Question

3. A short circuited line of length / with characteristic impedance 2, has an input load at 2 = 0 from the generator with EMF voltage V = Vo sin(wt) and zero internal resistance. Find: A) the phasors V(2), 7(2) for steady solutions for complex voltage and current along the line; B) voltage V(z,t) and current 1(z. t) along the line; C) The impedance Z of the line as a function of w and I.

Answer 1

(3 solution Given that I - - Zo iB 17:20 | ILE . NS-V 12=-V+ 1 Z=0 + for a shost circuit line [ =-lie Jo = -vo VCZ, t) = vot ē VBX + voet 183 (expression for volluse consufayon) =-16 évex tvo et 18x = völet j8X -28 N) = 6 2sjm (8x)] VEZ) = 206. j sim (820)] for loss Less transfronission line atx=z VCZ) = 2 Vor sin (up z) V(z) = ja vo sin (JP Z (OR) V(Z) = -12Votsin (iß Z) -JBZ No et v8z Zo expression for I (Z) = voité CSScanned with CamScanner current form

I (Z) = -vo zo [et j8z +ē 187] I(z) --VO [ 2008 &z] Zo IE) = -20 cosciBZ) Zo (for loss Less transmussin (OR) line) IZ) = + 2 vot cos CjBZ) Zo (9 v (z it) and Ilzit) (time domain representatio Vezit) - [vo ēUBZ + Vo ē VB7] evwt VCZ,t) = völ-ējßz evcut tesz elut] v (Zit) = völ-ejl wt B2) tev (wt +B7) vizit) = = vö Cev (BZ -wt) _ē J(BZ -ct) Vzit) = vo (2isin (BZ twt)] vizit) - 12 vo- sin(BZ twt) OR) V(zit) = -12 vot sin (BZ twt) similarly, :: Io(ICZ,t) = - 2V5 cos CB Etwt) OR) = +2V6 cos CBZ taut) CS Scanned with CamScanner Zo

(C). Impedence (E) of the line as a function of , :. Esc VCZ) IZ =-1 I(z) | Z = -1 Zo Zsc =j 2 Vo sint-ißl) =+ izotan(uße) -240 cos cjße) Esc = tizo tan cube:) OR) Esc = VCZ it)z I(Zitzal =) 2. Vo sin (BZ twz) -2x6 cos (BZ toz) zo => -17 Sin (BC-A +WE) cos CBC-l) + wz) ZSC =-j Zotan (wz-ßl) where, ß = phase constant - angular frequency. l = length of transmission line. CS Scanned with CamScanner