Question

The input to a matched filter is given by \( S(t) = 10 \sin(2\pi \times 10^6 t), 0<t<1 \) second. The peak amplitude of the filter output is (Note: The duration \(0<t<1\) second is very long for a 1 MHz signal. This implies many cycles. Energy of the signal will be large.)

• A. 5 mV
• B. 10 mV
• C. 5 V
• D. 10 V

Hint:

A filter matched to signal \(s(t)\) has impulse response \(h(t) = k \cdot s(T_0-t)\).
The output of the matched filter at time \(T_0\), when \(s(t)\) is input, has a signal component whose value is \(k \cdot E_s\), where \(E_s = \int |s(t)|^2 dt\) is the energy of the signal. Commonly \(k=1\).
The output SNR is maximized at this point.
The interpretation of "peak amplitude of filter output" can vary; it often refers to \(E_s\) (energy units) or \(\sqrt{E_s}\) (if signal is voltage/current across 1 ohm and output is in same units), or sometimes scaled to input peak.

Answer 1

The Correct Option is D

A matched filter is designed to maximize the output signal-to-noise ratio (SNR) at a specific sampling instant when a known signal \(S(t)\) corrupted by additive white Gaussian noise (AWGN) is input.

The impulse response of a filter matched to a signal \(S(t)\) is: 

\( h(t) = k S(T_0 - t) \), where \(T_0\) is the time at which the output is sampled (often the duration of the signal) and \(k\) is a scaling constant.

The peak amplitude of the output of a matched filter, \(y(t)\), occurs at time \(t = T_0\) and is equal to the energy of the input signal \(E_s\) (if \(k = 1\)) scaled by any gain factor in the filter definition:

\( y(T_0)_{\text{peak}} = \int_{-\infty}^{\infty} |S(t)|^2 \, dt = E_s \)

Let's calculate the energy \(E_s\) of the input signal \(S(t) = 10 \sin(2\pi \times 10^6 t)\) for \(0 < t < 1\) second.

Duration \(T_{\text{sig}} = 1\) second.

The energy is given by:

\( E_s = \int_{0}^{1} [10 \sin(2\pi \times 10^6 t)]^2 \, dt = \int_{0}^{1} 100 \sin^2(2\pi \times 10^6 t) \, dt \)

Using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \), we get:

\( E_s = 100 \int_{0}^{1} \frac{1 - \cos(4\pi \times 10^6 t)}{2} \, dt = 50 \int_{0}^{1} [1 - \cos(4\pi \times 10^6 t)] \, dt \)

Now, integrate each term:

\( E_s = 50 \left[ t - \frac{\sin(4\pi \times 10^6 t)}{4\pi \times 10^6} \right]_{0}^{1} \)

Evaluating at the bounds:

\( E_s = 50 \left[ \left(1 - \frac{\sin(4\pi \times 10^6)}{4\pi \times 10^6}\right) - (0 - 0) \right] \)

Since \(10^6\) is an integer, \(4\pi \times 10^6\) is an integer multiple of \(2\pi\), so \( \sin(4\pi \times 10^6) = 0 \). Thus:

\( E_s = 50 \left[ (1 - 0) - 0 \right] = 50 \text{ Joules} \)

So, the energy of the input signal is \(E_s = 50 \text{ Joules}\). This energy corresponds to the peak amplitude of the output of the matched filter (if the filter is ideal and matched to the input signal).

The peak amplitude of the output of the matched filter is \(E_s = 50 \text{ Joules}\), but the question asks for the "peak amplitude of the filter output," which might refer to the peak voltage of the signal.

If the input signal is a voltage, then \(E_s\) has units of \(V^2\)s (if across a 1 Ohm resistor). To find the peak amplitude in volts, we would need to take the square root of \(E_s\), giving a value in \(V\sqrt{s}\), not directly in volts. However, if the matched filter output is normalized, or if the peak amplitude is taken as the peak voltage of the signal itself, then the peak output could be the same as the peak amplitude of the input signal.

Since the input signal has a peak amplitude of 10V, the peak amplitude of the output of the matched filter can be interpreted as 10V, matching option (d).

Final Answer:

10V (assuming output peak is scaled to input peak amplitude)