Matlab: Symbolic Math Toolbox y Simulink

M.Sc. Luis Sánchez

Symbolic Math Toolbox

Permite:

• Calculus – integration, differentiation, Taylor series

expansion, …

• Linear Algebra – inverses, determinants, eigenvalues, …

• Simplification – algebraic and trigonometric expressions

• Equation Solutions – algebraic and differential equations

• Transforms – Fourier, Laplace, Z  transforms and inverse

transforms, … 

Objeto Symbolic Use sym to create a symbolic number, and double to convert to a normal number.

>> sqrt(2)ans = 1.4142

>> var = sqrt(sym(2))var = 2^(1/2)

>> double(var)ans = 1.4142

>> sym(2)/sym(5) + sym(1)/sym(3)ans = 11/15

Symbolic variables

Use syms to define symbolic variables.  (Or use sym to create an abbreviated symbol name.) >> syms m n b c x>> th = sym('theta')>> sin(th)ans = sin(theta)>> sin(th)^2 + cos(th)^2ans = cos(theta)^2 + sin(theta)^2>> y = m*x + by = b + m*x

Expresiones simbólicasThe subs function substitutes values or expressions for variables in a symbolic expression.  >> clear>> syms m x b>> y = m*x + b            → y = b + m*x>> subs(y,x,3)            → ans = b + 3*m>> subs(y, [m b], [2 3])  → ans = 2*x + 3>> subs(y, [b m x], [3 2 4])→ ans = 11

The symbolic expression itself is unchanged.

>> y                      → y = b + m*x

Substitutions, continued

Variables can hold symbolic expressions.  >> syms th z>> f = cos(th)   → f = cos(th)>> subs(f,pi)    → ans = -1 Expressions can be substituted into variables. >> subs(f, z*pi) → ans = cos(pi*z)

Diferenciación

Use diff to do symbolic differentiation.  >> clear>> syms m x b th n y

>> y = m*x + b;>> diff(y, x)     → ans = m>> diff(y, b)     → ans = 1

>> p = sin(th)^n  → p = sin(th)^n>> diff(p, th)    → ans = n*cos(th)*sin(th)^(n - 1)

Integración

Indefinite integrals

>> int(y, x)             →  ans = (m*x^2)/2 + b*x>> int(y, b)             →  ans = (b + m*x)^2/2>> int(1/(1+x^2))        →  ans = atan(x) Definite integrals

>> int(y,x,2,5)          →  ans = 3*b + (21*m)/2>> int(1/(1+x^2),x,0,1)  →  ans = pi/4

>> clear>> syms m b x >> y = m*x + b;

Graficando expresiones simbólicas

The ezplot function will plot symbolic expressions. >> clear; syms x y>> ezplot( 1 / (5 + 4*cos(x)) );>> hold on;  axis equal>> g = x^2 + y^2 - 3;>> ezplot(g);

Ejemplo>> clear; syms x>> digits(20)>> [x0, y0] = solve(' x^2 + y^2 - 3 = 0', ...                               'y = 1 / (5 + 4*cos(x)) ') → x0 = -1.7171874987452662214   y0 = 0.22642237997374799957>> plot(x0,y0,'o')>> hold on >> ezplot( diff( 1 / (5 + 4*cos(x)), x) )  

Solución de Ecuaciones diferenciales con Matlab

2

2

n

n

y

dy

dt

d y

dt

d y

dt

• y

• Dy

• D2y

• Dny

1 2(0) and '(0)y C y C

2

2 1 02sin

d y dyb b b y A atdt dt

• >> y = dsolve(‘b2*D2y+b1*D1y+b0*y=A*sin(a*t)’,• ‘y(0)=C1’, ‘Dy(0)=C2’)

• >> ezplot(y, [t1 t2])

Ejemplo. Resolver la ED usando Matlab

2 12dy

ydt

• >> y = dsolve('Dy + 2*y = 12', 'y(0)=10')

• y =

• 6+4*exp(-2*t)

• >> ezplot(y, [0 3])

• >> axis([0 3 0 10])

(0) 10y

• >> y = dsolve('Dy + 2*y = 12*sin(4*t)', 'y(0)=10')

• y =• -12/5*cos(4*t)+6/5*sin(4*t)+62/5*exp(-2*t)

• >> ezplot(y, [0 8])• >> axis([0 8 -3 10])

2 12sin 4dy

y tdt

(0) 10y

Ejemplo. Resolver la ED usando Matlab

• >> y = dsolve('D2y + 3*Dy + 2*y = 24', 'y(0)=10', 'Dy(0)=0')

• y =• 12+2*exp(-2*t)-4*exp(-t)

• >> ezplot(y, [0 6])

2

23 2 24

d y dyy

dt dt

(0) 10y '(0) 0y

Ejemplo. Resolver la ED usando Matlab

• >> y = dsolve('D2y + 2*Dy + 5*y = 20', • 'y(0) = 0', 'Dy(0) = 10')• y =• 4+3*exp(-t)*sin(2*t)-4*exp(-t)*cos(2*t)

• >>ezplot(y, [0 5]}

2

22 5 20

d y dyy

dt dt

(0) 0y '(0) 10y

Ejemplo. Resolver la ED usando Matlab

La trasformada de Laplace simbólica con Matlab

• Establezca s y t como variables simbólicas.•>> syms t s•La trasformada de laplace de una función f(t) se obtiene como:•>> F = laplace(f)•Algunas simplificaciones utilies son:•>> pretty(F)•>> simplify(F)

•Establish t and s as symbolic variables.

•>> syms t s

•The Laplace function F is then formed and the inverse Laplace transform command is

•>> f = ilaplace(F)

•The simplification operations may also be useful for inverse transforms.

La trasformada inversa de Laplace simbólica con Matlab

Ejemplo. Determine la trasformada de Laplace de f(t)=5t usando Matlab• >>syms t s• >> f = 5*t • f =• 5*t• >> F = laplace(f)• F =• 5/s^2

2 2( ) 3 sin 5 4 cos5t tv t e t e t • >> syms t s• >> v = 3*exp(-2*t)*sin(5*t) • + 4*exp(-2*t)*cos(5*t)• v =• 3*exp(-2*t)*sin(5*t)+4*exp(-2*t)*cos(5*t)

Ejemplo. Determine la trasformada de Laplace de v(t) usando Matlab

Ejemplo. Continuación…

• >> V = laplace(v) • V =• 15/((s+2)^2+25)+4*(s+2)/((s+2)^2+25)

• >> V=simplify(V) • V =• (23+4*s)/(s^2+4*s+29)•

• >> syms t s

• >> F=100*(s+3)/((s+1)*(s+2)*(s^2+2*s+5))

• F =• (100*s+300)/(s+1)/(s+2)/(s^2+2*s+5)

2

100( 3)( )

( 1)( 2)( 2 5)

sF s

s s s s

Ejemplo. Determine la trasformada Inversa de Laplace de F(s) usando Matlab

• >> f = ilaplace(F)

• f =• 50*exp(-t)-20*exp(-2*t)-30*exp(-t)*cos(2*t)-

10*exp(-t)*sin(2*t)

• >> pretty(f)• 50 exp(-t) - 20 exp(-2 t) - 30 exp(-t) cos(2 t)

- 10 exp(-t) sin(2 t)

Ejemplo. Continuación…

2

10 48( )

2 ( 2)( 16)Y s

s s s

• >> syms t s• >> Y = 10/(s+2) + 48/((s+2)*(s^2+16)) • Y =• 10/(s+2)+48/(s+2)/(s^2+16)

Ejemplo. Determine la trasformada Inversa de Laplace de F(s) usando Matlab

• >> y = ilaplace(Y)

• y =• 62/5*exp(-2*t)-12/5*cos(16^(1/2)*t)

+3/10*16^(1/2)*sin(16^(1/2)*t)

• >> y=simplify(y)

• y =• 62/5*exp(-2*t)-12/5*cos(4*t)+6/5*sin(4*t)

Ejemplo. Continuación…

Simulink Basics

click the Simulink button

the Simulink window

click the “new” button

the simulink model window

Simulink Basics

create a new model or open an existing one

• Build a Simulink model that solves the differential equation

• Initial condition

• First, sketch a simulation diagram of this mathematical model (equation)

(3 min.)

tx 2sin3

.1)0( x

Ejemplo. Modelo Simple

Diagrama de la Simulación

• Input is the forcing function 3sin(2t)• Output is the solution of the differential

equation x(t)

• Now build this model in Simulink

xxs1

3sin(2t)(input)

x(t)(output)

1)0( x

integrator

Select an input block

Drag a Sine Wave block from the Sources library to the model window

Select an operator block

Drag an Integrator block from the Continuous library to the model window

Select an output block

Drag a Scope block from the Sinks library to the model window

Connect blocks with signals

• Place your cursor on the output port (>) of the Sine Wave block

• Drag from the Sine Wave output to the Integrator input

• Drag from the Integrator output to the Scope input Arrows indicate the

direction of the signal flow.

Select simulation parameters

Double-click on the Sine Wave block to set amplitude = 3 and freq = 2.

This produces the desired input of 3sin(2t)

Select simulation parameters

Double-click on the Integrator block to set initial condition = -1.

This sets our IC x(0) = -1.

Select simulation parameters

Double-click on the Scope to view the simulation results

Run the simulation

In the model window, from the Simulation pull-down menu, select Start

View the output x(t) in the Scope window.

Simulation results

To verify that this plot represents the solution to the problem, solve the equation analytically.

The analytical result,

matches the plot (the simulation result) exactly.

ttx 2cos)( 23

21

Ejemplo. EDO 2 orden

• Build a Simulink model that solves the following differential equation– 2nd-order mass-spring-damper system– zero ICs– input f(t) is a step with magnitude 3– parameters: m = 0.25, c = 0.5, k = 1

)(tfkxxcxm

Create the simulation diagram

• On the following slides:– The simulation diagram for solving the ODE is

created step by step.– After each step, elements are added to the

Simulink model.

• Optional exercise: first, sketch the complete diagram (5 min.)

)(tfkxxcxm

(continue)

• First, solve for the term with highest-order derivative

• Make the left-hand side of this equation the output of a summing block

kxxctfxm )(

xm

summing block

Drag a Sum block from the Math library

Double-click to change the block parameters to rectangular and + - -

(continue)

• Add a gain (multiplier) block to eliminate the coefficient and produce the highest-derivative alone

xm m1 x

summing block

Drag a Gain block from the Math library

Double-click to change the block parameters.Add a title.

The gain is 4 since 1/m=4.

(continue)

• Add integrators to obtain the desired output variable

xm m1

summing block

s1

s1x xx

Drag Integrator blocks from the Continuous library

Add a scope from the Sinks library.Connect output ports to input ports.Label the signals by double-clicking on the leader line.

ICs on the integrators are zero.

(continue)

• Connect to the integrated signals with gain blocks to create the terms on the right-hand side of the EOM

xm m1

summing block

s1

s1x x x

c

k

xc

kx

Drag new Gain blocks from the Math library

Double-click on gain blocks to set parameters

Connect from the gain block input backwards up to the branch point.

Re-title the gain blocks.

To flip the gain block, select it and choose Flip Block in the Format pull-down menu.

c=0.5

k=1.0

Complete the model• Bring all the signals and inputs to the

summing block.• Check signs on the summer.

xm m1

s1

s1x x

c

k

xc

kx

f(t)input

+-

-x

x

xx(t)

output

Double-click on Step block to set parameters. For a step input of magnitude 3, set Final value to 3

Final Simulink model

Run the simulation

Results