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Stability

Module by: Carlos E. Davila. E-mail the author

Summary: Looks at bounded input bounded output stability.

Filter Stability

We define a filter as being stable if a bounded input produces a bounded output. This is called BIBO stability. Consider the convolution integral

y ( t ) = - h ( τ ) x ( t - τ ) d τ y ( t ) = - h ( τ ) x ( t - τ ) d τ
(1)

We wish to find a condition that is necessary for the filter output y(t)y(t) to be bounded whenever the input x(t)x(t) is bounded. Let's take the absolute value of both sides of Equation 1

y ( t ) = - h ( τ ) x ( t - τ ) d τ - h ( τ ) x ( t - τ ) d τ x m a x - h ( τ ) d τ < y ( t ) = - h ( τ ) x ( t - τ ) d τ - h ( τ ) x ( t - τ ) d τ x m a x - h ( τ ) d τ <
(2)

where the first inequality is due to the triangle inequality and the second inequality results from replacing |x(t-τ)||x(t-τ)| by its upper bound xmaxxmax. Therefore, the condition for BIBO stability is that the impulse response the absolutely integrable.

- h ( t ) d t < - h ( t ) d t <
(3)

Example 3.1 The impulse response of a filter is h(t)=cos(5t)u(t)h(t)=cos(5t)u(t). If the input to this filter is x(t)=cos(5t)u(t)x(t)=cos(5t)u(t), then the output is given by

y ( t ) = 0 cos ( 5 τ ) cos ( 5 ( t - τ ) ) u ( t - τ ) d τ = 0 t cos ( 5 τ ) cos ( 5 ( t - τ ) ) d τ = 1 2 0 t cos ( 10 τ - 5 t ) d τ + 1 2 cos ( 5 t ) 0 t d τ = 1 2 0 t cos ( 10 τ - 5 t ) d τ + 1 2 t cos ( 5 t ) y ( t ) = 0 cos ( 5 τ ) cos ( 5 ( t - τ ) ) u ( t - τ ) d τ = 0 t cos ( 5 τ ) cos ( 5 ( t - τ ) ) d τ = 1 2 0 t cos ( 10 τ - 5 t ) d τ + 1 2 cos ( 5 t ) 0 t d τ = 1 2 0 t cos ( 10 τ - 5 t ) d τ + 1 2 t cos ( 5 t )
(4)

where in the third line, we have used the trigonometric identity

cos ( θ 1 ) cos ( θ 2 ) = 1 2 cos ( θ 1 - θ 2 ) + cos ( θ 1 + θ 2 ) cos ( θ 1 ) cos ( θ 2 ) = 1 2 cos ( θ 1 - θ 2 ) + cos ( θ 1 + θ 2 )
(5)

As tt approaches infinity, then it is clear that y(t)y(t) also becomes unbounded. Therefore the filter is unstable. Moreover, it is clear that h(t)h(t) is not absolutely integrable.

If a transfer function is rational, it can be expressed as a sum of any direct term that may be present plus a proper rational function (q<pq<p).

H ( s ) = c m s m + c m - 1 s m - 1 + + c 1 s + c 0 + b q s q + b q - 1 s q - 1 + + b 1 s + b 0 a p s p + a p - 1 s p - 1 + + a 1 s + a 0 H ( s ) = c m s m + c m - 1 s m - 1 + + c 1 s + c 0 + b q s q + b q - 1 s q - 1 + + b 1 s + b 0 a p s p + a p - 1 s p - 1 + + a 1 s + a 0
(6)

The direct terms can be shown to produce unbounded outputs when the input is a step function (which of course, is bounded). The proper rational function produces output terms that depend on whether poles are distinct or repeated and whether these are real or complex:

  • Distinct real poles, s=σks=σk lead to an impulse response with terms:
    Keαktu(t)Keαktu(t)
    (7)
  • Distinct complex conjugate poles s=σk±jΩks=σk±jΩk produce impulse response terms:
    Keσktcos(Ωkt+θ)u(t)Keσktcos(Ωkt+θ)u(t)
    (8)
  • Repeated real poles s=σks=σk produce impulse response terms having the form:
    Ktneαktu(t)Ktneαktu(t)
    (9)

The quantities KK and θθ are constants while nn is a positive integer. In all of these cases the filter impulse response dies out with time if the poles have negative real parts, i.e. it is absolutely integrable and therefore leads to a stable filter. If the poles have zero or positive real parts, then the impulse response terms either oscillate or grow with time, and are not absolutely integrable. When this happens then one can always find a bounded input that produces an unbounded output. Therefore, in order for a filter with a rational transfer function to have BIBO stability, the transfer function should be proper and the poles of the transfer function should have negative real parts.

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