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Problem 2: (25 pts) Suppose the open-loop unstable plant, $qquad P(s) = frac{s+1}{s^2-9}$ and the controller, $qquad C(s) = k(s+2)$ are structured in a unity feedback control scheme. (a) Find the closed-loop transfer function from $r$ to $y$ as a ratio of polynomials in $s$: $qquad G(s) = frac{Y(s)}{R(s)} = frac{b_ms^m + b_{m-1}s^{m-1} + dots + b_1s + b_0}{a_ns^n + a_{n-1}s^{n-1} + dots + a_1s + a_0}$ (b) Find the range of $k$ such that the closed-loop system $G(s)$ is stable. (c) Regarding the steady-state error for tracking, what is the System TYPE? (d) Assume $k = 5$. Can this system track a step input perfectly? If so, prove it mathematically. If not, determine the steady-state error $e(infty)$ when the reference $r(t)$ is a unit step function. (e) Suppose the above controller is replaced by a controller of the form: $qquad C_1(s) = frac{k}{s}$ Determine whether a controller of this form can produce a closed-loop system capable of track- ing a step reference with zero steady-state-error $e(infty)_{step} = 0.$

          Problem 2: (25 pts)
Suppose the open-loop unstable plant,
$qquad P(s) = frac{s+1}{s^2-9}$
and the controller,
$qquad C(s) = k(s+2)$
are structured in a unity feedback control scheme.
(a) Find the closed-loop transfer function from $r$ to $y$ as a ratio of polynomials in $s$:
$qquad G(s) = frac{Y(s)}{R(s)} = frac{b_ms^m + b_{m-1}s^{m-1} + dots + b_1s + b_0}{a_ns^n + a_{n-1}s^{n-1} + dots + a_1s + a_0}$
(b) Find the range of $k$ such that the closed-loop system $G(s)$ is stable.
(c) Regarding the steady-state error for tracking, what is the System TYPE?
(d) Assume $k = 5$. Can this system track a step input perfectly? If so, prove it mathematically. If
not, determine the steady-state error $e(infty)$ when the reference $r(t)$ is a unit step function.
(e) Suppose the above controller is replaced by a controller of the form:
$qquad C_1(s) = frac{k}{s}$
Determine whether a controller of this form can produce a closed-loop system capable of track-
ing a step reference with zero steady-state-error $e(infty)_{step} = 0.$
        
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Problem 2: (25 pts)
Suppose the open-loop unstable plant,
qquad P(s) = fracs+1s^2-9
and the controller,
qquad C(s) = k(s+2)
are structured in a unity feedback control scheme.
(a) Find the closed-loop transfer function from r to y as a ratio of polynomials in s:
qquad G(s) = fracY(s)R(s) = fracbms^m + bm-1s^m-1 + dots + b1s + b0ans^n + an-1s^n-1 + dots + a1s + a0
(b) Find the range of k such that the closed-loop system G(s) is stable.
(c) Regarding the steady-state error for tracking, what is the System TYPE?
(d) Assume k = 5. Can this system track a step input perfectly? If so, prove it mathematically. If
not, determine the steady-state error e(infty) when the reference r(t) is a unit step function.
(e) Suppose the above controller is replaced by a controller of the form:
qquad C1(s) = fracks
Determine whether a controller of this form can produce a closed-loop system capable of track-
ing a step reference with zero steady-state-error e(infty)step = 0.

Added by Julia W.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Problem 2: (25 pts) Suppose the open-loop unstable plant, P(s) = (s+1)/(s^2-9), and the controller C(s) = k(s + 2) are structured in a unity feedback control scheme. (a) Find the closed-loop transfer function from r to y as a ratio of polynomials in s: G(s) = Y(s)/R(s) = (b_ms^m + b_{m-1}s^{m-1} + ... + b_1s + b_0) / (a_ns^n + a_{n-1}s^{n-1} + ... + a_1s + a_0) (b) Find the range of k such that the closed-loop system G(s) is stable. (c) Regarding the steady-state error for tracking, what is the System TYPE? (d) Assume k = 5. Can this system track a step input perfectly? If so, prove it mathematically. If not, determine the steady-state error e(infinity) when the reference r(t) is a unit step function. (e) Suppose the above controller is replaced by a controller of the form: C1(s) = k/s Determine whether a controller of this form can produce a closed-loop system capable of tracking a step reference with zero steady-state-error e(infinity)_step = 0.
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Transcript

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00:01 Characteristic equation is given as s squared multiplied by k plus 1 plus 3ks plus k minus 9.
00:12 We can rewrite this as s squared multiplied by k plus 1 plus s multiplied by 3k plus s to the power 0 multiplied by k minus 9.
00:27 Right we have this now s to the power zero is one actually so this is the same thing so now from here we have s square s and s to the power zero so now we are going to write the coefficient of all this so for s square you have k plus one for s you have 3k and for s to the power zero you have k minus 9 now see for the stability all the numbers in this column should be non -negative so therefore you'll have that your k plus one would be greater than or equals to zero so therefore from here you'll get that your k is greater than or equals to minus one from this over here you will have that your three k should be greater than or equals to zero so from here you'll have your k should be greater than or equals to zero and from the last one you'll have that your k minus nine should be greater than or equals to zero so therefore you'll have your k should be greater than or equals to 9 over here...
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