bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege
DESCRIPTION
Engineering 43. Oscilloscope Phase-Angle Measurement. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. Oscope Summarized. An Oscope does ONE thing:. Draws a PLOT of VOLTAGE vs TIME. And That’s IT!. These are Easy Check the VOLTS/DIV setting on the Scope - PowerPoint PPT PresentationTRANSCRIPT
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
OscilloscopePhase-Angle
Measurement
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Oscope Summarized An Oscope does ONE thing:
Draws a PLOT of
VOLTAGE vs TIME And That’s IT!
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Amplitude Measurements These are Easy1. Check the
VOLTS/DIV setting on the Scope• FILL screen
vertically
2. Count VERTICAL Deflection Divisions• i.e; Count
Squares
3. Multiply DIVs times VOLTS/DIV
5.1 Div
High
VVV
VDIVVDIVV
ppM
pp
28.12
55.25.01.5
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Vertical (V) Scale for Digital Scope
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Phase Angle, The Equation for a Phase-SHIFTED
Sinusoidal Electrical-Potential Signal
)cos( tVtv XMX
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
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
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 Signal2. Get ± from shifted-Signal LEAD or LAG 3. Get -Magnitude from TIME-SHIFT, td
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1. Define the BaseLine Signal For ANY Steady-State AC Signal
(SS-AC) We, as Ckt Analysts, get to PICK ONE Node-Voltage exOR Branch-Current 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
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1. Define the BaseLine Signal The BaseLine Signal is USUALLY (not
Always) the +Side of the Supply
VVt
VtVtv SMSMS
505secrads377cos5V , e.g.
0)0cos(
SV
On the Scope The BaseLine Signal is typically • The “A” or CH1 Trace• The Trigger Source
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
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-
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
This is the BASELINE Signal
The X-Signal LAGS the BASELINE; its PEAK occurs LATER in Time
vS(ωt) vX(ωt±||)
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
2. Lead or Lab = +/− by MATLAB
0 1 2 3 4 5 6 7 8-10
-8
-6
-4
-2
0
2
4
6
8
10Vx LEADS by 53°
time (mS)
Ele
ctric
al P
oten
ial (
V)
Vs(t)Vx(t)
0 1 2 3 4 5 6 7 8-10
-8
-6
-4
-2
0
2
4
6
8
10Vx LAGS by 53°
time (mS)
Ele
ctric
al P
oten
ial (
V)
Vs(t)Vx(t)
)53cos( tVtv XMX )53cos( tVtv XMX LEADING →
POSITIVE LAGGING →
NEGATIVE
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
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
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
T = 4.1DIV
vX Lagging
T = 360°
td = 1.6DIV
VXpp = 4.6D
IV
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Horizontal (t) Scale for digital scope
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
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 & HzfmS
DIVmSDIVT 48805.25.01.4
mSDIVmSDIVtd 8.05.06.1
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
3. -Magnitude Now since td/T is a Fraction of a Period
Multiply td/T by 360° to Find
5.1403601.46.1360
05.218.0
DIVDIV
PeriodmSPeriodmS
In this Case 360
Ttd
Use the LAGGING observation to apply the sign of as NEGATIVE
rads452.25.140Lagging- dt
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complete The Example From The Scope Voltage-
Measurements on the on the “” Slide Find• VXpp = 4.6 DIV • VOLTS/DIV = 0.5 V/Div
Calc VXM
VVV
VDivVDIVV
XppXM
Xpp
15.12
3.25.06.4
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complete the Example Now Can Fully Characterize the
Unknown Sinusoid Relative to the BaseLine
vX
Using The Results of the Phase and Amplitude Calcs
452.2sec
3066Re15.1
452.24882cos15.1
tradsj
X
eV
tVtv
• Note that ω = 2πf Alternatively in Std
Phasor Form 5.14015.1 VXV
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Find H(f) = VC/VS
Find Vc in the Scope-Measured Series RC Circuit
9.7V0°9.7V0° Vc
SCO
PE
BaseLine
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Series Ckt: GND => Vs => R => C => GND
-10
-8
-6
-4
-2
0
2
4
6
8
10
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Time (mS)
Pote
ntia
l to
GN
D (V
)
Vs (V)
Vc (V)
file = CR_RC_Phase-Difference_0601.xls
PARAMETERS• Vs = (9.7V)? 0°• R = 6.8 kΩ• C = 22 nF• f = 1300 Hz
T = 0.77 mS
Vc LAGS
td = 0.11 mSVc
m =
6.15
V
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The RC Series Ckt Phasor Calc The Frequency Parameters
seckRad17.8
seckCycle3.12
Cycle1Rad2
kHz3.1mS77.0cycle 1cycle 1
f
Tf
Calc noting that Vc LAGSrads89.051360
7711360
Ttd
Then Vcby 6.15VAmplitude
5115.6
89.0sec
8170 cos15.6
V
tVtvC
CV
9.7V0°9.7V0° Vc
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The RC Transfer Function The Transfer Function for the R→C
Circuit at 1.3 kHz
516833.0kHz3.1
or07.95115.6kHz3.1
S
S
VV
VV
C
C
fH
VVfH
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Swap C↔R for H(f) Find Vr in the Scope-Measured Series
CR CircuitSC
OPE
BaseLine
9.7V0°9.7V0° Vr
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Series Ckt: GND => Vs => C => R => GND
-10
-8
-6
-4
-2
0
2
4
6
8
10
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Time (mS)
Pote
ntia
l to
GN
D (V
)
Vs (V)
Vr (V)
PARAMETERS• Vs = (9.7V)? 0°• R = 6.8 kΩ• C = 22 nF• f = 1300 Hz
file = CR_RC_Phase-Difference_0601.xls
T = 0.77 mS
Vr LEADS
td = 0.084 mS
Vrm = 7.5V
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The CR Series Ckt Phasor Calc The Frequency Parameters
Calc noting that Vr LEADSrads 68.039360
77084360
T
td
Then Vrby 7.5VAmplitude
39V5.7
68.0sec
8170cosV5.7
RV
ttvR
9.7V0°9.7V0° Vr
seckRad17.8
seckCycle3.12
Cycle1Rad2
kHz3.1mS77.0cycle 1cycle 1
f
Tf
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[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt26
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The CR Transfer Function The Transfer Function for the C→R
Circuit at 1.3 kHz
516833.0kHz3.1
or07.95115.6kHz3.1
S
S
VV
VV
C
C
fH
VVfH
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
All Done with the Tutorial
PhasErson
Stun... A phaser RIFLE
(often referred to as a type-3 phaser)
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
MATLAB Script-Code% B. Mayer % ENGR43 * 19Jan06% Phase-Shift Lag Plot%% Parametersw = 1500; % Angular Freqency in rad/secVsa = 9.7; % Voltage Source Amplitude in VoltsAR = .73; % Attenuation Ratiophi = -0.925; % phase Angle in Radsphi_deg = 180*phi/pi % degrees%%% Calc periodT = 2*pi/w % seconds%% Define t vector over 1.2 periodst = linspace(0, 2.2*T, 200);% % Calc Vs & Vc over 1.2 periodsVs = Vsa*cos(w*t);Vx = AR*Vsa*cos(w*t + phi);%% Plot bothplot(1000*t, Vs, 1000*t, Vx, '--'), xlabel('time (mS)'),... ylabel('Electrical Potenial (V)'),... legend('Vs(t)', 'Vx(t)'), title('Vx LAGS by 53°')
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
More Scope Traces