An analog input signal x(t) = 1.5 + 1.5sin(5t) is sampled with 1 kHz sampling frequency fs. Assuming that sampling starts at t = 0 compute the value of the sample x[6]. What is the answer?

Answer 1

To compute x[6], we need to find the value of x(t) at t=6/fs.

Given sampling frequency fs = 1 kHz, we have a sampling interval of Ts = 1/fs = 1/1000 = 0.001 seconds.

So, t = 6*Ts = 0.006 seconds.

Substituting this value in x(t), we get:

x(0.006) = 1.5 + 1.5sin(5*0.006)
= 1.5 + 1.5sin(0.03)
≈ 1.53

Now, x[6] is the value of the sample of x(t) at the nearest integer multiple of Ts to 0.006, which is 6*Ts = 0.006 seconds. Since sampling starts at t=0, the index of the sample corresponding to 0.006 seconds is 6.

Therefore, x[6] = x(0.006) ≈ 1.53.