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]. show your work
To compute the value of the sample x[6], we need to find the value of the input signal x(t) at t = 6/fs.
Given:
x(t) = 1.5 + 1.5sin(5t)
fs = 1 kHz
To find x[6], we substitute t = 6/fs into the expression for x(t):
x[6] = 1.5 + 1.5sin(5(6/fs))
Simplifying the expression:
x[6] = 1.5 + 1.5sin(30/fs)
Since fs is given as 1 kHz (fs = 1,000 Hz), we substitute fs = 1,000 Hz into the expression:
x[6] = 1.5 + 1.5sin(30/1,000)
Using a calculator, we can evaluate the sine function:
x[6] ≈ 1.5 + 1.5sin(0.03)
x[6] ≈ 1.5 + 1.5(0.0299955)
x[6] ≈ 1.5 + 0.04499325
x[6] ≈ 1.54499325
Therefore, the value of the sample x[6] is approximately 1.545.