Staning Wav Intrfrnc btwn th incint & rflct wavs Staning wav A string with on n fix on a wall Incint: y, t) Y cos( t ) 1( Y 1 ( ) Y (St th incint wav s phas to b, i.., Y + ral & positiv.) Rflct: y, t) Y cos( t ) 2( Y 2 ( ) Y, whr Y Y Y Th total isplacmnt Y ( ) Y1 ( ) Y2 ( ) Y Y W must hav () Y Y Y i.. Y Y Y ( ) Y ( )
Y ( ) Y Rcall that ( ) 2 sin Y ( ) 2 Y sin( ) y(, t) R[ 2 Y sin( )] 2Y sin( )sin( ) t Why sin? S Wikipia Staning Wav animation to gt visual pictur: Harmonic oscillation at ach, with amplitu following sin() This is Homwork 1 Problm 4. Hr w ust us th phasor tool to o it th asy way. Rviw Homwork 1 Problm 4 (an also Qui 2), rlat th physical quantitis to th phasors.
Y ( ) Y ( y(, t) R[ 2 Y sin( )] 2Y ) sin( )sin( ) t L = L L 1 Similarly, short transmission lin: By finition of short circuit, () 1 ( ) 2 sin( ) v(, t) R[ 2 sin( )] 2 sin( )sin( ) t Lik a mirror. What proprty of a mirror maks it a mirror? I But, I I 1 I Fin I() an i(, t) on your own.
What if th transmission lin is trminat in opn circuit? Not: opn n opn circuit for high frquncis! (You will s how to mak an opn circuit latr.) ( ) ( ) 2 L L 1 cos( ) L = Fin v(, t) on your own. I I Fin I() an i(, t) on your own. (sinc total currnt is, by finition of opn circuit) In all th abov xampls, = 1. Compltly rflct.
At vry high frquncis, w oftn can only masur th amplitu or powr ( amplitu squar), but not th instantanous valus or th wavform. Th following xampl is for th short circuit. Th opn circuit is similar (ust with a shift of origin). Th amplitu of th voltag wav v(, t) at position is ( ) ( ) ( ) 2 ( ) * ( ) Th complx amplitu containing th phas ( ) * ( ) 2 sin( ) 2 sin( ) 4 2 sin 2 ( ) 2 () 2 2 2 [1 2cos(2)] () Qustion: What s th spatial prio of th staning wav? Th opn circuit: Just a shift of th origin () 2 () ( ) 2 cos( ) 2 2 2 ( ) 4 cos ( ) 2 2 [1 2cos(2)]
For both th short circuit (SC) an opn circuit (OC), ( ) 2 ( ) min 1 max Constructiv Dstructiv Complt rflction. Compltly a staning wav. Thr ar cass whr = 1 but 1. Also complt rflction. W ll talk about thos cass latr. What if 1? Partially staning, partially travling. Now, lt s look at th maxima an minima of this combination of a staning wav an a travling wav. Rcall that, in gnral, is a complx numbr: r
r ( ) Incint Rflct Notic that + is a complx amplitu Similarly, It s mor convnint to plot () 2 an I() 2 than th amplitus. Intrfrnc trm () 2 2 I() 2 2
Intrfrnc trm () 2 Constructiv intrfrnc 2 Pay attntion to th max, min, an avrag valus Dstructiv intrfrnc In this plot, w hav assum a spcial cas r =. Can you think of a kin of loa that las to r =? Qustion: In gnral, what s th conition for r =?
W stat that it s mor convnint to plot th amplitus squar than th amplitus thmslvs. But how os th plot of () look? It looks lik this: Work out th max an min valus. Notic th important iffrnc btwn its shap an that of ()
() 2 2 Now w fin th voltag staning wav ratio (SWR), or simply staning wav ratio (SWR) Spcial (xtrm) cass: All staning wav. (Rcall short & opn. Othr such cass to b iscuss) All travling wav. No rflction. (What s th conition for this? How os th plot look?)
Slott lin A tool to masur impanc. S in th txtbook, Fig. 2-16 (pp. 74 in 7/E, pp. 73 in 6/E, or pp. 6 in 5/E). Bas on th on-to-on mapping btwn L an. Th tctor masurs th local fil (proportional to voltag) as a function of longituinal position. Sliing th tctor, you fin th voltag maxima an minima. Th istanc btwn aacnt minima is? You also gt th max/min ratio min (You only car about th ratio, not th actual valus.) Solving, you gt. But this is not yt! Th hop is: If you know, you gt L using th on-to-on mapping btwn th two. (Rcall that.) You know, thus you can fin L.
Th hop is: If you know, you gt L using th on-to-on mapping btwn th two. (Rcall that.) You know, thus you can fin L. r W alray know. Just n to fin r. W know W know min So you fin r. r L L Qustion: In principl, w can also obtain th rsult by masuring positions of maxima. But w prfr minima. Why?
W hav so far always alt with ngativ, bcaus w raw th transmission lin to th lft of th loa. W on t lik to always carry th ngativ sign. So w fin =, th istanc from th loa. ) ( Incint Rflct ) ( ) ( I I I ) ( I I I ) ( Pay attntion to signs.
) ( I I I ) ( ) ( ) ( I I Compar to 1 1 L, thus th finition. Now lt s consir th quivalnt impanc looking into th transmission lin at a istanc from th loa:
, How to intrprt this? Say, th incint wav voltag is ( ) at. Whn it arrivs at th loa, it bcoms ( ) -- ust a phas shift. (quivalnt rflction cofficint at ) At th loa, th rflction wav is ( ) This rflction wav travls a istanc to back. At istanc, it bcoms 2 ( ) ( ) -- ust anothr phas shift. Thus th quivalnt rflction cofficint at is -- ust imagin th intrfac is at. () corrspons to in xactly th sam mannr as any L to. Thrfor th quivalnt circuit.
() corrspons to in xactly th sam mannr as any L to. Thrfor th quivalnt circuit. You can hav such an quivalnt circuit at any, all th way up to l for th ntir transmission lin: At th input n of th transmission lin, This way, you turn th transmission lin problm in to a simpl circuit problm. Qustion: Givn an, how o you fin?
All cass of = 1 Thr ar cass whr = 1 but 1. Also complt rflction, but nithr short nor opn. W mntion that, a quartr wavlngth away from a short, th quivalnt circuit is an opn. What is th quivalnt impanc anywhr in btwn? Lt s hav a closr look at th short circuit. Pay attntion to signs. W ar using now. OC SC OC SC
Unrstan this from a physics point of viw: Ractiv loas on t issipat powr. Thus complt rflction. Th iffrnc is ust in th phas. Equivalnt impanc Notic frquncy pnnc. OC SC OC SC
Th cas of opn circuit trmination Now that w alray know th cas of short circuit trmination, what s th asist way to work out th opn circuit trmination cas? With a short circuit, you can mak an opn circuit. (For complt solution, s Fig. 2-21 in txtbook, pp. 81 in 7/E or pp. 82 in 6/E)
Now lt s go back to th gnral cas an look at th quivalnt input impanc. (mak sur you unrstan how this is arriv at) Insrt L L an you gt in L cos l cos l L sin l sin l or in th normali form: in L cos l sin l cos l sin l L Rcall that L L Qustion: What is th unit of L?
Do not confus input with incint i or in is th voltag at th input n. It is th sum of incint an rflct wavs thr. incint rflct Th incint wav voltag at th input n is
Th quartr wavlngth magic in L cos l cos l L sin l sin l Prioic. This is for gnric L. For th spcial cas of purly ractiv loas, s sli 17.? min,
, or Qustion: What ar th quivalnt normali forms?? Th quartr wavlngth transformr -- a mtho for impanc matching min
Th quartr wavlngth transformr To bttr unrstan why it works, lt s look at its optical analog. Anti-rflction coating
Th quartr wavlngth magic xplain in th multipl rflction point of viw Fils sum to Optical analog: th AR coating Air AR coating Glass oltags (or fils) sum to
Tak-hom mssags Staning wavs ar simply u to intrfrnc btwn incint & rflct wavs. Th variations of ral positiv amplitus (of voltag & currnt) an moulus squars of 2 amplitus ( ( ) an I ( ) 2 ) with position (i.. istanc from loa) ar prioic, analogous to intrfrnc strips in optics an ar in on-imnsional intrfrnc pattrns. Th prio of th pattrns ar half wavlngth. Th rflction wav amplitu is a fraction ( 1) of th incint, an its phas is shift rlativ to th incint, right upon rflction. Thus th rflction cofficint is complx. In gnral voltag an currnt of a transmission lin ar combinations of a travling wav an a staning wav. Whn th loa is purly ractiv (incluing short an opn) complt rflction. Ths ar th sam in trms of absnc of nrgy issipation. Thus th you can obtain any sir ractanc valu by trminating a transmission lin in any ractiv componnt; you only n to hav th right istanc from th loa. At any istanc from th loa, you hav an quivalnt impanc (), such that you fl as if th transmission lin is trminat in () right thr. (/4) an L [or mor gnrally ( + /4) an ()] hav a spcial rlation, which is us as a mtho for impanc matching. This mtho liminats rflction bcaus multipl rflctions sum up to.