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| − | <html></html> | + | <html> |
| − | | + | <iframe src="https://trinket.io/embed/python/6224491962" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe> |
| − | | + | </html> |
| − | <center><math> | |
| − | \begin{align}
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| − | H&=\frac{j\omega+8}{(j\omega)^2+4*j\omega+13}\\
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| − | H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2}
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| − | \end{align}
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| − | </math></center>
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| − | From the <math>s</math> in the numerator, you can see that <math>A</math> is 1. That means the numerator so far is <math>1(s+2)</math> Therefore, there is already a constant 2 up top. To get the total of 8, then <math>B</math> needs to be 2. That is:
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| − | <center><math>
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| − | \begin{align}
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| − | H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2}=\frac{1(s+2)+2(3)}{(s+2)^2+(3)^2}
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| − | \end{align}
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| − | </math></center>
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| − | meaning
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| − | <center><math>
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| − | h(t):=e^{-2t}\left(1*\cos(3t)+2*\sin(3t)\right)~u(t)
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| − | </math></center> | |
| − | | |
| − | It is now possible to do inline math such as $$c=\sqrt{a^2+b^2}$$ with double dollar signs around the math.
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