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What will be the nature of time response if the roots of the characteristic equation are located on the s-plane imaginary axis?
- a)Oscillations
- b)Damped oscillations
- c)No oscillations
- d)Under damped oscilaations
Correct answer is option 'C'. Can you explain this answer?
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What will be the nature of time response if the roots of the character...
The nature of the time response is determined by the location of the roots of the characteristic equation in the s-plane. When the roots are located on the imaginary axis of the s-plane, the response is characterized by no oscillations.
Explanation:
Characteristic Equation:
The characteristic equation of a linear time-invariant system is obtained by setting the denominator of the transfer function to zero. It is given by:
\(D(s) = 1 + G(s)H(s) = 0\)
where G(s) is the transfer function of the system and H(s) is the transfer function of the feedback.
Roots on the Imaginary Axis:
When the roots of the characteristic equation are located on the imaginary axis of the s-plane, it means that the poles of the transfer function are purely imaginary.
The general form of a pole on the imaginary axis is given by:
\(s = j\omega\)
where \(j\) is the imaginary unit and \(\omega\) is the frequency.
No Oscillations:
When the poles of the transfer function are purely imaginary, it implies that there are no exponentially growing or decaying terms in the time response. Therefore, the response does not exhibit oscillations.
This can be understood by considering the Laplace transform of a sinusoidal input signal. When the poles are on the imaginary axis, the Laplace transform of a sinusoid will not have any exponential terms that grow or decay with time. As a result, the output will not exhibit oscillatory behavior.
In summary, when the roots of the characteristic equation are located on the imaginary axis of the s-plane, the system response will not exhibit oscillations. This is because there are no exponentially growing or decaying terms in the response.
Explanation:
Characteristic Equation:
The characteristic equation of a linear time-invariant system is obtained by setting the denominator of the transfer function to zero. It is given by:
\(D(s) = 1 + G(s)H(s) = 0\)
where G(s) is the transfer function of the system and H(s) is the transfer function of the feedback.
Roots on the Imaginary Axis:
When the roots of the characteristic equation are located on the imaginary axis of the s-plane, it means that the poles of the transfer function are purely imaginary.
The general form of a pole on the imaginary axis is given by:
\(s = j\omega\)
where \(j\) is the imaginary unit and \(\omega\) is the frequency.
No Oscillations:
When the poles of the transfer function are purely imaginary, it implies that there are no exponentially growing or decaying terms in the time response. Therefore, the response does not exhibit oscillations.
This can be understood by considering the Laplace transform of a sinusoidal input signal. When the poles are on the imaginary axis, the Laplace transform of a sinusoid will not have any exponential terms that grow or decay with time. As a result, the output will not exhibit oscillatory behavior.
In summary, when the roots of the characteristic equation are located on the imaginary axis of the s-plane, the system response will not exhibit oscillations. This is because there are no exponentially growing or decaying terms in the response.
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Community Answer
What will be the nature of time response if the roots of the character...
omplex conjugate (non-multiple): oscillatory (sustained oscillations)
Complex conjugate (multiple): unstable (growing oscillations).
Complex conjugate (multiple): unstable (growing oscillations).
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What will be the nature of time response if the roots of the characteristic equation are located on the s-plane imaginary axis?a)Oscillationsb)Damped oscillationsc)No oscillationsd)Under damped oscilaationsCorrect answer is option 'C'. Can you explain this answer?
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