Question

The convolution of \(x(-t)\) with impulse function \(\delta(-t - t_0)\) is equal to

• A. \( x(t+t_0) \)
• B. \( x(t-t_0) \)
• C. \( x(-t+t_0) \)
• D. \( x(-t-t_0) \)

Hint:

Remember two key properties of the delta function: 1. Scaling: \( \delta(at+b) = \frac{1}{|a|}\delta(t + b/a) \). 2. Sifting (Convolution): \( f(t) \delta(t-T) = f(t-T) \). First, simplify the delta function, then apply the sifting property.

Answer 1

The Correct Option is D

Step 1: Simplify the impulse function. The Dirac delta function \(\delta(t)\) has a scaling property: \( \delta(at) = \frac{1}{|a|}\delta(t) \). Let's apply this to the given impulse \( \delta(-t-t_0) \). \[ \delta(-t-t_0) = \delta(-(t+t_0)) \] Here, the scaling factor is \(a = -1\). \[ \delta(-(t+t_0)) = \frac{1}{|-1|}\delta(t+t_0) = \delta(t+t_0) \]
Step 2: Perform the convolution. The convolution is now \( x(-t) \delta(t+t_0) \). We use the sifting property of convolution with a shifted impulse: \( f(t) \delta(t-T) = f(t-T) \). In our case, the function is \( f(t) = x(-t) \) and the shift is \( T = -t_0 \). So, we replace \(t\) in \(x(-t)\) with \(t - (-t_0)\), which is \(t+t_0\). \[ x(-t) \delta(t+t_0) = x(-(t+t_0)) \] \[ = x(-t - t_0) \]