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u(t -a) 1 0 a bt u(t -b) 1 t b 1 b(t) 0 a t C Piecewise discontinuous functions. Example Graph of the function f (t) = eat u(t 1) u(t 2). Solution: a t 1 2 t 1 y f ( t ) = e [ u ( t -1 ) - u ( t -2 ) ] [ u ( t -1 ) - u ( t -2 ) ] e a t Notation: The function values u(t c) are denoted in the textbook as u c(t).

u(t) = 1 2 1 2 sgn(t) as can be seen from the plots 0 t 1!1 sgn (t)u The Fourier transform of the unit step is then F[u(t)] = F 1 2 1 sgn(t) = 1 2 (f) 1 2 1 jˇf : Cu (Lecture 7) ELE 301: Signals and Systems Fall 23 / 37 The transform pair is then u(t), 1 2 (f) 1 j2ˇf: 1 j!!! () 1 j! Cu (Lecture 7) ELE 301: Signals and

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$ begingroup$ @zdub You have two $u(t-1)$ in your question, or $2u(t-1)$. This means the derivative is a $2 delta(t-1)$ that is represented by an upward arrow with

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