Line 26: Line 26:
 
===Answer ===
 
===Answer ===
 
For a response to be critically damped we know that:<br />
 
For a response to be critically damped we know that:<br />
<math>b^2 - 4c = 0\\</math>
+
<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:
+
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}    and    c = \frac{1}{LC}\\</math>
 
<math> b = \frac{1}{RC}    and    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\\ so we get: \frac{-b}{2} = -4\\ therefore b = 8</math><br />
+
<math> \frac{-b}{2} = s<\math> so we get: <math>\frac{-b}{2} = -4\\ therefore b = 8</math><br />
Once we know b we can use the critically damped equation to solve for C.
+
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//

Revision as of 14:46, 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<\math> so we get: <math>\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

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