Question

A time domain signal is given by \[ x(t) = 20 \sin(100\pi t) + 36 \sin(150\pi t) - 2 \sin(300\pi t), \] where time \( t \) is in seconds. Among the following options, what is the frequency (in Hz) at which \( x(t) \) should be sampled for accurate signal reconstruction?

• A. 500
• B. 150
• C. 200
• D. 250

Hint:

To avoid aliasing, always sample at a frequency that is at least twice the highest frequency component of the signal, as per the Nyquist-Shannon theorem.

Answer 1

The Correct Option is A

Step 1: Identifying the frequency components of the signal
The given signal consists of three sinusoidal components: \[ x(t) = 20 \sin(100\pi t) + 36 \sin(150\pi t) - 2 \sin(300\pi t), \] where the angular frequencies are \( 100\pi \), \( 150\pi \), and \( 300\pi \), respectively. To find the highest frequency component, we calculate the frequencies by dividing the angular frequency by \( 2\pi \): - For \( 20 \sin(100\pi t) \), the frequency is: \[ f_1 = \frac{100\pi}{2\pi} = 50 { Hz}. \] - For \( 36 \sin(150\pi t) \), the frequency is: \[ f_2 = \frac{150\pi}{2\pi} = 75 { Hz}. \] - For \( -2 \sin(300\pi t) \), the frequency is: \[ f_3 = \frac{300\pi}{2\pi} = 150 { Hz}. \] Step 2: Applying the Nyquist-Shannon theorem
The highest frequency component is \( f_3 = 150 { Hz} \). According to the Nyquist-Shannon sampling theorem, the sampling frequency should be at least twice the highest frequency to ensure accurate signal reconstruction. Thus, the minimum sampling frequency is: \[ f_s = 2 \times 150 = 300 { Hz}. \] Step 3: Choosing the correct sampling frequency
Among the given options, the closest value greater than or equal to 300 Hz is 500 Hz. Thus, the correct answer is 500 Hz.