SYSTEM FUNCTIONS FOR SYSTEMS CHARACTERIZED BY LINEAR CONSTANT COEFFICIENT DIFFERENCE EQUATIONS

DIAGRAM UNITS: 3 BASICS

where z^-1 represents the transfer function of a unit delay:

so:

One implementation for: y(n) = x(n) + a y(n-1)

EXAMPLE

NICE FEATURE OF DIFFERENCE EQUATIONS -- WE CAN ALWAYS GET ANSWERS BY HAND:

x(n) = d(n)

y(-2)	= x(-2) +.5 y(-3)	= 0 + 0	= 0
y(-1)	= x(-1) +.5 y(-2)	= 0 + 0	= 0
y(0)	= x(0) +.5 y(-1)	= 1 + 0	= 1
y(1)	= x(1) +.5 y(0)	= 0 + .5 (1)	= .5
y(2)	= x(2) +.5 y(1)	= 0 + .5 (.5)	= (.5)2
y(3)	= x(3) +.5 y(2)	= 0 + .5 (.5)2	= (.5)3
y(4)	= x(4) +.5 y(-3)	= 0 + .5 (.5)3	= (.5)4

.....

y(n) = (.5)ⁿ u(n)

THIS IS AN INFINITE IMPULSE RESPONSE (IIR) OR RECURSIVE SYSTEM. OCCURS WHEN THE SYSTEM HAS A NON-ZERO POLE

CONTRAST TO FINITE IMPULSE RESPONSE (FIR) OR NONRECURSIVE SYSTEM: OCCURS WHEN SYSTEM HAS NO NON-ZERO POLES

y(n) = x(n) + 3x(n-2); x(n) = d(n) -----> h(n)

y(-2)	= x(-2) + 3x(-4)	= 0 + 0	= 0
y(-1)	= x(-1) + 3x (-3)	= 0 + 0	= 0
y(0)	= x(0) + 3x(-2)	= 1 + 0	= 1
y(1)	= x(1) + 3x(-1)	= 0 + 0	= 0
y(2)	= x(2) + 3x(0)	= 0 + 0	= 0
y(3)	= x(3) + 3x(1)	= 0 + 0	= 0
y(4)	= x(4) + 3x(2)	= 0 + 0	= 0

h(n) = d(n) + 3d(n-2)

LOOK AT x(n) = u(n):

y(-2)	= x(-2) + 3x(-4)	= 0 + 0	= 0
y(-1)	= x(-1) + 3x (-3)	= 0 + 0	= 0
y(0)	= x(0) + 3x(-2)	= 1 + 0	= 1
y(1)	= x(1) + 3x(-1)	= 1 + 0	= 1
y(2)	= x(2) + 3x(0)	= 1 + 3	= 4
y(3)	= x(3) + 3x(1)	= 1 + 3	= 4
y(4)	=x(4) + 3x(2)	= 1 + 3	= 4

Each: z-z_m = B_m exp(j q_m) and each z-p_k = A_kexp(j f_k)

GRAPHING TECHNIQUE

MULTIPLE POLES AND ZEROS

If H(z) is a rational system with real coefficients, then, by polynomial theory, if we have a zero or pole that has a complex component, then the complex conjugate of must also be zero or a pole, e.g.

Assume that: p_k = r e^jf

Then: 1-p_k z^-1= 1 - r e^jf z^-1 and 1-p_k^* z^-1= 1 - r e^-jf z^-1

or

The difference equation that implements this set of poles is:

and the impulse response is:

ANOTHER LOOK AT MOVING AVERAGE:

: