6 MatLab Tutorial Problems

Tutorial problems Introduction to MATLAB Problem1 A simple sine plot: plot y = sin x, 0 > tspan=[0 2]; x0=0; >> [t x] =ode23('simpode',tspan,x0); >> plot(t,x) >> xlabel('t'),ylabel('x') Problem 25: Solve the equation of motion of a nonlinear pendulum d 2θ d 2θ 2 + ω sin( θ ) = 0 ⇒ = −ω 2 sin(θ ) with initial conditions θ(0) = 1, dθ/dt (0) = 0. dt 2 dt 2

Solution 25: Step 1 Wrtie equations in terms of first order equations ⎧⎪ ∗ ⎫⎪ ⎧ Z2 ⎫ Let Z1 = θ and Z2 = dθ / dt, equation can be written in matrix form as, ⎨ Z∗ 1 ⎬ = ⎨ ⎬ 2 ⎪⎩Z 2 ⎪⎭ ⎩− ω sin(θ )⎭ Step 2 write a function to compute new state derivative function zdot = pend(t,z); % Call syntax zdot = pend(t,z) % inputs are t = time % z = [z(1); z(2)] = [theta; thetadot] % output is : zdot = [z(2); w.^2sin(z(1))] wsq = 1.56; % specify a value of w^2; zdot = [z(2); -wsq.*sin(z(1))];

Note that z(1) and z(2) refer to the first and second order elements of vector z. Step 3 use ode23 or ode45 for solution % tutorial 25 tspan=[0 50]; z0=[1;0]; % assign values of tspan [t z] =ode23('pend',tspan,z0); % run ode23 x = z(:,1);y=z(:,2); % x = 1st column of z, y = 2nd column figure(1) plot(t,x,t,y) % plot x and y xlabel('t'),ylabel('x and y') legend('Displacement','Velocity') figure(2) plot(x,y) % plot of phase potrait xlabel('Displacement'),ylabel('Velocity') title('Phase plane of nonlinear pendulum')

Step 4 Extract results and plot Problem 26:

Consider solving the following initial value problem: Solution 26: Create the following .m file called ex1.m:

function xprime=ex1(t,x) xprime=sin(tx); Save it in the workspace. In the Matlab command window, type: > > [t,x] = ode45(‘ex1’,[0 10], 1); >> plot(t,x); Problem 27:

Solution 27: Create a .m file called ex2.m with the following content:

function yprime=ex2(t,y) yprime=[cos(y(2))+sin(t); sin(y(1))-cos(t)]; Now y and yprime are both column vectors. In the command window, type the following: >> tspan=[0, 10]; >> y0=[5.1, 6.7]; >> [t, y]=ode45(‘ex2’, tspan, y0); >> plot(t, y); Of course, you can save the above commands to a .m file and type the name of the file in the command window to run it. Problem 28:

Let

, the problem is written as:

Solution 28: Create the function ex3.m as follows:

In the command window, type: >> z0=[0;1]; >> tspan=[0 5]; >> [t,z]=ode45(‘ex3’, tspan,z0); >> plot(t,z); We get this graph: Problem 29 Using MATLAB find the solution of Ax = y ⎡− 1⎤ ⎡3 2 ⎤ A=⎢ , y=⎢ ⎥ ⎥ ⎣1⎦ ⎣1 − 1⎦ Solution 29: A = [ 3 2; 1 -1]; Y= [-1 ; 1]; X=A\Y X= 0.2000 -0.800

Problem 30: Calculate C = A + B, D = A – B, E = AB Where, ⎡1 2 3 ⎤ ⎡ 4 1 2⎤ A = ⎢⎢0 1 4⎥⎥ B = ⎢⎢3 2 1 ⎥⎥ ⎢⎣3 0 2⎥⎦ ⎢⎣0 1 2⎥⎦ Solution 30: Problem 30: Two material properties of carbon monoxide gas are given below T Beta Alpha 300 3.33e3 0.212e4 400 2.5e3 0.3605e4 500 2.00e3 0.5324e4 600 1.67e3 0.719e4

Where T is temperature in Kelvin, Beta is thermal expansion coefficient, and alpha is thermal diffisivity. Find by MATLAB the properties for T = 321 440, 571 respectively Solution 30: % Tutorial 30 clear all clc Temp=[300 400 500 600]; Beta=1000*[3.33 2.5 2.0 1.67]; Alpha=10000*[0.2128 0.3605 0.5324 0.7190]; Ti = [321 440 571 581]; Propty1 = interp1(Temp,Beta,Ti,'linear'); Propty2 = interp1(Temp,Alpha,Ti,'linear'); [Ti; Propty1; Propty2]

Problem 31: Suppose a functional relation y = y(x) given in a tabular form as Y 0 0.25 0.50 0.75 1.00

Y(x) 0.9162 0.8109 0.6931 0.5596 0.4055

Where y(x) is a monotonically increasing function of x, Find values of x that satisfies y = 0.9, 0.7, 0.6 and 0.5, respectively by MATLAB % Tutorial 31 clear all clc x=[0 0.25 0.5 0.75 1.0];

y=[0.9162 0.8109 0.6931 0.5596 0.4055]; yi = [0.9 0.7 0.6 0.5]; xi = interp1(y,x,yi,'linear'); [yi; xi]

Problem 32: Desities of sodium for three temperatures are given as I 1 2 3

Temperature 94 205 371

Density 929 902 860

Write the Largrange interpolation formula that fits the three data points. Find density for T = 251 deg. C by Lagrange Interpolation Solution 32: % Tutorial 32 clear all clc x=[94 205 371]; y=[929 902 860]; xi = [251]; yi = lagrange(y,x,xi); [yi; xi] Problem 33: Matrix A is given by

⎡ 3 4 −2 ⎤ A = ⎢⎢ 3 −1 1 ⎥⎥ Find eigen values directly by eig, expand A into its characteristics polynomial, and ⎢⎣ 2 0 5 ⎥⎦ find the roots of the characteristics of equations Solution 33: % Tutorial 33 clear all clc A = [3 4 -2; 3 -1 1; 2 0 5]; A_eig = eig(A) A_poly = poly(A) A_roots = roots(A_poly) Output A_eig = -2.7503 4.8751 + 1.4314i 4.8751 - 1.4314i A_poly = 1.0000 -7.0000 -1.0000 71.0000

A_roots = 4.8751 + 1.4314i 4.8751 - 1.4314i -2.7503 Problem 34: A set of four data points is given by X = [ 1.1 2.3 3.9 5.1 ] Y = [ 3.887 4.276 4.651 2.117 ] Find the coeffificents of the interpolation polynomial fitted to the data set, determine value of y for x = 2.101 and plot the polynomial with data points Problem 35 An automobile of mass M = 2000 kg is crusing at speed of 30 m/s. The engine is suddenly disengaged at t = 0 sec. Assume that equation of crusing after t = 0 sec is given by du 2000u = −8.1u 2 − 1200 dx Where u is velocity and x is the linear distance of the car measured from the location at t = 0. The left side is the force of acceleration, the first term on right side is aerodynamic resistance. And second term is rolling resistance. Calculate how far the car moves before the speed reduces to 15 m/s. Solution 35 30

x

2000udu 2000udu = dx integrating equation ∫ = ∫ dx = x where sign on 2 2 −8.1u − 1200 8.1 u + 1200 15 0 left side is changed for switching the limits of integration, by applying a trapezoidal rule, ⎡ 16 ⎤ x ≈ Δu ⎢ ∑ fi − 0.5( f1 + f16 ) ⎥ where f is integrand function ⎣ i =1 ⎦ Programming Step 1 : Develop a trapezoidal function which evaluates integral function I = trapez_v(f, h); I = h*(sum(f) - (f(1)+f(length(f)))/2); Step 2 : Write a script file for calculation of distance as below, % Tutorial 35 clear all, clc, n_points = 16; i = 1:n_points; h = (30-15) / n_points; u = 15+(i-1)*h; f = 2000*u./(8.1*u.^2+1200); x = trapez_v(f,h) Rewrite equation as,

Problem 36 Knowing that the exact answer is I = 4.006994, analyze the effect of the number of intervals n on the errors of the trapezoidal rule applied to the following integral 2

I = ∫ 1 + exp( x)dx 0

Solutrion 36 Step 1 : Develop a trapezoidal function which evaluates integral

function I = trapez_v(f, h); I = h*(sum(f) - (f(1)+f(length(f)))/2); Step 2 Write a script file % Tutorial 36 clear all, clc, Iexact = 4.006994; a = 0; b = 2; fprintf('\n Extended Trapezoidal Rule \n'); fprintf('\n n I Error \n'); n = 1; for k=1:6 n = 2*n; h = (b-1)/n; i = 1:n+1; x = a + (i-1)*h; f = sqrt(1+exp(x)); I = trapez_v(f,h); nn(k)=n; I(k)=I; Error(k)=Iexact(1)-I(k); end out_data=[nn; I; Error;]' Problem 37 A spherical water reservior of radius 5 m is full to the top. Water is to be drained from the hole of radius b = 0.1 m at bottom, starting at t = 0 s. If there is no friction. How much time does it take to drain the water untill the water level reaches to 0.5m measured from bottom. ? Solve by simpsons rule Solution 37

u2 where u is 2 the velocity , z is level of water measured from the spherical center, R is the radius of the tank, and g is gravity acceleration. Consider a change of water level dz during time interval dt. The volume of water in dz is Πx2dz, where x is the radius of circular water surface at elevation z. Flow continuity relation can be written as, uAdt = - Πx2dz , where A is the cross section of exit hole and is given by A = Πb2 Radius of water surface x is related to z by R2 = z2 +x2. Eliminating u, x, and A from previous equation R2 − z 2 dt = 2 dz , Notice that the level of water at the top of the tank is z =R, while that at 0.5 m b 2g(z + R from bottom is z = 0.9R. Integrating from z = R to 0.9R, we get 0.9 R R2 − z 2 t= ∫ 2 dz 2 g ( z + R) R b Velocity of water draining from hole is determined by the energy equation, g ( z + R) =

Matlab Implementation Step 1 Develop a function for simpson’s rule

function I=Simps_v(f,h) n=length(f)-1; if n==1 fprintf('\nData has only one data point') return; end if n==2 I=h/3*(f(1)+4*f(2)+f(3)); return end if n==3 I=3/8*h*(f(1)+3*f(2)+3*f(3)+f(4)); return end I=0; if 2*floor(n/2)~=n I=3/8*h*(f(n-2)+3*f(n-1) ... +3*f(n)+f(n+1)); m=n-3; else m=n; end I = I+(h/3)*(f(1)+4*sum(f(2:2:m))+f(m+1)); if m>2 I = I + (h/3)*2*sum(f(3:2:m)); End Step 2 Write a script program as below % Tutorial 37 clear R=5; g=9.81; b=0.1; z1=-R*.90; z2=R; h=(z2-z1)/20; z=z1:h:z2; f=(R^2-z.^2)./(b^2*sqrt(2*g*(z+R))); I=Simps_v(f,h)/60/60 Problem 38 Motion of an electron in an uniform electromagnetic field is given by, dV m = eV + B + eE where V is the velocity vector, B is the magnetic field vector, E is the electric field dt vector, and m is mass of the electron, and e is the charge of electron. Initial condition V = (-10, 2, 0.1) X 105 m/s Initial Position of electron R = (0, 0, 0) m Magnetic Field Vector B = ( 0, 0, 0.1 ) T Electric Field Vector E = (0, 2, 0) X 104 V/m Mass of electron m =9.1 X 10-31 kg Charge of electron e = 1.6 X 10-19 c

Solve equation by Runge Kutta method with h = 0.5 X 10-11 s for 0 < t < 2 X 10-9 s, and determine the locus of the electron. Plot a trajectory of the electron in a three dimensional plane, and velocity components as functions of time as well as in the three dimensional phase space. Solution 38 Step 1 Develop a function which carry out a vector product function c = vxv_(a,b) % c = [a] X [b] % a b c are vectors c=[a(2)*b(3)-a(3)*b(2); -a(1)*b(3)+a(3)*b(1); a(1)*b(2)-a(2)*b(1)]; Step 2 Write a scrip file as below % Tutorial 38 clear all clf, clc m =9.1e-31; e = 1.6e-19; B = [0; 0; 0.1]*e/m; E = [0; 2e4; 0]*e/m; h=0.5e-11; v(:,1)=1e5*[-10;2;0.1];

t(1)=0; xyz(:,1)=[0;0;0]; epm=e/m; for i=2:400 t(i)=h*i; k1=h*(vxv_(v(:,i-1),B)+E); k2=h*(vxv_(v(:,i-1)+k1,B)+E); v(:,i)=v(:,i-1)+0.5*(k1+k2); xyz(:,i)=xyz(:,i-1)+0.5*(v(:,i-1)+v(:,i))*h; end figure(1) plot3(xyz(1,:),xyz(2,:),xyz(3,:)) axis([-5 5 -1 2 0 3]*1e-4) xlabel('X');ylabel('Y');zlabel('Z'); figure(2) plot3(xyz(1,:),xyz(2,:),xyz(3,:)) xlabel('X');ylabel('Y');zlabel('Z'); view([0,0,1]) figure(3) plot3(v(1,:),v(2,:),v(3,:)) xlabel('V_x');ylabel('V_y');zlabel('V_y'); figure(4) plot(t,v(1,:),t,v(2,:),t,v(3,:)) xlabel('t');ylabel('Velocities'); legend('V_x','V_y','V_z')

Problem 39 Write a program for lobe of oscillating liquid jet % Tutorial 39 % plots a lobe of liquid jet clear, clf hold on dth=2*pi/40; th=0:dth:2*pi; r =ones(size(th)); colormap gray for n=1:51 b=n-1; x(n,:)=cos(th)*(1-0.25*cos(b*0.3)); y(n,:)=sin(th)*(1+0.25*cos(b*0.3)); z(n,:)=n*0.3*ones(size(th)); m=n+9; if floor(m/2)*2 ==m plot3(z(n,:),x(n,:),y(n,:)-5) end end hold off figure(2) surfl(z,x,y+2,[-10,60]) axis([0 13 -5 15 -10 5])

Control System MATLAB Tutorials Problem 1 Find a poles of the follwing transfer functions and also compute step response characteristics of following systems 1 2 s 2 + 5s + 1 2 4s + 2 , T2 = 2 , T3 = 3 , T1 = 4 T1 = 2 2 s + 3s 2 + 3s + 1 s + 2s + 2 s + 2s + 2 s + 2 s 3 + 5s 2 + 2 s + 2 Solution 1: % Control Syst tut 01 % Poles of tranfer functions and % computation of step responses for all TFs num1=[2]; den1=[1 2 2]; sys1=tf(num1,den1) P1 = pole(sys1);Z1 = zero(sys1); figure(1) subplot(2,2,1), step(sys1) num2=[4 2]; den2=[1 2 2]; sys2=tf(num2,den2) P2 = pole(sys2);Z2 = zero(sys2); subplot(2,2,2), step(sys2) num3=[1]; den3=[2 3 3 1]; sys3=tf(num3,den3) P3 = pole(sys3);Z3 = zero(sys3); subplot(2,2,3), step(sys3) num4=[2 5 1]; den4=[1 2 5 2 2]; sys4=tf(num4,den4) P4 = pole(sys4);Z4 = zero(sys4); subplot(2,2,4), step(sys4)

Problem 2: Find each of second order systems below find ζ, ωn, Ts, Tp, Tr, % overshoot, and plot step response using MATLAB 130 0.045 10 3 T (s) = 2 , T ( s) = 2 , T ( s) = 2 s + 15s + 130 s + 0.025s + 0.045 s + 1.325 × 10 4 s + 10 8 Solution 2: % Control tut 02 % % first tranfer function num1=[130]; den1=[1 15 130]; Ta1=tf(num1,den1) wn1=sqrt(den1(3)) zeta1=den1(2)/(2*wn1) Tsa1=4/(zeta1*wn1) Tpa1=pi/(wn1*sqrt(1-zeta1^2)) Tra1=(1.76*zeta1^30.417*zeta1^2+1.039*zeta1+1)/wn1 Per1=exp(-zeta1*pi/sqrt(1-zeta1^2))*100 figure(1) subplot(221), step(Ta1)

num2=[0.045]; den2=[1 0.025 0.045]; Ta2=tf(num2,den2) wn2=sqrt(den2(3)) zeta2=den2(2)/(2*wn2) Tsa2=4/(zeta2*wn2) Tpa2=pi/(wn2*sqrt(1-zeta2^2)) Tra2=(1.76*zeta2^3-0.417*zeta2^2+1.039*zeta2+1)/wn2 Per2=exp(-zeta2*pi/sqrt(1-zeta2^2))*100 figure(1) subplot(222), step(Ta2) num3=[10E8]; den3=[1 1.325E4 10E8]; Ta3=tf(num3,den3) wn3=sqrt(den3(3)) zeta3=den3(2)/(2*wn3) Tsa3=4/(zeta3*wn3) Tpa3=pi/(wn3*sqrt(1-zeta3^2)) Tra3=(1.76*zeta3^3-0.417*zeta3^2+1.039*zeta3+1)/wn3 Per3=exp(-zeta3*pi/sqrt(1-zeta3^2))*100 figure(1) subplot(223), step(Ta3)

Problem 3:

C (s) 1 , = 2 R( s ) s + 2ξs + 1 plot the unit step response curves for ζ = 0.0, 0.1, 0.2, 0.4, 0.5, 0.6, 0.8, and 1.0. ωn is normalized to 1, plot a three dimensional coruve for step reponse as above

For a closed loop system defined by,

Solution 3: % Control system tut 03 % 2D and 3D plots for unit step response zeta = [0 0.1 0.2 0.4 0.5 0.6 0.8 1.0]; t = 0:0.2:15; figure(1) for n = 1:8; num = [0 0 1]; den = [1 2*zeta(n) 1]; sys(n) = tf(num,den) step(sys(n),t); y(:,n) = step(sys(n),t); hold on end figure(2) mesh(t,zeta,y') xlabel('t sec'), ylabel('\zeta'), zlabel('Amplitude')

Problem 4: Write a equations of motion for the simple pendulum where all mass is concentrated at the endpoint. Use MATLAB to determine the time history of θ to step input of Tc of 1 N-m. Assume l =1 m, m = 0.5 kg, and g = 9.81 m/s2. Solution 4:

Equation of motion of pendulum is •• T g θ + θ = c 2 and transfer function is given l ml 1 2 Θ( s ) = ml by Tc ( s ) s 2 + g l Script Program % Control tut 04 t=0:0.02:10; m=0.5; g=9.81; l=1; tc =1; num=[1/m*l^2]; den=[1 0 g/l]; sys = tf(num,den); y =step(sys,t); plot(t,y) Problem 5 Solve following state space equation by SIMULINK

Problem 6 Consider the double integrator system G(s), which is open loop unstable. To stabilize this plant, we use a lead compensator, K(s). Prepare a simulink model for this s +1 2.5 simulate this model upto 100 seconds G(s) = 2 , K (s) = s + 10s + 1 s+5 Solution 6

Problem 7 The Lorenz system is set of three first order nonlinear differential equations as given below, Simulate using Simulink Modeling •

x = a ( y − x) •

y = x(b − z ) − y where a =10, b = 28, c = -2.67, and Initial conditions are x(0) =1, y(0)=1, and z(0)=1 •

z = cz + xy Solution 7

Problem 8 Simulate a Two degree freedom system given by following relations ••

•

•

M 1 x1 + (k1 + k2 ) x1 + (c1 + c2 ) x1 − k2 x2 − c2 x2 = 0 ••

•

•

M 2 x2 + k2 x2 + c2 x2 − c2 x1 − k2 x1 = F sin(20t ) Where M1 =10kg, M2=20kg, k1 = 1000 N/m, k2=200 N/m, c1=c2=120 Ns/m, F = 100;

Solution

Problem 9 Simulate following Duffing System

•

x= y

x(0) = 5

•

y = −( x3 + x + y ) + cos θ •

θ =ω

y (0) = 5

θ (0) = 3

Problem 10 Simulate following Rossler System of equations •

x = −y − z x(0) = 5 • 1 y = x+ y y (0) = 5 5 • 1 z = + zx − 5.7 z z (0) = 5 5 Problem 11 Simple pendulum nonlinear equation is given by following relation, determine a velocity and displacement plot with respect to time using SIMULINK and ode solver also •

x1 = x2 •

x2 = −

T g sin x1 + c 2 l ml

Solution 11

ODE solution global Wn2 Uo Wn2=9.82; Uo=5; t0=0; tf=10;

tspan=[t0,tf]; X0=[0 0]; [t,x]=ode23(@pendot,tspan,X0) function xdot=pendot(t,x) global Wn2 Uo if t