STATE EQUATIONS

s₁(n+1) = - a₁ s₁(n) - a₂ s₂(n) + x(n)

s₂(n+1) = s₁(n)

OUTPUT EQUATION

y(n) = c₁s₁(n) + c₂s₂(n) + d x(n)

VECTOR EQUATIONS

s(n+1) = A s(n) + b x(n)

y(n) = c^t s(n) + d x(n)

STATE EQUATIONS

s₁(n+1) = - a1 s₁(n) - a₂ s₂(n) + x(n)

s₂(n+1) = s₁(n)

OUTPUT EQUATION

y(n) = g₀s₁(n+1) + g₁s₁(n) + g₂s₂(n)

y(n) = g₀(- a₁ s₁(n) - a₂ s₂(n) + x(n)) + g₁s₁(n) + g₂s₂(n)

y(n) = (g₁- a₁g₀) s₁(n) + (g₂- a₂g₀) s₂(n) + g₀ x(n)

Therefore: c^t = [ (g₁- a₁g₀) (g₂- a₂g₀)] d = g₀

CAN ALSO WRITE y(n) directly by following paths from state variables.

STATE TRANSFORMATIONS

^s(n) = T s(n) or T^-1^s(n) = s(n)

^s(n+1) = T s(n+1) = TA s(n) + T bx(n)

= TAT^-1+ T bx(n)

y(n) = c^t T^-1^s(n) + d x(n)

define: ^A= TAT^-1= T b; ^c= ct T^-1 ; ^d= d

SYSTEM FUNCTION

x(n) = d(n) with zero initial conditions:

s(n+1)	= A s(n) + b x(n)
s(n)	= A s(n-1) + b x(n -1)
s(0)	= A s(0-1) + b d(0 -1)	= 0 + 0	= 0
s(1)	= A s(0) + b d(0)	=A(0) + b	= b
s(2)	= A s(1) + b d(1)	= Ab	=Ab
s(3)	= A s(2) + b d(2)	= A2b	=A2b
.....
s(n)	= An-1b

y(n) = h(n)

y(n) = c^t s(n) + d x(n)

h(n) = c^t s(n) + d d(n)

h(0) = d

h(n) = c^t A^n-1b u(n-1) for n > 0

Therefore: h(n) = d d(n) + c^t A^n-1 b u(n-1)

which converges to one of the forms below for certain eigenvalues:

s(n+1) = A s(n) + b x(n)

z¹S(z) = A S(z) + b X(z)

z¹S(z) - A S(z) = (zI - A ) S(z) = b X(z)

S(z) =(zI - A )^-1 b X(z)

y(n) = c^t s(n) + d x(n)

Y(z) = c^t S(z) + d X(z)

Y(z) = c^t (zI - A )^-1 b X(z) + d X(z) = (c^t (zI - A )^-1 b +d )X(z)

EXAMPLE:

For an nth Order System

GENERAL STATE EQUATIONS

s₁(n+1) = - a₁ s₁(n) - a₂ s₂(n) - a₃ s₃(n) - a₄ s₄(n) ....- a_N s_N(n) + x(n)

s₂(n+1) = s₁(n)

s₃(n+1) = s₂(n)

s₄(n+1) = s₃(n)

....

s_N(n+1) = s_N-1(n)

OUTPUT EQUATION

y(n) = c₁s₁(n) + c₂s₂(n) + c₃s₃(n) + c₄s₄(n) ....+ c_Ns_N(n) + d x(n)

VECTOR EQUATIONS

Poles are the roots of the characteristic equation:

D(z) = |zI-A|

The eigenvalues of a matrix A are given by:

Ax = lx

(A-lI) x = 0 which has a nontrivial solution iff |A-lI| = 0

Thus the eigenvalues must equal the poles of the system function.

Our little 2x2 example:

Caley-Hamilton theorem:

D(z) = z^N + d_N-1 z^N-1+ ... + d₁z + d₀ = Characteristic Equation

Then: D(A) = A^N + d^N-1 A^N-1+ ... + d₁A + d₀ = 0

Diagonal Decomposition:

If A has distinct eigenvalues lk then A = P L P^-1

where P is composed of the eigenvectors: P = [v₁ | v₂| .... |v_N|]

i.e. Av_k = l_kv_k and AP = [ l₁v₁ | l₂v₂| .... | l_Nv_N|] = PL

Aⁿ = (P L P^-1) (P L P^-1) (P L P^-1) .....(P L P^-1)

= (P L (P^-1 P) L (P^-1P) L P-1) .....(P L P^-1)

Aⁿ = P Lⁿ P^-1

\

Therefore, letting T = P^-1

define: ^A= TAT^-1 = P^-1 P L P^-1 P -= L

Which is a partial fraction expansion of N decoupled equations:

If h(n) is real, all complex-values poles (and zeros) come in complex conjugate pairs. So assuming we only wish to work with real values in our diagrams, we pair up complex poles:

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