You are here: Home » Content » STABILITY OF SYSTEMS
Content Actions

STABILITY OF SYSTEMS

Module by: Nguyen Huu Phuong

Stability is perhaps the most important property of real systems. When a system is unstable a number of its operating parameters may change freely or go without bound, or (for computer pregrammes) give inconsistent results.
For DSP (or DTSP) systems the definition of stability is as follows: The system is stable when with respect to a bounded input it gives a bounded output. This stability criterion is called bounded-input bounded-output (BIBO). Mathematically:
∣ x ( n ) ∣ ≤ M X < ∞ ⇒ ∣ y ( n ) ≤ M y < ∞ ∣ ∣ x ( n ) ∣ ≤ M X < ∞ ⇒ ∣ y ( n ) ≤ M y < ∞ ∣ size 12{ lline x \( n \) rline <= M rSub { size 8{X} } < infinity matrix { {} # {} # {} } drarrow matrix { {} # {} # {} } \lline y \( n \) <= M rSub { size 8{y} } < infinity \lline } {}
Now we derive the condition of stability imposed on impulse response. Starting from the convolution summation
y ( n ) = x ( n ) ∗ h ( n ) = ∑ k = − ∞ + ∞ x ( n ) h ( n − k ) y ( n ) = x ( n ) ∗ h ( n ) = ∑ k = − ∞ + ∞ x ( n ) h ( n − k ) size 12{y \( n \) =x \( n \) * h \( n \) = Sum cSub { size 8{k= - infinity } } cSup { size 8{+ infinity } } {x \( n \) h \( n - k \) } } {}
Take the absolute value of both sides:
∣ y ( n ) ∣ = ∣ ∑ k = − ∞ ∞ x ( n ) h ( n − k ) ∣ ≤ ∑ k = − ∞ ∞ ∣ x ( n ) ∣ ∣ h ( n − k ) ∣ = ∣ x ( n ) ∣ ∑ k = − ∞ ∞ ∣ h ( n − k ) ∣ ∣ y ( n ) ∣ = ∣ ∑ k = − ∞ ∞ x ( n ) h ( n − k ) ∣ ≤ ∑ k = − ∞ ∞ ∣ x ( n ) ∣ ∣ h ( n − k ) ∣ = ∣ x ( n ) ∣ ∑ k = − ∞ ∞ ∣ h ( n − k ) ∣ size 12{ lline y \( n \) rline = lline Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {x \( n \) h \( n - k \) } rline <= Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } { lline x \( n \) rline } lline h \( n - k \) rline = lline x \( n \) rline Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } { lline h \( n - k \) rline } } {}
then for finite ∣x(n)∣,∣y(n)∣∣x(n)∣,∣y(n)∣ size 12{ lline x \( n \) rline , lline y \( n \) rline } {}is finite if
∑ k = − ∞ ∞ ∣ h ( k ) ∣ < ∞ ∑ k = − ∞ ∞ ∣ h ( k ) ∣ < ∞ size 12{ Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } { lline h \( k \) rline < infinity } } {}
Since k is a dummy variable we can change it to n and write the condition as
∑n=−∞∞∣h(n)∣<∞(ConditionofstabililtyBIBO)∑n=−∞∞∣h(n)∣<∞(ConditionofstabililtyBIBO) size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { lline h \( n \) rline < infinity } matrix { {} # {} # {} } \( matrix { ital "Condition" {} # ital "of" {} # ital "stabililty" {} # ital "BIBO"{} } \) } {} (1)
Here ∣h(n)∣∣h(n)∣ size 12{ lline h \( n \) rline } {} is the alsolute value if h(n)h(n) size 12{h \( n \) } {} is real-valued, and the magnitude if h(n)h(n) size 12{h \( n \) } {}is complex.
The condition for an LTI (or LSI) system to be stable is the total absolute values of its impulse response must be bounded. FIR systems are mostly stable , whereas as for IIR systems the stablity requires the impulse response decays fast enough with time.
Example 1 
LTI system has impulse response h ( n ) = a n n ≥ 0 b n n < 0 h ( n ) = a n n ≥ 0 b n n < 0 alignl { stack { size 12{h \( n \) =a rSup { size 8{n} } matrix { {} # {} # {} } n >= 0} {} # matrix { {} # {} # =b rSup { size 8{n} } {} } matrix { {} # {} # n<0{} } {} } } {} Find the condition for stability.
Solution
The overall impulse response consists of a causal part and a noncausal one. The condition of stability is
∑ n = − ∞ + ∞ ∣ h ( n ) ∣ = ∑ n = 0 + ∞ ∣ a ∣ n + ∑ n = − ∞ − 1 ∣ b ∣ n < ∞ ∑ n = − ∞ + ∞ ∣ h ( n ) ∣ = ∑ n = 0 + ∞ ∣ a ∣ n + ∑ n = − ∞ − 1 ∣ b ∣ n < ∞ size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{+ infinity } } { lline h \( n \) rline = Sum cSub { size 8{n=0} } cSup { size 8{+ infinity } } { lline a rline rSup { size 8{n} } + Sum cSub { size 8{n= - infinity } } cSup { size 8{ - 1} } { lline b rline rSup { size 8{n} } < infinity } } } } {}
First
∑ n = 0 ∞ ∣ a n ∣ = 1 + ∣ a ∣ + ∣ a ∣ 2 + . . . < ∞ ∑ n = 0 ∞ ∣ a n ∣ = 1 + ∣ a ∣ + ∣ a ∣ 2 + . . . < ∞ size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { lline a rSup { size 8{n} } rline =1+ lline a rline + lline a rline rSup { size 8{2} } + "." "." "." < infinity } } {}
Applying the formula of infinite geometric series (Equation (2.8)) will lead to the condition ∣a∣<1∣a∣<1 size 12{ lline a rline <1} {}.
Now
∑ n = − ∞ − 1 ∣ b ∣ n = ∑ n = 1 ∞ 1 ∣ b ∣ n = 1 ∣ b ∣ ( 1 + 1 ∣ b ∣ + 1 ∣ b ∣ 2 + . . . ) = 1 ∣ b ∣ 1 1 − 1 ∣ b ∣ , 1 ∣ b ∣ < 1 ∑ n = − ∞ − 1 ∣ b ∣ n = ∑ n = 1 ∞ 1 ∣ b ∣ n = 1 ∣ b ∣ ( 1 + 1 ∣ b ∣ + 1 ∣ b ∣ 2 + . . . ) = 1 ∣ b ∣ 1 1 − 1 ∣ b ∣ , 1 ∣ b ∣ < 1 alignl { stack { size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ - 1} } { \lline b \lline rSup { size 8{n} } } = Sum cSub {n=1} cSup { infinity } { { { size 8{1} } over { size 8{ \lline b \lline rSup { size 6{n} } } } } } size 12{ {}= { {1} over { \lline b \lline } } \( 1+ { {1} over { \lline b \lline } } } size 12{+ { {1} over { \lline b \lline rSup { size 6{2} } } } } size 12{+ "." "." "." \) }} {} # size 12{ matrix { matrix { matrix { {} # {} } {} # {} # {} } {} # {} # {} } matrix { {} # {} } = { {1} over { \lline b \lline } } { {1} over {1 - { { size 8{1} } over { size 8{ \lline b \lline } } } } } , matrix { {} # {} # {} } { {1} over { \lline b \lline } } <1} {} } } {}
The condition is 1∣b∣<11∣b∣<1 size 12{ { {1} over { lline b rline } } <1} {} or ∣b∣>1∣b∣>1 size 12{ lline b rline >1} {}. The overall condition is ∣a∣<1∣a∣<1 size 12{ lline a rline <1} {} and ∣b∣>1∣b∣>1 size 12{ lline b rline >1} {}

Comments, questions, feedback, criticisms?

Send feedback