Question

Consider the signal x(t) shown in the figure. Let h(t) denote the impulse response of the filter matched to x(t), with h(t) being non-zero only in the interval 0 to 4 sec. The slope of h(t) in the interval 3 greater than t greater than 4 sec is

• A. 0.5 sec\(^{-1}\)
• B. -0.5 sec\(^{-1}\)
• C. 1 sec\(^{-1}\)
• D. -1 sec\(^{-1}\)

Hint:

The impulse response of a matched filter, \(h(t) = x(T-t)\), is graphically obtained by taking the original signal \(x(t)\), flipping it horizontally (time-reversal to get \(x(-t)\)), and then shifting it to the right by T (to get \(x(-(t-T)) = x(T-t)\)). The slope of any segment in \(h(t)\) will be the negative of the slope of the corresponding original segment in \(x(t)\).

Answer 1

The Correct Option is D

Step 1: Recall the definition of a matched filter's impulse response.

The impulse response \(h(t)\) of a filter matched to a signal \(x(t)\), non-zero over \([0, T]\), is: \[ h(t) = x(T - t) \] Here, \(T = 4\) s. So: \[ h(t) = x(4 - t) \]

Step 2: Determine the corresponding interval in \(x(t)\).

We need the slope of \(h(t)\) in \(3 < t < 4\). Let \(\tau = 4 - t\). Then: \[ t = 3 \Rightarrow \tau = 1, \quad t = 4 \Rightarrow \tau = 0 \] So the interval \(3 < t < 4\) for \(h(t)\) corresponds to \(0 < \tau < 1\) for \(x(\tau)\).

Step 3: Slope of \(x(t)\) in the interval \(0 < t < 1\).

From the signal graph, \(x(t)\) rises from 0 to 1 in this interval. Therefore: \[ \text{Slope of } x(t) = \frac{1 - 0}{1 - 0} = 1 \]

Step 4: Relate the slope of \(h(t)\) to \(x(t)\).

Using the chain rule: \[ h(t) = x(4 - t) \Rightarrow \frac{d}{dt} h(t) = x'(4 - t) \cdot (-1) \] Thus: \[ m_h(t) = - m_x(4 - t) \] For \(t \in (3, 4)\), \(m_x(4 - t) = m_x(\tau) = 1\). Therefore: \[ m_h = -1 \]

Answer: The slope of \(h(t)\) in the interval \(3 < t < 4\) is \(-1\), which corresponds to option (D).