4.1DIV
T = 360°
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Horizontal (t) Scale for digital scope
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
3. -Magnitude
 From The Scope Time-Measurements
on the on the Last Slide Find
• T = 4.1 DIV = 360°
• td = 1.6 DIV, Lagging
• SEC/DIV = 0.5 millisec/Div
 Calc T & 
0.5mS
T  4.1DIV 
 2.05mS
DIV
0.5mS
t d  1.6 DIV 
 0.8mS
DIV
Engineering-43: Engineering Circuit Analysis
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f
 488 Hz 
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
3. -Magnitude
 Now since td/T is a Fraction of a Period
td
Multiply td/T by 360° to Find 
   360
T
 In this Case
1Period
360
1.6 DIV
  0.8mS 


 360  140.5
2.05mS Period 4.1DIV
 Use the LAGGING observation to
apply the sign of  as NEGATIVE
td - Lagging    140.5  2.452rads
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Complete The Example
 From The Scope VoltageMeasurements on the on the “” Slide
Find
• VXpp = 4.6 DIV
• VOLTS/DIV = 0.5 V/Div
 Calc VXM
0.5V
VXpp  4.6 DIV 
 2.3V
Div
VXM  VXpp 2  1.15V
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Complete the Example
 Now Can Fully
Characterize the
Unknown Sinusoid
Relative to the
BaseLine
vX
 Using The Results
of the Phase and
Amplitude Calcs
v X t   1.15V cos2  488t  2.452 
t  2.452 
 j 3066rads
sec
 1.15V  Re e



• Note that ω = 2πf
 Alternatively in Std
Phasor Form
VX  1.15V 140.5
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Example: Find H(f) = VC/VS
 Find Vc in the Scope-Measured Series
RC Circuit
9.7V0°
Engineering-43: Engineering Circuit Analysis
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SCOPE
BaseLine
Vc
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Series Ckt: GND => Vs => R => C => GND
10
8
6
Vc LAGS
2
T = 0.77 mS
0
Vcm =6.15V
Potential to GND (V)
4
-2
-4
PARAMETERS
• Vs = (9.7V)? 0°
• R = 6.8 kΩ
-6
-8
•
C
•
f
=
=
2
1
2
3
n
0
0
td = 0.11 mS
Vs (V)
F
H
z
Vc (V)
-10
0.0
0.1
0.2
file = CR_RC_Phase-Difference_0601.xls
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0.3
0.4
0.5
0.6
0.7
Time (mS)
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
0.8
0.9
The RC Series Ckt Phasor
 Calc The Frequency Parameters
f  1 cycle  T  1 cycle  0.77mS  1.3kHz
2 Rad
kCycle
kRad

f  2 1.3
 8.17
1 Cycle
sec
sec
9.7V0°
Vc
 Calc  noting that Vc LAGS
 td
11

360   360  51  0.89rads
T
77
 Then Vc
by 6.15V
Amplitude
Engineering-43: Engineering Circuit Analysis
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  8170 

vC t   6.15V cos 
t  0.89 

  sec 

VC  6.15V  51
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
The RC Transfer Function
 The Transfer Function for the R→C
Circuit at 1.3 kHz
VC 6.15V  51
H  f  1.3kHz  

VS
9.7V0
or
VC
H  f  1.3kHz  
 0.6833  51
VS
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Example: Swap C↔R for H(f)
 Find Vr in the Scope-Measured Series
CR Circuit
9.7V0°
Engineering-43: Engineering Circuit Analysis
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SCOPE
BaseLine
Vr
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Series Ckt: GND => Vs => C => R => GND
10
Vr LEADS
8
6
td = 0.084 mS
2
T = 0.77 mS
0
Vrm = 7.5V
Potential to GND (V)
4
-2
-4
PARAMETERS
• Vs = (9.7V)? 0°
• R = 6.8 kΩ
-6
•
C
•
f
=
2
2
n
Vs (V)
F
-8
=
1
3
0
0
H
z
Vr (V)
-10
0.0
0.1
0.2
file = CR_RC_Phase-Difference_0601.xls
Engineering-43: Engineering Circuit Analysis
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0.3
0.4
0.5
0.6
0.7
Time (mS)
Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
0.8
0.9
The CR Series Ckt Phasor
 Calc The Frequency Parameters
f  1 cycle  T  1 cycle  0.77mS  1.3kHz
2 Rad
kCycle
kRad

f  2 1.3
 8.17
1 Cycle
sec
sec
 Calc  noting that Vr LEADS
9.7V0°
360 84
  t d

360  39  0.68 rads
T
770
 Then Vr
 8170

vR t   7.5V cos
t  0.68 
by 7.5V
 sec

Amplitude V  7.5V39
R
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
Vr
The CR Transfer Function
 The Transfer Function for the C→R
Circuit at 1.3 kHz
VC 6.15V  51
H  f  1.3kHz  

VS
9.7V0
or
VC
H  f  1.3kHz  
 0.6833  51
VS
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
All Done with the Tutorial
PhasErs
on
Stun...
 A phaser RIFLE
(often referred to as a
type-3 phaser)
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
MATLAB Script-Code
% B. Mayer
% ENGR43 * 19Jan06
% Phase-Shift Lag Plot
%
% Parameters
w = 1500; % Angular Freqency in rad/sec
Vsa = 9.7; % Voltage Source Amplitude in Volts
AR = .73; % Attenuation Ratio
phi = -0.925; % phase Angle in Rads
phi_deg = 180*phi/pi % degrees
%
%
% Calc period
T = 2*pi/w % seconds
%
% Define t vector over 1.2 periods
t = linspace(0, 2.2*T, 200);
%
% Calc Vs & Vc over 1.2 periods
Vs = Vsa*cos(w*t);
Vx = AR*Vsa*cos(w*t + phi);
%
% Plot both
plot(1000*t, Vs, 1000*t, Vx, '--'), xlabel('time (mS)'),...
ylabel('Electrical Potenial (V)'),...
legend('Vs(t)', 'Vx(t)'), title('Vx LAGS by 53°')
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt
More Scope Traces
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt