LTI System Response using Laplace Transform

This solution details how to find the response of a Linear Time-Invariant (LTI) system to a sinusoidal input using the Laplace transform, focusing on the steady-state behavior indicated by the options.

System and Input Identification

• Unit Impulse Response: $h(t) = e^{-t} \sin(2t)u(t)$
• Input Signal: $x(t) = \sin(t)$

Laplace Transforms Calculation

• Laplace transform of the unit impulse response $h(t)$:

  $H(s) = L\{e^{-t} \sin(2t)u(t)\}$

  Using the frequency shifting property and the transform of $\sin(at)$, we get:

  $H(s) = \frac{2}{(s+1)^2 + 2^2} = \frac{2}{s^2 + 2s + 5}$

• Laplace transform of the input signal $x(t)$:

  $X(s) = L\{\sin(t)\} = \frac{1}{s^2 + 1^2} = \frac{1}{s^2 + 1}$

Steady-State Response Determination

The options provided are in the form of a scaled and phase-shifted sinusoid, suggesting the steady-state response is required. This is found using the system's frequency response, $H(j\omega)$, evaluated at the input signal's frequency.

• Input frequency: For $x(t) = \sin(t)$, the angular frequency is $\omega = 1$ rad/s.
• Calculate the frequency response $H(j\omega)$ by substituting $s = j\omega$ into $H(s)$:

  $H(j\omega) = \frac{2}{(j\omega)^2 + 2(j\omega) + 5} = \frac{2}{5 - \omega^2 + j2\omega}$

• Evaluate $H(j\omega)$ at $\omega = 1$:

  $H(j1) = \frac{2}{5 - 1^2 + j2(1)} = \frac{2}{4 + j2}$

• Simplify $H(j1)$ into rectangular form (a + jb):

  $H(j1) = \frac{2(4 - j2)}{(4 + j2)(4 - j2)} = \frac{8 - j4}{16 - (-4)} = \frac{8 - j4}{20} = 0.4 - j0.2$

• Determine the magnitude $|H(j1)|$ and phase $\arg(H(j1))$:

  Magnitude: $|H(j1)| = \sqrt{(0.4)^2 + (-0.2)^2} = \sqrt{0.16 + 0.04} = \sqrt{0.20} \approx 0.447$

  Phase: $\arg(H(j1)) = \arctan\left(\frac{-0.2}{0.4}\right) = \arctan(-0.5) \approx -26.6^\circ$

Output Response Calculation

For a sinusoidal input $x(t) = A \sin(\omega t)$, the steady-state output is $y_{ss}(t) = A |H(j\omega)| \sin(\omega t + \arg(H(j\omega)))$.

• With $A=1$ and $\omega=1$ for $x(t) = \sin(t)$:

  $y_{ss}(t) = 1 \times 0.447 \sin(1t + (-26.6^\circ))$

  $y_{ss}(t) = 0.447 \sin(t - 26.6^\circ)$

This result matches Option D. The transient part of the response, derived from the full $Y(s) = H(s)X(s)$ inverse transform, decays over time due to the $e^{-t}$ term in $h(t)$.