Consider a continuous time periodic signal𝑦(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 places)(c) Consider the following quantizer𝑄(𝑥) = {6,2,−2,−6,𝑥 ≥ 40 ≤ 𝑥 < 4−4 ≤ 𝑥 < 0𝑥 < −4Write down the quantized samples v[n] = Q(y[n])(d) Compute the quantization error q[n] = v[n] − y[n].(e) Under the conditions set above, determine the length of bit stream required to store y(t) for 8 seconds.

(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 is twice this frequency, which is 46π.

(b) Assuming a sampling frequency of fs = 5Hz, we can calculate the sampled data y[n] for n = 0, 1, ..., 9. We can use the formula y[n] = y(n/fs). Plugging in the values, we get: y[0] = y(0) = 6 cos(0) + 3 sin(0) = 6 y[1] = y(1/5) = 6 cos(23π/5) + 3 sin(16π/5) ≈ 4.47 y[2] = y(2/5) = 6 cos(46π/5) + 3 sin(32π/5) ≈ -2.12 y[3] = y(3/5) = 6 cos(69π/5) + 3 sin(48π/5) ≈ -6 y[4] = y(4/5) = 6 cos(92π/5) + 3 sin(64π/5) ≈ -2.12 y[5] = y(1) = 6 cos(23π) + 3 sin(16π) ≈ 4.47 y[6] = y(6/5) = 6 cos(138π/5) + 3 sin(96π/5) ≈ -2.12 y[7] = y(7/5) = 6 cos(161π/5) + 3 sin(112π/5) ≈ -6 y[8] = y(8/5) = 6 cos(184π/5) + 3 sin(128π/5) ≈ -2.12 y[9] = y(9/5) = 6 cos(207π/5) + 3 sin(144π/5) ≈ 4.47

(c) To obtain the quantized samples v[n] = Q(y[n]), we use the given quantizer 𝑄(𝑥). Plugging in the values of y[n], we get: v[0] = Q(y[0]) = Q(6) = 6 v[1] = Q(y[1]) = Q(4.47) = 6 v[2] = Q(y[2]) = Q(-2.12) = -2 v[3] = Q(y[3]) = Q(-6) = -6 v[4] = Q(y[4]) = Q(-2.12) = -2 v[5] = Q(y[5]) = Q(4.47) = 6 v[6] = Q(y[6]) = Q(-2.12) = -2 v[7] = Q(y[7]) = Q(-6) = -6 v[8] = Q(y[8]) = Q(-2.12) = -2 v[9] = Q(y[9]) = Q(4.47) = 6

(d) The quantization error q[n] can be computed as q[n] = v[n] - y[n]. Plugging in the values, we get: q[0] = v[0] - y[0] = 6 - 6 = 0 q[1] = v[1] - y[1] = 6 - 4.47 ≈ 1.53 q[2] = v[2] - y[2] = -2 - (-2.12) ≈ 0.12 q[3] = v[3] - y[3] = -6 - (-6) = 0 q[4] = v[4] - y[4] = -2 - (-2.12) ≈ 0.12 q[5] = v[5] - y[5] = 6 - 4.47 ≈ 1.53 q[6] = v[6] - y[6] = -2 - (-2.12) ≈ 0.12 q[7] = v[7] - y[7] = -6 - (-6) = 0 q[8] = v[8] - y[8] = -2 - (-2.12) ≈ 0.12 q[9] = v[9] - y[9] = 6 - 4.47 ≈ 1.53

(e) To determine the length of the bit stream required to store y(t) for 8 seconds, we need to calculate the total number of samples taken in 8 seconds. Since the sampling frequency is fs = 5Hz, the total number of samples is given by 8 * fs = 8 * 5 = 40. Therefore, the length of the bit stream required to store y(t) for 8 seconds is 40.