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Chapter 9: Problem 25

A \(10-\mathrm{kHz}\) sinewave is sampled. Determine the apparent frequency of the samples. Has aliasing occurred? The sampling frequency is a. \(11 \mathrm{kHz} ;\) b. \(8 \mathrm{kHz} ;\) c. \(40 \mathrm{kHz}\)

Short Answer

Expert verified
a. 1 kHz with aliasing; b. 2 kHz with aliasing; c. 10 kHz with no aliasing.

Step by step solution

01

Understanding the Nyquist Theorem

The Nyquist Theorem states that to avoid aliasing, a signal should be sampled at a rate that is at least twice its maximum frequency. For a sinewave with a frequency of \(10\, \mathrm{kHz}\), the minimum sampling rate required is \(20\, \mathrm{kHz}\).
02

Case a: Sampling Frequency is 11 kHz

The 11 kHz sampling frequency is below the Nyquist rate of 20 kHz. The apparent frequency can be calculated using the formula of aliasing: \(f' = |f_s - n \, f_{signal}|\) where \(f_s\) is the sampling frequency and \(n\) is an integer. Let's set \(n = 1\):\[f' = |11 - 10| = 1\, \mathrm{kHz}\]This implies that the apparent frequency of the samples is 1 kHz. Aliasing occurs because 11 kHz is below 20 kHz.
03

Case b: Sampling Frequency is 8 kHz

The 8 kHz sampling frequency is also below the Nyquist rate of 20 kHz. Using the same aliasing formula:\[f' = |8 - 10| = 2 \, \mathrm{kHz}\]This means the apparent frequency is 2 kHz. Aliasing occurs because 8 kHz is below 20 kHz.
04

Case c: Sampling Frequency is 40 kHz

The 40 kHz sampling frequency is above the Nyquist rate of 20 kHz. Therefore, the apparent frequency of the samples matches the original frequency as no aliasing occurs, producing a 10 kHz sinewave.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aliasing
Aliasing happens when a signal is sampled at a rate that is too low to accurately capture its frequency components. This distortion occurs because the sampling rate fails to meet the Nyquist Criterion, which states that a signal should be sampled at least twice as fast as its highest frequency.
  • When aliasing occurs, higher frequency components of the signal appear as lower frequencies. This can significantly distort the signal's representation.
  • To calculate the apparent frequency when aliasing occurs, the formula is: \(f' = |f_s - n \, f_{signal}|\), where \(f_s\) is the sampling frequency and \(n\) is an integer.
For instance, a 10 kHz signal sampled at 11 kHz will appear as a 1 kHz signal, due to aliasing (as 11kHz is less than the Nyquist Rate of 20kHz). Thus, aliasing creates a challenge in accurately reproducing the original signal when it is reconstructed from the samples.
Sampling frequency
The sampling frequency, often referred to as the sample rate, is critical in digital signal processing. It's the number of samples per second taken from a continuous signal to make it discrete.
  • The sampling frequency must be at least twice the highest frequency contained in the signal to satisfy the Nyquist Theorem.
  • For a signal with a maximum frequency of 10 kHz, the sampling frequency should be at least 20 kHz to prevent aliasing.
However, if a sampling frequency is below this threshold, the signal cannot be accurately reconstructed. A sampling frequency of 40 kHz is well above the required rate for a 10 kHz signal, thus avoiding aliasing altogether.
Apparent Frequency
The apparent frequency is the frequency at which a signal appears after sampling. It can differ from the original frequency if the signal is not sampled correctly.
  • The apparent frequency can be calculated using the formula: \(f' = |f_s - n \, f_{signal}|\).
  • In cases where aliasing is present, the apparent frequency will not match the original frequency of the sinewave.
For example, at an 8 kHz sampling frequency, the apparent frequency of a 10 kHz signal becomes 2 kHz, showcasing how the original signal can appear as a different frequency in digital form. This illustrates why maintaining the correct sampling rate is crucial to ensure that the apparent frequency aligns with the actual frequency of the signal.

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Most popular questions from this chapter

Suppose that the data collected from a sensor is found to contain an objectionable \(60-\mathrm{Hz}\) ac component. What potential causes for this interference would you look for? What are potential solutions for each cause of the interference?

A sensor produces a differential signal of \(6 \mathrm{mV}\) dc and a \(2-\mathrm{V}-\mathrm{rms} 60-\mathrm{Hz}\) ac commonmode signal. Write expressions for the voltages between the sensor output terminals and ground.

We need an ADC that can accept input voltages ranging from \(-5\) to \(+5 \mathrm{~V}\). How many bits must the code word have for a resolution of \(0.01 \mathrm{~V}\) and \(0.02 \mathrm{~V}\) respectively.

What signal-conditioner input impedance is best if the sensor produces short- circuit (Norton) current that is proportional to the measurand?

A \(60-\) Hz sinewave \(x(t)=A \cos (120 \pi t+\phi)\) is sampled at a rate of \(360 \mathrm{~Hz}\). Thus, the sample values are \(x(n)=A \cos \left(120 \pi n T_{s}+\phi\right)\), in which \(n\) assumes integer values and \(T_{s}=\) \(1 / 360\) is the time interval between samples. A new signal is computed by the equation $$ y(n)=\frac{1}{2}[x(n)+x(n-3)] $$ a. Show that \(y(n)=0\) for all \(n\). b. Now suppose that \(x(t)=V_{\text {signal }}+A \cos\) \((120 \pi t+\phi)\), in which \(V_{\text {signal }}\) is constant with time and again find an expression for \(y(n)\). c. When we use the samples of an input \(x(n)\) to compute the samples for a new signal \(y(n)\), we have a digital filter. Describe a situation in which the filter of parts (a) and (b) could be useful.
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