Question

Let the Laplace transform of a function f(t) which exists for t>0 be F₁(s) and the Laplace transform of its delayed version f(t – τ) be F₂(s). Let F₁ * (s) be the complex conjugate of F₁(s) with the Laplace variable set s = σ + jω. If \(G\left( s \right) = \frac{{{F_2}\left( s \right){F_1}*\left( s \right)}}{{{{\left| {{F_1}\left( s \right)} \right|}^2}}}\), then the inverse Laplace transform of G(s) is
1. An ideal impulse δ (t)
2. An ideal delayed impulse δ (t – τ)
3. An ideal step function u (t)
4. An ideal delayed step function u (t – τ)

Answer 1

Correct Answer - Option 2 : An ideal delayed impulse δ (t – τ)

\(\begin{array}{l} f\left( t \right)\mathop \leftrightarrow \limits^Z {F_1}\left( s \right)\\ f\left( {t - \tau } \right)\mathop \leftrightarrow \limits^Z {e^{ - s\tau }}{F_1}\left( s \right) = {F_2}\left( s \right)\\ G\left( s \right) = \frac{{{F_2}\left( s \right){F_1}*\left( s \right)}}{{{{\left| {{F_1}\left( s \right)} \right|}^2}}}\\ = \frac{{{e^{ - s\tau }}{F_1}\left( s \right){F_1}*\left( s \right)}}{{{{\left| {{F_1}\left( s \right)} \right|}^2}}}\\ = \frac{{{e^{ - s\tau }}|{F_1}\left( s \right){|^2}}}{{|{F_1}\left( s \right){|^2}}}\\ \left( {because\ {F_1}\left( S \right)F_1^{*}\left( S \right) = |{F_1}\left( S \right){|^2}} \right) \end{array}\)

= e ^–sτ

Taking inverse Laplace,

g(t) = L^-1 [e^-sτ ] = δ (t – τ)