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Chapter 7: Problem 15

Impulse-train sampling of \(x[n]\) is used to obtain $$g(n]=\sum_{k=-\infty}^{\infty} x[n] \delta[n-k N]$$ If \(X\left(e^{j \omega}\right)=0\) for \(3 \pi / 7 \leq|\omega| \leq \pi,\) determine the largest value for the sampling interval \(N\) which ensures that no aliasing takes place while sampling \(x[n]\)

Short Answer

Expert verified
The largest value for the sampling interval \(N\) is 2.

Step by step solution

01

Analyze the Sampling Condition

Given that the Fourier transform of the original signal, \(X(e^{j\omega})\), is zero for \(\frac{3\pi}{7} \leq |\omega| \leq \pi\), check the non-zero frequency band. It is non-zero in the range \(-\frac{3\pi}{7}, \frac{3\pi}{7}\).
02

Determine the Nyquist Criterion for Sampling

To avoid aliasing, the Nyquist-Shannon sampling theorem needs to be satisfied. This requires that the sampling frequency \(\Omega_N\) should be greater than twice the highest frequency present in the signal. The highest frequency in \(x[n]\) is \(\frac{3\pi}{7}\), thus the criterion is \(\Omega_N = \frac{2\pi}{N} > 2 \times \frac{3\pi}{7}\).
03

Solve for N

From the condition \(\frac{2\pi}{N} > \frac{6\pi}{7}\), solve the inequality for \(N\). Dividing both sides by \(2\pi\) gives \(\frac{1}{N} > \frac{3}{7}\). Inverting the inequality gives \(N < \frac{7}{3}\). Since \(N\) is an integer, the largest possible value is \(N = 2\).
04

Conclusion

Thus, the largest value for the sampling interval \(N\) that ensures no aliasing is 2. This ensures the sampled signal does not overlap in the frequency domain, thereby preventing aliasing.

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Key Concepts

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

Impulse-Train Sampling
Impulse-train sampling is a process used for converting continuous signals into discrete signals by capturing values at specific intervals. During impulse-train sampling, an impulse, which is a short burst or spike, is used to sample the signal at regular spaced intervals. This is mathematically represented as a sequence of Dirac delta functions.

The process can be visualized as taking samples of a signal at discrete moments in time and ignoring the values in between. This is analogous to taking snapshots of a moving object at regular intervals.
  • Each sample represents an impulse that captures a value of the signal.
  • The result is a train of impulses, hence the name impulse-train sampling.
This method is essential for converting analog signals into digital signals, enabling their processing by digital systems. By using this method, a complete record of the signal is constructed through periodic sampling.
Nyquist Criterion
The Nyquist Criterion, fundamental in signal processing, sets the rules for proper sampling. Named after Harry Nyquist, this theorem defines how to choose a sampling rate to ensure a signal can be accurately reconstructed from its samples without introducing distortion.

For a signal to be sampled without aliasing:
  • The sampling rate must be at least twice the highest frequency present in the signal (known as the Nyquist rate).
  • Mathematically, if the highest frequency in a signal is \(f_{max}\), then the sampling rate \(f_s\) must satisfy \(f_s > 2f_{max}\).
Adhering to this criterion prevents different signal frequencies from becoming indistinguishable from each other, a problem known as aliasing. The result is clear and accurate digital representation of the original analog signal.
Fourier Transform
The Fourier Transform is a mathematical tool for transforming a time-domain signal into its frequency components. This allows us to analyze the different frequencies in a signal, which is crucial in various applications such as audio processing, communications, and signal analysis.

When a signal is transformed using Fourier Transform:
  • It converts the signal from a time-based representation to a frequency-based representation.
  • Provides insight into the frequency content and allows us to identify at which frequencies the signal has significant energy.
Understanding the frequency domain representation of a signal helps engineers design systems that process only the desired signal components, filter out noise, and thereby improve clarity and efficiency.
Aliasing
Aliasing can occur when a signal is sampled, but not at a fast enough rate. It's the distortion that results when high frequency signal components are incorrectly interpreted as lower frequency components due to inadequate sampling.

To prevent aliasing:
  • Ensure the sampling rate is greater than twice the highest frequency component of the original signal (following Nyquist Criterion).
  • Use filters to limit the bandwidth of the input signal before sampling. This removes frequencies higher than half of the sampling rate.
Aliasing results in overlapping spectrums, where different signals become indistinguishable from one another. Understanding and preventing aliasing is crucial in any digital sampling to accurately capture and reproduce signal characteristics.

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Most popular questions from this chapter

Consider the signal $$x(t)=\left(\frac{\sin 50 \pi t}{\pi t}\right)^{2}$$ which we wish to sample with a sampling frequency of \(\omega_{s}=150 \pi\) to obcain a signal \(g(t)\) with Fourier transform \(G(j \omega) .\) Determine the maximum value of \(\omega_{0}\) for which it is guaranteed that $$G(j \omega)=75 X(j \omega) \text { for }|\omega| \leq \omega_{0}$$ where \(X(j \omega)\) is the Fourier transform of \(x(t)\)

A continuous-time signal \(x(t)\) is obtained at the output of an ideal lowpass filter with cutoff frequency \(\omega_{c}=1,000 \pi\). If impulse-train sampling is performed on \(x(t)\) which of the following sampling periods would guarantee that \(x(t)\) can be recovered from its sampled version using an appropriate lowpass filter? (a) \(T=0.5 \times 10^{-3}\) (b) \(T=2 \times 10^{-3}\) (c) \(T=10^{-4}\)

Let \(x_{c}(t)\) be a continuous-time signal whose Fourier transform has the property that \(X_{c}(j \omega)=0\) for \(|\omega| \geq 2,000 \pi .\) A discrete- time signal $$x_{d}[n]=x_{c}\left(n\left(0.5 \times 10^{-3}\right)\right)$$ is obtained. For each of the following constraints on the Fourier transform \(X_{d}\left(e^{j \omega}\right)\) of \(x_{d}[n],\) determine the corresponding constraint on \(X_{c}(j \omega)\) (a) \(X_{d}\left(e^{j \omega}\right)\) is real (b) The maximum value of \(X_{d}\left(e^{j \omega}\right)\) over all \(\omega\) is 1 (c) \(X_{d}\left(e^{j \omega}\right)=0\) for \(\frac{3 \pi}{4} \leq|\omega| \leq \pi\) (d) \(X_{d}\left(e^{j \omega}\right)=X_{d}\left(e^{f(\omega-\pi)}\right)\)

Let \(x(f)\) be a band-limited signal such that \(X(j \omega)=0\) for \(|\omega| \geq \frac{\pi}{T}\) (a) If \(x(t)\) is sampled using a sampling period \(T,\) determine an interpolating function \(g(t)\) such that $$\frac{d x(t)}{d t}=\sum_{n=-\infty}^{\infty} x(n T) g(t-n T)$$ (b) Is the function \(g(t)\) unique?

The frequency which, under the sampling theorem, must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals: (a) \(x(t)=1+\cos (2,000 \pi t)+\sin (4,000 \pi t)\) (b) \(x(r)=\frac{\sin (4,000 \pi t)}{\pi t}\) (c) \(x(t)=\left\\{\begin{array}{c}\left.\frac{\sin (4,000 \pi t)}{\pi t}\right)^{2} \\ \end{array}\right.\)
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