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Question $4(10$ points$)$

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Experss a parabolic train having a period of $6$ seconds where the fundamental portion of the signal is shown below, yet only utilize its positive portion $($i$.$e$.$ from the approximate x axis intercepes of $-5\mathrm{.}5$ to $0\mathrm{.}5).$ Minf: Note the following cartesian coordinates: $(1,-3)(-4,7)$ and $(-5,3).$ In other worls, this same signal should have its POSITIVE portion lasting $6$ seconds repeat to the left and right of $-5\mathrm{.}5$ and $0\mathrm{.}5.$

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$p(t)=(-2(t-3n{)}^2-5t-330n)[0[t5\mathrm{.}5-6n]-x(t-\mathrm{.}5-6n)]$

$p(t)=((t-6n{)}^2-5(t-6n)3)[t(t5-6n)-5(t-1-6n)]$

$x(t)=(-(2t-6n{)}^2-5(t-6n)30)[w(t5,5-6n)-w(t-\mathrm{.}5-6n)]$

$p(t)=(-2(t-3n{)}^2-4t-324n)[w(t5\mathrm{.}5-6n)-(t-\mathrm{.}5-6n)]$

$P(t)=((t-6n{)}^{2}5t-324n)|u(t5\mathrm{.}5-6n)-v(t-\mathrm{.}5-6n)|$

$p(t)=(-(t-6n{)}^2-5t330n)[v(t5,5-6n)-5(t-,5-6n)]$

$P(t)=-((t-6n{)}^{2}5t324n)[w(t5\mathrm{.}5-6n)-w(t-\mathrm{.}5-6n)]$

$p(t)=(-(t-6n{)}^2-5(t-6n)3)[v(t5-6n)-v(t-1-6n)]$

$p(t)=((t-6m{)}^{2}5t36n)[u(t5\mathrm{.}5-6m)-v(t-\mathrm{.}5-6n)]$

$F(t)=((t-6m{)}^2-4(t-6m)2)|(t5-6m)-(t-1-6n)|$

Question $5(10$ noints$)$