Sampling Let’s sample the signal Signal Sampling at a time interval of Δt

Analog Signal

Dr. Christopher M. Godfrey University of North Carolina at Asheville Δt Photo: C. Godfrey

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

Sampling Let’s sample the signal Sampling Reconstruct the curve at a time interval of Δt from the sampled points

Analog Signal Analog Signal Samples at Δt Samples at Δt

Δt Δt

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

Sampling High-frequency Sampling components in the curve are sampled poorly „ Given that there can be sampling problems, what Δt sampling interval should be selected in order to fully define the given input signal?

Low-frequency „ We’ll look at a version of the components in the curve are sampled well sampling theorem to answer

Analog Signal this… Samples at Δt ‰ …but first, let’s review some properties of sines and cosines Δt

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

1 Fourier Analysis Fourier Analysis

„ Fourier analysis shows that any analog signal can be represented perfectly by an infinite sum of sines and cosines

ATMS 320 – Fall 2011 D1

Great attributes of sine functions Sampling Theorem

„ It’s possible to define a sine wave, with unity amplitude, using only two points (more is better)! „ Let’s sample a signal periodically every Δts

„ For a given signal with multiple frequency ‰ Δts is the sampling time interval components, we must find the largest sampling time ‰ fs= 1/ Δts is the sampling frequency interval (Δt ), or lowest sampling frequency (f ) that s s „ Then each of the p samples strung together will allow recovery of the signal without error uniquely define the sampled signal, where ‰ Note that f = 1/(Δt) ‰ Think of f as the # complete cycles per unit time p(t)=p(n Δts) and n is an integer

„ If we know fmax (i.e., the highest frequency of the wave), we can appropriately sample the signal for error-free recovery!

‰ We can figure out fmax using Fourier analysis Δts

ATMS 320 – Fall 2011 D2 ATMS 320 – Fall 2011

Sampling Theorem The Nyquist Frequency For a limited bandwidth signal (bandwidth refers to the range of frequencies contained within a signal) with „ The Nyquist frequency is half the sampling maximum frequency fmax, the sampling frequency fs must be greater than twice the maximum frequency frequency: fmax in order to uniquely reconstruct the signal without aliasing: fN = fs/2=1/(2Δts)

fs > 2fmax If the sampled signal contains fmax > fN, aliasing (frequency folding) will occur. The sampled wave will The frequency 2fmax is called the Nyquist rate. Used in this context, the Nyquist rate is the lower bound for the appear to have a frequency that is folded to a lower sampling rate necessary for alias-free sampling. The frequency. To completely reconstruct a wave without Nyquist rate is a property of the signal. error, fN must be greater than fmax.

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

2 The Nyquist Frequency Sampling Example „ Assume that there is some signal with many „ The Nyquist rate is a property of a continuous signal frequency components associated with it ‰ If you want to sample a signal without aliasing, „ We know that the maximum frequency you must sample it with a sampling frequency component of the signal is 5 kHz that is greater than or equal to the Nyquist rate for that signal. „ To comppyletely reconstruct the si gnal, we would need to sample the signal at what „ The Nyquist frequency is a property of a discrete sampling system frequency and time interval?

‰ Any given sampling interval Δts chosen for a Answer: We would need to sample the wave particular sampling system (e.g., radar) at the Nyquist rate, or fs = 2fmax = 10 kHz. corresponds with a particular Nyquist frequency, This corresponds with a Δt of 0.0001 s. The regardless of the properties of the sampled s signal. Nyquist frequency in this case is fs/2 = 5 kHz, so no aliasing will occur. ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

Sampling Aliasing from inadequate sampling

„ Another question: What happens if we sample at a frequency that is less than the Nyquist rate?

‰ Recovery of the original signal is impossible /

‰ Funny things happen with the reconstructed signal…

If we sample close to the actual frequency of the input signal, the output will appear folded back to a lower frequency.

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

Aliasing from inadequate sampling Aliasing Example

Let’s sample a simple sine wave

If we sample with fs < 2fmax, those original frequencies that are above fN will appear as false low-frequency components of the sampled signal.

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

3 Aliasing Example Aliasing Example

If we sample it 1.5 times per cycle (fs = 1.5fmax), the re- If we sample it once per cycle (fs = fmax), constructed wave looks like a lower-frequency sine wave

we might think it is constant! Why? Because fN = ½ fs = ¾ fmax < fmax The original wave has a higher frequency than our Nyquist frequency.

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

Aliasing Example Aliasing Example

Sampled at twice the frequency of the input wave (i.e., fs = 2fmax), the reconstruction Oversampling results in an even better approximates the original sine wave (if we approximation to the original input signal sample in the right spot).

Why? Because fN = ½ fs = fmax ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

Aliasing from inadequate sampling Aliasing from inadequate sampling Frequency folding „ In general, the aliased output frequency is

± faliased = finput − 2mfN

where:

faliased is the aliased output frequency

finput is the input frequency of the original wave m is an integer

fN is the Nyquist frequency The x-axis shows the input frequency (i.e., the frequency of the original signal) in terms of the Nyquist frequency (which is ± f = f − 2mf determined by the sampling interval). aliased input N

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

4 Aliasing – Moiré Effect Aliasing – Velocity Folding

ATMS 320 – Fall 2011 ATMS 320 – Fall 2011

Aliasing – Stroboscopic Artifacts

Visit http://www.michaelbach.de/ot/mot_strob/index.html

ATMS 320 – Fall 2011

5