Laplace transform of cos (ωt) is $$\frac{{\text{s}}}{{{{\text{s}}^2} + {\omega ^2}}}.$$ The Laplace transform of e^-2t cos(4t) is

$$\frac{{{\text{s}} - 2}}{{{{\left( {{\text{s}} - 2} \right)}^2} + 16}}$$

$$\frac{{{\text{s}} + 2}}{{{{\left( {{\text{s}} - 2} \right)}^2} + 16}}$$

$$\frac{{{\text{s}} - 2}}{{{{\left( {{\text{s}} + 2} \right)}^2} + 16}}$$

$$\frac{{{\text{s}} + 2}}{{{{\left( {{\text{s}} + 2} \right)}^2} + 16}}$$

Laplace transform of cos (ωt) is $$\frac{{\text{s}}}{{{{\text{s}}^2} + {\omega ^2}}}.$$ The Laplace transform of e^-2t cos(4t) is Correct Answer $$\frac{{{\text{s}} + 2}}{{{{\left( {{\text{s}} + 2} \right)}^2} + 16}}$$

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Feb 20, 2025

Engineering Maths
Transform Theory

Laplace transform of cos (ωt) is $$\frac{{\text{s}}}{{{{\text{s}}^2} + {\omega ^2}}}.$$ The Laplace transform of e-2t cos(4t) is

Related Questions

Laplace transform of cos (ωt) is $$\frac{{\text{s}}}{{{{\text{s}}^2} + {\omega ^2}}}.$$ The Laplace transform of e^-2t cos(4t) is