Slide 1Slide 2Slide 3Slide 4Slide 5EE201 Lecture 34 P. 1SSS (cont.)A sinusoidal function:can be represented as,Phasor of x(t) = X = X = [Aej] = A For constant and known angular frequency, , x(t) = Acos(t+)x(t) = Re[Aej(t+] = Re[Aejt ej]Note: The ejωt term of , Anejωtej, is contained in all terms of the differential equation that describes the behavior of any first or second order circuit. Therefore, ejωt can be cancelled from all terms, but it is understood the frequency remains unchanged.x1(t) = A1cos(t+1)Summation of Phasorsx2(t) = A2cos(t+2)X1 = A1 1X2 = A2 2x1(t) + x2(t) = X1 + X2EE201 Lecture 34 P. 2Example: KCL to find current I3.I1 = 10 60oI2 = 10 -60oI3 = ?I3 = I1 + I2 = 10 I3 = 10 0oI1 = 10 60oI2 = 10 -60oImReX = X1(j) ÷ X2(j) = (X11)÷( X22) X = (1−2)EE201 Lecture 34 P. 3Multiplication of complex numbers Example:I = Y(j) V = (2.510o)(326o) I = 7.536o AX = X1(j) x X2(j) = (X11)( X22) X = X1X2(1+2)z = (a+jb)(c + jd) = (ac−bd) + j(ad+bc) Multiplication of PhasorsDivision of complex numbers z = = Division of PhasorsX2X1(a+jb)(c+jd) (a+jb)(c+jd) (c−jd) (c−jd) =(c2+d2)(ac+bd) + j(bc−ad)EE201 Lecture 34 P. 4Conversion: rectangular to polar coordinatesz = x + jy = (x2+y2) ejtan-1(y/x) Example: Find the SSS response for i1 and i2. 10 1000 0.5 H10 F100 cos(100 t) VZeq(j) = R10 + jL + Z(CR)ZL(j) = j50 ; YC(j) = jC = j0.00314 S Zeq(j) = 10 + j50 + (0.001 + j0.00314)-1Zeq(j) = 165-53.6o Step 1: find equivalent impedance ZeqEE201 Lecture 34 P. 5 I = VsZeq= 100 0 167 -52.3= 0.6 52.3 A Step 2: calculate input current I from ZeqStep 3: use current division to find I1 and I2I1 = YCYeq= 0.00314900.003373.1I = (0.9516.9 ) I II1 = (0.9516.9) (0.6 52.3) = 0.5769.2 A I2 = YRYeq= 0.00100.003373.1I = (0.30-73.1 ) I II2 = (0.30-73.1) (0.6 52.3) = 0.18-20.8 A i1(t) = 0.57 cos( t + ) Ai2(t) = 0.18 cos(t − )
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