(iv) Consider the lumped-elements circuit drawn in Figure 2 below, which contains the same generator and load as the cir

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899604
Joined: Mon Aug 02, 2021 8:13 am

(iv) Consider the lumped-elements circuit drawn in Figure 2 below, which contains the same generator and load as the cir

Post by answerhappygod »

Iv Consider The Lumped Elements Circuit Drawn In Figure 2 Below Which Contains The Same Generator And Load As The Cir 1
Iv Consider The Lumped Elements Circuit Drawn In Figure 2 Below Which Contains The Same Generator And Load As The Cir 1 (62.24 KiB) Viewed 37 times
(iv) Consider the lumped-elements circuit drawn in Figure 2 below, which contains the same generator and load as the circuit in Figure 2, and replaces the coaxial cable with a capacitor of the same value Ccable as that of the coaxial cable described before. Draw the voltage Vl(t) for t = 0 – 40 ns. RG = 50 = Vi = Ccable RL 200 22 Vg = 1 V (S + Figure 2
Note: you should have learned the time evolution of RC circuits in 2nd year electronics. As a reminder, the general formula is of the form: v(t) = v(00) + [v(0) – v(0)] exp(-t/RC) = You may use a plotting software like Matlab to produce the graph, or calculate a few points and draw a line that goes smoothly through them. [6 marks] (v) Comment on the similarity between vi(t) obtained with the lumped-elements model above, and the graph of vi(t) you have obtained at point (iii) when properly describing the propagation through the cable. Are the two graphs approximately similar? Why? Clearly explain your reasoning. [2 marks]
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply