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an analog input signal x(t) = 1.5 +0.5 sin(100t) is sampled with 2 kHz sampling frequency fs. Assuming that sampling starts at t = 0 compute the value of the sample x [20]. Show your work.

x[20] = 1.5 + 0.5 sin(2000π)

x[20] = 1.5 + 0.5 sin(6283.185)

x[20] = 1.5 + 0.5(0.9659)

x[20] = 2.4659

Did you know?

Did you know that when an analog input signal is sampled with a certain frequency, we can obtain individual samples of the signal at specific time points? In this case, the analog input signal x(t) = 1.5 + 0.5sin(100t) is sampled with a frequency of 2 kHz (2,000 samples per second). To compute the value of the sample x[20], we start sampling at t = 0, and each sample is taken every 1/fs seconds. Given that fs = 2 kHz, each sample is taken every 1/2,000 seconds or 0.0005 seconds. So, to find the value of x[20], we substitute t = 20 * 0.0005 seconds into the expression for x(t): x(20 * 0.0005) = 1.5 + 0.5sin(100 * 20 * 0.0005) = 1.5 + 0.5sin(0.1) = 1.5 + 0.5 * 0.0998 = 1.5 + 0.0499 = 1.5499 Therefore, the value of the sample x[20] is approximately 1.5499.