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Solutions Manual For Aircraft Dynamics From Modeling to Simulation 1st Edition By Napolitano

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Solutions Manual For Aircraft Dynamics From Modeling to Simulation 1st Edition By Napolitano Solutions Manual For Aircraft Dynamics From Modeling to Simulation 1st Edition By Napolitano Solutions Manual For Aircraft Dynamics From Modeling to Simulation 1st Edition By Napolitano

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Aircraft Dynamics From Modeling to Simulation 1e N
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Aircraft Dynamics From Modeling to Simulation 1e N

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December 11, 2025
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Problems for Chapter VIII


Problem 8.1
Consider a linear dynamic system described by the following Differential Equation (DE)
with constant coefficients:
y (t )  5 y(t )  7 y(t )  3 y(t )  u(t )
Find the state variable model associated with this system. Verify that the eigenvalues of
the state matrix A are coincident with the roots of the characteristic equation of the above
system.


Solution of Problem 8.1.
The system is a 3rd order system with 1 input and 1 output  n  3, m  1, l  1 . A suitable

selection of state variables is given by:
x1 (t )  y (t )
d y (t )
x2 (t )   y (t )
dt
d 2 y (t )
x3 (t )   y (t )  x3 (t )  y (t )
dt 2
Using the system DE equation,
x3 (t )  y (t )  5 y(t )  7 y(t )  3 y(t )  u(t )
Rewriting the above equations using the state variables will lead to the following:
x1 (t )  y (t )
x1 (t )  x2 (t )
x2 (t )  x3 (t )
x3 (t )  5 x3 (t )  7 x2 (t )  3 x1 (t )  u (t )


Thus, the state variable model in a matrix format is given by:
 x1 (t )   0 1 0   x1 (t )  0
    
 x2 (t )    0 0 1   x2 (t )   0  u (t )
 x (t )   3 7 5  x (t )  1 
 3    3   

,The characteristic equation is given by:

s3  5s 2  7s  3  0
with roots :
1  1, 1  1, 1  3
The eigenvalues of the system matrix are,


 0 1 0 
 
eig   0 0 1    1, 1, 3
  3 7 5 
 
The SV output equation is given by,


 x1 (t ) 
 
y  1 0 0  x2 (t )   0u (t )
 x (t ) 
 3 

,Problem 8.2
Consider a linear dynamic system described by the following set of Differential
Equations (DEs) with constant coefficients:

 y1 (t )  5 y2 (t )  u1 (t )

 y2 (t )  3 y1 (t )  2 y2 (t )  u2 (t )
- Find the state variable model of the above system.

- The system is subjected to 2 inputs  u1 (t ), u2 (t )  and has two measurable outputs

 y1 (t ), y2 (t )  . Therefore, a transfer function-based modeling of the system will require
the evaluation of a (2 x 2) matrix of transfer functions. Find the matrix of transfer
functions under the following conditions:
- Derivation directly from the DEs
- Derivation from the state variable model.


Solution of Problem 8.2
State Variable Model
The system is a 2nd order system with 2 inputs and 2 outputs  n  2, m  2, l  2  . A

suitable selection of state variables is given by:
x1 (t )  y1 (t )  x1 (t )  y1 (t )
x2 (t )  y2 (t )  x2 (t )  y2 (t )
Using the system DE equation,

 x1 (t )  5 x2 (t )  u1 (t )

 x2 (t )  3x1 (t )  2 x2 (t )  u2 (t )
Rearranging:


 x1 (t )   5 x2 (t )  u1 (t )

 x2 (t )  3x1 (t )  2 x2 (t )  u2 (t )
Thus, the SV state equation is given by:

,  x1 (t )  0 5  x1 (t )  1 0  u1 (t ) 
      
 x2 (t )  3 2  x2 (t )  0 1  u2 (t ) 
The SV output equation is given by,
 y1 (t )  1 0  x1 (t )  0 0  u1 (t ) 
      
 y2 (t )  0 1   x2 (t )  0 0 u2 (t ) 


Derivation of Transfer Function (TF) matrix DIRECTLY from the DEs
Following the application of the Laplace transformation to the above system we have:

 sY1 ( s)  5Y2 ( s)  U1 ( s)

 sY2 ( s)  3Y1 ( s)  2Y2 ( s)  U 2 ( s)
leading to the following matrix format:

s 5  Y1 ( s)  U1 ( s) 
 3 ( s  2)  Y ( s)   U ( s) 
  2   2 
By invoking the „superimposition of effect‟ property of linear systems, we can solve the

above system first considering u2 (t )  0  U 2 (s)  0 and, next, considering

u1 (t )  0  U1 ( s)  0 . The goal is to find all the Gij ( s) transfer functions in the
relationship:
Y1 ( s)  G11 ( s) U1 ( s)  G12 ( s) U 2 ( s)
Y2 ( s)  G21 ( s) U1 ( s)  G22 ( s) U 2 ( s)
Therefore, consider the system:

s 5  Y1 ( s)  U1 ( s) 
 3 ( s  2)  Y ( s)    0 
  2   


Using Cramer‟s rule we have:

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