0008.qti

<item ident="D2-6682" title="D2 | Laplace transforms from formula and definition | ver. 6682">
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  <presentation>
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        <div class="exercise-statement">
          <p>
            <strong>D2.</strong>
          </p>
          <p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" alt="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p>
          <p> Then show how the integral definition of the Laplace transform to obtains same result. </p>
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      <mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

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  <itemfeedback ident="general_fb">
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            <h4>Partial Answer:</h4>
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              <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}"/>
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        <mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D%20=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

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