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Old 02-Feb-2005, 04:44 PM
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homework help

got a question here

see if you can figure it out

A TTL waveform exists between 0 and 5 V. The frequency is 31.5 Khz. The duty cycle is 86 %. The waveform is fed into an integrator with an RC = 50ƒÝsec. At t = 0, the TTL signal is high. Calculate the following:

a. The time period for the TTL waveform.
=31.7us
b. The time period for the high of the TTL waveform.
=27.26us
c. The time period for the low of the TTL waveform.
4.438us
d. The voltage across the capacitor at the end of the first high.
=2.101v
e. The voltage across the capacitor at the end of the first low.
=1.92v
f. The voltage across the capacitor at the end of the second high.


???? for the life of me i can't figure out F.) the answer in the book says 3.22v but i have no idea how to get it
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Old 02-Feb-2005, 05:11 PM
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after the first cycle, you have residual voltage left. use superposition. figure our how much voltage you will have after he second cycle IF you started with zero, and since it is a linear system, add it to the residual voltage, and the sum of the two is your answer.
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Old 02-Feb-2005, 05:49 PM
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what he said^^^^^^
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Old 02-Feb-2005, 06:10 PM
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ERTW has it right somewhat...

you know the amount of charge left at the end of the first high (d) now calculate how much would be left after 31.7 us (assuming there is no more TL pulses) that value added to amount left after 1 high will give you the amount left after the second high. super position therom is fun. RC circuits suck ***.



Geez talk about the hard math questions.
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Old 02-Feb-2005, 06:52 PM
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can anyone sub the values into the equation, its kinda hard following just by what you guys are saying
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Old 02-Feb-2005, 07:52 PM
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Oh f&#@ am I glad college is over.

10 years later and I use about 3% of what I studied today.
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Old 02-Feb-2005, 07:52 PM
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it's been awhile. post the equation you used for D and E... and I'll redo them for F
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Old 02-Feb-2005, 07:57 PM
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Also a circuit diagram would be helpful
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Old 02-Feb-2005, 09:24 PM
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Jay no circuit diagram

d.) Vc = V (1-e ^-t/rc)

= 5v (1- e ^ 27.26us/50us)
=2.101

e.) Discharge

Vc = Vo e ^ -t/rc

=2.101 e ^ -4.438us/50us
=1.9216
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Old 03-Feb-2005, 12:44 AM
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ok so for F

use equation from question E with t = 31.7 us

then add to that the answer from D

now since I can't find one of my rcc calculators with e on it and I can't remember what it approximately equals I can't do the actual math for you but this should give you the answer thats int he book
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Old 03-Feb-2005, 07:24 AM
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oh man.thank god i'm in nursing....lol..

"shove my finger where?!?!? " hahah...
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Old 03-Feb-2005, 08:31 AM
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ok I found online that e = 2.71828.....

so I plugged that in with the 31.7us and got 1.1145 V adding to that the 2.101 from D and I get ~ 3.22 (I got 3.2155 but it's close enough)
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Old 03-Feb-2005, 11:40 AM
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thanks Jay

much appreciated
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Old 03-Feb-2005, 12:50 PM
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your welcome
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Old 04-Feb-2005, 11:53 AM
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WOW
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Old 04-Feb-2005, 12:01 PM
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wow what?
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