Exams › GATE › Technical

The phase of the above lead compensator is maximum at

1. √2 rad/s
2. √3 rad/s
3. √6 rad/s
4. 1/√3 rad/s

Correct answer: √2 rad/s

Solution

For a lead compensator, the phase is maximum at the frequency $\omega_m=\sqrt{ab}$, where $a$ and $b$ are the zero and pole parameters. Using the lead compensator from the previous question, the peak phase frequency evaluates to $\sqrt{2}$ rad/s.

Related GATE Technical questions

• If the closed-loop transfer function of a control system is given by \(T(s)=\dfrac{s-5}{(s+2)(s+3)}\), then it is
• A control system with a PD controller is shown in the figure. If the velocity error constant $K_v=1000$ and the damping ratio $\zeta=0.5$, then the values of $K_P$ and $K_D$ are
• The transfer function of a plant is $T(s)=\frac{5}{(s+5)(s^2+s+1)}$. The second-order approximation of $T(s)$ using the dominant pole concept is
• The open-loop transfer function of a plant is given by $G(s)=\frac{1}{s^2-1}$. If the plant is operated in a unity-feedback configuration, then the lead compensator that can stabilize this control system is
• The asymptotic Bode plot of a transfer function is shown in the figure. The transfer function $G(s)$ corresponding to the Bode plot is
• The state-space representation of a separately excited DC servo motor dynamics is given by \[ \begin{bmatrix} \dfrac{d\omega}{dt} \\ \dfrac{di_a}{dt} \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -1 & -10 \end{bmatrix} \begin{bmatrix} \omega \\ i_a \end{bmatrix} + \begin{bmatrix} 0 \\ 10 \end{bmatrix}u \] where \(\omega\) is the speed of the motor, \(i_a\) is the armature current, and \(u\) is the armature voltage. The transfer function \(\omega(s)/U(s)\) of the motor is

⚔️ Practice GATE Technical free + battle 1v1 →