Transcript Response of First Order Systems to Sinusoidal Input

Response of First Order Systems to Sinusoidal
Input
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First Order System - Sinusoidal Input
Consider the following first order system:

dO (t )
 O(t )  I (t ) O(s)(s  1)  I (s)
dt
O( s )
1
G( s) 

I ( s) s  1
Determine the response of the system if input is a sinusoidal:
I (t )  A sin(t )
Which may be transformed to:
I ( s) 
A
s2   2
the system response, O(s), is then:
O( s ) 
A


1
( s  )(s 2   2 )


a
s
1


b
c

s  i s  i
Solving then for a,b, and c:
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First Order System - Sinusoidal Input
To solve for a, multiply by s + 1/ , and let s = -1/
A
1

2


2
a
A
1   2 2
To solve for b, multiply by s + i, and let s = -i
A


1
(i  )(i  i )


A


2i 2 2 
2i


A
i 2  2i
2

A
A
 (  i )
   i 
2
b

  22 2
  i    i 
  1

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First Order System - Sinusoidal Input
c will be the complex conjugate of b and is:
A
(  i )
c  22 2
  1

Using the solution in the S&C for a sinusoidal function, the
solution becomes:
B  iC 
1  B  iC
2
2 rt
L 

2
B

C
e sin t   

 s  r  i s  r  i 
  A    ,r = 0
C
A
1




B

 2 2  
Where:   tan   C   2 2  
    1  2 
 B
    1  2 
1
 
A  1  2 2

2 2 2      1 e0t sin t   
    1  2 
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  tan1  
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First Order System - Sinusoidal Input
The complete solution is then:
t

A  
A

 sin t   
O(t )  2 2 e  

2 2
  1



1


It is important to note that the solution is made-up of a transient
(the first term) and a non-transient part (the second term).
Consider simply substuting i for s in the original transfer
function and solving for G(i ) G(i )
G (i ) 
1
i   1
G (i ) 
1
i  1

1  i
1  i


i 2 2 2  1
 2 2  1
1
1
    

2
2
G (i )  Im(G (i ))  Re(G (i ))   2 2    2 2  
  1   1
 2 2  1
Note, this gives us the non Im(G(i )) 
  tan1    transient solution
G(i )  tan1 
 Re(G(i )) 
2
2
for a unit sine input
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Plotting Frequency response
G(i ) 
1
 2 2  1
G(i )  tan1  
│G(i )│ 20log(│G(i )│)j

0.01/
1
0
0.5729
0.1/
1
0
5.7106
1/
0.7071
-3
45
10/
0.1
-20
84.289
100/
0.01
-40
89.427
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Plotting Frequency response
Magnitude Ratio (db)
0.01
Frequency/ ( / )
0.1
1
10
100
0
-5
-10
-15
-20
-25
-30
-35
-40
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Plotting Frequency response
Frequency/ (/)
0.01
0.1
1
10
100
Phase Lag (degrees)
0
-10
-20
-30
-40
-50
-60
-70
-80
-90
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Adding terms of Frequency Response
G1 i 
G2 i 
20log G1 i  G2 i    20log G1 i    20log G2 i  
G1 i G2 i   G1 i   G2 i 
We can simply add terms on the Bode Magnitude plot
and on the Bode Phase plot to get total response
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