The asymptotic Bode plot for the gain magnitude of a minimum phase system G(s) is shown in the figure. The transfer function of the system G(s) is

\frac{100}{(1 + s / 10) (1 + s / 250)}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{40}{s (s + 250)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{100}{(s + 10) (s + 250)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{100 s}{(s + 10) (s + 250)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{100}{(1 + s / 10) (1 + s / 250)}$
The initial slope is 0 dB/decade, thus it is a type zero system
$∴ 20 l o g k = 40$
$l o g k = \frac{40}{20}$
$k = 10^{2} = 100$

$∴ T . F . = \frac{100}{(1 + s / 10) (1 + s / 250)}$

Suggest Corrections

Similar questions

The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure.

If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input is

Q. Bode plot of a stable system is shown in figure The transfer function of the system is

Q. The asymptotic Bode plot of the transfer function

K / [1 + (s / a)]

is given in figure. The error in phase angle and dB gain at a frequency of

ω = 0.5

are respectively.

Q. The asymptotic Bode magnitude plot of minimum phase transfer function G(s) is shown below.

Consider the following statements:
Statement-1: Transfer function G(s) has three poles and one zero.

Statement -2: At very high frequency

(ω \to \infty)

, the phase angle

∠ G (j ω) = - \frac{3 π}{2}

Which of the following option is correct ?

Q. The system having the Bode magnitude plot show in figure below has the transfer fucntion(s).

Join BYJU'S Learning Program

The asymptotic Bode plot for the gain magnitude of a minimum phase system G(s) is shown in the figure. The transfer function of the system G(s) is

The correct option is A 100(1+s/10)(1+s/250)The initial slope is 0 dB/decade, thus it is a type zero system ∴20 log k=40 logk=4020 k=102=100 ∴T.F.=100(1+s/10)(1+s/250)

The correct option is A $\frac{100}{(1 + s / 10) (1 + s / 250)}$
The initial slope is 0 dB/decade, thus it is a type zero system
$∴ 20 l o g k = 40$
$l o g k = \frac{40}{20}$
$k = 10^{2} = 100$

$∴ T . F . = \frac{100}{(1 + s / 10) (1 + s / 250)}$