EECS 216 LECTURE NOTES
BASIC SYSTEM PROPERTIES
What: Input x[n]→ |SYSTEM| → y[n] output.
Why: Design the system to filter input x[n].
DEF: A system is LINEAR if these two properties hold:
1. Scaling: If x[n]→ |SYS| → y[n], then ax[n]→ |SYS| → ay[n]
for: any constant a. NOT true if avaries with time (i.e., a[n]).
2. Superposition: If x1[n]→ |SYS| → y1[n] and x2[n]→ |SYS| → y2[n],
Then: (ax1[n] + bx2[n]) → |SYS| → (ay1[n] + by2[n])
for: any constants a, b. NOT true if aor bvary with time (i.e., a[n], b[n]).
EX: y[n] = 3x[n−2]; y[n] = x[n+1]−nx[n]+2x[n−1]; y[n] = sin(n)x[n].
NOT: y[n] = x2[n]; y[n] = sin(x[n]); y[n] = |x[n]|;y[n] = x[n]/x[n−1].
NOT: y[n] = x[n] + 1 (try it). This is called an affine system.
HOW: If any nonlinear function of x[n], not linear. Nonlinear of just nOK.
DEF: A system is TIME-INVARIANT if this property holds:
If x[n]→ |SYS| → y[n], then x[n−N]→ |SYS| → y[n−N]
for: any integer time delay N. NOT true if Nvaries with time (e.g., N(n)).
EX: y[n] = 3x[n−2]; y[n] = sin(x[n]); y[n] = x[n]/x[n−1].
NOT: y[n] = nx[n]; y[n] = x[n2]; y[n] = x[2n]; y[n] = x[−n].
HOW: If nappears anywhere other than in x[n], not time-invariant. Else OK.
DEF: A system is CAUSAL if it has this form for some function F(·):
y[n] = F(x[n], x[n−1], x[n−2] . . .) (present and past input only).
Note: Physical systems must be causal. But DSP filters need not be causal!
DEF: A system is MEMORYLESS if y[n] = F(x[n]) (present input only).
DEF: A system is (BIBO) STABLE iff: Let x[n]→ |SYSTEM| → y[n].
If |x[n]|< M for some constant M, then |y[n]|< N for some N.
i.e.: “Every bounded input (BI) produces a bounded output (BO).”
HOW: BIBO stable ⇔P∞
n=−∞ |h[n]|< L for some constant L
where: Impulse δ[n]→ |SYS| → h[n]=impulse response.
EX: A time-invariant system is observed to have these two responses:
{0,0,3} → |SYS| → {0,1,0,2}and {0,0,0,1} → |SYS| → {1,2,1}.
Prove: The system is nonlinear.
Proof: By contradiction. Suppose the system is linear. But then:
{0,0,0,1} → |SYS| → {1,2,1}implies {0,0,3} → |SYS| → {3,6,3}
since we know it is time-invariant. Then {0,0,3}produces two outputs!
