Sinusoidal Steady State Response of Linear Circuits. The circuit shown on Figure 1 is driven by a sinusoidal voltage source v s (t) of the form

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1 Sinusidal Steady State espnse f inear Circuits The circuit shwn n Figure 1 is driven by a sinusidal ltage surce v s (t) f the frm v () t = v cs( ωt) (1.1) s i(t) + v (t) - + v (t) s v c (t) - C Figure 1. Series C circuit driven by a sinusidal frcing functin Our gal is t determine the ltages v c (t) and the current i(t) which will cmpletely characterize the Steady State respnse f the circuit. The equatin that describes the behavir f this circuit is btained by applying KV arund the mesh. v () t + v () t = v () t (1.2) c s Using the current ltage relatinship f the resistr and the capacitr, Equatin (1.2) becmes dvc () t C + vc() t = cs( ωt) (1.3) dt Nte that the cefficient C has the unit f time. ( Ohm)(Farad) secnds Befre prceeding with the slutin f this differential equatin let s explre its physical significance. This linear circuit is driven (frced) by an independent sinusidal surce, v s (t). We may view its respnse (its effect n the circuit) as the superpsitin f the respnse with the surce set equal t zer (surce-free r natural respnse (v ch (t)) ) and the frced respnse (v cp (t)). v () t = v () t + v () t (1.4) c ch cp 6.071/ Spring 2006, Chanitakis and Cry 1

2 Schematically the superpsitin is shwn n Figure 2(a) and (b). + + v ch (t) C v (t) s v cp (t) C - - (a) (b) Figure 2 In mathematical language we call these tw respnses the hmgeneus slutin (v ch (t)) ) and the particular slutin (v cp (t)) f the equatin that characterizes the system (Equatin (1.3) in ur case) The hmgeneus slutin crrespnds t the differential equatin And the particular slutin t the equatin dvch() t C + vch() t = 0 (1.5) dt dvcp () t C + vcp () t = cs( ωt) (1.6) dt The hmgeneus slutin (r the natural respnse f the system) has the frm t vch () t = Bexp C (1.7) The particular slutin (r the frced respnse f the system) is the csine functin f amplitude A, frequency ω, and phase φ. v () t = Acs( ωt+ φ) (1.8) cp And s the general frm f the system respnse is t vc () t = Acs( ωt+ φ) + Bexp C (1.9) 6.071/ Spring 2006, Chanitakis and Cry 2

3 At this time let s recall the prblem statement which says that we are interested in btaining the Steady State respnse f the system. This is equivalent t saying that the surce v s (t) was cnnected t the system lng time ag and all transient phenmena are gne. In rder t be mre precise, if the time t C then the expnential term in Equatin (1.9) wuld g t zer. In this case the bservable and thus the imprtant respnse f the system is the Steady State respnse which is given by v () t = Acs( ωt+ φ) (1.10) c We may nw prceed t determine the details f the slutin by calculating the amplitude A and the phase φ. T d this we substitute the frm f the slutin (Equatin (1.10) ) int the differential Equatin (1.3). Befre we make that substitutin lets use the trignmetric identities t expand Equatin (1.10). v () t = Acs( ωt+ φ) c = Acsφ cs( ωt) Asinφsin( ωt) (1.11) And the crrespnding derivative dv c dt = Aω csφsin( ωt) Aωsinφcs( ωt) (1.12) Substituting Equatins (1.11) and (1.12) int Equatin (1.3) we have Aω csφsin( ωt) Aωsinφcs( ωt) + 1 C v Acs cs( t) Asin sin( t) = cs( ωt) C 0 [ φ ω φ ω ] (1.13) Cllecting the cefficients f cs( ω t) and sin( ωt) we have the fllwing equatins fr the unknwns A and φ. A Aωcsφ sinφ = 0 (1.14) C A C v 0 ωsinφ + csφ = (1.15) C A These equatins are independent and thus they may be slved simultaneusly fr A and φ / Spring 2006, Chanitakis and Cry 3

4 Frm Equatin (1.14) we btain the phase and A becmes ( C ω) φ = arctan (1.16) v A = csφ ωc sinφ (1.17) Therefre, the Steady State respnse f the system is vc () t = cs( ωt+ φ ) (1.18) csφ ωc sinφ ( ) where φ = arctan ωc Figure 3 shws a plt f the phase and the amplitude rati A/ as a functin f the dimensinless parameter ω C. Observe the frequency selectivity f this system. It passes lw frequencies while it attenuates higher frequencies ωc ωc Figure / Spring 2006, Chanitakis and Cry 4

5 The ltage acrss the resistr, v ( t ) i(t) and multiplying it by., may als be determined by calculating the current dvc() t Cω π it () = C = cs ωt+ φ+ dt csφ ωc sinφ 2 (1.19) ωc π v () t = i() t = cs ωt+ φ+ csφ ωc sinφ 2 (1.20) Figure 4 shws the plt f amplitude rati v / as a functin f the dimensinless parameterω C. Here we see the cmplementary behavir t that shwn n Figure 3. Observe again the frequency selectivity f this system. If the utput is thus taken acrss the resistr it passes high frequencies while it attenuates lwer frequencies ωc 10 Figure / Spring 2006, Chanitakis and Cry 5

6 Nw let s lk at a few frequencies f interest 1. Fr ω = 0 (dc signal) the phase φ = 0 and the amplitude A=. The ltage acrss the capacitr is cnstant. N current flws in the circuit The capacitr behaves as an pen circuit. The circuit equivalent when ω = 0 is shwn n Figure 5. v C v C Figure 5 2. Fr ω C = 1 0 The phase φ = 45 and the amplitude A = = 1/ 2 + 1/ 2 2 cs( ω t) v C cs( ωt π / 4) 2 Here current is flwing and thus sme pwer is dissipated in. Als the capacitr stres sme energy / Spring 2006, Chanitakis and Cry 6

7 3. Fr ωc 1 0 The phase φ = 90 and the amplitude A = 0 0 ωc( 1) = ωc As ω increases mre pwer is dissipated in the resistr. When ω the capacitr acts as a shrt circuit and all pwer is dissipated in. The circuit equivalent when ω is shwn n Figure 6. cs( ω t) C ω cs( ω t) C Figure / Spring 2006, Chanitakis and Cry 7

8 Nw let s cnsider the circuit i(t) v (t) s + v (t) - We wuld like t characterize this circuit by btaining the current i(t) and the ltage v (t). Apply KV arund the mesh di() t vs () t = i() t + (1.21) dt With v s (t) f the frm v ( t) = v cs( ωt), Equatin (1.21) becmes s di () t v0 + it () = cs( ω t ) (1.22) dt The cefficient Henry is a time cnstant. secnds Ohm Here again we are interested in the behavir f the system fr times that are lng cmpared t the time cnstant. In this scenari the nly cntributin t the slutin is again the ne frced by the sinusidal surce ltage v ( t) = v cs( ωt). And the frm f the frced respnse slutin is s it () = Acs( ωt+ φ) = Acsφ cs( ωt) Asinφsin( ωt) (1.23) di() t = Aω csφsin( ωt) Aωsin φcs( ωt) (1.24) dt Substituting back int the differential equatin (1.22) we get 6.071/ Spring 2006, Chanitakis and Cry 8

9 Aω csφsin( ωt) Aωsinφcs( ωt) + v0 [ A csφ cs( ω t ) A sinφsin( ω t )] = cs( ωt) (1.25) We nw separate the csine and sine functinal frms f the slutin since they are independent cntributins. The cefficients f the sin( ω t) terms are: And the cefficients f the cs( ω t) terms are: A Aωcsφ sinφ = 0 (1.26) A Equatin (1.26) gives us the expressin fr the phase φ. v 0 ωsinφ + csφ = (1.27) A And the amplitude A becmes φ tan 1 = ω (1.28) / A = ω csφ sinφ (1.29) And the slutin fr i(t) becmes / it () = cs( ωt+ φ ) (1.30) csφ ω sinφ 1 ω Where the phase φ = tan Figure 7 (a) and (b) shw the plts f the phase and the amplitude as a functin f the ω parameter / Spring 2006, Chanitakis and Cry 9

10 ω/ 10 (a) ω/ 10 (b) Figure / Spring 2006, Chanitakis and Cry 10

11 et s lk at a few imprtant frequencies as befre 1. Fr ω = 0 (dc signal) the phase φ = 0 and the amplitude A= v /. The ltage acrss the inductr is zer and the current flwing in the circuit is v /. All pwer is dissipated in. A magnetic field is generated in the inductr but it des nt change ver time (n change in the magnetic field n ltage) The inductr behaves as a shrt circuit as indicated n Figure 8. v / v v Figure 8 2. Fr ω / = 1 0 The phase φ = 45 / 1 and the amplitude f the current is A = = 1/ 2 + 1/ 2 2 cs( ω t) v 1 i = cs( ωt π / 4) 2 Here current is flwing and thus sme pwer is dissipated in. Als the inductr stres sme energy / Spring 2006, Chanitakis and Cry 11

12 3. Fr ω/ 1 0 The phase φ = 90 and the amplitude / / A = 0 ω = ω 0 ( 1) As ω increases the current decreases and as ω the inductr acts as an pen circuit (see Figure 9). cs( ω t) ω cs( ω t) Figure / Spring 2006, Chanitakis and Cry 12

13 Using the cmplex frcing functin Our gal is t be able t analyze C and circuits withut having t every time emply the differential equatin methd, which can be cumbersme. If we draw upn ur current understanding f C and netwrks and the fact that they represent linear systems we will be able t cnsiderably simplify the mathematical steps inlved in the cmputatin. This simplificatin will require the use f fundamental cmplex arithmetic and will in the end reduce the differential equatins int simple algebraic equatins. (indeed a simplificatin using cmplex numbers!) The linearity f the system implies that if we use a frcing functin f the frm cs( ωt + θ ) then the utput will have the same frequency but with a different phase Acs( ωt + φ ). Als if we were t scale the surce by a factr k then the utput will be scaled by the same factr. Figure 10 graphically demnstrates these tw statements. inear inear cs( ωt+ θ ) Acs( ωt+ φ) kacs( ωt+ θ ) kacs( ωt+ φ) syestem syestem Figure 10 If the factr k is an imaginary number like j = 1, then linearity still hlds as we graphically demnstrate n Figure 11. j sin( ωt+ θ ) inear syestem jasin( ωt+ φ) Figure 11 Therefre by superpsitin a frcing functin f the frm v cs( ωt) + jv sin( ωt) (1.31) Will prduce a respnse f the frm Acs( ωt+ φ) + jasin( ωt+ φ) (1.32) 6.071/ Spring 2006, Chanitakis and Cry 13

14 By using Euler s identity Equatins (1.31) and (1.32), the frcing functin and the crrespnding respnse, becme respectively and ( t) j ve ω (1.33) j Ae ω ( t φ ) + (1.34) Therefre we culd emply the cmplex frm f the frcing functin, prceed with the develpment f the slutin and then extract the desirable part f the respnse depending n whether the frcing functin was a csine r a sine functin. S what is the advantage f this methd? 1. Very easy t perfrm algebra with the expnentials 2. educes differential equatins t algebraic equatins. et s explre this methd by cnsidering again the circuit analyzed previusly and shwn again n Figure 12. The surce has the frm v ( t) = v cs( ωt). s i(t) vs(t) cs( ω t) + v (t) - Figure 12 j t Since cs( ω t) is the real part f ve ω we will prceed with the analysis and at the end we will simply extract the real part f the slutin. The equatin characterizing the system is If we use the cmplex surce di() t vs + it () = (1.35) dt j t ve ω (1.36) 6.071/ Spring 2006, Chanitakis and Cry 14

15 Then the crrespnding cmplex respnse is Ae j( ωt+ φ ) (1.37) Equatin (1.35) becmes which upn simplificatin becmes j Ae A e Cntinue with mre simplificatin we btain v jφ jφ ω + = (1.38) ( ω ) jφ Ae j v And by writing the cmplex number in plar frm we have Ae jφ + = (1.39) / = (1.40) 1 + jω/ Ae jφ = / ω e 1 ω j tan (1.41) Where the amplitude f the current is A = / ω (1.42) And the phase is φ tan 1 = The cmplete cmplex respnse f the system is ω (1.43) 6.071/ Spring 2006, Chanitakis and Cry 15

16 / e 2 2 ω ω j tan j t e ω (1.44) Since ur frcing surce was the csine functin, all we need t d is extract the real part f the functin given by Equatin (1.44) which is / 1 ω it ( ) = cs ωt tan ω (1.45) 1+ 2 Which is the same as Equatin (1.30) / Spring 2006, Chanitakis and Cry 16