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July 17th, 2015, 07:45 AM   #1
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step functions

hi guys, trying to get back to studying, I am doing some basic step functions and all the examples were along the lines of

u(t-1)

now all that is obviously straightforward but they then asked me to plot the graph

u(t+1)

it is the first I have seen a pos+, do I just move to the left on the x axis

sitting alone trying to learn, the simple things drive you nuts. all help appreciated.

anyone know any sites to plot functions? thanks

stevie
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July 17th, 2015, 09:51 AM   #2
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Yes, u(t- 1) is the same as u(x) if x= t- 1 so going from "u(x)" to "u(t-1)" is the same as shifting the graph one place to the right and going the other way, from u(t- 1) to u(x) shifts the graph one place to the left. Then u(t+ 1) is the same as u(x) if x= t+ 1 shifting the graph one place to the left. So going from u(t- 1) to u(t+ 1) is shifting the graph two places to the left.

(That was for general function, u. Specifically, the step function, u(t- 1), is 0 for x< 1 and 1 for x> 1. The step function, u(t+ 1), is 0 for x< -1, 1 for x> -1, so is u(t- 1) shifted two places to the left, shifting x= 1 to x= -1.)
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Last edited by Country Boy; July 17th, 2015 at 10:05 AM.
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July 17th, 2015, 10:11 AM   #3
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Country Boy,

thanks very much for your help, I thought so but couldn't confirm. I could not find one reference to a pos on all my google searches, was starting to get to me. Now to really pester you the very next question was u(t-3)-u(t-2). When graphing it I am slightly confused as with the 3 on the left hand side (of formula) it looks like I am going neg on the 2 (right hand side of formula) before I have gone pos on the 3. Again probably straightforward but when sitting alone the simple things can drive you insane.

thanks again for your help

Stevie

Last edited by stevie1710; July 17th, 2015 at 10:17 AM.
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