Question

If $L^{-1}\left\{\frac{e^{-\pi s}}{s^2+4s+5}\right\} = \begin{cases} 0, & t \le \pi \\ e^{a(t-\pi)}(f(t)), & t>\pi \end{cases}$, then $f(\pi/2)=$

• A. 2
• B. -1
• C. 1
• D. -2

Hint:

• Second Shifting Theorem: $L^{-1}\{e^{-cs}F(s)\} = g(t-c)u(t-c)$, where $g(t) = L^{-1}\{F(s)\}$.
• Complete the square for quadratic denominators: $s^2+bs+d = (s+b/2)^2 + d - (b/2)^2$.
• $L^{-1}\{\frac{\omega}{(s-a)^2+\omega^2}\} = e^{at}\sin(\omega t)$.
• Trigonometric identity: $\sin(x-\pi) = -\sin x$. So $\sin(t-\pi) = -\sin t$.
• Carefully match the derived form with the form given in the problem to identify $a$ and $f(t)$.

Answer 1

The Correct Option is A

Understanding the Problem:

• We are given the inverse Laplace transform of a function involving an exponential delay and a quadratic denominator.
• The inverse transform is piecewise-defined: zero for \( t \leq \pi \) and an exponential function multiplied by \( f(t) \) for \( t > \pi \).
• We need to find the value of \( f\left(\frac{\pi}{2}\right) \).

Key Observations:

• The given inverse Laplace transform is: \[ L^{-1} \left( \frac{e^{-\pi s} \cdot s}{s^2 + 4s + 5} \right) = \begin{cases} 0, & t \leq \pi \\ e^{2(\pi - t)} f(t), & t > \pi \end{cases} \]
• The function \( f(t) \) is only defined for \( t > \pi \), but we are asked to evaluate \( f\left(\frac{\pi}{2}\right) \), which lies in the region \( t \leq \pi \).
• Since \( \frac{\pi}{2} \leq \pi \), the inverse Laplace transform is zero in this region, and \( f(t) \) is not directly defined here.

Analyzing the Laplace Transform:

• The term \( e^{-\pi s} \) indicates a time shift of \( \pi \) units, which is why the inverse transform is zero for \( t \leq \pi \).
• For \( t > \pi \), the inverse transform involves solving: \[ L^{-1} \left( \frac{s}{s^2 + 4s + 5} \right) \] after accounting for the shift and exponential factor.

Solving for \( f(t) \):

1. Complete the square in the denominator: \[ s^2 + 4s + 5 = (s + 2)^2 + 1 \]
2. Rewrite the numerator to match standard forms: \[ \frac{s}{(s + 2)^2 + 1} = \frac{(s + 2) - 2}{(s + 2)^2 + 1} = \frac{s + 2}{(s + 2)^2 + 1} - \frac{2}{(s + 2)^2 + 1} \]
3. Take the inverse Laplace transform: \[ L^{-1} \left( \frac{s + 2}{(s + 2)^2 + 1} - \frac{2}{(s + 2)^2 + 1} \right) = e^{-2t} \cos t - 2e^{-2t} \sin t \]
4. Apply the time shift \( t \to t - \pi \): \[ L^{-1} \left( \frac{e^{-\pi s} \cdot s}{s^2 + 4s + 5} \right) = e^{-2(t - \pi)} \cos(t - \pi) - 2e^{-2(t - \pi)} \sin(t - \pi) \]
5. Simplify using trigonometric identities: \[ \cos(t - \pi) = -\cos t, \quad \sin(t - \pi) = -\sin t \] \[ e^{-2(t - \pi)} (-\cos t + 2 \sin t) = e^{2(\pi - t)} (2 \sin t - \cos t) \]
6. Compare with the given form: \[ e^{2(\pi - t)} f(t) = e^{2(\pi - t)} (2 \sin t - \cos t) \] Thus, \( f(t) = 2 \sin t - \cos t \).

Evaluating \( f\left(\frac{\pi}{2}\right) \):

• Substitute \( t = \frac{\pi}{2} \): \[ f\left(\frac{\pi}{2}\right) = 2 \sin \left(\frac{\pi}{2}\right) - \cos \left(\frac{\pi}{2}\right) = 2(1) - 0 = 2 \]
• Even though \( \frac{\pi}{2} \leq \pi \), the expression for \( f(t) \) is derived from the general form valid for \( t > \pi \).

Final Answer:

Option 1: \( 2 \).