Sinusoidal Steady-State Analysis

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1 Sinusoidal Steady-State Analysis Mauro Forti October 27, 2018

2 Constitutive Relations in the Frequency Domain

3 Consider a network with independent voltage and current sources at the same angular frequency ω. After the transient has vanished, all currents and voltages in the network are sinusoids at the same angular frequency ω. We say the network is in a steady-state sinusoidal regime at angular frequency ω phasor analysis method is based on writing the constitutive relations (CRs) of circuits elements and Kirchhoff laws directly in the frequency domain given a two-terminal element with voltage v(t) = V M cos(ωt + ϕ v ) and current i(t) = I M cos(ωt + ϕ i ), let V = V M e jϕv and Ī = I M e jϕ i be the corresponding phasors CRs of basic circuit elements in the frequency domain. Each two-terminal element establishes a link between the phasor current V and phasor current Ī at its terminal once they are coordinated via for example the passive convention

4 Ideal Resistor in the time domain, using the passive convention, the CR is given by Ohm s law v(t) = Ri(t) hence due to the linearity of the phasor transform we obtain the CR in the frequency domain V = RĪ i.e., the phasor voltage is simply the phasor current times the resistance

5 symbolic circuit of resistor in the frequency domain

6 the phasor information is twofold. Since V = V M e jϕv = RI M e jϕ i we obtain V M = RI M i.e., there holds Ohm s law between the amplitude of voltage and current, and ϕ v = ϕ i i.e., the voltage and current in an ideal resistor are in phase

7 since R is positive, the phasor current and phasor voltage are parallel to each other

8 Ideal inductor in the time domain, using the passive convention, the CR is given by v(t) = L di(t) dt hence due to the linearity and differentiation rule of the phasor transform we obtain the CR in the frequency domain V = jωlī i.e., the phasor voltage is the phasor current times jωl

9 the inductor is a memoryless element in the frequency domain obeying the symbolic Ohm s law Ī = V jωl = V Z L where Z L = jωl is the complex impedance of the inductor we have Z L = jx L where X L = ωl has dimension of Ohm, is called inductive reactance, and represents physically an apparent resistance offered by L to an alternate current

10 symbolic circuit of the inductor in the frequency domain

11 the phasor information is twofold. Since V = V M e jϕv = jωlī = e j π 2 ωlim e jϕ i we obtain V M = ωli M = X L I M i.e., there holds Ohm s law between the amplitude of voltage and current, with R replaced by X L, and ϕ v = ϕ i + π 2 i.e., the voltage phasor leads the current phasor by π/2

12 phasor diagram for an ideal inductor. Geometric interpretation of jωl

13 Ideal capacitor in the time domain, using the passive convention, the CR is given by i(t) = C dv(t) dt hence due to the linearity and differentiation rule of the phasor transform we obtain the CR in the frequency domain Ī = jωc V i.e., the phasor current is the phasor voltage times jωc

14 the capacitor is a memoryless element in the frequency domain obeying the symbolic Ohm s law Ī = V j ωc = V Z C where Z C = j ωc is the complex impedance of the capacitor we have Z C = jx C where X C = 1 ωc has dimension of Ohm, is called capacitive reactance, and represents physically an apparent resistance offered by C to an alternate current

15 symbolic circuit of the capacitor in the frequency domain

16 the phasor information is twofold. Since Ī = I M e jϕ i = jωc V = e j π 2 ωcvm e jϕv we obtain I M = ωcv M = V M X C i.e., there holds Ohm s law between the amplitude of voltage and current, and ϕ i = ϕ v + π 2 i.e., the current phasor leads the voltage phasor by π/2

17 phasor diagram for an ideal capacitor. Geometric interpretation of jωc

18 Kirchhoff Laws in the Frequency Domain

19 Kirchhoff Voltage Law (KVL) and Kirchhoff Current Law (KCL). Consider a circuit in sinusoidal steady-state at angular frequency ω. Starting from the KVL in the time domain, and using the linearity of phasor transform it can be shown that: phasor voltages of two-terminal elements obey KVL phasor currents of two-terminal elements obey KCL

20 Analysis of Elementary R L and R C Circuits

21 Series R L circuit simple circuit given by the series of a sinusoidal voltage source v g (t) = V M cos(ωt + ϕ v ), a resistor R and an inductor L

22 to find the current in sinusoidal steady-state, consider the symbolic circuit in the frequency domain. This is a memoryless circuit that can be analyzed using the methods of resistive circuits

23 KCL yields while from KVL we obtain Ī = Ī R = Ī L V g = V R + V L the CRs of R and L are where we let V g = V M e jϕv V R = RĪ R, V L = jωlī L these yield the phasor current in the circuit Ī = V g jωl + R

24 using the exponential form we obtain Ī = I M e jϕ i = V g jωl + R = V M ωl arctan( R 2 + ω 2 ej(ϕv R )) L2 hence the steady-state sinusoidal current in the time-domain is obtained as i(t) = I M cos(ωt + ϕ i ) = V M R 2 + ω 2 L 2 cos(ωt + ϕ v arctan( ωl R ))

25 the impedance (apparent resistance) offered by a series R L circuit to a sinusoidal current is given by Z = R 2 + XL 2 in a series R L circuit the phasor current lags the phasor voltage by an angle arctan( ωl R ) (0, π/2)

26 draw a phasor diagram that interprets the KVL V g = V R + V L = RĪ + jωlī and find geometrically V R and V L, in the case ϕ v = 0

27 Series R C circuit simple circuit given by the series of a sinusoidal voltage source v g (t) = V M cos(ωt + ϕ v ), a resistor R and a capacitor C repeat the previous arguments and show that hence Ī = I M e jϕ i = Ī = V g R j 1 ωc V M R ω 2 C 2 e j(ϕv +arctan( 1 ωcr )) and the sinusoidal current in the time-domain is obtained as i(t) = I M cos(ωt +ϕ i ) = V M R ω 2 C 2 1 cos(ωt +ϕ v +arctan( ωcr ))

28 the impedance (apparent resistance) offered by a series R C circuit to a sinusoidal current is given by Z = R 2 + XC 2 in a series R C circuit the phasor current leads the phasor voltage by an angle arctan( 1 ωcr ) (0, π/2) Phasor Diagram construct as before a phasor diagram assuming ϕ v = 0

29 Impedance and Admittance

30 Consider a two-terminal element in the frequency domain. Let V be the phasor voltage and Ī the phasor current coordinated via the passive convention. The impedance is defined by and the admittance by Z = V Ī Ȳ = Ī V note that Z and Ȳ are complex numbers related by Ȳ = 1/ Z the rules for computing impedances and admittances are the same as those for computing resistances and conductances. For example, the impedance of the series of two-terminal elements is the sum of the impedances

31 we write Z(ω) = R(ω) + jx (ω) where R(ω) is called resistance and X (ω) reactance. Moreover Ȳ(ω) = G(ω) + jb(ω) where G(ω) is called conductance and B(ω) susceptance we have hence Ȳ = 1 Z = 1 R + jx = R jx R 2 + X 2 R G = R 2 + X 2, B = X R 2 + X 2 also find R and X as a function of G and B

32 exponential form. We have Z = Ze jϕz where and ϕ z is such that Z = R 2 + X 2 cos ϕ z = R R 2 + X 2, sin ϕ z = X R 2 + X 2 moreover where Ȳ = Ye jϕy Y = 1 Z, ϕ y = ϕ z find the other links between different representations

33 we have Z = Ze jϕz = V Ī = V Me jϕv I M e jϕ i hence we obtain the useful relationships and Z = V M I M ϕ z = ϕ v ϕ i this means that the modulus of the impedance coincides with the ratio of voltage amplitude and current amplitude. Moreover, the argument of the impedance coincides with the phase shift between voltage and current

34 similarly, we have Ȳ = Ye jϕy = Ī V = I Me jϕ i V M e jϕv hence and Y = I M V M ϕ y = ϕ i ϕ v

35 consider again the series R L circuit. We have Z = Ze jϕz = R + jωl = R 2 + ω 2 L 2 ωl j arctan( e R ) hence we obtain and Z = R 2 + ω 2 L 2, ϕ z = arctan( ωl R ) I M = V M Z = V M R 2 + ω 2 L 2 ϕ i = ϕ v ϕ z = ϕ v arctan( ωl R )

36 Ohmic-inductive and Ohmic-capacitive Loads

37 Suppose that R > 0 (passive loads). If X > 0 (inductive load), from ϕ i = ϕ v ϕ z = ϕ v arctan( X R ) we obtain that the current lags the voltage by an angle arctan( X R ) (0, π/2) this is true in the simplest example of a series R L circuit If X < 0 (capacitive), we obtain instead that the current leads the voltage by an angle arctan( X R ) (0, π/2) this is true in the simplest example of a series R C circuit.

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