Question

Consider a continuous-time signal \[ x(t) = -t^2 \left[ u(t+4) - u(t-4) \right] \] where \( u(t) \) is the continuous-time unit step function. Let \( \delta(t) \) be the continuous-time unit impulse function. The value of \[ \int_{-\infty}^{\infty} x(t) \delta(t+3) \, dt \text{ is:} \]

• A. 9
• B. 3
• C. -9
• D. -3

Hint:

To evaluate integrals involving \( \delta(t) \), simply evaluate the function at the location where the impulse is centered.

Answer 1

The Correct Option is C

Step 1: The signal is given by: \[ x(t) = -t^2 \left[ u(t+4) - u(t-4) \right] \] This represents a function that is nonzero only between \( t = -4 \) and \( t = 4 \), as the unit step functions effectively "turn on" and "turn off" the signal at these points.
Step 2: The expression \( \int_{-\infty}^{\infty} x(t) \delta(t+3) \, dt \) is the value of \( x(t) \) at \( t = -3 \), because the delta function \( \delta(t+3) \) is centered at \( t = -3 \).
Substitute \( t = -3 \) into \( x(t) \): \[ x(-3) = -(-3)^2 \left[ u(-3+4) - u(-3-4) \right] \] \[ x(-3) = -9 \left[ u(1) - u(-7) \right] \] Since \( u(1) = 1 \) and \( u(-7) = 0 \), we have: \[ x(-3) = -9 \]
Thus, the value of the integral is \( -9 \).