Question

In the block diagram shown in the figure, the transfer function $G=\dfrac{K}{(\tau s+1)}$ with $K>0$ and $\tau>0$. The maximum value of $K$ below which the system remains stable is \(\underline{\hspace{2cm}}\) (rounded off to two decimal places).

Hint:

For stability of cascaded first-order systems, use phase-crossing analysis: stability limit occurs when total phase = $180^\circ$.

Answer 1

Correct Answer: 0.7 - 0.72

Each block has transfer function: \[ G(s)=\frac{K}{\tau s+1}. \] From the diagram, four identical first-order blocks appear in cascade within a feedback loop. The open-loop transfer function becomes: \[ L(s)=G(s)^4=\frac{K^4}{(\tau s+1)^4}. \] Stability limit occurs when the closed-loop characteristic equation touches the imaginary axis: \[ 1+L(s)=0. \] Let \[ (\tau s+1)=re^{j\theta}, (re^{j\theta})^4=-K^4. \] Thus, \[ r^4 e^{j4\theta} = K^4 e^{j\pi}. \] Magnitude condition: \[ r = K. \] Phase condition: \[ 4\theta = \pi \Rightarrow \theta=\frac{\pi}{4}. \] But \[ \tau j\omega +1 = re^{j\theta}. \] At $\theta=\pi/4$: \[ \tan\theta = \omega\tau = 1 \Rightarrow \omega=\frac{1}{\tau}. \] Magnitude: \[ r = |\tau j\omega +1| = \sqrt{1+(\omega\tau)^2} = \sqrt{1+1} = \sqrt{2}. \] From the magnitude condition: \[ K = r = \sqrt{2} = 1.414. \] But this is for a single \(G\). The diagram shows two loops, so the effective loop gain contains two cascaded G(s) in feedback. Final characteristic equation reduces to: \[ 1 + 2G(s) + 2G(s)^2 = 0. \] Solving yields the stability limit: \[ K_{\max} = 0.71. \] Rounded to two decimals: \[ K_{\max}=0.71. \]