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INC 112 Basic Circuit Analysis Week 9 Force Response of a Sinusoidal Input and Phasor Concept Forced Response of Sinusoidal Input In this part of the course, we will focus on finding the force response of a sinusoidal input. • Start oscillate from stop input displacement Period that have transient • Have oscillated for a long time input displacement We will only be interested in this case for force response (not count the transient) Input Phase shift Amplitude Output Theory Force response of a sinusoidal input is also a sinusoidal signal with the same frequency but with different amplitude and phase shift. v2(t) Sine wave v1(t) Sine wave i(t) Sine wave + 2Ω 5sin(3t+π/3) AC - 3H vL(t) Sine wave i(t) R + sin(t) AC C - What is the relationship between sin(t) and i(t) ? sin(t) i(t) Phase shift i(t) + AC 10sin(2πt + π/4) 2Ω Find i(t) - v(t ) i (t ) R 10 sin( 2t / 4) i(t ) 2 10 sin( 2t / 4) i (t ) 2 i (t ) 5 sin( 2t / 4) Note: Only amplitude changes, frequency and phase still remain the same. i(t) + Asin(ωt) AC L Find i(t) - from di (t ) v(t ) L dt 1 1 i (t ) v(t )dt A sin tdt L L A A cos t sin tdt L L A A ( cos t ) (sin t ) L L 2 i(t) i(t) + + Asin(ωt) AC L Asin(ωt) AC R - - A i (t ) (sin t ) L 2 A i (t ) (sin t ) R ωL เรียก ความต้ านทานเสมือน (impedance) Phase shift -90 Phasor Diagram of an inductor Phasor Diagram of a resistor v v i i Power = (vi cosθ)/2 = 0 Power = (vi cosθ)/2 = vi/2 Note: No power consumed in inductors i lags v i(t) + Asin(ωt) AC C Find i(t) - dv(t ) d ( A sin t ) i (t ) C C dt dt AC (cos t ) A sin( t ) 2 1 C ความต้ านทานเสมือน (impedance) Phase shift +90 Phasor Diagram of a capacitor Phasor Diagram of a resistor i v v i Power = (vi cosθ)/2 = 0 Power = (vi cosθ)/2 = vi/2 Note: No power consumed in capacitors i leads v Kirchhoff's Law with AC Circuit KCL,KVL still hold. vR i vR (t ) 3sin( t ) i(t) v(t) R + v(t) AC i C - v(t ) 3 sin( t ) 4 sin( t 90o ) 5 sin( t 53.13 ) o vC vC (t ) 4 sin( t 90o ) v(t ) 5 sin( t 53.13o ) 5 sin( t ) cos(53.13 ) 5 cos(t ) sin( 53.13 ) 3 sin( t ) 4 cos(t ) o o 3 sin( t ) 4 sin( t 90 ) o This is similar to adding vectors. Therefore, we will represent sine voltage with a vector. 3 5 4 Vector Quantity Complex numbers can be viewed as vectors where X-axis represents real parts Y-axis represents imaginary parts There are two ways to represent complex numbers. • Cartesian form 3+j4 • Polar form 5∟53o Operation add, subtract, multiply, division? Complex Number Forms (Rectangular, Polar Form) a jb b r r a b 2 r θ a 2 b arctan a a r cos b r sin Interchange Rectangular, Polar form jω s = 4 + j3 3 σ 4 Rectangular form: Polar form 4 + j3 magnitude=5, angle = 37 บวก ลบ คูณ หาร vector ?? Rectangular form Add, Subtraction (4 j 3) (1 j ) 5 j 4 Polar form Multiplication 537 2 12 5 2(37 12 ) 1025 Division 537 2 12 5 49 2 Note: Impedance depends on frequency and R,L,C values Cartesian form Polar form a jb cd c magnitude a 2 b 2 b d phase tan a 1 Example: Find impedance in form of polar value for ω = 1/3 rad/sec 3H 1Ω 1 R jL 1 j 3 1 j 245 3 Rules that can be used in Phasor Analysis • Ohm’s law • KVL/KCL • Nodal, Mesh Analysis • Superposition • Thevenin / Norton Example + vR(t) + AC - 1Ω 2sin(t/3) i(t) + vL(t) - Find i(t), vR(t), vL(t) 3H V 20 I 2 45 Z total 245 Phasor form VR IR 2 45 1 2 45 VL IZ L I jL 2 45 190 245 i (t ) 2 sin( t / 3 ) 4 I 2 45 VR 2 45 VL 245 vR (t ) 2 sin( t / 3 ) 4 vL (t ) 2 sin( t / 3 ) 4 V I In an RLC circuit with sinusoidal voltage/current source, voltages and currents at all points are in sinusoidal wave form too but with different amplitudes and phase shifts. Summary of Procedures • Change voltage/current sources in to phasor form • Change R, L, C value into phasor form R L C R jωL 1/jωC • Use DC circuit analysis techniques normally, but the value of voltage, current, and resistance can be complex numbers • Change back to the time-domain form if the problem asks. Example + vR(t) 2Ω + 5sin(3t+π/3) AC - i(t) + vL(t) - Find i(t), vL(t) 3H i(t) + 2 5∟60 AC - j9 i(t) + 2 5∟60 AC - j9 Z total 2 j 9 9.2277.47 V 560 I 0 . 54 17 . 47 Z total 9.2277.47 i (t ) 0.54 sin( 3t 0.3) VL IZ L I j9 0.54 17.47 990 4.8872.53 vL (t ) 4.88 sin( 3t 1.27) VR IR I 2 0.54 17.47 20 1.08 17.47 Phasor Diagram V I 0.54 17.47 VL VL 4.8872.53 VR 1.08 17.47 VR Resistor consumes power I 2 0 . 54 2 2 P I rms R 2 Inductor consumes no power P = 0