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posted by  Pooley on 7/23/2008 12:43:02 AM  |  status: Live  

Second Order Lag Process - Please verify my work- Will rate - Urgent!!

Course Textbook Chapter Problem
N/A Intro. to Control System Technology - by Robert N. Bateson (Prentice Hall 2002) N/A N/A
Question Details:
An electrical interacting second-order circuit has the following component values:
R1 = 100 Ω
R2 = 300 Ω
RL = 50 Ω
C1 = 0.1 μF
C2 = 0.8 μF
 
Determine τ1 (Tau1 ), τ2 (Tau2) , the time-domain equation, and the transfer function.
PLEASE NOTE: DUE TO A FORMATTING ERROR, T = TAU
 
My solution:
τ1 ( i.e Tau1 ) = R1C1 = τ1 = 100 Ω x 10 μF = 10s
τ2 (i.e. Tau2 ) = R2C2  = 300 Ω x 0.8 μF = 240s
Time domain equation given by:

 
 
 
= 0.11 [250 + 60 + 80]
= 43
 
A2 = Gτ1τ2 = GTau1Tau2 = 0.11 x 10 x 240 = 264
 
Time domain equation = 264 d2y/dt2 + 43 dy/dt + y = 0.11x
Transfer Function = Y(s)/X(s) = 0.11/1 + 43s + 264s2
Need some direction. Please help.

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(Cramster SME)
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posted by shann(EE-SME) on 7/25/2008 6:07:49 AM  |  status: Live
Asker's Rating: None Provided    Moderator's Rating: N/A
Response Details:
Given differential equation
............. (1)
Here
Now equation (1) becomes
In s-domain it can be represented as below
Transfer function
In time domain we can write
Transfer function in s-domain
Magnitude and phase plots using Matlab
Code:

>> y=tf([416666666.7],[1,162500,3750003750])
>> bode(y);
>> grid
 
Plot:
 
 
I hope this helps you..................
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