The steady state solution of stable systems is due to simple poles in the jOhm a

Q: The steady state solution of stable systems is due to simple poles in the jOhm a

final value theorem states that sF(s) = f(t) s->inf t->0 H(s) = 1/([(s+1)^2]+4)) = 1/( s^2 + 2s + 5 ) x(t) = tu(t) X(s) = 1/s^2 Y(s) = H(s) * X(s) = 1/( s^2 * ( s^2 + 2s + 5 ) ) lim(t approaches infinity)y(t) = lim(t approaches zero)sY(s) = lim(t approaches zero) 1/( s^2 * ( s^2 + 2s + 5 ) ) = lim(t approaches zero) 1/( s^4 + 2s^3 + 5^2 ) ) = 8 as the limit is tending towards infinity for t--->8 , the system is BIBO unstable.

ID: 1923529 • Letter: T

Question

The steady state solution of stable systems is due to simple poles in the jOhm axis of the s-plane coming from the input. Suppose the transfer function of the system is:
H_ (s)= (Y(s))/(X(s)) = 1/([(s+1)^2]+4))
To explain the behavior in the case above, consider the following: Is the input x (t) = tu (t) bounded? That is, is there some finite value M such that |x (t)| < M for all times? So what would you expect the output to be knowing that the system is stable?

Explanation / Answer

final value theorem states that sF(s) = f(t) s->inf t->0 H(s) = 1/([(s+1)^2]+4)) = 1/( s^2 + 2s + 5 ) x(t) = tu(t) X(s) = 1/s^2 Y(s) = H(s) * X(s) = 1/( s^2 * ( s^2 + 2s + 5 ) ) lim(t approaches infinity)y(t) = lim(t approaches zero)sY(s) = lim(t approaches zero) 1/( s^2 * ( s^2 + 2s + 5 ) ) = lim(t approaches zero) 1/( s^4 + 2s^3 + 5^2 ) ) = 8 as the limit is tending towards infinity for t--->8 , the system is BIBO unstable.

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The steady state solution of stable systems is due to simple poles in the jOhm a

The steady-state concentration of a substance in a bucket (see diagram below) is

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The steady state solution of stable systems is due to simple poles in the jOhm a

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