Sampling
Aliasing
Learning Objectives:
- Understand the concept of aliasing in the context of signal processing, data acquisition, and imaging, where it refers to the misrepresentation of a high-frequency signal as a lower frequency signal due to insufficient sampling.
- Aliasing happens when the waveform is sampled at low frequency.
- Aliasing can be avoided if sampling rate is at least twice the max frequency of the measured signal.
- The wagon wheel effect: Continously varying images are being discretely sampled at a rate of 24 frames per second. This means aliasing will happen at any point in the image there are frequncy components, or light to dark transitions, that occur faster than f$_s/2$, whichm, in this case is, 12 frames per second, such as wagon wheel or propeller rotating at high speed.
Solved Example: 9964-01
Calculate the minimum sampling rate to avoid aliasing when a continuous-time signal is given by $x(t) = 5 \cos 400 \pi t$
A. 100 Hz
B. 250 Hz
C. 400 Hz
D. 20 Hz
Correct Answer: C
Solved Example: 9964-02
When aliasing takes place:
A. Sampling signals less than Nyquist rate
B. Sampling signals more than Nyquist rate
C. Sampling signals equal to Nyquist rate
D. Sampling signals at a rate which is twice of Nyquist rate
Correct Answer: A
Nyquist Theorem
Learning Objectives:
- Determine the minimum sampling rate of an analog signal.
- Determine if a sampled signal contains aliased signals.
The Nyquist sampling theorem provides a prescription for the nominal sampling interval required.
Nyquist theorem: A periodic signal must be sampled at more than twice the highest frequency component of the signal.
Or in mathematical terms:
\[f_s \geq 2f_c\] where,f$_s$ is the sampling frequency (how often the samples are taken per unit time or space) and f$_c$ is the highest frequency contained in the signal.
Solved Example: 9973-01
A system has a sampling rate of 50,000 samples per second. The maximum frequency of the signal it can acquire to reconstruct is:
A. 25 kHz
B. 50 kHz
C. 100 kHz
D. 10 kHz
Correct Answer: A
Solved Example: 9973-02
The Nyquist sampling rate for the signal \(x\left( t \right) = \dfrac{{\sin \left( {500\pi t} \right)}}{{\pi t}}\; \times \dfrac{{\sin \left( {700\pi t} \right)}}{{\pi t}}\)
A. 1200 Hz
B. 1400 Hz
C. 600 Hz
D. 400 Hz
Correct Answer: A
Solved Example: 9973-03
Arrange the Nyquist sampling interval of the signal in descending order: (A) $\sin c(300t)$ (B) $\sin c(300t) + \sin c^2(300t)$ (C) $\sin c(200t)$ (D) $\sin c(200t) + \sin c^2(200t)$ (E) $\sin c(200t) + \sin c(500t)$
A. (A), (C), (B), (D), (E)
B. (C), (D), (E), (A), (B)
C. (C), (A), (D), (E), (B)
D. (A), (E), (D), (C), (B)
Correct Answer: C
Solved Example: 9973-04
To satisfy the sampling theorem, a 100 Hz sine wave should be sampled at:
A. 10 Hz
B. 100 Hz
C. 200 Hz
D. 50 Hz
Correct Answer: C