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Learn how to measure phase angle using an oscilloscope in electrical engineering. Discover the steps to determine phase shift, magnitude, and sign. Obtain valuable insights on analyzing AC signals.
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Engineering 43 OscilloscopePhase-Angle Measurement Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu
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 FILL screen vertically Count VERTICAL Deflection Divisions i.e; Count Squares Multiply DIVs times VOLTS/DIV Amplitude Measurements 5.1 DivHigh
Phase Angle, • The Equation for a Phase-SHIFTED Sinusoidal Electrical-Potential Signal • 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 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 • Define (pick) a BASELINE Signal • Get ± from shifted-Signal LEAD or LAG • 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 • 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 vX(ωt±||) vS(ωt)
2. Lead or Lab = +/− by MATLAB • 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 tdby SEC/DIV, Multiply by 360°/T
td= 1.6DIV vX Lagging VXpp = 4.6DIV T = 4.1DIV T = 360°
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 &
3. -Magnitude • Now since td/T is a Fraction of a Period Multiply td/T by 360° to Find • In this Case • Use the LAGGING observation to apply the sign of as NEGATIVE
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
Now Can Fully Characterize the Unknown Sinusoid Relative to the BaseLine Complete the Example • Using The Results of the Phase and Amplitude Calcs vX • Note that ω = 2πf • Alternatively in Std Phasor Form
Example: Find H(f) = VC/VS • Find Vc in the Scope-Measured Series RC Circuit SCOPE BaseLine
Vc LAGS T = 0.77 mS td= 0.11 mS Vcm =6.15V
The RC Series Ckt Phasor • Calc The Frequency Parameters • Calc noting that Vc LAGS • Then Vcby 6.15VAmplitude
The RC Transfer Function • The Transfer Function for the R→C Circuit at 1.3 kHz
Example: Swap C↔R for H(f) • Find Vr in the Scope-Measured Series CR Circuit SCOPE BaseLine
Vr LEADS td= 0.084 mS T = 0.77 mS Vrm = 7.5V
The CR Series Ckt Phasor • Calc The Frequency Parameters • Calc noting that Vr LEADS • Then Vrby 7.5VAmplitude
The CR Transfer Function • The Transfer Function for the C→R Circuit at 1.3 kHz
All Done with the Tutorial PhasErsonStun... • A phaser RIFLE (often referred to as a type-3 phaser)
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°')