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EENG 2610: Circuit Analysis Class 14: Sinusoidal Forcing Functions, Phasors, Impedance and Admittance Oluwayomi Adamo Department of Electrical Engineering College of Engineering, University of North Texas AC Steady-State Analysis Sinusoidal forcing function (f(t) is forcing function) dx(t ) ax(t ) f (t ), dt x(t ) x p (t ) xc (t ) The natural response xc(t) is a characteristics of the circuit network and it is independent of the forcing function. The forced response xp(t) depends on the type of forcing function. Why study sinusoidal forcing function? This is the dominant waveform in electric power industry. Any periodic signal can be represented by a sum of sinusoids (you will learn it in Fourier analysis) We will only concentrate on the steady-state forced response of networks with sinusoidal forcing functions. We will ignore the initial conditions and the transient or natural response d 2 x(t ) dx(t ) a a2 x(t ) f (t ), 1 2 dt dt Sinusoids Definition x(t ) X M sin( t ) XM is the amplitude ω is radian or angular frequency (unit: radian/second) θ is phase angle (unit: radian), radian 180 T=2π/ω is period (unit: second) 2 f=1/T is frequency (unit: Hertz), 2 f T In-phase and out-of-phase Any point on the waveform XMsin(ωt+θ) occurs θ radians earlier in time than the corresponding point on the waveform XMsin(ωt). We say XMsin(ωt+θ) leads XMsin(ωt) by θ radians, or XMsin(ωt) lags XMsin(ωt+θ) by θ radians. In the more general situation, if Then, x1 (t ) X M sin( t 1 ) x2 (t ) X M sin( t 2 ) x1(t) leads x2(t) by (θ1 - θ2) radians, or, x2(t) lags x1(t) by (θ1 - θ2) radians. If θ1 = θ2 ,the waveforms are identical and the functions are said in phase; otherwise, it is said out of phase. Important Trigonometric Identities cos(t ) sin( t 2 ) sin( t ) cos(t ) 2 radian 180 Rectangular form cos(t ) cos(t ) sin( t ) sin( t ) cos( ) cos( ) sin( ) sin( ) sin( ) sin cos cos sin sin( ) sin cos cos sin cos( ) cos cos sin sin cos( ) cos cos sin sin Polar form x jy re j r x2 y2 y tan 1 x x r cos y r sin 1 e j j e sin( 2 ) 2 sin cos 2 2 cos( 2 ) cos sin 2 2 cos 1 2 1 2 sin Sinusoidal and Complex Forcing Functions Forcing function and circuit response If we apply a constant forcing function (i.e., source function) to a network, the steady state circuit response is also a constant. If we apply a sinusoidal forcing function to a linear network, the steady state circuit response will also be sinusoidal. Sinusoidal source function If the input source is v(t)=Asin(ωt+θ), then the output will be in the same sinusoidal form. For example, i(t)=Bsin(ωt+φ). That means: if the input source is sinusoidal function, we know the form of the output response, and therefore the solution involves simply determining the values of the two parameters B and φ. Learning Example di (t ) Ri (t ) v (t ) dt In steady state i (t ) A cos( t ), or i (t ) A1 cos t A2 sin t KVL : L di (t ) A1 sin t A2 cos t dt ( LA1 RA2 ) sin t ( LA2 RA1 ) cos t VM cos t LA1 RA2 0 algebraic problem LA2 RA1 VM A1 RVM LVM , A 2 R 2 (L) 2 R 2 (L) 2 Determining the steady state solution can be accomplished with only algebraic tools! FURTHER ANALYSIS OF THE SOLUTION The solution is i (t ) A1 cos t A2 sin t The applied voltage is v (t ) VM cos t For comparison purposes one can write i (t ) A cos( t ) A1 A cos , A2 Asin A1 RVM LVM , A 2 R 2 (L) 2 R 2 (L) 2 A i (t ) A A12 A22 , tan A2 A1 VM 1 L , tan R R 2 (L) 2 VM 1 L cos( t tan ) 2 2 R R (L) For L 0 the current ALWAYS lags the voltage If R 0 (pure inductor) the current lags the voltage by 90 Phasors If the forcing function for a linear network is of the form v(t)=VMejωt, Then every steady-state voltage or current in the network will have the same form and the same frequency ω; for example, a current will be of the form i(t)=IMej(ωt+φ). In our circuit analysis, we can drop the factor ejωt, since it is common to every term in the describing equations. Phasors are defined as: Sinusoidal signal: V VM VM e j VM (cos j sin ) v(t ) VM cos(t ) Re{VM e j ( t ) } Re{VM e j e j t } Re{VM e j t } I I M I M e j I M (cos j sin ) The magnitude of phasors are positive ! V VM VM ( ) Phasor Analysis (or Frequency Domain Analysis) The circuit analysis after dropping ejωt term is called phasor analysis or frequency domain analysis. By phasor analysis, we have transformed a set of differential equations with sinusoidal forcing functions in the time domain into a set of algebraic equations containing complex numbers in the frequency domain. The phasors are then simply transformed back to the time domain to yield the solution of the original set of differential equations. Phasor representation: Time Domain Frequency Domain A cos(t ) A A sin( t ) A( 90) Phasor Relationships for Circuit Elements We will establish the phasor relationships between voltage and current for the three passive elements R, L, C. v(t ) Ri (t ) VM e j ( t v ) RI M e j ( t i ) VM e j v RI M e j i V RI V VM e j v VM v I I M e j i I M i Phasor diagram di (t ) dt d L {I M e j ( t i ) } dt jLI M e ji v(t ) L VM e j ( t v ) VM e j v V j L I V VM e j v VM v I I M e j i I M i j 1 1e j 90 190 Voltage leads current by 90° dv(t ) dt d C {VM e j ( t v ) } dt jCVM e j v i (t ) C I M e j ( t i ) I M e j i I jCV V VM e j v VM v I I M e j i I M i j 1 1e j 90 190 Current leads voltage by 90° Definition of Impedance (unit: ohms): Impedance is defined as the ratio of the phasor voltage V to the phasor current I at the two terminals of the element related to one another by the passive sign convention: Z Z z V VM v VM ( v i ) I I M i I M Z( ) R( ) jX ( ) R : resistance , Z R 2 X 2 X : reactance z tan 1 ( X / R) R Z cos z X Z sin z It’s important to note that: Resistance R and reactance X are real function of the frequency of the forcing function ω, thus Z(ω) is frequency dependent. Impedance Z is a complex number; however, it is not a phasor, since phasors denote sinusoidal functions. Passive element impedance: Element Phasor Equation Impedance R L C V RI V jLI 1 V I jC ZR Z jL 1 Z jC Equivalent impedance if impedances are connected in series: Z s Z1 Z 2 ... Z n Equivalent impedance if impedances are connected in parallel: 1 1 1 1 ... Z p Z1 Z 2 Zn Two terminal input admittance: KVL and KCL are both valid in frequency domain 1 I (unit : siemens) Z V Y( ) G ( ) jB( ) Y G : conductanc e, B : susceptanc e Y G jB 1 1 Z R jX G R X , B R2 X 2 R2 X 2 In general, R and G are not reciprocals of one another. The purely resistive case is an exception. Example 8.9: Determine the equivalent impedance of the network. Then compute i(t) for f = 60 Hz and f = 400 Hz. v(t ) 50 cos( t 30) V SPECIAL APPLICATION: IMPEDANCES CAN BE COMBINED USING THE SAME RULES DEVELOPED FOR RESISTORS I V1 I V2 I Zs Z1 Z2 Z2 Z1 Z1 Z2 V V 1 1 k Zp Zk Z s k Zk LEARNING EXAMPLE I Zp Z1Z 2 Z1 Z 2 f 60 Hz, v (t ) 50 cos( t 30) Compute equivalent impedance and current 120 , V 5030, Z R 25 ZR R Z L jL ZC 1 jC 1 j120 50 106 Z L j 7.54, ZC j53.05 Z L j120 20 103 , ZC Z s Z R Z L ZC 25 j 45.51 I V 5030 5030 ( A) ( A) Z s 25 j 45.51 51.93 61.22 I 0.9691.22( A) i (t ) 0.96 cos(120 t 91.22)( A) LEARNING EXAMPLE SERIES-PARALLEL REDUCTIONS Z3 4 j 2 1 2 j4 2 2 j 4 (2) (4) 2 1 4 j2 Y34 4 j2 20 Y2 Y4 j 0.25 j 0.5 j 0.25 Z 4 1 / Y4 j 4 1 ( j 2) 1 j2 1 Z1 1 j 0.5 Z1 1 j 0.5 Z1 1 (0.5) 2 Z1 0.8 j 0.4() Z4 j 4 ( j 2) 8 j4 j2 j2 Y2 0.1 j 0.2( S ) Y34 0.2 j 0.1 Z2 2 j 6 j 2 2 j 4 Z34 4 j 2 Z 234 Y234 0.3 j 0.1( S ) Z 234 1 1 0.3 j 0.1 Y234 0.3 j 0.1 0.1 Z 2 Z34 3 j1 Z 2 Z34 Zeq Z1 Z234 3.8 j 0.6 3.8478.973 AC Steady-State Analysis For relatively simple circuits (e.g., those with single source), use: Ohm’s law for AC analysis, i.e., V=IZ The rules for combining impedance Z (or admittance Y) KCL and KVL Current division and voltage division For more complicated circuits with multiple sources, use: Nodal analysis Loop or mesh analysis Superposition Source exchange Thevenin’s and Norton’s theorem Software tools: MATLAB, PSPICE, … LEARNING EXAMPLE COMPUTE ALL THE VOLTAGES AND CURRENTS Compute I1 Use current divider for I2 , I3 Ohm' s law for V1 , V2 V1 690 I 2 Zeq 4 ( j 6 || 8 j 4) V1 16.2678.42(V ) 24 j 48 32 j8 24 j 48 Zeq 4 8 j2 8 j2 V2 7.2815(V ) 56 j56 79.19645 9.60430.964() 8 j2 8.24614.036 V 2460 I1 S 2.49829.036( A) Zeq 9.60430.964 j6 690 I3 I1 2.49829.036( A) 8 j2 8.24614.036 Zeq I2 8 j4 8.944 26.565 I1 2.49829.036( A) 8 j2 8.24614.036 I1 2.529.06 I 2 2.71 11.58 V2 4 90 I3 I3 1.82105 Steady-State Power Analysis Here we study powers in AC circuits: Instantaneous power Average power Maximum power transfer, Power factor, Complex power. Device power ratings: Typically, electrical and electronic devices have peak power or maximum instantaneous power ratings that cannot be exceeded without damaging the devices. Instantaneous Power Steady-state voltage and current: v(t ) VM cos(t v ) i (t ) I M cos(t i ) Instantaneous power: With passive sign convention p(t ) v(t )i (t ) VM cos(t v ) I M cos(t i ) VM I M [cos( v i ) cos( 2t v i )] 2 independent of time a function of time LEARNING EXAMPLE Instantaneous Power Supplied to Impedance p( t ) v ( t ) i ( t ) p(t ) 4 cos 30 4 cos(2 t 90) i (t ) I M cos( t i ) p(t ) VM I M cos( t v ) cos( t i ) p( t ) 1 cos(1 2 ) cos(1 2 ) 2 VM I M cos( v i ) cos(2 t v i ) 2 constant V 460 230( A) Z 230 i (t ) 2 cos( t 30)( A) VM 4, v 60 I I M 2, i 30 In steady State v (t ) VM cos( t v ) cos1 cos2 Assume : v (t ) 4 cos( t 60), Z 230 Find : i (t ), p(t ) Twice the frequency Average Power Since p(t) is a periodic function of time, the average power: 1 t 0 T P p(t )dt t 0 T 1 t 0 T VM cos(t v ) I M cos(t i )dt T t0 1 t0 T VM I M [cos( v i ) cos( 2t v i )]dt t 0 T 2 For passive sign convention. 1 VM I M cos( v i ) 2 In the equation, t0 is arbitrary, T=2π/ω is the period of the voltage or current. 2 1 1 1 1 V 2 M For purely resistive circuit (i.e., Z = R+j0 ): P V I cos( ) V I I R M M v i M M M 2 2 2 2 R For purely reactive circuit (i.e., Z = 0 + jX ): 1 P VM I M cos(90) 0 2 That’s why reactive elements are called lossless elements LEARNING EXAMPLE Determine the average power absorbed by each resistor, the total average power absorbed and the average power supplied by the source Inductors and capacitors do not absorb power in the average Ptotal 18 28.7W Psupplied Pabsorbed Psupplied 46.7W If voltage and current are in phase 2 Verification 1 2 1 1 V M v i P VM I M RI1M 2 2 2 R I I1 I 2 345 5.3671.57 1245 I1 345( A) 4 I 8.1562.10( A) 1 VM I M P4 12 3 18W P cos( v i ) 2 2 1245 1245 I2 5.3671.57( A) 1 2 j1 5 26.37 Psupplied 12 8.15 cos(45 62.10) 2 1 P2 2 5.362 (W ) 28.7W 2 Maximum Average Power Transfer Average power at the load: 1 1 P VL I L cos( vL iL ) I L2 RL 2 2 Voc I L ZTh Z L V Voc Z L L ZTh Z L Voc2 RL 1 PL 2 ( RTh RL ) 2 ( X Th X L ) 2 If the load is purely resistive (i.e., XL = 0): dPL 0 dRL ZTh RTh jX Th Z L RL jX L For maximum average power transfer: I L Voc /( 2 RTh ) * Z L ZTh RTh jX Th Voc2 1 2 PL,max 2 I L RTh 8R Th ZL RL jX L RL For maximum average power transfer: RL R X 2 Th 2 Th PL ,max 1 2 I L RL 2 Effective or RMS Values The effective value of a periodic current (or voltage) is defined as a constant or DC value, which would deliver the same average power to a resistor R. The 120 V AC electrical outlets in our 120 2 170 Vrms 120 V home is the rms value of the voltage: v(t ) 170 cos(377t ) 2 60 377 f 60 Hz It is common practice to specify the voltage rating of AC electrical devices (such as light bulb) in terms of the rms voltage. The average power delivered to a resistor by DC effective current: P The average power delivered to a resistor by a periodic current: P On using the rms values for the sinusoidal voltage and current, the average power: Effective (or rms) value of a periodic current: I eff2 R I eff I rms 1 T Vrms t 0 T t0 VM , 2 1 t0 T 2 i (t )dt t 0 T i 2 (t ) Rdt I rms IM 2 P Vrms I rms cos( v i ) The power absorbed by a resistor: 2 2 P I rms R Vrms R LEARNING EXAMPLE Compute the rms value of the voltage waveform T 3 X rms 1 T 4t 0 t 1 v (t ) 0 1 t 2 4( t 2) 2 t 3 T v 1 2 3 (t )dt (4t ) dt (4(t 2))2 dt 0 2 0 2 1 3 16 3 32 v ( t ) dt 2 3 t 3 0 0 2 Vrms 1 32 1.89(V ) 3 3 t 0 T x t0 2 (t )dt