Consider the triangular Voltage Signal (V) as a function of time in ms. Each coo
ID: 1832036 • Letter: C
Question
Consider the triangular Voltage Signal (V) as a function of time in ms.
Each coordinate is from 0 to .1, .2, .3 .... to .9.
a. Identify th end points in time where the function is discontinuous.
b. What is the period of the signal?
c. What is the voltage as t approaches .2 ms from the left?
d. What is the voltage as t apporoaches .2 ms from the right?
e. Plot the derivative of this function in the interval 0<.8 ms.
f. What is the derivative at t=.2 ms?
From this graph I know that the function is discontinuous at each point where there is a kink in the graph. ie., .2, .4, and .8
REPOST
For the answer posted on this question, "GURU" says that this function is also discontinuous at 0 and .8, could someone please expain this? I thought that a function is only discontinuous where there is a kink or point in the function, not at 0 and the endpoint of the function?
Also the plot for the derivative of this function was explained by GURU and I'm not sure it is right or how he came up with the answer, please give a more detailed explanation.
Explanation / Answer
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a) Yep, the function would be discontinuous wherever there's a kink. 0, .2, .4, .6, .8, etc. More exactly, it would be 0.2*n where n is an integer.
b) The period would be whenever the signal repeats. In this case, it repeats from 0 to 0.4 and from 0.4 to 0.8. This shows a period of = 0.4 ms.
c) This could be done in 2 ways. One way would be to determine the equations of the lines from 0 < t < 0.2 ms and for 0.2 < t < 0.4 ms and then plug into those equations. However, there's an easier way. Even though the graph is discontinuous, we see that it isn't disjunct (the graph never "jumps"). For this reason, the value as t approaches 0.2 ms from the left will equal the value as approached from the right, and we can see visually that that value is 0.7 V.
d) As above, V = 0.7 V.
e) The slope of these lines is 0.7 / 0.2 = 3.5. And it will be alternating between positive and negative when the direction changes. For some reason the Cramster diagram creator isn't working correctly for me, so here's a link to an image of the derivative graph: http://tinypic.com/r/mk7qzb/7
f) Because the signal is discontinuous at t = 0.2 ms, there is no derivative at that point. The derivative at t = .2 ms does not exist.