Download Phasor Analysis in Electrical Engineering: Understanding Sinusoidal Steady-state Response and more Lecture notes Structures and Materials in PDF only on Docsity! 3-1 Sinusoidal steady-state analysis Phasor analysis is a technique to find the steady-state response when the system input is a sinusoid. That is, phasor analysis is sinusoidal analysis. Phasor analysis is a powerful technique with which to find the steady-state portion of the complete response. Phasor analysis does not find the transient response. Phasor analysis does not find the complete response. R Original circuit Acos (wt + 5) C) 8 L R Add an imaginary sine source to obtain A cos (at + @,) @) ¢ Sl jAsin (at + @,) @) R : felat Use Euler's relation to obtain Ait @) L Aelat+ Os) = agl@sgiat = gajat lecture 3 outline 3-2 The differential equation becomes Rhe™+Ld(he™)ydt = Ve Rice + jal Le =v, ei Rh + joLh = Vs The complex currents and voltages in the equations above are called phasors—phasor currents and phasor voltages. Many use the convention of using RMS values when using phasor analysis in electrical circuits. That’s what we'll do in this course from now on. Notice that, in the equation above, the inductance appears as a "resistance" of jaL. This quantity is referred to as the inductance's impedance. Impedance The algebraic relationship between a phasor voltage and a phasor current is a generalization of resistance and is termed an element's impedance. The unit of impedance is the ohm. Resistance Let's assume that all the voltages are of the form Ve! and all the currents are of the form /e!". Let's look at the resistance's element relation, ohm's law. Ve = Rie V=RI The impedance of the resistance Z, is just its resistance. That is, Z=V/I=R lecture 3 outline Perform nodal 3-5 analysis on this circuit: 10 cos(3t + 20°) VG) 1H OD 4 & x Ve 6 F j3a (2) 40 Q 7.071 Z20°V @) Vy 2 Q lecture 3 outline 3-6 Using Maple Phasor analysis example > restart: > alias(l="I’,j=sqrt(-1)): > eqns:={v1=7.071*exp(j*20*Pi/180), > (V2-V1)/(j*3)+V2/(-[*2)+(v2-v3)/4=0, > (V3-v2)/4+v3/5=0}; > soln:=solve(eqns): > assign(soln): > ve:=evalf(polar(v2),4); eqns := {9/20 v3- 1/4 v2 = 0, v1 = (10N2) exp(1/9j Pi), - 1/3 j (v2- v1) + 1/2 j v2 + 1/4 v2- 1/4 v3 = 0} ve := polar(11.77, -2.205) > evalf(-2.205*180/Pi,4); -126.3 check your work: Vvo(t) = 11.77 cos (3t- 126.3°) V, ix(t) = 1.31 cos (3t- 126.3°)A Discuss Again, compare with the big gun technique. How many variables would have been required? How many equations? Really good advice on phasor analysis Far and away the best tool for phasor analysis is a good engineering calculator. Experience shows that you will avoid pain, work less, and learn more by taking this simple step. lecture 3 outline
