Given the differential equation model of a physical system, determine the time constant of the system:

\(40 \frac{dx}{dt}+2x=f(t)\)

Option 2 : 20

**Concept: **

Time constant \(\tau = \frac{{ - 1}}{{{\rm{real\ part\ of\ Dominant\ pole}}}}\)

**Calculation:**

\(40\frac{{dx}}{{dt}} + 2x = f\left( t \right)\)

Taking Laplace transform, we get

40 s X(s) + 2X(s) = 12(s)

\(\frac{{X\left( s \right)}}{{F\left( s \right)}} = \frac{1}{{40s + 2}}\)

\( = \frac{1}{{40\left( {s + \frac{1}{{20}}} \right)}}\)

Pole will be at -1/20.

Time constant \( = \frac{1}{{pole}} = 20\)

A second order underdamped system is cascaded with a first order system as shown below.

Based on dominant pole concept match the nature of third order system (**List-2**) with the nature of poles of both systems (**List-1**)

**List-1**

A. Poles of second order system are dominant poles

B. Both the system poles are dominant poles

C. Pole of 1^{st} order system is dominant pole

**List-2**

1. Over damped

2. Critically damped

3. Underdamped

Option 2 : A – 3, B – 2, C – 1

**Dominant pole concept: **The pole which is closer to the imaginary axis is called as the dominant pole as this pole contributes more in the response of the system.

**Cascading of a second order underdamped system with a first order system:**

**Case 1:** When the poles of the second order system are the dominant poles

- In this case the response of the third order system
**will remain underdamped** - In this case the percentage overshoot of the third order system will be less than the percentage overshoot of the pure second order system.
- Rise time of the third order system will be greater than the rise time of the pure second order system.
- Settling time of the third order system will be less than the settling time of the pure second order system.

**Case 2:** When both the system poles are the dominant poles

In this case the output response of the third order system will be same as the output response of the pure second order **critically damped system**

**Case 3:** When the pole of the first order system is the dominant pole

A plant has an open-loop transfer function

\({G_P}\left( s \right) = \frac{{20}}{{\left( {s + 0.1} \right)\left( {s + 2} \right)\left( {s + 100} \right)}}\)

The approximate model obtained by retaining only one of the above poles, which is closest to the frequency response of the original transfer function at low frequency isOption 1 : \(\frac{{0.1}}{{s + 0.1}}\)

According to dominant pole concept, we can neglect the poles which are far from origin but the DC gain should be constant.

Approximate transfer function \(= \frac{{20}}{{\left( {s + 0.1} \right)\left( 2 \right)\left( {100} \right)}} = \frac{{0.1}}{{\left( {s + 0.1} \right)}}\)Given the differential equation model of a physical system, determine the time constant of the system:

\(40 \frac{dx}{dt}+2x=f(t)\)

Option 2 : 20

**Concept: **

Time constant \(\tau = \frac{{ - 1}}{{{\rm{real\ part\ of\ Dominant\ pole}}}}\)

**Calculation:**

\(40\frac{{dx}}{{dt}} + 2x = f\left( t \right)\)

Taking Laplace transform, we get

40 s X(s) + 2X(s) = 12(s)

\(\frac{{X\left( s \right)}}{{F\left( s \right)}} = \frac{1}{{40s + 2}}\)

\( = \frac{1}{{40\left( {s + \frac{1}{{20}}} \right)}}\)

Pole will be at -1/20.

Time constant \( = \frac{1}{{pole}} = 20\)