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Hello, I have a homework where I have to do multiple convolutions . The first signal (base) is in the attachment ( h1 ) . I have to do z(t) =( m(t)*m(t) ) * m(t) . m(t) is the M(t) signal from the picture and I will call it this way from now on.
I wrote my signal as this m(t) = u(t+0.5) - u(t-0.5) where u(t) is the heavyside step function ( t>= 0 : value 1 ; t<0 : value 0 )
So what I taught of doing was use the laplace transform this way : writing [tex]Ms=\frac{1}{s}\cdot e^{0.5s} - \frac{1}{s}\cdot e^{-0.5s}[/tex]
Alright, using the property that if a(t) = M(t)*M(t) => A(s) = M(s) x M(s) where M(s) is the laplace transform of m(t) and A(s) the laplace transform of a(t) it results that Z(s) = ( M(s)*M(s) ) * M(s) = M(s) * M(s) * M(s) right ?
=> [tex]Z(s) = ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{s}}\cdot e^{-0.5s} )\cdot ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{s}}\cdot e^{-0.5s} )\cdot ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{2}}\cdot e^{-0.5s} )[/tex] .
=> [tex]Z(s) = ( \frac{1}{s^{2}}\cdot e^{s} - \frac{2}{s^{2}} + \frac{1}{s^{2}}\cdot e^{-s} ) \cdot ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{2}}\cdot e^{-0.5s} )[/tex]
=> [tex] Z(s) = \frac{1}{s^{3}}\cdot e^{1.5s} - \frac{1}{s^{3}}\cdot e^{0.5s} - \frac{2}{s^{3}}\cdot e^{0.5s} + \frac{2}{s^{3}}\cdot e^{-0.5s} + \frac{1}{s^{3}}\cdot e^{-0.5s} - \frac{1}{s^{3}}\cdot e^{-1.5s} [/tex]
=> [tex] Z(s) = \frac{1}{s^{3}}\cdot e^{1.5s} - 3\cdot \frac{1}{s^{3}}\cdot e^{0.5s} + 3\cdot\frac{1}{s^{3}}\cdot e^{-0.5s} - \frac{1}{s^{3}}\cdot e^{-1.5s} [/tex]
And now using the inverse laplace transform it would be something like this : [tex] 2 \cdot z(t) = (t+1.5)^2 \cdot u(t+1.5) - 3 \cdot (t+0.5)^{2} \cdot u(t+0.5) + 3 \cdot (t-0.5)^{2} \cdot u(t-0.5 - (t-1.5)^{2{ \dot u(t-1.5) [/tex][/plan]
The signal would be : [plain][tex] 2z(t) = \begin{cases}0 & t <= -1.5 \\-t^{2}+3\cdot t - 2.25 & -1.5<t<-0.25 \\ 2\cdot t^{2} - 1.5 & -0.5<t<0.5 \\ -t^{2} - 3\cdot t - 2.25 & 0.5<t<1.5 \\ 0 & t >= 1.5\end{cases} [/tex]
Now, if i will try and represent this it doesnt get me to the answer which would be in the pic number 2. Can someone help me represent this signal ?
Edit: what's wrong with the AAC code? I inserted pictures after each equation since AAC code isnt working..
I wrote my signal as this m(t) = u(t+0.5) - u(t-0.5) where u(t) is the heavyside step function ( t>= 0 : value 1 ; t<0 : value 0 )
So what I taught of doing was use the laplace transform this way : writing [tex]Ms=\frac{1}{s}\cdot e^{0.5s} - \frac{1}{s}\cdot e^{-0.5s}[/tex]
Alright, using the property that if a(t) = M(t)*M(t) => A(s) = M(s) x M(s) where M(s) is the laplace transform of m(t) and A(s) the laplace transform of a(t) it results that Z(s) = ( M(s)*M(s) ) * M(s) = M(s) * M(s) * M(s) right ?
=> [tex]Z(s) = ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{s}}\cdot e^{-0.5s} )\cdot ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{s}}\cdot e^{-0.5s} )\cdot ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{2}}\cdot e^{-0.5s} )[/tex] .
=> [tex]Z(s) = ( \frac{1}{s^{2}}\cdot e^{s} - \frac{2}{s^{2}} + \frac{1}{s^{2}}\cdot e^{-s} ) \cdot ( \frac{1}{s^{2}}\cdot e^{0.5s} - \frac{1}{s^{2}}\cdot e^{-0.5s} )[/tex]
=> [tex] Z(s) = \frac{1}{s^{3}}\cdot e^{1.5s} - \frac{1}{s^{3}}\cdot e^{0.5s} - \frac{2}{s^{3}}\cdot e^{0.5s} + \frac{2}{s^{3}}\cdot e^{-0.5s} + \frac{1}{s^{3}}\cdot e^{-0.5s} - \frac{1}{s^{3}}\cdot e^{-1.5s} [/tex]
=> [tex] Z(s) = \frac{1}{s^{3}}\cdot e^{1.5s} - 3\cdot \frac{1}{s^{3}}\cdot e^{0.5s} + 3\cdot\frac{1}{s^{3}}\cdot e^{-0.5s} - \frac{1}{s^{3}}\cdot e^{-1.5s} [/tex]
And now using the inverse laplace transform it would be something like this : [tex] 2 \cdot z(t) = (t+1.5)^2 \cdot u(t+1.5) - 3 \cdot (t+0.5)^{2} \cdot u(t+0.5) + 3 \cdot (t-0.5)^{2} \cdot u(t-0.5 - (t-1.5)^{2{ \dot u(t-1.5) [/tex][/plan]
The signal would be : [plain][tex] 2z(t) = \begin{cases}0 & t <= -1.5 \\-t^{2}+3\cdot t - 2.25 & -1.5<t<-0.25 \\ 2\cdot t^{2} - 1.5 & -0.5<t<0.5 \\ -t^{2} - 3\cdot t - 2.25 & 0.5<t<1.5 \\ 0 & t >= 1.5\end{cases} [/tex]
Now, if i will try and represent this it doesnt get me to the answer which would be in the pic number 2. Can someone help me represent this signal ?
Edit: what's wrong with the AAC code? I inserted pictures after each equation since AAC code isnt working..
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