Question

This is seismic data processing course. Explains the following: 1. Sampling in time domain produces duplicated...

This is seismic data processing course.

Explains the following:

1. Sampling in time domain produces duplicated versions of the signal in frequency domain.?

2. under - sampling of a signal will result in a new signal with low frequency.?

3. Multiples are characterized by their non - linear, which makes them difficult to recognize and eliminate in t - x domain and F-K domain.?

thank you. ??

Homework Answers

Answer #1

Frequency domain=

In physics, electronics, control systems engineering, and statistics, the frequency domain refers to the analysis of mathematical functions or signals with respect to frequency, rather than time.[1] Put simply, a time-domain graph shows how a signal changes over time, whereas a frequency-domain graph shows how much of the signal lies within each given frequency band over a range of frequencies. A frequency-domain representation can also include information on the phase shift that must be applied to each sinusoid in order to be able to recombine the frequency components to recover the original time signal.

A given function or signal can be converted between the time and frequency domains with a pair of mathematical operators called transforms. An example is the Fourier transform, which converts a time function into a sum or integral of sine waves of different frequencies, each of which represents a frequency component. The "spectrum" of frequency components is the frequency-domain representation of the signal. The inverse Fourier transform converts the frequency-domain function back to the time function. A spectrum analyzer is a tool commonly used to visualize electronic signals in the frequency domain.

Some specialized signal processing techniques use transforms that result in a joint time–frequency domain, with the instantaneous frequency being a key link between the time domain and the frequency domain.

The Fourier transform converts the function's time-domain representation, shown in red, to the function's frequency-domain representation, shown in blue. The component frequencies, spread across the frequency spectrum, are represented as peaks in the frequency domain.

Undersampling=

The Fourier transforms of real-valued functions are symmetrical around the 0 Hz axis. After sampling, only a periodic summation of the Fourier transform (called discrete-time Fourier transform) is still available. The individual frequency-shifted copies of the original transform are called aliases. The frequency offset between adjacent aliases is the sampling-rate, denoted by fs. When the aliases are mutually exclusive (spectrally), the original transform and the original continuous function, or a frequency-shifted version of it (if desired), can be recovered from the samples. The first and third graphs of Figure 1 depict a baseband spectrum before and after being sampled at a rate that completely separates the aliases.

The second graph of Figure 1 depicts the frequency profile of a bandpass function occupying the band (A, A+B) (shaded blue) and its mirror image (shaded beige). The condition for a non-destructive sample rate is that the aliases of both bands do not overlap when shifted by all integer multiples of fs. The fourth graph depicts the spectral result of sampling at the same rate as the baseband function. The rate was chosen by finding the lowest rate that is an integer sub-multiple of A and also satisfies the baseband Nyquist criterion: fs > 2B. Consequently, the bandpass function has effectively been converted to baseband. All the other rates that avoid overlap are given by these more general criteria, where A and A+B are replaced by fL and fH, respectively:[2][3]

{\displaystyle {\frac {2f_{H}}{n}}\leq f_{s}\leq {\frac {2f_{L}}{n-1}}}, for any integer n satisfying: {\displaystyle 1\leq n\leq \left\lfloor {\frac {f_{H}}{f_{H}-f_{L}}}\right\rfloor }

If the sampling theorem is interpreted as requiring twice the highest frequency, then the required sampling rate would be assumed to be greater than the Nyquist rate 216 MHz. While this does satisfy the last condition on the sampling rate, it is grossly oversampled.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions