1 16.202: PHASORS Consider sinusoidal source i(t) = Acos(ωt + φ) Using Eulers Notation: Acos(ωt + φ) = Re[Ae j(ωt+φ) ] Phasor Representation of i(t): = Ae jφ = A φ f v(t) = Bsin(ωt + ψ) First convert the function to a cosine: sin(x) = cos(x π/2) Therefore v(t) = Bcos(ωt + ψ π/2) Phasor: V = B ψ π/2 Term e jωt is implicit and suppressed and ω is a known constant. Phasor Representation is a Frequency Domain Representation, since the response depends on ω.
2 Phasor Diagram mag. Axis Complex Plane A φ A φ θ B Leading Direction Lagging Direction Real Axis B θ Figure 1: Phasor Diagram Note: f v(t) = Bsin(ωt + ψ) Phasor representation requires the form v(t) = Bcos(ωt + φ) (No negative amplitudes) Therefore: v(t) =Bcos(ωt + ψ + π/2) Phasor : V = B (ψ + π/2) f v(t) = Acos(ωt + φ) Convert to v(t) = Acos(ωt + φ + π) Phasor V = A (φ + π) Applying: cos(π) = 1 and sin(π/2) = 1
3 Phasors Represenation of Derivatives Consider i(t) = Acos(ωt + φ) di dt = Aωsin(ωt + φ) = Aωcos(ωt + φ + π/2) = Re[ωAe jωt e jφ e jπ/2 ] = Re[ωAe jωt e jφ j] Where e jπ/2 = cos(π/2) + jsin(π/2) = 0 + j di dt = Re[jωAe jωt ] Phasor Rep. of di dt : jω = Aejφ
4 Phasors Represenation of ntegrals Consider i(t) = Acos(ωt + φ) idt = A ωsin(ωt + φ) idt Phasor Rep of idt : jω = A ωcos(ωt + φ π/2) = Re[ A ω ejωt e jφ e jπ/2 )] = Re[ A ω ejωt e jφ ( j))] = Re[ A jω ejωt e jφ ] = Ae jφ
5 Phasors Summary Time Domain Representation: v(t) and i(t) Real Functions of Time Phasor or Frequency Domain Representation: Vand are Complex Variables Analysis assumes that frequency ω is fixed.
6 Phasors for Circuit Elements f i(t) =Acos(ωt + φ) and = A φ Resistor: v(t) = i(t)r V = R Voltage and Current are in Phase nductor: v(t) = L di dt V = jωl V Leads the current by 90 o. Capacitor: v(t) = 1 C V = jωc idt Voltage V Lags the current by 90 o. Current through a capacitor Leads Voltage across it by 90 o
7 i(t) + v(t) R + V = R - R m φ V Re i(t) L L V m + v(t) V =jωl φ Re + i(t) m v(t) C V C φ Re = jωcv V Figure 2: V- Relationships
8 mpedance and Admittance mpedance Z of a circuit element is the ratio of the phasor voltage V to the phasor current Unit of mpedance: Ohms : Measures the resistance offered by the element to the flow of current Note: Z is Complex but it is NOT a Phasor The e jωt variation is not associated with Z Resistor: Z R = R nductor : Z L = jωl Capacitor : Z C = 1 jωc Note: For DC: ω > 0 nductor: Zero mpedance: Short Circuit Capacitor: nfinite mpedance: Open Circuit
9 n Polar Form : Z = Z θ = R + jx R = Re[Z] is termed Resistance (ohms) X = m[z] is termed Reactance (ohms) mpedance is nductive if X is positive mpedance is Capacitive if X is negative Z = R 2 + X 2 θ = tan 1 X R R = Z cos(θ) X = Z sin(θ)
10 Admittance Y = Z 1 = V Resistor: Y R =1/R nductor : Y L = 1 jωl Capacitor : Y C = jωc Y = G + JB G = Re[Y] is termed Conductance (Siemens or mhos ) B = m[y] is termed Susceptance (Siemens or mhos ) From G + jb = 1 R+jX, equating real and imag. parts G = R B = X R 2 +X 2 R 2 + X 2
11 Verify KVL and KCL in Frequency Domain + V 1 - Z 1 +V 2 - Z 2 + V 1 - Z 3 V v(t) Z eq i(t) Figure 3: Kirchoff s Voltage Law v(t) = v 1 (t)+v 2 (t)+v 3 (t) Vcos(ωt + φ) = V 1 cos(ωt + φ 1 ) + V 2 cos(ωt + φ 2 ) + V 3 cos(ωt + φ 3 ) Re[Ve jφ e jωt ] = Re[V 1 e jφ 1 e jωt ]+Re[V 2 e jφ 2 e jωt ]+Re[V 3 e jφ 3 e jωt ] Re[(Ve jφ V 1 e jφ 1 V 2 e jφ 2 V 3 e jφ 3 )e jωt ] = 0 Re[(V V 1 V 2 V 3 )e jωt ] = 0 V = V 1 + V 2 + V 3
12 1 2 3 i(t) Z 1 Z 2 Z 3 Z eq Figure 4: Kirchoff s Current Law i(t) = i 1 (t)+i 2 (t)+i 3 (t) cos(ωt + φ) = 1 cos(ωt + φ 1 ) + 2 cos(ωt + φ 2 ) + 3 cos(ωt + φ 3 ) Re[e jφ e jωt ] = Re[ 1 e jφ 1 e jωt ]+Re[ 2 e jφ 2 e jωt ]+Re[ 3 e jφ 3 e jωt ] Re[(e jφ 1 e jφ 1 2 e jφ 2 3 e jφ 3 )e jωt ] = 0 Re[( 1 2 3 )e jωt ] = 0 = 1 + 2 + 3
13 Equivalent mpedance and Admittance KVL: V = (Z 1 + Z 2 + Z 3 ) = V = Z 1 + Z 2 + Z 3 Voltage Division: V 1 = V Z2 +Z Z 1 3 Z eq KCL : = V(Y 1 + Y 2 + Y 3 ) Y eq = V = Y 1 + Y 2 + Y 3 Current Division: 1 = (Z 1 +(Z 2 Z 3 )) Z 2 Z 3