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1. Use combinations of STEP FUNCTIONS to describe each continuous-time signal shown below. f(t) 0 2 4 6 0 1 2 3 0 1 2 3 4 2.

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Answer #1

f (t) can be writen as 15. 04 f (t) = 58 (t) - 2.5 8(+-2) - 58 (+-2) +2.5 r(+-6) fH) = 5861)-7.5764-2) +2.5 t-26 where 8 A) i*4)=+. (44+2) 42-)] u(++2) = t3,-2 + u A-1)= uct+2) - (t-1) 9 + [u (t +2) – u(t-1)] (t) 2061)= telt! - 1 tl*.4) - rid = -tet to - tet +3,0 Izol t sol 2 ti 30+ .ut-). 44-2). U(1-3 d) 264) A744) - 4(4-2) 443)@ g (t) = x (34) Linear day, 4) + by El) = an, (4) + bx, (t) | 44) = T[X] y (t) = x(34) is a linear system Tim Invarient YET)Y(t) = a 2x, (1-t) + b2% Clot) - aytt byt) 193 (t) = ay, (t) + by 4) | Hence Linear system TIME invarient 4Ct)- 23(-+) y (17)@ )+(43) 6 4) 7 (4) = + G+2) + (1+() 8 According to the propesties of impulse - - signal. (H) 8+) = xco&A) 4, (t) = 62) & (+2

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