Question

Consider a causal LTI system characterized by differential equation \(\frac{{dy\left( t \right)}}{{dt}} + \frac{1}{6}y\left( t \right) = 3x\left( t \right)\). The response of the system to the input \(x\left( t \right) = 3{e^{ - \frac{t}{3}u\left( t \right)}}\). Where u(t) denotes the unit step function is
1. \(9{e^{ - \frac{t}{3}}}u\left( t \right)\)
2. \(9{e^{ - \frac{t}{6}}}u\left( t \right)\)
3. \(9{e^{ - \frac{t}{3}}}u\left( t \right) - 6{e^{ - \frac{t}{6}}}u\left( t \right)\)
4. \({54^{ - \frac{t}{6}}}u\left( t \right) - 54{e^{ - \frac{t}{3}}}u\left( t \right)\)

Answer 1

Correct Answer - Option 4 : \({54^{ - \frac{t}{6}}}u\left( t \right) - 54{e^{ - \frac{t}{3}}}u\left( t \right)\)

Concept:

Laplace transform is an important tool for converting differential equations in the time domain to algebraic equations in the frequency domain.

This analysis involves three important processes:

1) The transformation from the time domain to the frequency domain.

2) Manipulate the algebraic equations to form a solution.

3) The inverse transformation from the frequency to the time domain.

Mathematically, the Laplace transform of a function f(t) is defined as:

\(L[f(t)]=F(s)=\mathop \smallint \limits_0^\infty f\left( t \right){e^{ - st}}dt\)

s = σ + jω is the complex frequency 

Applications of the Laplace Transform:

1) Solve differential equations (both ordinary and partial)

2) Application in RLC circuit analysis.

Calculation:

Given \(\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{y}}\left( {\rm{t}} \right) + \frac{1}{6}{\rm{y}}\left( {\rm{t}} \right) = 3{\rm{x}}\left( {\rm{t}} \right)\)

Apply Laplace transform we get

\({\rm{sY}}\left( {\rm{s}} \right) + \frac{1}{6}{\rm{Y}}\left( {\rm{s}} \right) = 3{\rm{X}}\left( {\rm{s}} \right)\) But \({\rm{X}}\left( {\rm{s}} \right) = \frac{3}{{\left( {{\rm{s}} + \frac{1}{3}} \right)}}\) 

\(\begin{array}{l} \therefore {\rm{Y}}\left( {\rm{s}} \right) = \frac{{3{\rm{X}}\left( {\rm{s}} \right)}}{{\left( {{\rm{s}} + \frac{1}{6}} \right)}} = \frac{9}{{\left( {{\rm{s}} + \frac{1}{6}} \right)\left( {{\rm{s}} + \frac{1}{3}} \right)}}\\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{9}{{\left( {{\rm{s}} + \frac{1}{6}} \right)\left( {\frac{{ - 1}}{6} + \frac{1}{3}} \right)}} + \frac{9}{{\left( {{\rm{s}} + \frac{1}{3}} \right)\left( {\frac{{ - 1}}{3} + \frac{1}{6}} \right)}}\\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{{54}}{{{\rm{s}} + \frac{1}{6}}} - \frac{{54}}{{\left( {{\rm{s}} + \frac{1}{3}} \right)}} \end{array}\)

Apply Inverse Laplace transform we get

\({\rm{y}}\left( {\rm{t}} \right) = 54\left[ {{{\rm{e}}^{ - \frac{1}{6}{\rm{t}}}} - {{\rm{e}}^{ - \frac{1}{3}{\rm{t}}}}} \right]{\rm{u}}\left( {\rm{t}} \right)\)