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A-tutorial.pdf
1.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 1 Complex Numbers, Phasors and Circuits Complex numbers are defined by points or vectors in the complex plane, and can be represented in Cartesian coordinates 1 z a jb j = + = − or in polar (exponential) form ( ) ( ) ( ) ( ) exp( ) cos sin cos sin z A j A jA a A b A = φ = φ + φ = φ = φ r imagina ea ry lpart part where 2 2 1 tan b A a b a − = + φ =
2.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 2 exp( ) exp( 2 ) z A j A j j n = φ = φ ± π Note : φ z Re Im A b a
3.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 3 Every complex number has a complex conjugate ( ) * * z a jb a jb = + = − so that 2 2 2 2 * ( ) ( ) z z a jb a jb a b z A ⋅ = + ⋅ − = + = = In polar form we have ( ) ( ) ( ) ( ) * exp( ) * exp( ) exp 2 cos sin z A j A j A j j A jA = φ = − φ = π − φ = φ − φ
4.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 4 The polar form is more useful in some cases. For instance, when raising a complex number to a power, the Cartesian form ( ) ( ) ( ) n z a jb a jb a jb = + ⋅ + + … is cumbersome, and impractical for non−integer exponents. In polar form, instead, the result is immediate [ ] ( ) exp( ) exp n n n z A j A jn = φ = φ In the case of roots, one should remember to consider φ + 2kπ as argument of the exponential, with k = integer, otherwise possible roots are skipped: ( ) 2 exp 2 exp n n n k z A j j k A j j n n φ π = φ + π = + The results corresponding to angles up to 2π are solutions of the root operation.
5.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 5 In electromagnetic problems it is often convenient to keep in mind the following simple identities exp exp 2 2 j j j j π π = − = − It is also useful to remember the following expressions for trigonometric functions ( ) ( ) exp( ) exp( ) exp( ) exp( ) cos ; sin 2 2 jz jz jz jz z z j + − − − = = resulting from Euler’s identity exp( ) cos( ) sin( ) jz z j z ± = ±
6.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 6 Complex representation is very useful for time-harmonic functions of the form ( ) ( ) [ ] ( ) ( ) [ ] ( ) [ ] cos Re exp Re exp exp Re exp A t A j t j A j j t A j t ω + φ = ω + φ = φ ω = ω The complex quantity ( ) exp A A j = φ contains all the information about amplitude and phase of the signal and is called the phasor of ( ) cos A t ω + φ If it is known that the signal is time-harmonic with frequency ω, the phasor completely characterizes its behavior.
7.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 7 Often, a time-harmonic signal may be of the form: ( ) sin A t ω + φ and we have the following complex representation ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( ) [ ] sin Re cos sin Re exp Re exp / 2 exp exp Re exp / 2 exp Re exp A t jA t j t jA j t j A j j j t A j j t A j t ω + φ = − ω + φ + ω + φ = − ω + φ = − π φ ω = φ − π ω = ω with phasor ( ) ( ) exp / 2 A A j = φ − π This result is not surprising, since cos( / 2) sin( ) t t ω + φ − π = ω + φ
8.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 8 Time differentiation can be greatly simplified by the use of phasors. Consider for instance the signal ( ) ( ) 0 0 ( ) cos exp V t V t V V j = ω + φ = φ withphasor The time derivative can be expressed as ( ) ( ) ( ) { } ( ) 0 0 0 ( ) sin Re exp exp ( ) exp V t V t t j V j j t V t j V j j V t ∂ = −ω ω + φ ∂ = ω φ ω ∂ ⇒ ω φ = ω ∂ is the phasor of
9.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 9 With phasors, time-differential equations for time harmonic signals can be transformed into algebraic equations. Consider the simple circuit below, realized with lumped elements This circuit is described by the integro-differential equation ( ) 1 ( ) ( ) t d i t v t L Ri i t dt dt C −∞ = + + ∫ C R L v (t) i (t)
10.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 10 Upon time-differentiation we can eliminate the integral as ( ) 2 2 ( ) 1 ( ) d i t d v t d i L R i t dt dt C dt = + + If we assume a time-harmonic excitation, we know that voltage and current should have the form 0 0 0 0 ( ) cos( ) exp( ) ( ) cos( ) e pha ph xp( ) o a or s r s V V I I v t V t V V j i t I t I I j = ω + α = α = ω + α = α ⇒ ⇒ If V0 and αV are given, ⇒ I0 and αI are the unknowns of the problem.
11.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 11 The differential equation can be rewritten using phasors ( ) { } ( ) { } ( ) { } ( ) { } ω ω ω ω ω ω ω 2 Re exp Re exp 1 Re exp Re exp − + + = L I j t R j I j t I j t j V j t C Finally, the transform phasor equation is obtained as 1 V R j L j I Z I C = + ω − = ω where 1 Z R j L C = + ω − ω Impedance Resistance Reactance
12.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 12 The result for the phasor current is simply obtained as ( ) 0 exp 1 I V V I I j Z R j L j C = = = α + ω − ω which readily yields the unknowns I0 and αI . The time dependent current is then obtained from ( ) ( ) { } ( ) 0 0 ( ) Re exp exp cos I I i t I j j t I t = α ω = ω + α
13.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 13 The phasor formalism provides a convenient way to solve time- harmonic problems in steady state, without having to solve directly a differential equation. The key to the success of phasors is that with the exponential representation one can immediately separate frequency and phase information. Direct solution of the time- dependent differential equation is only necessary for transients. Anti- Transform Direct Solution ( Transients ) Solution Transform Integro-differential equations i ( t ) = ? i ( t ) Algebraic equations based on phasors I = ? I
14.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 14 The phasor representation of the circuit example above has introduced the concept of impedance. Note that the resistance is not explicitly a function of frequency. The reactance components are instead linear functions of frequency: Inductive component ⇒ proportional to ω Capacitive component ⇒ inversely proportional to ω Because of this frequency dependence, for specified values of L and C , one can always find a frequency at which the magnitudes of the inductive and capacitive terms are equal 1 1 r r r L C LC ω = ⇒ ω = ω This is a resonance condition. The reactance cancels out and the impedance becomes purely resistive.
15.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 15 The peak value of the current phasor is maximum at resonance 0 0 2 2 1 V I R L C = + ω − ω IM ωr | I0| ω
16.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 16 Consider now the circuit below where an inductor and a capacitor are in parallel The input impedance of the circuit is 1 2 1 1 in j L Z R j C R j L LC − ω = + + ω = + ω − ω C V I L R
17.
Transmission Lines © Amanogawa,
2006 – Digital Maestro Series 17 When 0 1 in in in Z R Z LC Z R ω = = ω = → ∞ ω → ∞ = At the resonance condition 1 r LC ω = the part of the circuit containing the reactance components behaves like an open circuit, and no current can flow. The voltage at the terminals of the parallel circuit is the same as the input voltage V.
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