Initial Value Theorem MCQ Quiz in తెలుగు - Objective Question with Answer for Initial Value Theorem - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Apr 4, 2025

పొందండి Initial Value Theorem సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Initial Value Theorem MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest Initial Value Theorem MCQ Objective Questions

Top Initial Value Theorem MCQ Objective Questions

Initial Value Theorem Question 1:

If dydt(t)+2y(t)=3u(t),y(0)=3, then the Forced and natural response of the system are

  1. 32e2t,3(1e2t)
  2. 32(1e2t),3e2t
  3. 3(1 – e-2t), 2e-2t
  4. 32(1+et),e2t

Answer (Detailed Solution Below)

Option 2 : 32(1e2t),3e2t

Initial Value Theorem Question 1 Detailed Solution

Taking Laplace transform, we have

Sy(s) – y(0-) + 2y(s) = 3U(s)

Sy(s) – 3 + 2y(s) = 3U(s)

(s + 2) y(s) = 3U(s) + 3

y(s)={3s+2U(s)}+3(s+2)3s+2Responseduetointialvalue  

3s+2U(s)Responseduetoinput

Now taking Inverse laplace

y(t)=32(1e2t)+3e2t

Initial Value Theorem Question 2:

The poles of a causal system (zeroes not shown) are shown below:

Gate EC Signals Jitendra Ch.4 13

It is also known that output is zero when input is e3t and the impulse response of the system at t=0+ is 8. Then the dc gain of the system is _______.

Answer (Detailed Solution Below) 3

Initial Value Theorem Question 2 Detailed Solution

We have transfer function, H(s)=z(s)(s+2)(s4)

where z(s) is polynomial in s whose roots determine the zeroes of the system. It is given that y(t) = 0 when x(t) = e3t  and at frequency s=3

⇒ y (t) = H(3)⋅ e3t

⇒ z(s) = 0 at s = 3

⇒ z(s) has a root at s = 3

⇒ (s-3) is a zero

H(s)=(s3)w(s)(s+2)(s4)

Now, sH(s)=s(53)w(s)(s+2)(s4)

Now lim s→∞ will be non-zero finite only when the degree of numerator and degree of denominator of sH(s) is same.

⇒ w(s) is a constant

Let the constant be K, then using the initial value theorem, we have

limt0+x(t)=x(0+)=limss(s3)K(s+2)(s4)=8

⇒ K = 8

Thus, H(s)=8(s3)(s+2)(s4)

Writing in standard form we have

H(s)=8.3(1s3)2.4(1+s2)(1s4)H(s)=3(1s3)(1+s2)(1s4)

Thus dc gain = 3

Initial Value Theorem Question 3:

If F(s) = L[f(t)] = K(s+1)(s2+4)then limtf(t)is given by

  1. K/4

  2. Zero

  3. 0

  4. 5

Answer (Detailed Solution Below)

Option 2 :

Zero

Initial Value Theorem Question 3 Detailed Solution

By final value theorem we have, limx0SF(s)=lims0SK(s+1)(s2+4)=0

Initial Value Theorem Question 4:

If F(s) =2(s+1)s2+2s+5then f(0) and f(∞) are

  1. 0, 2

  2. 0, 1

  3. 2, 0

  4. 2/5, 0

Answer (Detailed Solution Below)

Option 3 :

2, 0

Initial Value Theorem Question 4 Detailed Solution

f(0+)=limssF(s)=2

f()=lims0sF(s)=0

Initial Value Theorem Question 5:

Initial value of signal x(t) for which Laplace transform X(s) is 1s2+7s4, assuming x(t) to be a causal signal is

  1. -1/4

  2. 1

  3. 0

  4. Does Not exist

Answer (Detailed Solution Below)

Option 3 :

0

Initial Value Theorem Question 5 Detailed Solution

We have from intial value theorem,

lim t0x(t)=lim ssX(s)

lim ss1s2+7s4=lim s1s1+7s4s2=0

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