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The state space representation of a first-order system is given as
x = -x + u
y = x
where, x is the state variable, u is the control input and y is the controlled output. Let u = -Kx be the control law, where K is the controller gain. To place a closed-loop pole at -2 , the value of K i s _______ .
Correct answer is '1'. Can you explain this answer?
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The state space representation of a first-order system is given asx = ...
Characteristic equation,
|sI + KI + I| = 0
|(S + 1 + K)I| = 0
S + 1 + K = 0
S = -1 - K
-2 = -1 - K
K = 1
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The state space representation of a first-order system is given asx = ...
State Space Representation:
The given first-order system can be represented in state space form as follows:
\[\dot{x} = -x\]
\[y = x\]
where \(\dot{x}\) represents the derivative of \(x\) with respect to time, and \(y\) is the controlled output.
Control Law:
The control law is given as \(u = -Kx\), where \(K\) is the controller gain. Substituting this control law into the state equation, we get:
\[\dot{x} = -x\]
\[u = -Kx\]
Closed-Loop Pole:
To place a closed-loop pole at -2, we need to choose a value of \(K\) that satisfies this condition. The closed-loop transfer function for the system is given by:
\[G(s) = \frac{Y(s)}{U(s)} = \frac{1}{s + 1}\]
where \(s\) is the Laplace variable.
The pole of the closed-loop transfer function is the value of \(s\) that makes the denominator zero. Setting \(s + 1 = 0\), we find that the pole is at \(s = -1\).
Pole-Placement:
To place a closed-loop pole at -2, we need to choose a value of \(K\) such that the characteristic equation of the closed-loop system becomes \(s + 2 = 0\).
The characteristic equation is obtained by substituting the control law into the state equation and equating the resulting characteristic polynomial to zero:
\[\det(sI - A + BK) = 0\]
where \(A\) is the system matrix and \(B\) is the input matrix.
For the given first-order system, the system matrix \(A\) is -1 and the input matrix \(B\) is 1. Substituting these values into the characteristic equation, we get:
\[\det(s - (-1) + 1(-K)) = 0\]
\[\det(s + 1 + K) = 0\]
To place a pole at -2, we set \(s + 1 + K = 0\) and solve for \(K\):
\[-2 + 1 + K = 0\]
\[K = 1\]
Conclusion:
Therefore, to place a closed-loop pole at -2, the value of \(K\) should be 1.
The given first-order system can be represented in state space form as follows:
\[\dot{x} = -x\]
\[y = x\]
where \(\dot{x}\) represents the derivative of \(x\) with respect to time, and \(y\) is the controlled output.
Control Law:
The control law is given as \(u = -Kx\), where \(K\) is the controller gain. Substituting this control law into the state equation, we get:
\[\dot{x} = -x\]
\[u = -Kx\]
Closed-Loop Pole:
To place a closed-loop pole at -2, we need to choose a value of \(K\) that satisfies this condition. The closed-loop transfer function for the system is given by:
\[G(s) = \frac{Y(s)}{U(s)} = \frac{1}{s + 1}\]
where \(s\) is the Laplace variable.
The pole of the closed-loop transfer function is the value of \(s\) that makes the denominator zero. Setting \(s + 1 = 0\), we find that the pole is at \(s = -1\).
Pole-Placement:
To place a closed-loop pole at -2, we need to choose a value of \(K\) such that the characteristic equation of the closed-loop system becomes \(s + 2 = 0\).
The characteristic equation is obtained by substituting the control law into the state equation and equating the resulting characteristic polynomial to zero:
\[\det(sI - A + BK) = 0\]
where \(A\) is the system matrix and \(B\) is the input matrix.
For the given first-order system, the system matrix \(A\) is -1 and the input matrix \(B\) is 1. Substituting these values into the characteristic equation, we get:
\[\det(s - (-1) + 1(-K)) = 0\]
\[\det(s + 1 + K) = 0\]
To place a pole at -2, we set \(s + 1 + K = 0\) and solve for \(K\):
\[-2 + 1 + K = 0\]
\[K = 1\]
Conclusion:
Therefore, to place a closed-loop pole at -2, the value of \(K\) should be 1.
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