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

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

Answer 1

[KALYAN ACHARJYA]

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.

Answer 2

[Nagul Meera Shaik]

CAN YOU SHOW WHAT MATRIX FORM HAVE YOU USED FOR SAME QUESTION FOR LATERAL DYNAMICS