Final Answer:
The transfer function is stable for 0 < k < 3. This range ensures the system's stability based on the Routh stability test.
Step-by-step explanation:
The stability of a system with a given transfer function can be determined using the Routh-Hurwitz stability criterion. For the given transfer function, let's denote the characteristic equation as
. The coefficients of the characteristic equation are [1, -3k, -2k]. Now, we construct the Routh array:
![\[ \begin{array}{cccc} s\textsuperscript{2} & 1 & -2k \\ s\textsuperscript{1} & -3k & 0 \\ s\textsuperscript{0} & -2k & 0 \end{array} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/h1cakgcoy5lrfdi5lh4z8cfs9gq6ykx89y.png)
For the system to be stable, all the elements in the first column of the Routh array must be positive. In this case, for stability, we require -3k > 0 and -2k > 0, leading to k < 0. However, since k is squared in the second row of the characteristic equation, we need
, which implies
. Combining these conditions, we get
for the system to be stable.
In conclusion, the Routh-Hurwitz stability test determines that the system is stable for values of k within the range 0 < k < 3. This means that the stability of the system is guaranteed when the proportional gain k falls within this specified interval. Beyond this range, the system may become unstable, leading to undesirable behaviors such as oscillations or divergence in response to input signals.