𝑦(𝑡) = 6 cos(23𝜋𝑡) + 3 sin(16𝜋𝑡)(a) State the Nyquist sampling rate of y(t).(b) Regardless of your answer in (a), assume sampling frequency fs = 5Hz. Write down the sampled data y[n] for n =0, 1, . . . , 9. Suppose y[0] = y(0). (Give your answers correct to two decimal place

(a) The Nyquist sampling rate of y(t) can be determined by considering the highest frequency component in the signal. In this case, the highest frequency component is 23π, so the Nyquist sampling rate would be twice that value, which is 46π.

(b) Assuming a sampling frequency of fs = 5Hz, we can calculate the sampled data y[n] for n = 0, 1, ..., 9. To do this, we substitute the values of t = n/fs into the equation for y(t).

For n = 0: y[0] = 6 cos(23π(0/5)) + 3 sin(16π(0/5)) = 6 cos(0) + 3 sin(0) = 6(1) + 3(0) = 6

For n = 1: y[1] = 6 cos(23π(1/5)) + 3 sin(16π(1/5)) ≈ 6 cos(4.6π) + 3 sin(3.2π) ≈ 6(-0.9998) + 3(0) ≈ -5.999

For n = 2: y[2] = 6 cos(23π(2/5)) + 3 sin(16π(2/5)) ≈ 6 cos(9.2π) + 3 sin(6.4π) ≈ 6(0.9998) + 3(0) ≈ 5.999

Continuing this process, we can calculate the values of y[n] for n = 3, 4, 5, 6, 7, 8, and 9.