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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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y(t) = ln(t1)u(t). Example 3: pulse input, unit step response. Let x(t) = u(t) u(t 2), h(t) = u(t). The integrand is zero when 0 and 2, so that at most the integrand is non-zero when 0 2. It is also zero when t. This sets up three intervals for t. First, when t 0 there is no way that 0 2 and t.

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Recall `u(t)` is the unit-step function. 1. ℒ`{u(t)}=1/s` 2. ℒ`{u(t-a)}=e^(-as)/s` 3. Time Displacement Theorem: If `F(s)=` ℒ`{f(t)}` then ℒ`{u(t-a)g(t-a)}=e^(-as)G(s)` [You can see what the left hand side of this expression means in the section Products Involving Unit Step Functions.]

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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