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ECE 3144 Lecture 30 Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University 1 Reminder from Lecture 29 • Representation of sinusoidal functions Acos(t+) • Steady state response of Sinusoid functions v(t)=VM cost i(t ) VM R L 2 2 2 cos(t tan 1 L R i(t) 2 ) Complex Number • • A complex number a+jb: – a = Re(a+jb) is the real part of the complex number a+jb – b=Im(a+jb) is the imaginary part of the complex number a+bj Complex number relationships a+bj = rej r • a2 b2 tan 1 b a a=rcos, b=rsin Now we want to establish a relationship between time-varying sinusoidal functions and complex numbers through Euler’s equation e jt exp( jt ) cos t j sin t cost = Re(ejt) sint = Im(ejt) 3 Complex forcing function • Think about applying a complex forcing function v(t) = VMcos(t+) + jVMsin(t+ ) to a linear circuit network. – We expect a complex forcing function to produce a complex response; – The real part of the forcing function, VMcos(t+ ), will produce the real part of the response IMcos(t+) – The imaginary part of the forcing function, jVMsin(t+ ), will result in the imaginary portion of the response jIMsin(t+) – Based on the superposition, the response to the complex forcing function v(t) = VMcos(t+ ) + jVMsin(t+ ) is i(t) = IMcos(t+) + jIMsin(t+) • Applying Euler’s identity – v(t) = VMcos(t+) + jVMsin(t+ )= VMej(t+) – i(t) =IMcos(t+) + jIMsin(t+) = IMej(t+) – The complex source and response are illustrated as v(t) = VMej( t+) i(t) =IMej( t+) 4 Apply complex forcing function to the problem as shown in Lecture 29 Once gain, let us determine the current in the RL circuit. However, rather than applying VMcost, we will apply VMejt as the input signal v(t)=VM cost The forced response of input VMejt will be of form i(t) = Imej(t+) , where IM and need to be decided. i(t) The KVL equation for the circuit is L di (t ) Ri (t ) VM e jt dt Substituting i(t) = Imej(t+) into the above differential equation d ( I M e j (t ) ) VM e jt dt jLI M e j (t ) VM e jt => RI M e j (t ) L RI M e j (t ) => RI M e j jLI M e j VM => VM I M e j , converting the right side to exponential form => R jL I M e j VM R L 2 2 2 e j ( tan 1 (L / R )) IM => VM R 2 2 L2 and tan 1 L R Since the actual forcing function was VMcost, the actual response is the real part of the complex response i (t ) I M cos(t ) VM R L 2 2 2 cos(t tan 1 L R ) 5 • • • • • • Phasor By introducing in complex forcing function, the differential equations becomes simple algebraic equations with coefficients as complex numbers. The actual response is the real part of the complex response from the complex forcing function. For the given example, the forcing function is v(t) = VMej t and steady sate response is i(t) = Imej( t+) . The steady state response in the network has the same form and the same frequency with the forcing function. So from now on, we will simply drop the frequency ej t since each term in the equation will contain it. Thus every voltage and current can be fully described by a magnitude and phase. This abbreviated complex representation is called a phasor. For a voltage v(t)=VM cos(t+), a current i(t)=IM cos(t+), – Their exponential notations are v(t) = Re[VMei(t+ )], i(t) = Re[IMei(t+ )] – Their phasor notations are V VM I I M • Phasor operation: • v(t) represents a voltage in time domain while phasor V represents the voltage in the frequency domain. A11 A 1 1 2 A2 2 A2 Time Domain Acos(t) Asin(t) Frequency Domain A A 90o 6 Phasor relationship for a resistor R The voltage-current relationship is known as v(t) = Ri(t) (1) j (t v ) Applying the complex voltage source VM e results in the j (t ) complex current I M e , the above equation becomes: i VM e j (t v ) RI M e j (t i ) VM e j v RI M e j i => (2) The above equation can be written in phasor form as V=RI Where V VM e j v VM v (3) I I M e j i I M i From equation (2) we see that v=I. Thus the current through and voltage across a resistor are in phase. 7 Phasor relationship for an inductor L The voltage-current relationship for an inductor is di (t ) dt v (t ) L j (t v ) Applying the complex voltage source VM e results in the j (t ) complex current I M e , the above equation becomes: i d I M e j (t i ) dt => VM e j (t v ) jLI M e j (t i ) => VM e j (t v ) L VM e j v jLI M e j i (1) Equation (1) in phasor notation is V = jLI Note that (2) j 1e j 90 1900 o Thus the voltage and current for an inductor are 90o out of phase. Actually I lags V by 90o for an inductor. 8 Phasor relationship for a capacitor C The voltage-current relationship for an inductor is i (t ) C dv (t ) dt Applying the complex voltage source VM e j (t ) results in the complex current I M e j (t ) , the above equation becomes: v i I M e j (t i ) C d VM e j (t v ) dt I M e j (t i ) jCVM e j (t v ) I M e j i jCVM e j v => => (1) Equation (1) in phasor notation is I = jCV Note that (2) j 1e j 90 1900 o Thus the voltage and current for a capacitor are 90o out of phase. Actually V lags I by 90o for a capacitor. 9 Homework for lecture 30 • Problems 7.5, 7.6, 7.7 • Due April 8 10