cos t sin(t 90 ) sin t cos(t 90 ) sin t sin(t 180 ) cos t cos(t 180 ) sin(t ) sin t cos cos t sin cos(t ) cos t cos sin t sin sin( ) sin cos( ) cos 1 2 2 2 2 x (t ) sin 2 t sin 6 t sin 10 t sin 14 t ... 2 3 5 7 2 6 10 14 The VI Relationship for a Resistor R v (t ) v (t ) R i (t ) i (t) Let the current through the resistor be a sinusoidal given as i (t ) I m cos(t i ) v (t ) R i (t ) R I m cos(t i ) R I m cos(t i ) v (t ) R I m cos(t i ) voltage phase Is also sinusoidal with amplitude amplitude V m RI m And phase v i The sinusoidal voltage and current in a resistor are in phase The VI Relationship for an Inductor L v (t ) v (t ) L di (t ) dt i (t) Let the current through the resistor be a sinusoidal given as i (t ) I m cos(t i ) v (t ) L di (t ) LI m sin(t i ) dt Now we rewrite the sin function as a cosine function v (t ) LI m cos(t i 90o ) The sinusoidal voltage and current in an inductor are out of phase by 90o The voltage lead the current by 90o or the current lagging the voltage by 90o The VI Relationship for a Capacitor C v (t ) i (t ) C dv (t ) dt i (t) Let the voltage across the capacitor be a sinusoidal given as v (t ) V m cos(t v ) i (t ) C dv (t ) CV m sin(t v ) dt Now we rewrite the sin function as a cosine function i (t ) CI m cos(t i 90o ) The sinusoidal voltage and current in an inductor are out of phase by 90o The voltage lag the current by 90o or the current leading the voltage by 90o The Sinusoidal Response i (t) R v S (t ) L KVL v s (t ) V m cos(t ) v s (t ) Ri L di dt L di Ri V m cos(t ) dt This is first order differential equations which has the following solution i (t ) Vm R 2 2L 2 cos(t L were tan 1 R ) We notice that the solution is also sinusoidal of the same frequency However they differ in amplitude and phase Complex Numbers A = a + jb j= 1 j = 2 Rectangular Representation j = j j= ()j= j j = j j = ()() 1 3 2 4 2 2 Imaginary A b a Re(A) = Re(a + jb) =a Im(A) = Im(a + jb) = b Real Complex Numbers (Polar form) A = a + jb j= 1 Rectangular Representation Imaginary A b 2 | A|= a b 2 |A| a = |A| cos(θ) θ a b 1 θ tan a b = | A| sin(θ) Real A = a + jb = |A| cos(θ) + j|A| sin(θ) = | A| cos(θ) + j sin(θ) Euler’s Identity e j cos( ) j sin( ) Euler’s identity relates the complex exponential function to the trigonometric function e j cos( ) j sin( ) e j cos( ) j sin( ) Adding e j e j 2cos( ) e j cos( ) j sin( ) e j cos( ) j sin( ) Subtracting e j e j 2 j sin( ) j j e e cos( ) 2 j j e e sin( ) 2j Euler’s Identity e j cos( ) j sin( ) The left side is complex function cos( ) e j e 2 j The left side is real function The right side is complex function j j e e sin( ) 2j The right side is real function Complex Numbers (Polar form) A = a + jb j= 1 Rectangular Representation Imaginary A b 2 | A|= a b 2 |A| a = |A| cos(θ) θ a b 1 θ tan a b = | A| sin(θ) Real A = a + jb = |A| cos(θ) + j|A| sin(θ) = | A| cos(θ) + j sin(θ) = | A| e j =| A| Short notation j e Real Numbers A =1 Rectangular Representation Imaginary Polar Representation A =1 = 1e j0 = 1 0 θ=0 | A|= 1 Real 1 OR θ = 360 A =1 = 1e j 360 = 1 360 A = 1 Rectangular Representation Imaginary Polar Representation A = 1 = 1e j 180 = 1 180 θ = 180 | A|= 1 1 OR θ = 180 Real A = 1 = 1e j 180 = 1 180 Imaginary Numbers A = j= 1 Imaginary Rectangular Representation Polar Representation A = j= 1e j= 1 j 90 = 1 90 | A|= 1 θ = 90 Real OR θ = 270 A = 1A = j = 1e j 270 = 1 270 A = j Rectangular Representation Imaginary Real | A|= 1 θ = 90 Polar Representation A = j= 1e j= 1 1 1 j j j j 2 (1) j jj j 1 1 j 90 OR j 1e j 90 = 1e =1 90 j 90 = 1 90 Complex Conjugate A = a + jb = | A| e j A =| A| A Complex Conjugate is defined as A = a jb * = | A| e j A =| A| A Complex Numbers (Addition) A = a + jb B = c + jd A+B =(a + c)+ j(b + d) Imaginary B b+d b Re(A + B) = a +c = Re(A) +Re(B) Im(A + B) = b+d = Im(A) +Im(B) A+B A B d a c Real a +c Complex Numbers (Subtraction) A = a + jb Imaginary B = c + jd d A B =(a c)+ j(b d) B A b Re(A B) = a c = Re(A) Re(B) Im(A B) = b d = Im(A) Im(B) a c c a bd B A B Real Complex Numbers (Multiplication) A = a + jb = | A| e j B = c + jd = |B| e A j B =| A| A 2 | A|= a b =|B| B |B|= c d 2 2 2 b 1 A tan a d 1 B tan c Multiplication in Rectangular Form AB = (a + jb)(c+ jd) = ac + jad + jbc + j2bd = ac + jad + jbc bd = (ac bd)+ j(ad +bc) Multiplication in Polar Form AB = ( | A| e j A )( |B| e j B ) = | A| |B|e j A e j B = | A| |B|e Multiplication in Polar Form is easier than in Rectangular form j (A B ) Complex Numbers (Division) A = a + jb = | A| e j =| A| A A 2 | A|= a b 2 B = c + jd = |B| e j =|B| B |B|= c d 2 B 2 b 1 A tan a d 1 B tan c Division in Rectangular Form A (a + jb) (a + jb)(c+ jd) (a + jb)(c jd) (ac bd) j(bc ad) = 2 2 * ( c + j d)(c jd) B (c+ jd) (c+ jd)(c+ jd) (c d ) * (ac bd) 2 2 j(bc ad) 2 2 (c d ) (c d ) Complex Numbers (Division) A = a + jb = | A| e j B = c + jd = |B| e A j B =| A| A 2 | A|= a b =|B| B |B|= c d 2 2 2 b 1 A tan a d 1 B tan c Division in Polar Form j A ) = | A| e j ( = = ( | A| e j |B| B ( |B| e ) A A B ) B Division in Polar Form is easier than in Rectangular form Complex Conjugate Identities ( can be proven) * 2 2 AA* = (a + jb)(a + jb) =(a + jb)(a jb) = a jab + jab b 2 2 2 = a +b =| A | OR AA* = ( | A| e j A 2 =| A| e )( | A| e j (0) j A 2 ) =| A|2 e j e j =| A| e A A =| A |2 Other Complex Conjugate Identities ( can be proven) * * * (A ) = A * * * (AB) = A B (A +B) = A* +B* * A A* B = B* j (A A ) + i R (t ) di (t ) v (t ) L R R dt Let the current through the resistor be a sinusoidal given as i (t ) I m cos(t i ) R v R (t ) L di (t ) v (t ) L R R dt + i I (t ) v (t ) L I di (t ) I dt v (t ) LI m cos(t i 90o ) R Let the current through the resistor be a sinusoidal given as i (t ) I m sin(t i ) I v I (t ) L v (t ) L From Linearity if I di (t ) I dt v (t ) LI m cos(t i ) I i (t ) j I m sin(t i ) I then v I (t ) j LI m cos(t i ) + V(t ) I(t) i R (t ) ji I (t ) V(t) L LI m cos(t i 90o) + j LI m cos(t i ) I(t ) I m cos(t i ) j I m sin(t i ) = |I| e j (t i ) = Ime j (t i ) V(t) v R (t ) jv I (t ) V(t) LI m cos(t i 90o) + j LI m cos(t i ) v (t ) LI m cos(t i 90o) The solution which was found earlier The VI Relationship for an Inductor L v (t ) v (t ) L di (t ) dt i (t) Let the current through the resistor be a sinusoidal given as i (t ) I m cos(t i ) v (t ) L di (t ) LI m sin(t i ) dt Now we rewrite the sin function as a cosine function v (t ) LI m cos(t i 90o ) The sinusoidal voltage and current in an inductor are out of phase by 90o The voltage lead the current by 90o or the current lagging the voltage by 90o From Linearity if + V(t ) I(t) i R (t ) ji I (t ) L I(t ) I m cos(t i ) j I m sin(t i ) = |I| e j (t i ) = Ime j (t i ) V(t) v R (t ) jv I (t ) V(t) LI m cos(t i 90o) + j LI m cos(t i ) v (t ) LI m cos(t i 90o) The solution which was found earlier v (t ) Re( V(t) ) + v (t ) i (t ) I m cos(t i ) L + V(t ) I(t)= Im e j (t i ) L The solution is the real part of V(t ) v (t ) Re( V(t) ) This will bring us to the PHASOR method in solving sinusoidal excitation of linear circuit + V(t ) I(t) i R (t ) ji I (t ) L I(t ) I m cos(t i ) j I m sin(t i ) = |I| e j (t i ) = Ime j (t i ) = Im e j (t ) e j (i ) V(t) v R (t ) jv I (t ) V(t) LI m cos(t i 90o) + j LI m cos(t i ) LI m cos(t i 90o) + j LI m sin(t i 90o) LI m [cos(t i 90o) + j sin(t i 90o)] = LI me j (t i 90o ) = LI m e j (t ) e j (i 90o ) i (t ) + v (t ) L i (t ) I m cos(t i ) v (t ) L di (t ) dt v (t ) LI m cos(t i 90o ) j (t i ) Re[ Im e I(t) j (t i +90o ) Re[ LI me V(t) ] ] The real part is the solution Now if you pass a complex current + V(t ) I(t)= Im e L j (t i ) = Im e j t e j i I = Ime Phasor j i You get a complex voltage V(t) LI me j (t i 90o ) j LI me LI me j t j i e e j 90o j j t j i e Phasor V j LI me j L I j i The phasor The phasor is a complex number that carries the amplitude and phase angle information of a sinusoidal function The phasor concept is rooted in Euler’s identity e j cos( ) j sin( ) We can think of the cosine function as the real part of the complex exponential and the sine function as the imaginary part cos( ) {e j } sin( ) {e j } Because we are going to use the cosine function on analyzing the sinusoidal steady-state we can apply j cos( ) {e } e j cos( ) j sin( ) cos( ) {e j } sin( ) {e j } v V m cos(t ) V m {e j (t )} {V me j e j t } Moving the coefficient Vm inside v V m {e j t e j } Phasor Transform V V me j P{V m cos(t )} Were the notation P{V m cos(t )} Is read “ the phasor transform of V m cos(t ) The VI Relationship for a Resistor R v (t ) v (t ) R i (t ) i (t) Let the current through the resistor be a sinusoidal given as i (t ) I m cos(t i ) v (t ) R i (t ) R I m cos(t i ) R I m cos(t i ) v (t ) R I m cos(t i ) voltage phase Is also sinusoidal with amplitude amplitude V m RI m And phase v i The sinusoidal voltage and current in a resistor are in phase R i (t ) I m cos(t i ) v (t ) i (t) v (t ) R I m cos(t i ) Now let us see the pharos domain representation or pharos transform of the current and voltage i I I me j I m i V m RI m V RI V RI me j and v i V I RI m Vm i v Which is Ohm’s law on the phasor ( or complex ) domain R i V RI R V RI V I The voltage and the current are in phase Imaginary V Vm Im I i v i Real The VI Relationship for an Inductor L v (t ) v (t ) L di (t ) dt i (t) Let the current through the resistor be a sinusoidal given as i (t ) I m cos(t i ) v (t ) L di (t ) LI m sin(t i ) dt The sinusoidal voltage and current in an inductor are out of phase by 90o The voltage lead the current by 90o or the current lagging the voltage by 90o You can express the voltage leading the current by T/4 or 1/4f seconds were T is the period and f is the frequency L v (t ) i (t ) I m cos(t i ) v (t ) LI m sin(t i ) i (t) Now we rewrite the sin function as a cosine function ( remember the phasor is defined in terms of a cosine function) v (t ) LI m cos(t i 90o ) The pharos representation or transform of the current and voltage I I me j I m i (t ) I m cos(t i ) i v (t ) LI m cos(t i But since Therefore 90o ) j 1e j 90 1 o i V L I me o i j 90o V j LI me j LI me j 90 e j LI me j ( V m LI m and L I me j e j 90 j LI me j j ( i 90o ) o i v i 90o i i 90o ) LI m Vm (i 90o ) v i j L V I V j L I V m LI m and v i 90o The voltage lead the current by 90o or the current lagging the voltage by 90o Imaginary V Vm v Im I i Real The VI Relationship for a Capacitor C v (t ) i (t ) C dv (t ) dt i (t) Let the voltage across the capacitor be a sinusoidal given as v (t ) V m cos(t v ) i (t ) C dv (t ) CV m sin(t v ) dt The sinusoidal voltage and current in an inductor are out of phase by 90o The voltage lag the current by 90o or the current leading the voltage by 90o The VI Relationship for a Capacitor C v (t ) V m cos(t v ) i (t ) CV m sin(t v ) v (t ) i (t) The pharos representation or transform of the voltage and current v V V me j V m v (t ) V m cos(t v ) v i (t ) CV m sin(t v ) CV m cos(t v 90o ) I CV me j e j 90 o v j V I j C V m Im C I me j i 1e j 90 C j and o j CV me j v j ( 90 ) I me C v i 90o i o I LV me j ( j CV Im C Vm (i 90o ) v v 90o ) 1 j C V V I j C V m Im C I and v i 90o The voltage lag the current by 90o or the current lead the voltage by 90o Imaginary V I i Im Vm v Real Phasor ( Complex or Frequency) Domain Time-Domain Imaginary R R i (t) v (t ) R i (t ) V Im V RI I i (t ) I m cos(t i ) Real Imaginary v Vm I Im i Real I i (t ) I m cos(t i ) v (t ) LI m sin(t i ) V j L I v (t ) L di (t ) dt Imaginary 1 j C C v (t ) V V i (t) V j L I j L L v (t ) I v i v (t ) R I m cos(t i ) V RI Vm v (t ) V V I i i (t) i (t ) C dv (t ) dt v (t ) V m cos(t v ) i (t ) CV m sin(t v ) V I V Im V Vm I j C v Real I j C Impedance and Reactance The relation between the voltage and current on the phasor domain (complex or frequency) for the three elements R, L, and C we have V RI V j L I V I j C 1 j C I When we compare the relation between the voltage and current , we note that they are all of form: V ZI Which the state that the phasor voltage is some complex constant ( Z ) times the phasor current This resemble ) ( شبهOhm’s law were the complex constant ( Z ) is called “Impedance” ) (أعاقه Recall on Ohm’s law previously defined , the proportionality content R was real and called “Resistant” ) (مقاومه V Z Solving for ( Z ) we have I The Impedance of a resistor is ZR R The Impedance of an indictor is ZL j L 1 ZC j C The Impedance of a capacitor is In all cases the impedance is measured in Ohm’s W V j L I V RI Z Impedance V 1 j C I V I The Impedance of a resistor is ZR R The Impedance of an indictor is ZL j L 1 ZC j C The Impedance of a capacitor is In all cases the impedance is measured in Ohm’s W The imaginary part of the impedance is called “reactance” The reactance of a resistor is XR 0 We note the “reactance” is associated with energy storage elements like the inductor and capacitor XL L The reactance of a capacitor is XC 1 C Note that the impedance in general (exception is the resistor) is a function of frequency The reactance of an inductor is At = 0 (DC), we have the following ZL j L j (0) L 0 ZC 1 j C 1 j (0) C short open 9.5 Kirchhoff’s Laws in the Frequency Domain ( Phasor or Complex Domain) Consider the following circuit R2 v 2 (t ) R1 Phasor Transformation + v 1 (t ) v 3 (t ) L V 1=V 1e v 2 (t ) V 2 cos(t 2 ) V 2 =V 2e v 3(t ) V 3 cos(t 3) V 3 =V 3e v 4 (t ) V 4 cos(t 4 ) v 4 (t ) j 1 v 1(t ) V 1 cos(t 1) j 2 j 3 V 4 =V 4e j 4 v 1(t ) v 2 (t ) v 3(t ) v 4 (t ) 0 KVL C V 1 cos(t 1) V 2 cos(t 2 ) V 3 cos(t 3) V 4 cos(t 4 ) 0 j Using Euler Identity we have {V 1e 1e j t } {V 2e j Which can be written as {V 1e 1e j t V 2e {(V 1e Factoring e j t j 1 V 2e V1 Phasor So in general V 1 +V 2 + +V n = 0 V2 j 2 j 2 j t e V 3e V3 j 3 j 2 j t e j } {V 3e 3e j t } {V 4e j V 3e 3e j t V 4e V 4e V4 j 4 j 4 j t e )e j t } 0 j 4 j t e } 0 } 0 V 1 +V 2 +V 3 +V 4 = 0 Can not be zero KVL on the phasor domain Kirchhoff’s Current Law A similar derivation applies to a set of sinusoidal current summing at a node i1(t ) I 1 cos(t 1) Phasor Transformation I 1=I 1e KCL j 1 i 2 (t ) I 2 cos(t 2 ) I 2 =I 2e i n (t ) I n cos(t n ) j 2 i1(t ) i 2 (t ) I 1 +I 2 + +I n = 0 I n =I ne j n i n (t ) 0 KCL on the phasor domain Example 9.6 for the circuit shown below the source voltage is sinusoidal v s (t ) 750cos(5000t 30o ) (a) Construct the frequency-domain (phasor, complex) equivalent circuit ? (b) Calculte the steady state current i(t) ? The source voltage pahsor transformation or equivalent o V s =750 e j 30 =75030 o The Impedance of the indictor is Z j L j (5000)(32 X 10) j 160 W The Impedance of the capacitor is ZC L 1 j C 1 j 40 W j (5000) (5 X 106 ) v s (t ) 750cos(5000t 30o ) To Calculate the phasor current I I Vs Z ab o j 30 750 e 90 j 160 j 40 o j 30 750 e 90 j 120 o 75030 15053.13o i (t ) 5cos(5000t 23.13o ) A 5 23.13o A Example 9.7 Combining Impedances in series and in Parallel i s (t ) 8cos(200,000t ) (a) Construct the frequency-domain (phasor, complex) equivalent circuit ? (b) Find the steady state expressions for v,i1, i2, and i3 ? ? (a) A Ex 6.4:Determine the voltage v(t) in the circuit Replace: source 2cos 5t 30 with 230 desired voltage v(t) with Vˆ Impedance of capacitor is 1 1 1 1 j j2 j C 2 j 5 A single-node pair circuit 1 1 j 1 1 ˆ 2 2 230 V j 230 2 1 1 2 j Parallel combination 2 2 1 90 4 230 0.707 15 1 2 45 2 Hence time-domain voltage becomes v (t ) 0.707 cos(5t 15 ) V Ex 6.5 Determine the current i(t) and voltage v(t) Single loop phasor circuit 230 The current Iˆ 230 0.185 26.31 6 j 12 j 3 6 j 9 By voltage division Vˆ j 12 j 3 990 230 230 1.6663.69 6 j 12 j 3 6 j9 The time-domain i (t ) 0.185sin(4t 26.31 ) A v (t ) 1.66sin(4t 63.69 ) V Ex 6.6 Determine the current i(t) The phasor circuit is Combine resistor and inductor (3)( j 3) 990 3 3 3 3 j3 45 j 3 j 3 3 245 2 2 2 3 3 3 j3 j 2 2 Use current division to obtain capacitor current 3 45 3 j 3 2 Iˆ 10 60 10 60 13.42 33.43 3 3 3 j 3 j1 j j1 2 2 Hence time-domain current is: i (t ) 13.42cos(3t 33.43 ) A 9.7 Source Transformations and Thevenin-Norton Equivalent Circuits Source Transformations Thevenin-Norton Equivalent Circuits Example 9.9 Ex 6.7 Determine i(t) using source transformation Phasor circuit Voltage of source: Hence the current In time-domain Transformed source j 6 690 575 30165 30165 ˆ I 6.71101.57 2 j6 j 2 i (t ) 6.71sin(2t 101.57 ) A Ex 6.9 Find voltage v(t) by reducing the phasor circuit at terminals a and b to a Thevenin Phasor circuit equivalent VˆOC 2 210 0.43 67.47 2 j9 4 ZˆTH j 2 j 9 2.11 25.52 1.91 j 0.91 3 Vˆ j6 j6 VˆOC 0.43 67.47 0.48 46.94 j 6 1.91 j 0.91 j 6 ZˆTH 1.10 20.53 v (t ) 0.48cos(3t 46.94 ) V v OC (t ) 0.43cos(3t 67.47 ) V The Thevenin impedance can be modeled as 1.19 resistor in series with a capacitor with value 1 0.91 ( 3)C or C 0.37 F