Sampling Techniques Quiz

According to the Nyquist-Shannon sampling theorem, the sampling frequency must be at least twice the maximum frequency component in the signal. Therefore, for a signal with maximum frequency 25 kHz, the Nyquist rate is 2 × 25 kHz = 50 kHz.

When a signal is undersampled, aliasing occurs. The aliased frequency f_alias = |f_signal - n × f_sampling|, where n is an integer. For f_signal = 10 kHz and f_sampling = 12 kHz, the closest n is 1: f_alias = |10 - 1×12| = 2 kHz. Alternatively, we can compute: 10 kHz > f_sampling/2 = 6 kHz, so it will alias to 12 - 10 = 2 kHz.

For a bandpass signal with bandwidth B = 30 - 20 = 10 kHz and highest frequency f_H = 30 kHz, the minimum sampling frequency according to bandpass sampling is 2B = 20 kHz, provided that f_s lies in the range [2B, 2f_H/(floor(f_H/B))]. Here, floor(f_H/B) = floor(30/10) = 3, so the valid range is [20, 2×30/3 = 20] kHz. Thus, f_s = 20 kHz is sufficient.

For an n-bit ADC with voltage range V_range, the quantization step size Δ = V_range / (2^n). Here, V_range = 5V, n = 12, so Δ = 5 / (2^12) = 5 / 4096 ≈ 0.0012207 V = 1.22 mV.

Bit rate = Sampling rate × Bits per sample. 256 quantization levels require log2(256) = 8 bits per sample. Therefore, bit rate = 8 kHz × 8 bits = 64 kbps.

For linear PCM, SQNR (in dB) ≈ 6.02n + 1.76, where n is the number of bits. For n = 14, SQNR ≈ 6.02×14 + 1.76 = 84.28 + 1.76 = 86.04 dB. This formula assumes a full-scale sinusoidal input.

Flat-top sampling (also called sample-and-hold) holds the sample value constant during the sampling period, resulting in a staircase-like waveform when reconstructed. Natural sampling multiplies the signal by a pulse train, while ideal sampling uses impulse trains (theoretically).

If the maximum acceptable aliasing power is -40 dB relative to the signal, the anti-aliasing filter must provide at least 40 dB of attenuation at frequencies above the Nyquist frequency (half the sampling rate) to ensure aliased components are attenuated to the acceptable level.

The signal contains frequency components at f1 = 2000π/(2π) = 1000 Hz and f2 = 5000π/(2π) = 2500 Hz. The maximum frequency is 2500 Hz. The Nyquist rate is 2 × f_max = 2 × 2500 = 5000 Hz = 5000 samples/s. However, note that the Nyquist rate is the minimum sampling rate to avoid aliasing, but the question asks for the Nyquist rate in samples per second, which is 5000. But wait, let's double-check: f_max = 2500 Hz, so Nyquist rate = 5000 samples/s. However, looking at the options, 5000 is option B. But let me re-examine: Actually, the correct answer should be 5000 samples/s. But looking at the options, 5000 is option B. However, I need to check if there's any trick: The signal has components at 1 kHz and 2.5 kHz. The Nyquist rate is twice the highest frequency, which is 5 kHz. So answer should be B. But let me check the answer key I provided: I said C (10000 samples/s). That was an error in my initial answer key. The correct answer is B (5000 samples/s).

Oversampling ratio (OSR) = f_sampling / (2 × Bandwidth). Here, f_sampling = 48 kHz, Bandwidth = 4 kHz, so OSR = 48000 / (2 × 4000) = 48000 / 8000 = 6. Alternatively, sometimes OSR is defined as f_sampling / f_Nyquist, where f_Nyquist = 2 × Bandwidth = 8 kHz, so OSR = 48/8 = 6. Wait, that gives 6, which is option A. But let me check: Actually, the correct answer is 6 (option A). I made an error in the answer key. OSR = f_s / (2×B) = 48/(2×4) = 48/8 = 6. So the correct answer is A.

Using the formula SQNR ≈ 6.02n + 1.76 dB, we need SQNR ≥ 50 dB. Solving: 6.02n + 1.76 ≥ 50 → 6.02n ≥ 48.24 → n ≥ 48.24/6.02 ≈ 8.01. Since n must be an integer, the minimum is 9 bits. With 8 bits, SQNR ≈ 6.02×8 + 1.76 = 48.16 + 1.76 = 49.92 dB, which is slightly below 50 dB.

The sampling period T_s must be greater than the acquisition time. Additionally, according to Nyquist, T_s ≤ 1/(2f_max) to avoid aliasing. But here, we're concerned with the sample-and-hold circuit's acquisition time. The minimum sampling period should be at least the acquisition time (10 ns) plus the hold time. However, for accurate sampling, the sampling period should be much larger than the acquisition time. For a 1 MHz signal, the period is 1 μs. A reasonable sampling period would be at least 1 μs to allow for proper acquisition and settling. Typically, the sampling period is chosen based on the signal frequency, not just the acquisition time.