Stability Analysis from Bode Plot

Q. The asymptotic Bode plot of the transfer function
$K/[1+(s/a\left)\right]$ is given in figure. The error in phase angle and dB gain at a frequency of $\omega =0.5$ are respectively.

1. ${4.9}^{\circ },0.97dB$
2. ${5.7}^{\circ },3dB$
3. ${4.9}^{\circ },3dB$
4. ${5.7}^{\circ },0.97dB$

Q.

The figure below shows the bode magnitude and phase plots of a stable transfer function
$G\left(s\right)=\frac{n_0}{s^3+d_2{s}^2+d_{1}s+d_0}$

Consider the negative unity feedback configuration with gain $k$ in the feedforward path. The closed loop in stable for $k<k_0$. The maximum value of $k_0$ is

1. 0.1

Q. The asymptotic magnitude Bode plot of an open loop system G(s) with K > 0 and all poles and zeros on the left hand side of the s-plane is shown in the figure. It is completetly sysmmetric about ${\omega }_c$. The minimum absolute angle contribution by G(s) is given by

1. ${78.6}^o$
2. ${90}^o$
3. ${101.4}^o$
4. ${180}^o$

Q. The asymptotic Bode magnitude plot of lead network with its pole and zero on the left half of the s-plane is shown in the adjoining figure. The frequency at which the phase angle of the network in maximum (in rad/s) is

1. $\frac{3}{\sqrt{10}}$
2. $\frac{1}{\sqrt{20}}$
3. $\frac{1}{20}$
4. $\frac{1}{30}$

Q. Bode plot of a plant transfer function with unity negative feedback is shown below


If a lead compensator is added, then system becomes.

1. Effect of lead compensator is similar to addition of zero. So, system becomes more stable.
2. less stable
3. marginal stable
4. unstable

Q. The approximate phase response of $\frac{{100}^2{e}^{-0.01s}}{s^2+0.2s+{100}^2}is$

1. 
2. 
3. 
4.