Ch07 Sinusoidal Steady-State Analysis 1s09

Ch07 Sinusoidal Steady-State Analysis 1s09 - Chapter 7 The...

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Sinusoidal Steady Sinusoidal Steady -State State Analysis Analysis Chapter 7 Artemio P. Magabo Artemio P. Magabo Professor of Electrical Engineering Professor of Electrical Engineering Department of Electrical and Electronics Engineering University of the Philippines - Diliman Department of Electrical and Electronics Engineering The Sinusoidal Function α T ) t cos( F ) t ( f m α + ω = where F m = amplitude or peak value ω = angular frequency, rad/sec α = phase angle at t=0, rad The sinusoid is described by the expression ω t, rad -F m F m π 2 π Department of Electrical and Electronics Engineering The sinusoid is generally plotted in terms of ω t, expressed either in radians or degrees. Consider the plot of the sinusoidal function f(t)=F m cos ω t. -F m F m ω t, deg 180 360 ω t, rad π 2 π T When ω t=2 π , t=T. Thus we get ω T=2 π or ω = . The function may also be written as T 2 π t T 2 cos F t cos F ) t ( f m m π = ω = Department of Electrical and Electronics Engineering The peak value of the voltage is V m =311 volts. The angular frequency is ω =377 rad/sec. The frequency is f=60 Hz. The period is T=16.67 msec. Define: The frequency of the sinusoid T 1 f = sec -1 or cycles/sec or Hertz (Hz) Then, the sinusoid may also be expressed as ft 2 cos F t cos F ) t ( f m m π = ω = Note : The nominal voltage in the Philippines is a sinusoid described by V ) t 377 cos( 311 ) t ( v α + = Department of Electrical and Electronics Engineering Leading and Lagging Sinusoids Note: We say either “f 1 (t) leads f 2 (t) by an angle of 30 ° ” or that “f 2 (t) lags f 1 (t) by an angle of 30 ° .” ) 30 t cos( F ) t ( f m 2 ° + ω = Consider the plot of the sinusoidal functions and ) 60 t cos( F ) t ( f m 1 ° + ω = 30 60 90 180 -30 -60 -90 ω t, deg 270 360 ) t ( f 1 ) t ( f 2 Department of Electrical and Electronics Engineering Note: The current is in phase with the voltage. From Ohm’s law, we get t cos RI Ri v m R R ω = = Consider a resistor. Let the current be described by t cos I i m R ω = R + - v R i R 90 180 -90 ω t, deg 270 360 v R i R The Resistor
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Department of Electrical and Electronics Engineering The Inductor Consider an inductor. Let the current be described by t cos I i m L ω = L i L + - v L From v L = , we get t sin LI v m L ω ω - = dt di L L 90 180 -90 ω t, deg 270 360 i L v L Note: The current lags the voltage by 90 o . Department of Electrical and Electronics Engineering The Capacitor i C v C + - C Consider a capacitor. Let the current be described by t cos I i m C ω = From v C = , we get t sin C I v m C ω ω = dt i C 1 C 90 180 -90 ω t, deg 270 360 i C Note: The current leads the voltage by 90 o . v C Department of Electrical and Electronics Engineering Note: We will show later that: Summary: In a resistor, i R and v R are in phase. 1. In an inductor, i
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This note was uploaded on 01/13/2011 for the course EEEI 33 taught by Professor Co during the Winter '10 term at University of the Philippines Diliman.

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Ch07 Sinusoidal Steady-State Analysis 1s09 - Chapter 7 The...

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