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

Explanation:

Final Value Theorem in Laplace Transform

Definition: The Final Value Theorem (FVT) is a fundamental concept in control theory and signal processing, which provides a method to determine the steady-state value of a time-varying function as time approaches infinity. It is particularly useful in analyzing the long-term behavior of systems described by differential equations.

Formula: The Final Value Theorem states that for a given function $f(t)$, if the Laplace transform $F(s)$ of $f(t)$ exists, then the final value of $f(t)$ as $t$ approaches infinity can be found using the formula:

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

This theorem applies under the condition that all poles of $sF(s)$ (except possibly at $s=0$) are in the left half of the complex plane. This ensures that the function $f(t)$ approaches a steady-state value as $t$ approaches infinity.

Working Principle: The Final Value Theorem provides a direct way to find the steady-state value of a function by analyzing its Laplace transform. The theorem essentially relates the behavior of the function in the time domain to its behavior in the frequency domain, allowing for easier computation of the final value.

Example: Consider a function $f(t)$ with the Laplace transform $F(s)=\frac{5}{s(s+2)}$. To find the final value of $f(t)$, we use the Final Value Theorem:

$\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to 0}{lim}s\left(\frac{5}{s(s+2)}\right)=\underset{s\to 0}{lim}\frac{5}{s+2}=\frac{5}{2}=2.5$

Therefore, the final value of $f(t)$ as $t$ approaches infinity is 2.5.

Advantages:

• Simplicity in finding the steady-state value without needing to perform inverse Laplace transforms.
• Useful for analyzing the long-term behavior of systems described by differential equations.
• Provides a direct relationship between the time domain and frequency domain representations of a function.

Disadvantages:

• Applicable only if the function $f(t)$ approaches a steady-state value as $t$ approaches infinity.
• Requires the existence of the Laplace transform $F(s)$ of the function $f(t)$.
• Not applicable if the function has poles on the right half of the complex plane or on the imaginary axis (except at $s=0$).

Correct Option Analysis:

The correct option is:

Option 3: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to 0}{lim}sF(s)$

This option correctly represents the Final Value Theorem. The formula $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to 0}{lim}sF(s)$ provides a method to find the final value of a time-varying function $f(t)$ by analyzing its Laplace transform $F(s)$.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to \mathrm{\infty }}{lim}sF(s)$

This option is incorrect because it suggests taking the limit as $s$ approaches infinity, which does not provide the final value of the function $f(t)$ as $t$ approaches infinity. The correct limit should be as $s$ approaches zero.

Option 2: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{t\to 0}{lim}tf(t)$

This option is incorrect because it involves the time domain limit as $t$ approaches zero, which does not relate to finding the final value as $t$ approaches infinity. The Final Value Theorem specifically deals with the limit as $t$ approaches infinity.

Option 4: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{t\to 0}{lim}f(t)$

This option is incorrect because it again involves the time domain limit as $t$ approaches zero, which does not help in finding the final value as $t$ approaches infinity. The correct approach involves the Laplace transform and the limit as $s$ approaches zero.

Conclusion:

Understanding the Final Value Theorem is essential for analyzing the steady-state behavior of time-varying functions. The correct application of the theorem allows for the determination of the final value of a function by evaluating its Laplace transform. This method simplifies the process and provides a direct relationship between the time domain and frequency domain representations of the function.

Concept:

The initial value theorem is given by:

In time domain:

$Lt\stackrel{t=0}{\to }x(t)$

In the Laplace domain:

$Lt\stackrel{s=\infty }{\to }sx(s)$

The final value theorem is given by:

In time domain:

$Lt\stackrel{t=\infty }{\to }x(t)$

In the Laplace domain:

Calculation:

Final value theorem is applicable to the system with poles in left half of s-plane i.e., stable system.

A/C to final value theorem

In time domain,

$\mathrm{l}{\mathrm{t}}_{\mathrm{t}\to \infty }\mathrm{f}(\mathrm{t})=\mathrm{f}(\infty )$

$\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to \infty }\mathrm{s}.\mathrm{F}(\mathrm{s})=\mathrm{F}(\infty )$

Given signal is f(t) = e^-2t sin(5t) u(t)

A/C to laplace theorem,

$e^{at}\sin btu(t)\leftrightarrow \frac{b}{(s-a{)}^2+b^2}$

$e^{-2t}sin5tu(t)\leftrightarrow \frac{5}{(s+2{)}^2+5^2}$

$\mathrm{F}(\mathrm{s})=\frac{5}{(\mathrm{s}+2{)}^2+25}$

∴ F(∞) = $\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to 0}\mathrm{s}\mathrm{F}(\mathrm{s})=\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to 0}\frac{\mathrm{s}\times 2}{(\mathrm{s}+2{)}^2+25}$

$=\frac{0\times 2}{2^2+5^2}=0$

Concept:

The steady-state output of signal y(t) is given as:

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(1-z^{-1}\right)Y\left(z\right)$

‘OR’

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(z-1\right)Y\left(z\right)$   ---(1)

Final value theorem is only applied when if X(z) has no poles outside the unit circle, and it has no higher poles on the unit circle centered at the origin of the z-plane

Calculation:

Given:

$H\left(z\right)=\frac{4z-1}{z^2+3z-1},$

$x\left(n\right)=8u\left(n\right)$

z-transform of x(n) is,

$X\left(z\right)\to \frac{8}{1-z^{-1}}=\frac{8z}{z-1}$

Y(z) = H(z) ⋅ X(z)

$Y\left(z\right)=\frac{4z-1}{z^2+3z-1}\cdot \frac{8z}{z-1}$

From equation (1);

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\frac{\left(z-1\right)\left(4z-1\right)\cdot 8}{\left(z^2+3z-1\right)\left(z-1\right)}$

$y\left(\mathrm{\infty }\right)=\frac{3\times 8}{3}$

Concept:

For a causal signal x(n), the final value theorem states that:

$x\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left[z-1\right]X\left(z\right)$

This can also be written as:

$\underset{k\to \mathrm{\infty }}{lim}x(k)=\underset{z\to 1}{lim}(z-1)X^{+}(z)$

Notes:

For a causal signal x(n), the initial value theorem states that:

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

Concept:

The initial value theorem is given by:

In time domain:

$Lt\stackrel{t=0}{\to }x(t)$

In the Laplace domain:

$Lt\stackrel{s=\infty }{\to }sx(s)$

The final value theorem is given by:

In time domain:

$Lt\stackrel{t=\infty }{\to }x(t)$

In the Laplace domain:

Calculation:

Final value theorem is applicable to the system with poles in left half of s-plane i.e., stable system.

A/C to final value theorem

In time domain,

$\mathrm{l}{\mathrm{t}}_{\mathrm{t}\to \infty }\mathrm{f}(\mathrm{t})=\mathrm{f}(\infty )$

$\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to \infty }\mathrm{s}.\mathrm{F}(\mathrm{s})=\mathrm{F}(\infty )$

Given signal is f(t) = e^-2t sin(5t) u(t)

A/C to laplace theorem,

$e^{at}\sin btu(t)\leftrightarrow \frac{b}{(s-a{)}^2+b^2}$

$e^{-2t}sin5tu(t)\leftrightarrow \frac{5}{(s+2{)}^2+5^2}$

$\mathrm{F}(\mathrm{s})=\frac{5}{(\mathrm{s}+2{)}^2+25}$

∴ F(∞) = $\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to 0}\mathrm{s}\mathrm{F}(\mathrm{s})=\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to 0}\frac{\mathrm{s}\times 2}{(\mathrm{s}+2{)}^2+25}$

$=\frac{0\times 2}{2^2+5^2}=0$

Explanation:

Final Value Theorem in Laplace Transform

Definition: The Final Value Theorem (FVT) is a fundamental concept in control theory and signal processing, which provides a method to determine the steady-state value of a time-varying function as time approaches infinity. It is particularly useful in analyzing the long-term behavior of systems described by differential equations.

Formula: The Final Value Theorem states that for a given function $f(t)$, if the Laplace transform $F(s)$ of $f(t)$ exists, then the final value of $f(t)$ as $t$ approaches infinity can be found using the formula:

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

This theorem applies under the condition that all poles of $sF(s)$ (except possibly at $s=0$) are in the left half of the complex plane. This ensures that the function $f(t)$ approaches a steady-state value as $t$ approaches infinity.

Working Principle: The Final Value Theorem provides a direct way to find the steady-state value of a function by analyzing its Laplace transform. The theorem essentially relates the behavior of the function in the time domain to its behavior in the frequency domain, allowing for easier computation of the final value.

Example: Consider a function $f(t)$ with the Laplace transform $F(s)=\frac{5}{s(s+2)}$. To find the final value of $f(t)$, we use the Final Value Theorem:

$\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to 0}{lim}s\left(\frac{5}{s(s+2)}\right)=\underset{s\to 0}{lim}\frac{5}{s+2}=\frac{5}{2}=2.5$

Therefore, the final value of $f(t)$ as $t$ approaches infinity is 2.5.

Advantages:

• Simplicity in finding the steady-state value without needing to perform inverse Laplace transforms.
• Useful for analyzing the long-term behavior of systems described by differential equations.
• Provides a direct relationship between the time domain and frequency domain representations of a function.

Disadvantages:

• Applicable only if the function $f(t)$ approaches a steady-state value as $t$ approaches infinity.
• Requires the existence of the Laplace transform $F(s)$ of the function $f(t)$.
• Not applicable if the function has poles on the right half of the complex plane or on the imaginary axis (except at $s=0$).

Correct Option Analysis:

The correct option is:

Option 3: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to 0}{lim}sF(s)$

This option correctly represents the Final Value Theorem. The formula $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to 0}{lim}sF(s)$ provides a method to find the final value of a time-varying function $f(t)$ by analyzing its Laplace transform $F(s)$.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{s\to \mathrm{\infty }}{lim}sF(s)$

This option is incorrect because it suggests taking the limit as $s$ approaches infinity, which does not provide the final value of the function $f(t)$ as $t$ approaches infinity. The correct limit should be as $s$ approaches zero.

Option 2: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{t\to 0}{lim}tf(t)$

This option is incorrect because it involves the time domain limit as $t$ approaches zero, which does not relate to finding the final value as $t$ approaches infinity. The Final Value Theorem specifically deals with the limit as $t$ approaches infinity.

Option 4: $\underset{t\to \mathrm{\infty }}{lim}f(t)=\underset{t\to 0}{lim}f(t)$

This option is incorrect because it again involves the time domain limit as $t$ approaches zero, which does not help in finding the final value as $t$ approaches infinity. The correct approach involves the Laplace transform and the limit as $s$ approaches zero.

Conclusion:

Understanding the Final Value Theorem is essential for analyzing the steady-state behavior of time-varying functions. The correct application of the theorem allows for the determination of the final value of a function by evaluating its Laplace transform. This method simplifies the process and provides a direct relationship between the time domain and frequency domain representations of the function.

Concept:

The initial value theorem is given by:

In time domain:

$Lt\stackrel{t=0}{\to }x(t)$

In the Laplace domain:

$Lt\stackrel{s=\infty }{\to }sx(s)$

The final value theorem is given by:

In time domain:

$Lt\stackrel{t=\infty }{\to }x(t)$

In the Laplace domain:

Calculation:

Concept:

The steady-state output of signal y(t) is given as:

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(1-z^{-1}\right)Y\left(z\right)$

‘OR’

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(z-1\right)Y\left(z\right)$   ---(1)

Final value theorem is only applied when if X(z) has no poles outside the unit circle, and it has no higher poles on the unit circle centered at the origin of the z-plane

Calculation:

Given:

$H\left(z\right)=\frac{4z-1}{z^2+3z-1},$

$x\left(n\right)=8u\left(n\right)$

z-transform of x(n) is,

$X\left(z\right)\to \frac{8}{1-z^{-1}}=\frac{8z}{z-1}$

Y(z) = H(z) ⋅ X(z)

$Y\left(z\right)=\frac{4z-1}{z^2+3z-1}\cdot \frac{8z}{z-1}$

From equation (1);

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\frac{\left(z-1\right)\left(4z-1\right)\cdot 8}{\left(z^2+3z-1\right)\left(z-1\right)}$

$y\left(\mathrm{\infty }\right)=\frac{3\times 8}{3}$

Final value theorem is applicable to the system with poles in left half of s-plane i.e., stable system.

A/C to final value theorem

In time domain,

$\mathrm{l}{\mathrm{t}}_{\mathrm{t}\to \infty }\mathrm{f}(\mathrm{t})=\mathrm{f}(\infty )$

$\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to \infty }\mathrm{s}.\mathrm{F}(\mathrm{s})=\mathrm{F}(\infty )$

Given signal is f(t) = e^-2t sin(5t) u(t)

A/C to laplace theorem,

$e^{at}\sin btu(t)\leftrightarrow \frac{b}{(s-a{)}^2+b^2}$

$e^{-2t}sin5tu(t)\leftrightarrow \frac{5}{(s+2{)}^2+5^2}$

$\mathrm{F}(\mathrm{s})=\frac{5}{(\mathrm{s}+2{)}^2+25}$

∴ F(∞) = $\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to 0}\mathrm{s}\mathrm{F}(\mathrm{s})=\mathrm{l}{\mathrm{t}}_{\mathrm{s}\to 0}\frac{\mathrm{s}\times 2}{(\mathrm{s}+2{)}^2+25}$

$=\frac{0\times 2}{2^2+5^2}=0$

Concept:

The steady-state output of signal y(t) is given as:

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(1-z^{-1}\right)Y\left(z\right)$ ---(1)

‘OR’

$y\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(z-1\right)Y\left(z\right)$   

Calculation:

Given:

$X(z)=\frac{2z^{-1}}{1-1.8z^{-1}+0.8z^{-2}}$

From equation (1),

$X\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(1-z^{-1}\right)X\left(z\right)$

$X\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(1-z^{-1}\right)\frac{2z^{-1}}{1-1.8z^{-1}+0.8z^{-2}}$

$X\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left(1-z^{-1}\right)\frac{2z^{-1}}{(1-z^{-1})(1-0.8z^{-1})}$

$X\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\frac{2z^{-1}}{1-0.8z^{-1}}$

$X(\mathrm{\infty })=\frac{2}{1-0.8}$

$X(\mathrm{\infty })=10$

Concept:

For a causal signal x(n), the initial value theorem states that:

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

For a causal signal x(n), the final value theorem states that:

$x\left(\mathrm{\infty }\right)=\underset{z\to 1}{lim}\left[z-1\right]X\left(z\right)$

Analysis:

$X(z)=\frac{z(8z-7)}{4z^2-7z+3}$

$=\frac{z(8z-7)}{(z-1)(4z-3)}$

Using the final value theorem:

$\underset{n\to \mathrm{\infty }}{lim}x(n)=\underset{z\to 1}{lim}(z-1)X(z)$

$=\underset{z\to 1}{lim}(z-1)\frac{z(8z-7)}{(z-1)(4z-3)}$

= 1

Concept:

The final value is given by

$\underset{n\to \mathrm{\infty }}{lim}x\left[n\right]=\underset{z\to 1}{lim}\left(\frac{z-1}{z}\right)X\left(z\right)$

Calculation:

Final value

$=\underset{z\to 1}{lim}\frac{(z-1)}{z}\left[\frac{0.792z^2}{(z-1)(z^2-0.416z+0.208)}\right]$

Final value theorem it states that

$\begin{array}{l}x\left(\mathrm{\infty }\right)=\begin{array}{c}lim\\ z\to 1\end{array}\left(1-z^{-1}\right)x\left(z\right)\\ =\begin{array}{c}lim\\ z\to 1\end{array}\frac{1}{8}\left(1-z^{-1}\right)\frac{z^{-1}\left(1+z^{-5}\right)}{\left(1-z^{-1}\right)\left(1+z^{-1}\right)}\\ =\begin{array}{c}lim\\ z\to 1\end{array}\frac{1}{8}\frac{z^{-1}\left(1+z^{-5}\right)}{\left(1+z^{-1}\right)}\\ =\begin{array}{c}lim\\ z\to 1\end{array}\frac{1}{8}\frac{\frac{1}{z}\left(1+\frac{1}{z^5}\right)}{\left(1+\frac{1}{z}\right)}\end{array}$

$\begin{array}{l}=\begin{array}{c}lim\\ z\to 1\end{array}\frac{1}{8}.\frac{z^5+1}{\left(z+1\right)}\\ =\frac{1}{8}=0.125\end{array}$