Generate a state space matrix of longitudinal dynamics of an aircraft and determine it's transfer function

Question

Kashmira Vinay 2022-11-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1845973-generate-a-state-space-matrix-of-longitudinal-dynamics-of-an-aircraft-and-determine-it-s-transfer-fu

回答： Nagul Meera Shaik 2023-11-14

For Boeing 747 @40,000 ft, certain parameters are assigned. Based on those, we have to find the state space matrix for Longitudinal Dynamics and then transfer that state space matrix into a transfer function.

I have written the following code:

%Aircraft Parameters
Al = 40,000; %Altitude; ft
M = 0.900; %Mach 
TAS = 871; %True Air Speed; ft/s
Pd = 222.8; %Dynamic Pressure; lb/ft^2
W = 636,636; %Weight; lb
s = 5,500; %Wing Area; ft^2
b = 196; %Wing Span; ft
c.bar = 27.3; %Wing Chord; ft
CG = 0.25; %x*c.bar
Alpha = 2.4; %Trim Angle of Attack; degree
Ixxs = 1.82 * 10^7; %slug-ft^2
Iyys = 3.31 * 10^7; %slug-ft^2
Izzs = 4.97 * 10^7; %slug-ft^2
Ixzs = -3.50 * 10^5; %slug-ft^2
U1 = 306.261; %m/s
%Longitudinal Derivatives
Xu = -0.0218; %1/s
Xalp = 1.2227; %X alpha; ft/s^2
Zu = -0.0569; %1/s
Zalp = -339.0036; %Z aplha; ft/s^2
Mu = -0.0001; %1/ft.s
Malp = -1.6165; %M alpha; 1/s^2
Malpdot = -0.1425; %M alpha dot; 1/s
Mq = -0.4038; %1/s
Xde = 0; %X delta e; ft/s^2
Zde = -18.3410; %Z delta e; ft/s^2
Mde = -1.2124; %M delta e; 1/s^2
%Lateral Derivatives
Yb = -55.7808; %Y beta; ft/s^2
Lb = -1.2555; %L beta; 1/s^2
Lp = -0.4758; %1/s
Lr = 0.2974; %1/s
Nb = 1.0143; %N beta; 1/s^2
Np = 0.0109; %1/s
Nr = -0.1793; %1/s
Ydr = 3.7187; %Y delta r; ft/s^2
Ldr = 0.2974; %L delta r; 1/s^2
Ndr = -0.4589; %N delta r; 1/s^2
Yda = 0; %Y delta a; ft/s^2
Lda = 0.1850; %L delta a; 1/s^2
Nda = -0.0135; %N delta a; 1/s^2
%'0' Parameters
Xtu = 0;
Zq = 0;
Mtu = 0;
Tetha1 = 0;
Ntb = 0;
Yp = 0;
Yr = 0;
Ntb = 0;
Zalpdot = 0;
%gravitational constant
g = 9.81;
%State Space Matrices for Longitudinal Dynamics
A = [Xu+Xtu Xalp 0 -g*cos(Tetha1); 
    (Zu/(U1-Zalpdot)) (Zalp/(U1-Zalpdot)) ((Zq+U1)/(U1-Zalpdot)) ((-g*sin(Tetha1))/(U1-Zalpdot));
    Mu+Mtu+((Malpdot*Zu)/(U1-Zalpdot)) Malp+((Malpdot*Zalp)/(U1-Zalpdot)) Mq+(Malpdot*(U1+Zq)/(U1-Zalpdot)) (Malpdot*g*sin(Tetha1)/(U1-Zalpdot));
    0 0 1 0]
B = [Xde;
    (Zde/(U1-Zalpdot));
    Mde+((Malpdot*Zde)/(U1-Zalpdot));
    0]
C = [1 0 0 0;
    0 1 0 0;
    0 0 1 0;
    0 0 0 1]
D = [0;0;0;0]
[Nu,Du] = ss2tf(A, B, C, D)
TF = tf(Nu,Du)

However, when I try to run the code I get an error saying

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Answer 1

KALYAN ACHARJYA 2022-11-8

编辑：KALYAN ACHARJYA 2022-11-8

在 MATLAB Online 中打开

If you check the Transfer fuction MATLAB docs, its clearly states that the values of the "Numerator" and "Denominator" properties must be row vectors or cell arrays of row vectors. The same has been reflected in the error message. In your case

clc
Nu=[0         0   -0.0732   10.2980   12.2155
0   -0.0599   -1.2379   -0.0270   -0.0022
0   -1.2039   -1.2715   -0.0274         0
0         0   -1.2039   -1.2715   -0.0274]
Nu = 4×5
         0         0   -0.0732   10.2980   12.2155
         0   -0.0599   -1.2379   -0.0270   -0.0022
         0   -1.2039   -1.2715   -0.0274         0
         0         0   -1.2039   -1.2715   -0.0274
Du =[1.0000    1.6750    2.0997    0.0445    0.0019]
Du = 1×5
    1.0000    1.6750    2.0997    0.0445    0.0019

Hence error

TF = tf(Nu,Du)
Error using tf
The values of the "Numerator" and "Denominator" properties must be row vectors or cell arrays of row vectors, where each vector is nonempty and containing numeric data. Type "help tf.num" or "help
tf.den" for more information.

For Testing case lets change the Nu to row vector

Nu=[0   0   -0.0732   10.2980   12.2155]
Du=[1.0000    1.6750    2.0997    0.0445    0.0019]

TF =

-0.0732 s^2 + 10.3 s + 12.22
-----------------------------------------------
s^4 + 1.675 s^3 + 2.1 s^2 + 0.04448 s + 0.00186
Continuous-time transfer function.