By V. Dlab, P. Gabriel
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12 Truncated-ramp voltage drive. (a) Circuit; (b) VG = f(t); (c) v= f(x) for indicated times, (ns), for forward wave (d) v =f(x) for indicated times for reverse wave Vrl; (e) v=f(x) for t = 26 ns, obtained by summing respective parts of (c) and (d); (f) VT(t) for 48 ns> t> 0; (g) Vrl and Vf2 att = 32 ns and t = 34 ns; (h) v =f(x) for indicated times; (i) Reflection chart for step voltage input; (j) VI(t), for 40 ns r;::,t> 0 ns, for a step input; (k) VI(t), for truncated-ramp voltage input. 12(a}, when Fig.
6(b). 2) Using the numerical data for our example gives: vI(O+) = +3 V; V OCF = 6 V; = 2 V. The value of VT(td ) agrees, of course, with that which can be obtained from the reflection chart in Fig. 5. As RT =I=- Ro there is a reflected wave v R which proceeds from RT to the battery and causes successive points on the line to take on the voltage value VT(td ) as it reaches them. 3) V OCR This equation is represented in Fig. 7 by a line of slope +Ro passing v Fig. 7 Counterpart of Fig. 3 for reverse wave for 2td> t;;t;td.
9 Input current waveform i(t) for (a) is shown in (b). Consider Fig. 6 of the text. Assume Sw opens at t = 40 ns. Sketch vl(t), vT(t) for 65 ns ~ t ~ 40 ns. Refer to Fig. 1. Let RT = 00. Show that: a Ll v I(2t ct ) = [Ro/(Ro + Rd] [I + PvaJ V (Ll indicates 'change in') b For t = r(2tct ), where r = 2, 3, 4 .... [Llv l [r(2t ct )]! od where t = n(2tct ), and n ~ 2. 9, let P~a = exp( -tlr). Then, taking logarithms and using the binomial expansion for the case Ra» R o, show that a T == RaCT, where CT = tota/line capacitance; b vI(t) == V[l - exp( - tI CTRdJ.