Initial Value Theorem MCQ Quiz - Objective Question with Answer for Initial Value Theorem - Download Free PDF

Concept:

The initial value theorem gives:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

Using the initial value theorem,

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(2\mathrm{s}+3)}{(\mathrm{s}+1)(\mathrm{s}+3)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}{\mathrm{s}}^2.\frac{(2+\frac{3}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{1}{\mathrm{s}})(1+\frac{3}{\mathrm{s}})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}.\frac{(2+\frac{3}{\mathrm{s}})}{(1+\frac{1}{\mathrm{s}})(1+\frac{3}{\mathrm{s}})}}$

∴ x(0+) = 2

Concept:

Final value theorem:

A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression

Final value theorem states that the final value of a system can be calculated by

$x\left(\mathrm{\infty }\right)=\underset{s\to 0}{lim}sX\left(s\right)$

 Where X(s) is the Laplace transform of the function.

For the final value theorem to be applicable, th system should be stable in steady-state and for that , the real part of the poles should lie in the left side of s plane.

Initial value theorem:

$x\left(0\right)=\underset{t\to 0}{lim}x\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sX\left(s\right)$

It is applicable only when the number of poles of X(s) is more than the number of zeros of X(s).

Concept:

The initial value theorem gives:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

Using the initial value theorem,

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(2\mathrm{s}+3)}{(\mathrm{s}+1)(\mathrm{s}+3)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}{\mathrm{s}}^2.\frac{(2+\frac{3}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{1}{\mathrm{s}})(1+\frac{3}{\mathrm{s}})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}.\frac{(2+\frac{3}{\mathrm{s}})}{(1+\frac{1}{\mathrm{s}})(1+\frac{3}{\mathrm{s}})}}$

∴ x(0+) = 2

The correct answer is option 2): (3)

Concept:

Using the Initial value theorem

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(3\mathrm{s}+5)}{({\mathrm{s}}^2+10\mathrm{s}+21)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{{\mathrm{s}}^2(3+\frac{5}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{3+\frac{5}{\mathrm{S}}}{1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2}}}$

x(0+) = 3

Concept:

Initial Value Theorem:

$\underset{\mathrm{t}\to 0}{lim}\mathrm{f}(\mathrm{t})=\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})$

where; f(t) is any function in the time domain

F(s) is Laplace Transform of f(t)

Calculation:

 F(s) = s × $\frac{2(\mathrm{s}+1)}{{\mathrm{s}}^2+2\mathrm{s}+5}$

  =  s × 

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\times \frac{2\mathrm{s}(1+\frac{1}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{2}{\mathrm{s}}+\frac{5}{{\mathrm{s}}^2})}$

= 

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\frac{2{\mathrm{s}}^2(1+\frac{1}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{2}{\mathrm{s}}+\frac{5}{{\mathrm{s}}^2})}$

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\frac{2(1+\frac{1}{\mathrm{\infty }})}{(1+\frac{2}{\mathrm{\infty }}+\frac{5}{\mathrm{\infty }})}$

Since; $\frac{1}{\mathrm{\infty }}$ = 0

= 2

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\frac{2(1+0)}{(1+0+0)}$

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=2$

Additional Information Final Value Theorem:

$\underset{\mathrm{t}\to \mathrm{\infty }}{lim}\mathrm{f}(\mathrm{t})=\underset{\mathrm{s}\to 0}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})$

Conditions where initial and final value theorem fails:

1. The initial value theorem does not hold good for impulse signals.
2. The final value theorem is not applicable for an unstable system, i.e. when poles lie on the right-hand side of the jω axis or imaginary axis.

Concept:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

Using the initial value theorem,

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(3\mathrm{s}+5)}{({\mathrm{s}}^2+10\mathrm{s}+21)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{{\mathrm{s}}^2(3+\frac{5}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{3+\frac{5}{\mathrm{S}}}{1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2}}}$

∴ x(0+) = 3

Alternate method:

using Inverse Laplace Transform method,

we have,

$\mathrm{X}(\mathrm{s})=\frac{3\mathrm{S}+5}{{\mathrm{S}}^2+10\mathrm{S}+21}=\frac{3\mathrm{S}+5}{{\mathrm{S}}^2+10\mathrm{S}+21+4-4}$

$\mathrm{X}(\mathrm{s})=\frac{3\mathrm{S}+5}{(\mathrm{S}+5{)}^2-2^2}=\frac{3\mathrm{S}+15-10}{(\mathrm{S}+5{)}^2-2^2}$

$\mathrm{X}(\mathrm{S})=\frac{3(\mathrm{S}+5)}{(\mathrm{S}+5{)}^2-2^2}-\frac{10}{(\mathrm{S}+5{)}^2-2^2}$

Taking inverse Laplace transform

x(t) = 3e-5t cosh2t - 5e-5t sinh2t

x(t) = e-5t(3 cosh2t - 5 sinh2t)

At t = 0+,

x(0+) = e0(3cosh 0 - 5 sinh0)

x(0+) = 1(3 - 0)

x(0+) = 3

Concept:

By initial value theorem,

Initial value of f(t) is given by,

$\underset{t\to 0}{lim}f\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sF\left(s\right)$      ---(1)

Final value of f(t) is given by,

$\underset{t\to \mathrm{\infty }}{lim}f\left(t\right)=\underset{s\to 0}{lim}sF\left(s\right)$      ---(2)

Calculation:

${\displaystyle \underset{t\to o}{lim}f(t)={\displaystyle \underset{(s\to \mathrm{\infty })}{lim}\frac{s(2s+1)}{s^2+4s+7}=2}}$

${\displaystyle \underset{t\to \mathrm{\infty }}{lim}f(t)={\displaystyle \underset{(s\to 0)}{lim}\frac{s(2s+1)}{s^2+4s+7}=0}}$

Concept:

Initial value: Value of the system at t = 0

It can be found as,

Initial value $=\underset{t\to 0}{lim}F(t)=\underset{s\to \mathrm{\infty }}{lim}sF(s)$

Calculation:

Initial value $=\underset{s\to \mathrm{\infty }}{lim}sY(s)$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{s(2s+1)}{(s+1+j)(s+1-j)}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{2s^2+s}{(s+1{)}^2+1}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{s^2(2+\frac{1}{s})}{s^2+2s+2}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{(2+\frac{1}{s})}{(1+\frac{2}{s}+\frac{2}{s^2})}=2$

The correct answer is option 2): (3)

Concept:

Using the Initial value theorem

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(3\mathrm{s}+5)}{({\mathrm{s}}^2+10\mathrm{s}+21)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{{\mathrm{s}}^2(3+\frac{5}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{3+\frac{5}{\mathrm{S}}}{1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2}}}$

x(0+) = 3

Concept:

Initial Value Theorem:

$\underset{\mathrm{t}\to 0}{lim}\mathrm{f}(\mathrm{t})=\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})$

where; f(t) is any function in the time domain

F(s) is Laplace Transform of f(t)

Calculation:

 F(s) = s × $\frac{2(\mathrm{s}+1)}{{\mathrm{s}}^2+2\mathrm{s}+5}$

  =  s × 

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\times \frac{2\mathrm{s}(1+\frac{1}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{2}{\mathrm{s}}+\frac{5}{{\mathrm{s}}^2})}$

= 

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\frac{2{\mathrm{s}}^2(1+\frac{1}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{2}{\mathrm{s}}+\frac{5}{{\mathrm{s}}^2})}$

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\frac{2(1+\frac{1}{\mathrm{\infty }})}{(1+\frac{2}{\mathrm{\infty }}+\frac{5}{\mathrm{\infty }})}$

Since; $\frac{1}{\mathrm{\infty }}$ = 0

= 2

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=\frac{2(1+0)}{(1+0+0)}$

$\underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})=2$

Additional Information Final Value Theorem:

$\underset{\mathrm{t}\to \mathrm{\infty }}{lim}\mathrm{f}(\mathrm{t})=\underset{\mathrm{s}\to 0}{lim}\mathrm{s}\mathrm{F}(\mathrm{s})$

Conditions where initial and final value theorem fails:

1. The initial value theorem does not hold good for impulse signals.
2. The final value theorem is not applicable for an unstable system, i.e. when poles lie on the right-hand side of the jω axis or imaginary axis.

Initial value theorem:

The initial value of a function with a given Laplace transform can be obtained using the Initial value theorem as:

$c\left(0\right)=\underset{t\to 0}{lim}c\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sC\left(s\right)$

It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

Application:

$F(s)=\frac{5}{s(s^2+3s+2)}$

Since the number of poles is greater than the number of zeros, we can apply the initial value theorem as:

$f\left(0\right)=\underset{t\to 0}{lim}f\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sF\left(s\right)$

$\underset{t\to 0}{lim}f\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}s.\frac{5}{s(s^2+3s+2)}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{5}{s^2+3s+2}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{5}{s^2(1+\frac{3}{s}+\frac{2}{s^2})}$

f(0) = 0

Final value theorem:

​The final value theorem states that the final value of a system can be calculated by

$f\left(\mathrm{\infty }\right)=\underset{s\to 0}{lim}sF\left(s\right)$

F(s) is the Laplace transform of the function.

For the final value theorem to be applicable, the system should be stable in steady-state and for that real part of poles should lie on the left side of s plane.

Initial value theorem:

The initial value of a function with a given Laplace transform can be obtained using Initial value theorem as:

$c\left(0\right)=\underset{t\to 0}{lim}c\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sC\left(s\right)$

It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

Application:

$X\left(s\right)=\frac{5s+10}{s\left(s+4\right)}$

Since the number of poles is greater than the number of zeros, we can apply the initial value theorem as:

$x\left(0\right)=\underset{t\to 0}{lim}x\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sX\left(s\right)$

$\underset{t\to 0}{lim}x\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}s.\frac{5s+10}{s\left(s+4\right)}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{5s+10}{\left(s+4\right)}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{5+10/s}{\left(1+4/s\right)}$

x(0) = 5

Final value theorem:

​Final value theorem states that the final value of a system can be calculated by

$f\left(\mathrm{\infty }\right)=\underset{s\to 0}{lim}sF\left(s\right)$

F(s) is the Laplace transform of the function.

For the final value theorem to be applicable, the system should be stable in steady-state and for that real part of poles should lie on the left side of s plane.

Concept:

Final value theorem:

A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression

Final value theorem states that the final value of a system can be calculated by

$x\left(\mathrm{\infty }\right)=\underset{s\to 0}{lim}sX\left(s\right)$

 Where X(s) is the Laplace transform of the function.

For the final value theorem to be applicable, th system should be stable in steady-state and for that , the real part of the poles should lie in the left side of s plane.

Initial value theorem:

$x\left(0\right)=\underset{t\to 0}{lim}x\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sX\left(s\right)$

It is applicable only when the number of poles of X(s) is more than the number of zeros of X(s).

Concept:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

Using the initial value theorem,

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(3\mathrm{s}+5)}{({\mathrm{s}}^2+10\mathrm{s}+21)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{{\mathrm{s}}^2(3+\frac{5}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{3+\frac{5}{\mathrm{S}}}{1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2}}}$

∴ x(0+) = 3

Alternate method:

using Inverse Laplace Transform method,

we have,

$\mathrm{X}(\mathrm{s})=\frac{3\mathrm{S}+5}{{\mathrm{S}}^2+10\mathrm{S}+21}=\frac{3\mathrm{S}+5}{{\mathrm{S}}^2+10\mathrm{S}+21+4-4}$

$\mathrm{X}(\mathrm{s})=\frac{3\mathrm{S}+5}{(\mathrm{S}+5{)}^2-2^2}=\frac{3\mathrm{S}+15-10}{(\mathrm{S}+5{)}^2-2^2}$

$\mathrm{X}(\mathrm{S})=\frac{3(\mathrm{S}+5)}{(\mathrm{S}+5{)}^2-2^2}-\frac{10}{(\mathrm{S}+5{)}^2-2^2}$

Taking inverse Laplace transform

x(t) = 3e-5t cosh2t - 5e-5t sinh2t

x(t) = e-5t(3 cosh2t - 5 sinh2t)

At t = 0+,

x(0+) = e0(3cosh 0 - 5 sinh0)

x(0+) = 1(3 - 0)

x(0+) = 3

Concept:

By initial value theorem,

Initial value of f(t) is given by,

$\underset{t\to 0}{lim}f\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sF\left(s\right)$      ---(1)

Final value of f(t) is given by,

$\underset{t\to \mathrm{\infty }}{lim}f\left(t\right)=\underset{s\to 0}{lim}sF\left(s\right)$      ---(2)

Calculation:

${\displaystyle \underset{t\to o}{lim}f(t)={\displaystyle \underset{(s\to \mathrm{\infty })}{lim}\frac{s(2s+1)}{s^2+4s+7}=2}}$

${\displaystyle \underset{t\to \mathrm{\infty }}{lim}f(t)={\displaystyle \underset{(s\to 0)}{lim}\frac{s(2s+1)}{s^2+4s+7}=0}}$