A casual system having the transfer function Hs - 1s-2 is excited with 10ut- The time at which the output reaches 99

H(s)=1s+2 r(t)= 10u(t) R(s)=10s C(s)=H(s).R(s)=1s+2.10s ⇒10s(s+2)=As+Bs+2 10=A(s+2)+Bs s=0,10=2A ⇒ A=5 s= 2, 10 = 2 B ⇒ B = 5 ∴ C(s) =5s 5s+2 c(t) =5[1 e^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Control Systems

Step Response of First Order System

A casual syst...

Question

A casual system having the transfer function $H (s) = \frac{1}{s + 2}$ is excited with $10 u (t)$ . The time at which the output reaches 99% of its steady state value is

2.7 s e c

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

2.5 s e c

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

2.3 s e c

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

2.1 s e c

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

Open in App

Solution

The correct option is C $2.3 s e c$
$H (s) = \frac{1}{s + 2}$
$r (t) = 10 u (t)$
$R (s) = \frac{10}{s}$
$C (s) = H (s) . R (s) = \frac{1}{s + 2} . \frac{10}{s}$
$\Rightarrow \frac{10}{s (s + 2)} = \frac{A}{s} + \frac{B}{s + 2}$
$10 = A (s + 2) + B s s = 0, 10 = 2 A$
$\Rightarrow A = 5$
$s = - 2, 10 = - 2 B$
$\Rightarrow B = - 5$
$∴ C (s) = \frac{5}{s} - \frac{5}{s + 2}$
$c (t) = 5 [1 - e^{- 2 t}] u (t)$
Steady state value when $t = \infty$ is 5 and steady state value is reached at
$5 [1 - e^{- 2 t}] = 0.99 \times 5$
$\Rightarrow 1 - e^{- 2 t} = 0.99$
$e^{- 2 t} = 0.1$
$\Rightarrow - 2 t = ln 0.1$
$\Rightarrow t = 2.3 s e c$

Suggest Corrections

Similar questions

Q. A system with transfer function:

G (s) = \frac{(s^{2} + 9) (s + 2)}{(9 s + 1) (s + 3) (s + 4)}

is excited

s i n (ω t)

. The steady-state output of the system is zero at

Q. A linear, time - invarient, casual continuous time system has a rational transfer function with simple poles at

s = - 2

and

s = - 4

, and one simple zero at

s = - 1

.A unit step

u (t)

is applied at the input of the system. At steady state, the output has constant value of

1

. The impulse response of this system is

Q. Consider a casual second-order system will be transfer function

G (s) = \frac{1}{1 + 2 s + s^{2}}

with a unit-step

R (s) = \frac{1}{s}

as an input. Let

C (s)

be the corresponding output. The time taken by the system output

c (t)

to reach

94 %

of its steady-state value

lim t \to \infty c (t)

, rounded off to two decimal places, is

Q. Consider an LTI system with transfer function,

H (s) = \frac{1}{s (s + 10)}

. If input to the system is cos 2t and the steady state output is

A s i n (2 t + a),

then the value of A is _______(Answer upto 2 decimal places)

Q. A system has a complex pole pair of (-1

\pm

j2) and a real zero of (-3). The steady state output to a unit step input is 2. The transfer function of the system is.

Join BYJU'S Learning Program

A casual system having the transfer function H(s)=1s+2 is excited with 10u(t). The time at which the output reaches 99% of its steady state value is

A casual system having the transfer function $H (s) = \frac{1}{s + 2}$ is excited with $10 u (t)$ . The time at which the output reaches 99% of its steady state value is