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<math>b^2 - 4c = 0</math><br />
 
<math>b^2 - 4c = 0</math><br />
 
The next step would be to simplify the circuit as shown in the image below. Once simplified it becomes a parallel RLC circuit where we know:<br />
 
The next step would be to simplify the circuit as shown in the image below. Once simplified it becomes a parallel RLC circuit where we know:<br />
<math> b = \frac{1}{RC}</math> and <math>c = \frac{1}{LC}\\</math>
+
<math> b = \frac{1}{RC} </math> and <math> c = \frac{1}{LC} </math>
 
[[File:ECE201P6_1.png|500px|center]]
 
[[File:ECE201P6_1.png|500px|center]]
  
 
Since the root was given to be -4 we can find b.<br />
 
Since the root was given to be -4 we can find b.<br />
<math> \frac{-b}{2} = s<\math> so we get:  <math>\frac{-b}{2} = -4\\ therefore b = 8</math><br />
+
<math> \frac{-b}{2} = s </math> so we get:  <math> \frac{-b}{2} = -4 </math> therefore b = 8. <br />
 
Once we know b we can use the critically damped equation to solve for C.<br />
 
Once we know b we can use the critically damped equation to solve for C.<br />
 
<math>\begin{align}
 
<math>\begin{align}
8^2 - \frac{4}{2C} = 0
+
8^2 - \frac{4}{2C} = 0\\
64 = \frac{2}{C}
+
64 = \frac{2}{C}\\
C = \frac{1}{32}
+
C = \frac{1}{32}\\
 
\end{align}
 
\end{align}
 
</math>
 
</math>

Latest revision as of 14:51, 2 May 2015


Critically Damped Practice

Practice question for ECE201: "Linear circuit analysis I"

By: Chinar Dhamija

Topic: Critically Damped Second Order Equation


Question

Find the value for C that will make the zero input response critically damped with roots at -4.

ECE201P6.png


Answer

For a response to be critically damped we know that:
$ b^2 - 4c = 0 $
The next step would be to simplify the circuit as shown in the image below. Once simplified it becomes a parallel RLC circuit where we know:
$ b = \frac{1}{RC} $ and $ c = \frac{1}{LC} $

ECE201P6 1.png

Since the root was given to be -4 we can find b.
$ \frac{-b}{2} = s $ so we get: $ \frac{-b}{2} = -4 $ therefore b = 8.
Once we know b we can use the critically damped equation to solve for C.
$ \begin{align} 8^2 - \frac{4}{2C} = 0\\ 64 = \frac{2}{C}\\ C = \frac{1}{32}\\ \end{align} $


Questions and comments

If you have any questions, comments, etc. please post them below

  • Comment 1
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