Question

Consider the following continuous-time signal

xa(t)=xa1(t)+xa2(t)+xa3(t),
where
xa1(t)=6cos(2πf1t+π/3)+sin(2πf2t+π/6),
xa2(t)=2cos(2πf3t−π/5)+sin(2πf4t+π/4)
xa3(t)=4cos(2πf5t)+5cos(2πf6t),
where f1=200 Hz,f2=1kHz,f3=3kHz,f4=4kHz,f5=6kHz it is required to design a digital signal processing-hased system to separated the three signals xe1(t) and xn2(t), and xe3(t).

Determine the minimum required sampling rate fsamp. Hence, use a value of 1.5 such a value.

Answer 1

The minimum required sampling rate fsamp is calculated to be 18 kHz.

Step-by-step explanation:

To design a digital signal processing (DSP) system to separate the three signals xa1(t), xa2(t), and xa3(t), we need to determine the minimum required sampling rate fsamp. The Nyquist-Shannon sampling theorem states that the sampling rate should be at least twice the highest frequency component in the signal. In this case, we have f1 = 200 Hz, f2 = 1 kHz, f3 = 3 kHz, f4 = 4 kHz, f5 = 6 kHz, and f6 = 6 kHz. Therefore, the highest frequency component in the signal is 6 kHz.

According to the Nyquist-Shannon sampling theorem, the minimum required sampling rate fsamp should be at least 2 times the highest frequency component, which is 2 * 6 kHz = 12 kHz.

Since the question mentions using a value of 1.5 times the minimum required sampling rate, we can calculate the sampling rate as 1.5 * 12 kHz = 18 kHz.