( ) = Q 0. ( ) R = R dq. ( t) = I t

Transcription

1 ircuis onceps The addiion of a simple capacior o a circui of resisors allows wo relaed phenomena o occur The observaion ha he ime-dependence of a complex waveform is alered by he circui is referred o as a ime-domain analysis On he oher hand, observaion ha a single-frequency wave undergoes an ampliude and phase shif upon passage hrough he circui is referred o as a frequency-domain analysis Time-Domain Analysis of he ircui The behavior of his "" circui can be analyzed in he ime-domain by solving an appropriae differenial equaion wih he appropriae boundary condiions Begin wih Kirchoff's Poenial Law, which is a consequence of conservaion of energy: = V +V = I + Q = dq d + Q onsider he case in which iniially he capacior is charged o hrough he horizonal swich while he verical swich is open The boundary or iniial condiion is ha a = 0 Q 0 = / Now he horizonal swich is opened and he verical swich is closed harge flows from one side of he capacior o he oher, and he differenial equaion o solve is simply The soluion is V +V = dq d + Q = 0 = Q 0 e dq d = Q Q wih Q 0 = The quaniy τ = is he ime consan or characerisic ime or /e ime, and i is he ime for he charge o decay from Q 0 o Q 0 / e The poenials across he resisor and capacior are and V = I = dq d = e V = Q / = e Now consider he case in which iniially is grounded hrough closure of he verical swich When his swich is opened and he horizonal swiched is closed, curren begins o flow and he capacior begins o charge The iniial or boundary condiion is ha a = 0, V +V =, and he equaion o solve is

2 = V +V = I + Q = dq d + Q The soluion is he soluion above (he soluion o he homogeneous equaion) plus whaever is necessary o saisfy he iniial condiion Thus, he soluion is Q = Q 0 e wih Q 0 = In ime τ =, Q rises from 0 o Q 0 ( / e) Noice ha he curren decreases wih ime as I = dq d = The poenials across he resisor and capacior are and V V = I = Q e = dq d = e / = e Insead of using manual swiches, i is easier o use a square wave from a funcion generaor o alernae he applied poenial beween 0 and This circui and he resuling V ou = V measured across he oupu erminals for a low frequency square wave are shown in he figure below, V ou As he frequency of he applied square wave increases, he oupu waveform is diminished because here is insufficien ime for he capacior o charge o he applied volage Examining he expression for V a imes small compared o τ, we find V = e! V 0 = V 0 using he expansion e x =! + x n=0 n! Thus V inegraor of a square wave, as shown below x n is linear in wih a slope A high frequencies, he circui acs as an 2

3 A V ou B Time-Domain Analysis of he ircui A If we swap he posiions of he resisor and capacior, we have a circui, where we measure he oupu volage across he resisor The Kirchoff loop analysis from above is sill valid here, he only change being ha V ou = V Using he resuls from above, he oupu signal response o a square wave inpu is shown below In his case, he inpu frequency is low and he circui behaves as a differeniaor B V ou Frequency Domain Analysis The behavior of hese circuis in he frequency-domain can be deermined by solving he same differenial equaion using a single-frequency applied poenial = e i As picured below, he resuling V ou has a smaller ampliude and is shifed in phase V ou 3

4 The analysis begins wih In seady sae, Q differen phase: Q or = V +V = I + Q ( ) = dq d + Q ( ) oscillaes a he same frequency as he applied signal bu wih a = Q 0 e i( +α ) Thus, V o e i = i + Q 0 ei +α = i + Q 0 eiα is he ampliude of he inpu poenial a frequency, and Q 0 / is he ampliude of he oupu signal across he capacior a frequency A his poin, we have wo unknowns, Q 0 and α, bu only one equaion However, here are acually wo equaions here is a real number and mus be equal o he real par of he righ hand side of he equaion, and he imaginary par of he righ hand side mus be zero ewriing his expression as = ( + i )e = e i an e iα Q 0 = e i ( α +an ) he fac ha he imaginary par of he righ hand side mus be zero yields α = an Then, V = Q Q 0 = The final resul for he oupu signal across he capacior is V V ( ) = 0 ( ) e i an The frequency-dependen ransmission funcion or response funcion for his circui is = V ou ( ) = V ( ) = A We define he corner, or 3dB, frequency as yielding he response funcion A( ) = c =, e i an + c e i an ( c ) Noice ha he phase difference α 0 as 0 and α π / 2 for c and ha he ampliude A( ) is a low frequency bu ends o zero a high frequency The conclusion is, 4

5 ha his circui is a low-pass filer, meaning ha i does no ransmi high frequencies very well oncep of Impedance The curren-poenial relaionship across he capacior is ineresing The curren in he circui is so, in he frequency domain, The ineresing resul is ha Thus, he poenial across a capacior is where he impedance of he capacior is I = dq d = iq ( 0 ei +α ) I ( ) = Q 0 ( ) I V i α +π /2 e = i V ( ) = I ( )Z ( ) Z ( ) = i looks like Ohm's law, bu i is no I simply saes ha The expression V ( ) = I ( )Z here is a linear relaionship beween he frequency-dependen complex poenial and he frequency-dependen complex curren apaciors are non-dissipaive elemens, so he ime-average power e V I = 0 For a resisor, he impedance is Z ( ) =, a real quaniy independen of frequency I is imporan o undersand he physical significance of he impedance Z ( ) of a capacior A low frequencies, he impedance Z and he capacior acs as an open circui A high frequencies, he impedance Z 0 and he capacior acs as an shor circui These wo limiing behaviors are useful in deermining he behavior of a circui wihou performing a deailed analysis oncep of Impedance Wih his definiion of he complex impedance of a capacior, i is now easy o analyze an circui in he frequency domain The expression for a poenial divider can be used o deermine he poenial across he capacior in he circui: Z V ( ) = ( ) ( ) i i + Z + = = ( ) + i = V e i an This expression is consisen wih he previous conclusion ha he circui is a low-pass filer For he circui, he poenial divider expression yields he volage across he resisor: 5

6 V ( ) = Z = ( ) i + = V 0 = ( ) i e i an 6