Designing Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 3A

Transcription

1 EECS 16B Desgnng Informaton Devces and Systems II Sprng 018 J. Roychowdhury and M. Maharbz Dscusson 3A 1 Phasors We consder snusodal voltages and currents of a specfc form: where, Voltage vt) = V 0 cosωt φ v ) Current t) = I 0 cosωt φ ), a) V 0 s the voltage ampltude and s the hghest value of voltage vt) wll attan at any tme. Smlarly, I 0 s the current ampltude. b) ω s the frequency of oscllaton. c) φ v and φ are the phase terms of the voltage and current respectvely. These capture a delay, or a shft n tme. We know from Euler s dentty that e jθ = cosθ) j snθ). Usng ths dentty, we can obtan an expresson for cosθ) n terms of an exponental: cosθ) = Ree jθ ) Extendng ths to our voltage sgnal from above: vt) = V 0 cosωt φ v ) = V 0 Ree jωt jφ v ) = V 0 Ree jφ v e jωt ) Now, snce we know that the crcut wll not change the frequency of the sgnal, we can drop the e jωt, as long as we remember that all sgnals related to the voltage wll be snusodal wth angular frequency ω. The result s called the phasor form of ths sgnal: Ṽ = V 0 e jφ v The phasor representaton contans the magntude and phase of the sgnal, but not the tme-varyng porton. Phasors let us handle snusodal sgnals much more easly, lettng us use crcut analyss technques that we already know to analyze AC crcuts. Note that we can only use ths f we know that our sgnal s a snusod. Wthn ths standard form, the phasor doman representaton s as follows. The general equaton that relates cosnes to phasors s below, where Ṽ s the phasor. V 0 cosωt φ v ) = ReṼ e jωt ) The standard forms for voltage and current phasors are gven below: Voltage Current Ṽ = V 0 e jφ v Ĩ = I 0 e jφ We defne the mpedance of a crcut component to be Z =, where Ṽ and Ĩ represent the voltage across ṼĨ and the current through the component, respectvely. EECS 16B, Sprng 018, Dscusson 3A 1

2 1.1 Phasor Relatonshp for Resstors R v Fgure 1: A smple resstor crcut Consder a smple resstor crcut as n Fgure 1, wth current beng t) = I 0 cosωt φ) By Ohm s law, vt) = t)r = I 0 Rcosωt φ) In phasor doman, Ṽ = RĨ 1. Phasor Relatonshp for Capactors C v Fgure : A smple capactor crcut Consder a capactor crcut as n Fgure, wth voltage beng By the capactor equaton, vt) = V 0 cosωt φ) t) = C dv dt t) = CV 0 ω snωt φ) = CV 0 ω cos ωt φ π ) ) = CV 0 ω cos ωt φ π ) = ωc)v 0 cos ωt φ π ) EECS 16B, Sprng 018, Dscusson 3A

3 In phasor doman, Ĩ = ωce j π Ṽ = Ṽ The mpedence of a capactor s an abstracton to model the capactor as a resstor n the phasor doman. Ths s denoted Z C. Z C = Ṽ Ĩ = 1 1. Proof of Inducton Gven the voltage-current relatonshp of an nductor V = L d dt, show that ts complex mpedance s Z L = jωl. L v Fgure 3: A smple nductor crcut Consder a smple resstor crcut as n Fgure 3, wth current beng By the nductor equaton, t) = I 0 cosωt φ) vt) = L d dt t) = LI 0 ω snωt φ) = LI 0 ω cos ωt φ π ) = ωl)i 0 cos ωt φ π ) In phasor doman, Ṽ = ωle j π Ĩ = jωlĩ The mpedence of an nductor s an abstracton to model the nductor as a resstor n the phasor doman. Ths s denoted Z L.. Phasor Analyss Z L = Ṽ Ĩ = jωl Any snusodal tme-varyng functon xt), representng a voltage or a current, can be expressed n the form xt) = Re[Xe jωt ], 1) EECS 16B, Sprng 018, Dscusson 3A 3

4 where X s a tme-ndependent functon called the phasor counterpart of xt). Thus, xt) s defned n the tme doman, whle ts counterpart X s defned n the phasor doman. The phasor analyss method conssts of fve steps. Consder the RC crcut below. R t) v s v C t) C The voltage source s gven by v s = 1sn ωt π ), ) 4 wth ω = rad s, R = 3kΩ, and C = 1µF. Our goal s to obtan a soluton for t) wth the snusodal voltage source v s. a) Step 1: Adopt cosne references All voltages and currents wth known snusodal functons should be expressed n the standard cosne format. Convert v s nto a cosne and wrte down ts phasor representaton V s. v s t) = 1cos ωt π 4 π ) = 1cos ωt 3π 4 ) The phasor s gven by V s = 1e j 3π 4 b) Step : Transform crcuts to phasor doman The voltage source s represented by ts phasor V s. The current t) s related to ts phasor counterpart I by t) = Re[Ie jωt ]. What are the phasor representatons of R and C? Z R = R Z C = 1 EECS 16B, Sprng 018, Dscusson 3A 4

5 c) Step 3: Cast KCL and/or KVL equatons n phasor doman Use Krchhoff s laws to wrte down a loop equaton that relates all phasors n Step. Z R I Z C I = V s R 1 ) I = 1e j 3π 4 d) Step 4: Solve for unknown varables Solve the equaton you derved n Step 3 for I and V C. What s the polar form of I Ae jθ, where A s a postve real number) and V C? To derve the polar form, I = 1e j 3π 4 R 1 V C = IZ C = j1ωce j 3π 4 1 jωrc = j1ωce j 3π 4 1 jωrc 1 = 1e 3π j 4 1 jωrc I = j1e j 3π j 3 = 1e j 3π 4 e j π 10 3 e j π 3 = 6e j 7π 1 ma. V = 1e j 3π 4 1 jωrc = 1e 3π j 4 1 j 3 = 1e 3π j 4 e j π 3 = 6e j 13π 1 V e) Step 5: Transform solutons back to tme doman To return to tme doman, we apply the fundamental relaton between a snusodal functon and ts phasor counterpart. What s t) and v C t)? What s the phase dfference between t) and v C t)? t) = Re[Ie jωt ] = Re[6e j 7π 1 e jωt ] = 6cos ωt 7π 1 ) ma v C t) = Re[V C e jωt ] = Re[6e j 13π 1 e jωt ] = 6cos ωt 13π 1 The phase dfference between the two, wth respect to t) s π. 3. RLC Crcut In AC We study a smple RLC crcut wth an AC voltage source gven by v s = Bcosωt φ) ) V EECS 16B, Sprng 018, Dscusson 3A 5

6 C R L v s a) Wrte out the phasor representaton of v s, R, C, and L. V s = Be jφ, Z R = R, Z C = 1, Z L = jωl b) Use Krchhoff s laws to wrte down a loop equaton relatng the phasors n the prevous part. Z R I Z C I Z L I = V s R 1 ) jωl I = Be jφ c) Solve the equaton n the prevous step for the current I. What s the polar form of I? The magntude of I s The phase of I s I = Be jφ R 1 jωl = Be jφ R j 1 ωc ωl) B I = R 1 ωc ωl) I = φ tan 1 1 R ωl 1 ) ωc Contrbutors: Sddharth Iyer. Yen-Sheng Ho. EECS 16B, Sprng 018, Dscusson 3A 6