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f CH)= u(k +2) - u Ct-3) * ultt2), odvances by & (left shift ed) odvances by & Cleff shifted) kt2=0. t%=D-2 -2 10 * +ult), delogued by 3 - uck-3) 3. -1 -1 f)=uCt +2) -u(t-3) = u(tta) +Eu(t- Aーu(は-3) +1 -> -4 1. 042-1=0 to

f CH)= u(k +2) - u Ct-3) * ultt2), odvances by & (left shift ed) odvances by & Cleff shifted) kt2=0. t%=D-2 -2 10 * +ult), delogued by 3 - uck-3) 3. -1 -1 f)=uCt +2) -u(t-3) = u(tta) +Eu(t- Aーu(は-3) +1 -> -4 1. 042-1=0 to

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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I added a problem that someone work it out already. I am confuse from u(t+2) is shifted to the left and -u(t-3) is shifted to the right for quadrant 4 ( x,y plane) how do we end up with a regtangule from -2 to 3 

f Ct)= u(k +2) - u Ct-3) U (t+2) odvances by & (Cieft shiftcdl) ultta) kt2=0.→t=-2. -2 10 7. * +ult), deloged by 3 deloyed by 3 u(t-3) 1 – 4(t-3) 3. -1 -1 f)=uct +2) -u(t-3) = u(tta) +Eu(t- ヘーu(は-3) +1 -> 1. 3 to
Transcribed Image Text:f Ct)= u(k +2) - u Ct-3) U (t+2) odvances by & (Cieft shiftcdl) ultta) kt2=0.→t=-2. -2 10 7. * +ult), deloged by 3 deloyed by 3 u(t-3) 1 – 4(t-3) 3. -1 -1 f)=uct +2) -u(t-3) = u(tta) +Eu(t- ヘーu(は-3) +1 -> 1. 3 to
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