Linear system equation Solve the following system of linear equations: 2 x + y – 4 z = 5 x - 2 y - 5 z = y + 2 z = 4 One possible solution is by computing the inverse of the matrix of coefficients: To do this exercise follow these steps:

Solving linear equations Solve the following system of linear equations: 2 x + y – 4 z = 5 x - 2 y - 5 z = y + 2 z = 4 One possible solution is by computing the inverse of the matrix coefficients: To do this exercise follow these steps:

In matrix form, the above equations can be written as

Define the matrix D and the vector M shown below using the same method used in step 1:

Solving linear equations_1 % y+4z+x = 10 % -2z+3x+4y = 0 % -2*z+3*x+4*y = 0' In general, you can use sym or syms to create symbolic variables M_File for solving equations is: clear, syms x y z; eq1 = '2*x-3*y+4*z = 5' eq2 = 'y+4*z+x = 10', eq3 = '-2*z+3*x+4*y = 0' [x,y,z] = solve(eq1,eq2,eq3,x,y,z)

Solving linear equations_2 % 2 x + y – 4 z = 5 % x - 2 y - 5 z = % y + 2 z = 4 % Multiply the inverse of the matrix D by the vector M by typing D=[2 1 4; ;0 1 2] M=[5; -6.5; 4] X=inv(D)*M

Differentiation diff_f =df/dx f = e -ax x 3b sin(cx) a, b and c are unspecified constants clear, syms a b c x; f=exp(-a*x)*x^(3*b)*sin(c*x), diff_f = diff(f,x)

Differentiation f = exp(-a*x)*x^(3*b)*sin(c*x) diff_f = -a*exp(-a*x)*x^(3*b)*sin(c*x)+3*exp(- a*x)*x^(3*b)*b/x*sin(c*x)+exp(-a*x)*x^(3*b)*cos(c*x)*c

derivatives of polynomial functions Examples of derivatives of polynomial functions: syms s n p = s^3 + 4*s^2 -7*s -10; >> d = diff(p) d = 3*s^2+8*s-7

Integral of equation int(x^3 +4*x^2 + 7*x + 10) ans = 1/4*x^4+4/3*x^3+7/2*x^2+10*x int(x^3,a,b) ans = 1/4*b^4-1/4*a^4

Integral int(1/x) ans = log(x) >> int(cos(x)) ans = sin(x) >> int(1/(1+x^2)) ans = atan(x) >> int(exp(-x^2)) ans = 1/2*pi^(1/2)*erf(x)

Limit integral f = inline(vectorize(sin(x^3 - 7*x)),'x') quad(f,2,4) f = Inline function: f(x) = sin(x.^3-7.*x) ans =

A sample evaluation of a double integral

Infinite integral We type in: >> f = inline(vectorize((1-u)*sin(u*v*(1- u))),’u’,’v’) >> dblquad(f,0,1,0,1)

Control Systems Closed-Loop Poles The closed-loop transfer function is: thus the poles of the closed loop system are values of s such that 1 + K H(s) = 0

Closed-Loop system If we write H(s) = b(s)/a(s), then this equation has the form: a(s)+kb(s)=0 A(s)/k +b(s)=0 Let n = order of a(s) and m = order of b(s) [the order of a polynomial is the highest power of s that appears in it]. We will consider all positive values of k. In the limit as k -> 0, the poles of the closed-loop system are a(s) = 0 or the poles of H(s). I n the limit as k -> infinity, the poles of the closed-loop system are b(s) = 0 or the zeros of H(s).

Plotting the root locus of a transfer function Consider an open loop system which has a transfer function of

Closed loop systems num=[1 7]; den=conv(conv([1 0],[1 5]),conv([1 15],[1 20])); rlocus(num,den) axis([ ])axis

gain and phase margins margin(50,[ ])

Bode Plots As noted above, a Bode plot is the representation of the magnitude and phase of G(j*w) (where the frequency vector w contains only positive frequencies). To see the Bode plot of a transfer function, you can use the Matlab bode command. For example, bode(50,[ ]) displays the Bode plots for the transfer function:

bode(50,[ ])

bode(1, [1 1])

bode([1 0], 1)

bode(1,[ 1, 0.1])

bode(1,[ 1, 10])

Plot all these bodes bode([1, 0.1],1) # S+0.1 bode([1, 10],1) # S+10 bode([1 1], 1) # S+1