Solved Examples | Nyquist Rate & Aliasing | Digital Signal Processing

Summary

In this video lecture on Digital Signal Processing (DSP), the instructor revisits aliasing and the sampling theorem by solving practical examples. Key topics covered include determining minimum sampling rates to avoid aliasing, deriving discrete-time signals after sampling at different rates, and exploring Nyquist rate calculations. The video also discusses potential issues with sampling at Nyquist frequency, such as zero-crossing problems and incorrect amplitude reconstruction. The instructor concludes with examples of reconstructing analog signals from discrete samples, emphasizing the importance of choosing appropriate sampling rates to avoid aliasing.

Takeaways

• 📘 Aliasing and the Sampling Theorem are the core focus of this lecture, discussed with solved examples.
• 🎯 The minimum sampling rate to avoid aliasing is determined by Nyquist theorem, which states the sampling frequency should be at least twice the highest input frequency.
• 📊 For the signal 3cos(100πt), the frequency is 50 Hz, so the minimum sampling rate required is 100 Hz.
• 📉 Sampling at 200 Hz yields the discrete-time signal 3cos(π/2 * n).
• 🔄 When sampled at 75 Hz, the discrete-time signal is 3cos(2π/3 * n), illustrating aliasing.
• 🔍 Aliasing causes signals at different frequencies (like 50 Hz and 25 Hz) to have identical discrete-time samples when sampled at 75 Hz.
• ⚠️ Sampling below the Nyquist rate can lead to missing signal components, as seen in an example where all sampled values are zero.
• 📏 The Nyquist rate for the signal 3cos(50πt) + 10sin(300πt) − cos(100πt) is 300 Hz, based on the highest frequency of 150 Hz.
• 🛠️ Sampling at 5000 Hz for the signal 3cos(2000πt) + 5sin(6000πt) + 10cos(12000πt) leads to aliasing, with a distorted reconstructed signal.
• 🎓 The lecture ends with a practical note on the importance of sampling above the Nyquist rate to avoid aliasing-related issues.

Q & A

What is the minimum sampling rate required to avoid aliasing for the signal x_a(t) = 3 cos(100πt)?

The minimum sampling rate required to avoid aliasing, according to the sampling theorem, is twice the highest frequency present in the signal. For this signal, the frequency is 50 Hz, so the minimum sampling rate is 100 Hz.

What is the discrete-time signal obtained when the analog signal x_a(t) = 3 cos(100πt) is sampled at 200 Hz?

The discrete-time signal is x[n] = 3 cos(π/2 * n) when the sampling frequency is 200 Hz.

What is the discrete-time signal obtained when the analog signal x_a(t) = 3 cos(100πt) is sampled at 75 Hz?

When sampled at 75 Hz, the discrete-time signal is x[n] = 3 cos(2π/3 * n).

What is the frequency of a sinusoid that yields the same samples as 3 cos(2π/3 * n) when sampled at 75 Hz?

The frequency of the sinusoid is 25 Hz. This is because 25 Hz is an alias of 50 Hz when sampled at 75 Hz.

What is the Nyquist rate for the signal x_a(t) = 3 cos(50πt) + 10 sin(300πt) - cos(100πt)?

The Nyquist rate for this signal is 300 Hz, as the highest frequency component is 150 Hz, and the Nyquist rate is twice that frequency.

What happens when you sample the signal 2 sin(200πt) at the Nyquist rate of 200 Hz?

Sampling the signal at the Nyquist rate (200 Hz) results in all zero-valued samples because the signal is sampled at its zero-crossing points.

What is the difference when you sample the signal 2 sin(200πt + θ) at 200 Hz compared to sampling 2 sin(200πt) at the same rate?

When sampling 2 sin(200πt + θ) at 200 Hz, the samples are not all zero, but the correct amplitude cannot be reconstructed. This highlights a problem with sampling exactly at the Nyquist rate.

Why is it recommended to sample at a rate higher than the Nyquist rate?

Sampling at a rate higher than the Nyquist rate avoids issues such as missing signal components at zero crossings and difficulties in reconstructing the correct amplitude of the original signal.

What is the Nyquist rate for the signal x_a(t) = 3 cos(2000πt) + 5 sin(6000πt) + 10 cos(12000πt)?

The Nyquist rate for this signal is 12,000 Hz, as the highest frequency component is 6,000 Hz, and the Nyquist rate is twice that frequency.

What is the discrete-time signal obtained when the signal x_a(t) = 3 cos(2000πt) + 5 sin(6000πt) + 10 cos(12000πt) is sampled at 5000 Hz?

The discrete-time signal obtained is x[n] = 13 cos(2π/5 * n) - 5 sin(4π/5 * n) after sampling at 5000 Hz.

What is the analog signal that can be reconstructed from the samples of x_a(t) = 3 cos(2000πt) + 5 sin(6000πt) + 10 cos(12000πt) if ideal interpolation is used?

The reconstructed analog signal is x(t) = 13 cos(2000πt) - 5 sin(4000πt), which is different from the original signal due to aliasing caused by sampling at a rate lower than the Nyquist rate.