Question

Sensitivity of the following open loop system is:

• A. 1
• B. 2/3
• C. 0.5
• D. 3/2

Hint:

An fundamental law of control engineering states that the sensitivity of any open-loop system with respect to variations in its forward path parameter is always equal to unity (\(1\)). This happens because there is no feedback loop present to compensate or damp down any parameter deviations.

Answer 1

The Correct Option is A

Concept: The sensitivity of a system's overall transfer function \(T(s)\) with respect to variations in a parameter \(P\) is defined mathematically as the ratio of the percentage change in \(T(s)\) to the percentage change in \(P\): \[ S^{T}_{P} = \frac{\partial T(s) / T(s)}{\partial P / P} = \frac{\partial T(s)}{\partial P} \cdot \frac{P}{T(s)} \] For any standard open-loop configuration where the total forward transfer function is simply \(T(s) = G(s)\), any fractional change in the system components reflects entirely and directly into the overall transmission. Let us evaluate this parameter-wise for confirmation.

Step 1: Determine the overall transmission function of the system. The given system is an open-loop block with a forward path gain function: \[ T(s) = G(s) = \frac{K}{1 + 0.5s} \]

Step 2: Calculate sensitivity with respect to the parameter matrix variable \(K\). Using the definition of sensitivity \(S^{T}_{K}\): \[ S^{T}_{K} = \frac{\partial T(s)}{\partial K} \cdot \frac{K}{T(s)} \] Differentiating \(T(s)\) with respect to \(K\): \[ \frac{\partial T(s)}{\partial K} = \frac{\partial}{\partial K}\left[\frac{K}{1 + 0.5s}\right] = \frac{1}{1 + 0.5s} \] Now substitute this back into the sensitivity expression: \[ S^{T}_{K} = \left(\frac{1}{1 + 0.5s}\right) \cdot \frac{K}{\left(\frac{K}{1 + 0.5s}\right)} \] Canceling out identical terms in the numerator and denominator: \[ S^{T}_{K} = \left(\frac{1}{1 + 0.5s}\right) \cdot \left(\frac{1 + 0.5s}{1}\right) = 1 \] Hence, the sensitivity of this open loop system is exactly 1, confirming option (A).