[email protected] engr-43_scope_phase-angle_tutorial.ppt 1 bruce mayer, pe engineering-43:...

[email protected] • ENGR-43_Scope_Phase-Angle_Tutorial.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

OscilloscopeOscilloscopePhase-Angle Phase-Angle

MeasurementMeasurement



Oscope SummarizedOscope Summarized

An Oscope does ONE thing:

Draws a Draws a PLOT of PLOT of

VOLTAGE vs VOLTAGE vs TIMETIME And That’s IT!



Amplitude MeasurementsAmplitude Measurements These are Easy

1. 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

VDIV

VDIVV

ppM

pp

28.12

55.25.0

1.5



Vertical (V) Scale for TDS-340Vertical (V) Scale for TDS-340



Phase Angle, Phase Angle, The Equation for a Phase-SHIFTED

Sinusoidal Electrical-Potential Signal

)cos( tVtv XMX

Where• VXM The AMPLITUDE (Max Value) of the

Sinusoid in Volts The PHASE Angle in DEGREES

– MAGNITUDE <180°

– SIGN can be POSITIVE or NEGATIVE



Scope Phase-AngleScope Phase-Angle

The Scope Trace Tells usNOTHING 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,



1. Define the BaseLine Signal1. 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



1. Define the BaseLine Signal1. Define the BaseLine Signal

The BaseLine Signal is USUALLY (not Always) the +Side of the Supply

VVt

VtVtv SMSMS

505sec

rads377cos5V , e.g.

0)0cos(

SV

On the Scope The BaseLine Signal is typically • The “A” or CH1 Trace

• The Trigger Source



2. Determine the Sign of 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-



This is the BASELINE Signal

The X-Signal LAGS the BASELINE; its PEAK occurs LATER in Time

vS(ωt) vX(ωt±||)



2. Lead or Lab = +/2. Lead or Lab = +/−− by MATLAB 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)E

lect

rical

Pot

enia

l (V

)

Vs(t)

Vx(t)

)53cos( tVtv XMX )53cos( tVtv XMX LEADING →

POSITIVE LAGGING →

NEGATIVE



3. 3. -Magnitude-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,

Now Realize that will be some FRACTION of a Period; Thus• Find by SEC/DIV, Multiply by 360°/T



T = 4DIV

vX Lagging

T = 360°

= 1.6DIV

VX

pp =

4.6DIV



Horizontal (t) Scale for TDS-340Horizontal (t) Scale for TDS-340




From The Scope Time-Measurements on the on the Last Slide Find• T = 4 DIV = 360° = 1.6 DIV, Lagging

• SEC/DIV = 0.5 millisec/Div

Calc T & HzfmS

DIV

mSDIVT 5002

5.04

mSDIV

mSDIV 8.0

5.06.1




Now since /T is a Fraction of a Period Multiply / T by 360° to Find

1443604

6.1360

2

18.0

DIV

DIV

PeriodmS

PeriodmS

In this Case 360

T

Use the LAGGING observation to apply the sign of as NEGATIVE

radsLagging 513.2144



Complete The ExampleComplete 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

VDiv

VDIVV

XppXM

Xpp

15.12

3.25.0

6.4



Complete the ExampleComplete the Example Now Can Fully

Characterize the Unknown Sinusoid Relative to the BaseLine

vX

Using The Results of the Phase and Amplitude Calcs

513.2sec

3142Re15.1

513.25002cos15.1

trads

j

X

eV

tVtv

• Note that ω = 2πf

Alternatively in Std Phasor Form

14415.1 VXV



Another ExampleAnother Example

Find Vc in the Scope-Measured Series RC Circuit

9.7V0°9.7V0° Vc

SC

OP

E

BaseLine



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)

Po

ten

tial

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

= 0.11 mS

Vc

m =

6.1

5V



The RC Series Ckt PhasorThe RC Series Ckt Phasor

Calc The Frequency Parameters

sec17.83.121

2

3.177.0cycle 1cycle 1

kradkfcycle

rads

kHzmSTf

Calc noting that Vc LAGS

radsT

89.05136077

11360

Then Vcby 6.15VAmplitude

5115.6

89.08170cos15.6

V

tVtvC

CV

9.7V0°9.7V0° Vc



Series CR ExampleSeries CR Example

Find Vr in the Scope-Measured Series CR Circuit

SC

OP

E

BaseLine

9.7V0°9.7V0° Vr



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)

Po

ten

tial

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 = 0.084 mS

Vrm

= 7

.5V



The CR Series Ckt PhasorThe CR Series Ckt Phasor

Calc The Frequency Parameters

sec17.83.121

2

3.177.0cycle 1cycle 1

kradkfHz

rads

kHzmSTf

Calc noting that Vr LEADS

radsT

68.039360770

84360

Then Vrby 7.5VAmplitude

395.7

68.08170cos5.7

V

tVtvR

RV

9.7V0°9.7V0° Vr



All Done with the TutorialAll Done with the Tutorial

PhasErson

Stun...

A phaser RIFLE (often referred to as a type-3 phaser)



MATLAB Script-CodeMATLAB 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°')



More Scope TracesMore Scope Traces

[email protected] engr-43_scope_phase-angle_tutorial.ppt 1 bruce mayer, pe engineering-43:...

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