bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

BMayer@ChabotCollege.edu • ENGR-43_Scope_Phase-Angle_Tutorial.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 43

OscilloscopePhase-Angle

Measurement

Oscope Summarized An Oscope does ONE thing:

Draws a PLOT of

VOLTAGE vs TIME And That’s IT!

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

VDIVVDIVV

55.25.01.5

Vertical (V) Scale for Digital Scope

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

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

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

1. Define the BaseLine Signal The BaseLine Signal is USUALLY (not

Always) the +Side of the Supply

VtVtv SMSMS

505secrads377cos5V , e.g.

0)0cos(

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

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 = +/− by MATLAB

0 1 2 3 4 5 6 7 8-10

10Vx LEADS by 53°

time (mS)

Vs(t)Vx(t)

0 1 2 3 4 5 6 7 8-10

10Vx LAGS by 53°

time (mS)

Vs(t)Vx(t)

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

POSITIVE LAGGING →

NEGATIVE

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

T = 4.1DIV

vX Lagging

T = 360°

td = 1.6DIV

VXpp = 4.6D

Horizontal (t) Scale for digital scope

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

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

Use the LAGGING observation to apply the sign of as NEGATIVE

rads452.25.140Lagging- dt

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

VDivVDIVV

3.25.06.4

Complete the Example Now Can Fully Characterize the

Unknown Sinusoid Relative to the BaseLine

Using The Results of the Phase and Amplitude Calcs

452.2sec

3066Re15.1

452.24882cos15.1

tradsj

• Note that ω = 2πf Alternatively in Std

Phasor Form 5.14015.1 VXV

Example: Find H(f) = VC/VS

Find Vc in the Scope-Measured Series RC Circuit

9.7V0°9.7V0° Vc

BaseLine

Series Ckt: GND => Vs => R => C => GND

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Time (mS)

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

The RC Series Ckt Phasor Calc The Frequency Parameters

seckRad17.8

seckCycle3.12

Cycle1Rad2

kHz3.1mS77.0cycle 1cycle 1

Calc noting that Vc LAGSrads89.051360

7711360

Then Vcby 6.15VAmplitude

5115.6

89.0sec

8170 cos15.6

9.7V0°9.7V0° Vc

The RC Transfer Function The Transfer Function for the R→C

Circuit at 1.3 kHz

516833.0kHz3.1

or07.95115.6kHz3.1

Example: Swap C↔R for H(f) Find Vr in the Scope-Measured Series

CR CircuitSC

BaseLine

9.7V0°9.7V0° Vr

Series Ckt: GND => Vs => C => R => GND

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Time (mS)

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

The CR Series Ckt Phasor Calc The Frequency Parameters

Calc noting that Vr LEADSrads 68.039360

77084360

Then Vrby 7.5VAmplitude

39V5.7

68.0sec

8170cosV5.7

9.7V0°9.7V0° Vr

seckRad17.8

seckCycle3.12

Cycle1Rad2

kHz3.1mS77.0cycle 1cycle 1

The CR Transfer Function The Transfer Function for the C→R

Circuit at 1.3 kHz

516833.0kHz3.1

or07.95115.6kHz3.1

All Done with the Tutorial

PhasErson

Stun... A phaser RIFLE

(often referred to as a type-3 phaser)

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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