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# Rotordynamics

Rotor dynamics is a specialized branch of applied mechanics concerned with the behavior and diagnosis of rotating structures. It is commonly used to analyze the behavior of structures ranging from jet engines and steam turbines to auto engines and computer disk storage. At its most basic level, rotor dynamics is concerned with one or more mechanical structures (rotors) supported by bearings and influenced by internal phenomena that rotate around a single axis. The supporting structure is called a stator. As the speed of rotation increases the amplitude of vibration often passes through a maximum that is called a critical speed. This amplitude is commonly excited by unbalance of the rotating structure; everyday examples include engine balance and tire balance. If the amplitude of vibration at these critical speeds is excessive, then catastrophic failure occurs. In addition to this, turbo machinery often develop instabilities which are related to the internal makeup of turbo machinery, and which must be corrected. This is the chief concern of engineers who design large rotors.

Rotating machinery produces vibrations depending upon the structure of the mechanism involved in the process. Any faults in the machine can increase or excite the vibration signatures. Vibration behavior of the machine due to imbalance is one of the main aspects of rotating machinery which must be studied in detail and considered while designing. All objects including rotating machinery exhibit natural frequency depending on the structure of the object. The critical speed of a rotating machine occurs when the rotational speed matches its natural frequency. The lowest speed at which the natural frequency is first encountered is called the first critical speed, but as the speed increases, additional critical speeds are seen. Hence, minimizing rotational unbalance and unnecessary external forces are very important to reducing the overall forces which initiate resonance. When the vibration is in resonance, it creates a destructive energy which should be the main concern when designing a rotating machine. The objective here should be to avoid operations that are close to the critical and pass safely through them when in acceleration or deceleration. If this aspect is ignored it might result in loss of the equipment, excessive wear and tear on the machinery, catastrophic breakage beyond repair or even human injury and loss of lives.

The real dynamics of the machine is difficult to model theoretically. The calculations are based on simplified models which resemble various structural components (lumped parameters models), equations obtained from solving models numerically (Rayleigh–Ritz method) and finally from the finite element method (FEM), which is another approach for modelling and analysis of the machine for natural frequencies. There are also some analytical methods, such as the distributed transfer function method,[1] which can generate analytical and closed-form natural frequencies, critical speeds and unbalanced mass response. On any machine prototype it is tested to confirm the precise frequencies of resonance and then redesigned to assure that resonance does not occur.

## Basic principles

The equation of motion, in generalized matrix form, for an axially symmetric rotor rotating at a constant spin speed Ω is

${\displaystyle {\begin{matrix}\mathbf {M} {\ddot {\mathbf {q} }}(t)+(\mathbf {C} +\mathbf {G} ){\dot {\mathbf {q} }}(t)+(\mathbf {K} +\mathbf {N} ){\mathbf {q} }(t)&=&\mathbf {f} (t)\\\end{matrix}}}$

where:

M is the symmetric Mass matrix

C is the symmetric damping matrix

G is the skew-symmetric gyroscopic matrix

K is the symmetric bearing or seal stiffness matrix

N is the gyroscopic matrix of deflection for inclusion of e.g., centrifugal elements.

in which q is the generalized coordinates of the rotor in inertial coordinates and f is a forcing function, usually including the unbalance.

The gyroscopic matrix G is proportional to spin speed Ω. The general solution to the above equation involves complex eigenvectors which are spin speed dependent. Engineering specialists in this field rely on the Campbell Diagram to explore these solutions.

An interesting feature of the rotordynamic system of equations are the off-diagonal terms of stiffness, damping, and mass. These terms are called cross-coupled stiffness, cross-coupled damping, and cross-coupled mass. When there is a positive cross-coupled stiffness, a deflection will cause a reaction force opposite the direction of deflection to react the load, and also a reaction force in the direction of positive whirl. If this force is large enough compared with the available direct damping and stiffness, the rotor will be unstable. When a rotor is unstable, it will typically require immediate shutdown of the machine to avoid catastrophic failure.

## Campbell diagram

Campbell Diagram for a Simple Rotor

The Campbell diagram, also known as "Whirl Speed Map" or a "Frequency Interference Diagram", of a simple rotor system is shown on the right. The pink and blue curves show the backward whirl (BW) and forward whirl (FW) modes, respectively, which diverge as the spin speed increases. When the BW frequency or the FW frequency equal the spin speed Ω, indicated by the intersections A and B with the synchronous spin speed line, the response of the rotor may show a peak. This is called a critical speed.

## Jeffcott rotor

The Jeffcott rotor (named after Henry Homan Jeffcott), also known as the de Laval rotor in Europe, is a simplified lumped parameter model used to solve these equations. The Jeffcott rotor is a mathematical idealization that may not reflect actual rotor mechanics.

## History

The history of rotordynamics is replete with the interplay of theory and practice. W. J. M. Rankine first performed an analysis of a spinning shaft in 1869, but his model was not adequate and he predicted that supercritical speeds could not be attained. In 1895, Dunkerley published an experimental paper describing supercritical speeds. Gustaf de Laval, a Swedish engineer, ran a steam turbine to supercritical speeds in 1889, and Kerr published a paper showing experimental evidence of a second critical speed in 1916.

Henry Jeffcott was commissioned by the Royal Society of London to resolve the conflict between theory and practice. He published a paper now considered classic in the Philosophical Magazine in 1919 in which he confirmed the existence of stable supercritical speeds. August Föppl published much the same conclusions in 1895, but history largely ignored his work.

Between the work of Jeffcott and the start of World War II there was much work in the area of instabilities and modeling techniques culminating in the work of Nils Otto Myklestad [2] and M. A. Prohl [3] which led to the transfer matrix method (TMM) for analyzing rotors. The most prevalent method used today for rotordynamics analysis is the finite element method.

Modern computer models have been commented on in a quote attributed to Dara Childs, "the quality of predictions from a computer code has more to do with the soundness of the basic model and the physical insight of the analyst. ... Superior algorithms or computer codes will not cure bad models or a lack of engineering judgment."

Prof. F. Nelson has written extensively on the history of rotordynamics and most of this section is based on his work.

## Software

There are many software packages that are capable of solving the rotor dynamic system of equations. Rotor dynamic specific codes are more versatile for design purposes. These codes make it easy to add bearing coefficients, side loads, and many other items only a rotordynamicist would need. The non-rotor dynamic specific codes are full featured FEA solvers, and have many years of development in their solving techniques. The non-rotor dynamic specific codes can also be used to calibrate a code designed for rotor dynamics.

Rotordynamic specific codes:

• Dynamics R4 (Alfa-Tranzit Co. Ltd) - Commercial software developed for design and analysis of spatial systems
• AxSTREAM RotorDynamics, ( SoftInWay) - Integrated Software platform for Rotor Dynamics, capable of lateral, torsional, and axial rotor dynamics for all widely used rotor types using the Finite Element Method on either beam or 2D-axisymmetric elements, and is capable of being automated.
• Rotortest, ( LAMAR - University of Campinas) - Finite Element Method based software, including different types of bearing solver. Developed by LAMAR (Laboratory of Rotating Machinery) - Unicamp (University of Campinas).
• SAMCEF ROTOR,(SAMCEF) - Software Platform for Rotors Simulation (LMS Samtech,A Siemens Business)
• MADYN (Consulting engineers Klement) - Commercial combined finite element lateral, torsional, axial and coupled solver for multiple rotors and gears, including foundation and housing.
• MADYN 2000 (DELTA JS Inc.) - Commercial combined finite element (3D Timoshenko beam) lateral, torsional, axial and coupled solver for multiple rotors and gears, foundations and casings (capability to import transfer functions and state space matrices from other sources), various bearings (fluid film, spring damper, magnetic, rolling element)
• iSTRDYN (DynaTech Software LLC) - Commercial 2-D Axis-symmetric finite element solver
• FEMRDYN (DynaTech Engineering, Inc.) - Commercial 1-D Axis-symmetric finite element solver
• DyRoBeS (Eigen Technologies, Inc.) - Commercial 1-D beam element solver
• RIMAP (RITEC) - Commercial 1-D beam element solver
• XLRotor (Rotating Machinery Analysis, Inc.) - Commercial 1-D beam element solver, including magnetic bearing control systems and coupled lateral-torsional analysis. A powerful, fast and easy to use tool for rotor dynamic modeling and analysis using Excel spreadsheets. Readily automated with VBA macros, plus a plugin for 3D CAD software.
• ARMD (Rotor Bearing Technology & Software, Inc.) - Commercial FEA-based software for rotordynamics, multi-branch torsional vibration, fluid-film bearings (hydrodynamic, hydrostatic, and hybrid) design, optimization, and performance evaluation, that is used worldwide by researchers, OEMs and end-users across all industries.
• XLTRC2 (Texas A&M) - Academic 1-D beam element solver
• ComboRotor (University of Virginia) - Combined finite element lateral, torsional, axial solver for multiple rotors evaluating critical speeds, stability and unbalance response extensively verified by industrial use
• MESWIR (Institute of Fluid-Flow Machinery, Polish Academy of Sciences) - Academic computer code package for analysis of rotor-bearing systems within the linear and non-linear range
• RoDAP (D&M Technology) - Commercial lateral, torsional, axial and coupled solver for multiple rotors, gears and flexible disks(HDD)
• ROTORINSA (ROTORINSA) - Commercial finite element software developed by a French engineering school (INSA-Lyon) for analysis of steady-state dynamic behavior of rotors in bending.
• COMSOL Multiphysics, Rotordynamics Module add-on (Rotordynamics Module)
• RAPPID - (Rotordynamics-Seal Research) Commercial finite element based software library (3D solid and beam elements) including rotordynamic coefficient solvers

Non-rotordynamic specific codes:

• Ansys - Version 11 workbench and classic is capable of solving the rotordynamic equations (3-D/2-D and beam element)
• Nastran - Finite element based (3-D/2-D and beam element)
• SAMCEF - Finite element based (3-D/2-D and beam element)
• eComputingX, Analytical solutions (Rotordynamics Toolbox)

## References

1. ^ Liu, Shibing; Yang, Bingen (2017-02-22). "Vibrations of Flexible Multistage Rotor Systems Supported by Water-Lubricated Rubber Bearings". Journal of Vibration and Acoustics. 139 (2): 021016–021016–12. doi:10.1115/1.4035136. ISSN 1048-9002.
2. ^ Myklestad, Nils (April 1944). "A New Method of Calculating Natural Modes of Uncoupled Bending Vibration of Airplane Wings and Other Types of Beams". Journal of the Aeronautical Sciences (Institute of the Aeronautical Sciences). 11 (2): 153–162. doi:10.2514/8.11116.
3. ^ Prohl, M. A. (1945), "A General Method for Calculating Critical Speeds of Flexible Rotors", Trans ASME, 66: A–142
• Chen, W. J., Gunter, E. J. (2005). Introduction to Dynamics of Rotor-Bearing Systems. Victoria, BC: Trafford. ISBN 978-1-4120-5190-3.CS1 maint: multiple names: authors list (link) uses DyRoBeS
• Childs, D. (1993). Turbomachinery Rotordynamics Phenomena, Modeling, & Analysis. Wiley. ISBN 978-0-471-53840-0.
• Fredric F. Ehrich (Editor) (1992). Handbook of Rotordynamics. McGraw-Hill. ISBN 978-0-07-019330-7.CS1 maint: extra text: authors list (link)
• Genta, G. (2005). Dynamics of Rotating Systems. Springer. ISBN 978-0-387-20936-4.
• Jeffcott, H. H. (1919). "The Lateral Vibration Loaded Shafts in the Neighborhood of a Whirling Speed. - The Effect of Want of Balance". Philosophical Magazine. 6. 37.
• Krämer, E. (1993). Dynamics of Rotors and Foundations. Springer-Verlag. ISBN 978-0-387-55725-0.
• Lalanne, M., Ferraris, G. (1998). Rotordynamics Prediction in Engineering, Second Edition. Wiley. ISBN 978-0-471-97288-4.CS1 maint: multiple names: authors list (link)
• Muszyńska, Agnieszka (2005). Rotordynamics. CRC Press. ISBN 978-0-8247-2399-6.
• Nelson, F. (June 2003). "A Brief History of Early Rotor Dynamics". Sound and Vibration.
• Nelson, F. (July 2007). "Rotordynamics without Equations". International Journal of COMADEM. 3. 10. ISSN 1363-7681.
• Nelson, F. (2011). An Introduction to Rotordynamics. SVM-19 [1].
• Lalanne, M., Ferraris, G. (1998). Rotordynamics Prediction in Engineering, Second Edition. Wiley. ISBN 978-0-471-97288-4.CS1 maint: multiple names: authors list (link)
• Vance, John M. (1988). Rotordynamics of Turbomachinery. Wiley. ISBN 978-0-471-80258-7.
• Tiwari, Rajiv (2017). Rotor Systems: Analysis and Identification. CRC Press. ISBN 9781138036284.
• Vance, John M., Murphy, B., Zeidan, F. (2010). Machinery Vibration and Rotordynamics. Wiley. ISBN 978-0-471-46213-2.CS1 maint: multiple names: authors list (link)
• Vollan, A., Komzsik, L. (2012). Computational Techniques of Rotor Dynamics with the Finite Element Method. CRC Press. ISBN 978-1-4398-4770-1.CS1 maint: multiple names: authors list (link)
• Yamamoto, T., Ishida, Y. (2001). Linear and Nonlinear Rotordynamics. Wiley. ISBN 978-0-471-18175-0.CS1 maint: multiple names: authors list (link)
• Ganeriwala, S., Mohsen N (2008). Rotordynamic Analysis using XLRotor. SQI03-02800-0811


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# -1 Solvers

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Apparently this problem has -1 solvers. This is because I posted the problem with an incorrect answer, then I reported it myself to change it to the correct answer, and a moderator confirmed it. Now my attempt was considered incorrect, so a -1 tally was added to the amount of solvers. This brought the number of solvers to -1.

Note by Daniel Liu
3 years, 2 months ago

Sort by:

Ok, the glitch is gone now. Did somebody fix it or was it just reset by itself? · 3 years, 2 months ago

The creator of a problem is an edge case that has to be dealt with separately. In this case, the code was present but took time to update. Staff · 3 years, 2 months ago

Also, is the red banner gone on my problem? It still displays a red banner (although not the usual one) when I look at it:

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Also, the report button is permanently used:

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· 3 years, 2 months ago

I've just removed the banner (not at my computer, so don't have easy access).

The red report button indicates that you have filed a report on the question. Staff · 3 years, 2 months ago

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text:
Corpus ID: 14257218

# The average singular value of a complex random matrix decreases with dimension

@article{Abreu2016TheAS,
title={The average singular value of a complex random matrix decreases with dimension},
author={L. D. Abreu},
journal={ArXiv},
year={2016},
volume={abs/1606.00494}
}
• L. D. Abreu
• Published 2016
• Mathematics, Computer Science
• ArXiv
• We obtain a recurrence relation in $d$ for the average singular value $% \alpha (d)$ of a complex valued $d\times d$\ matrix $\frac{1}{\sqrt{d}}X$ with random i.i.d., N( 0,1) entries, and use it to show that $\alpha (d)$ decreases monotonically with $d$ to the limit given by the Marchenko-Pastur distribution.\ The monotonicity of $\alpha (d)$ has been recently conjectured by Bandeira, Kennedy and Singer in their study of the Little Grothendieck problem over the unitary group \$\mathcal{U}_{d… CONTINUE READING
1 Citations

#### Paper Mentions

Multitaper spectral estimation and off-grid compressive sensing: MSE estimates
• Mathematics
• 2017 International Conference on Sampling Theory and Applications (SampTA)
• 2017

#### References

SHOWING 1-10 OF 20 REFERENCES
Approximating the little Grothendieck problem over the orthogonal and unitary groups
• Mathematics, Computer Science
• Math. Program.
• 2016
• 32
• Highly Influential
• PDF
Zeros of Gaussian Analytic Functions and Determinantal Point Processes
• Mathematics, Computer Science
• University Lecture Series
• 2009
• 378
• PDF
Random Matrix Theory and Wireless Communications
• Mathematics, Computer Science
• Found. Trends Commun. Inf. Theory
• 2004
• 1,778
• PDF
Random Matrices
• Physics
• 2000
• 3,018
• PDF


text:
# Numerical Quadrature of Oscillating Integral With Non-oscillating part

As you will know there are different numerical integrals (I believe Levin's method is the most popular one) for the numerical quadrature of oscillating integrands which may roughly speaking be written as a product of oscillating part and non-oscillating amplitude $exp(i \omega t) A(t)$ (the form can of course be slightly more general). Integrands of the form $exp(i \omega t) A(t)+B(t)$ may be treated using a method for oscillatory and one for non-oscillatory for the first and second part, respectively. I wonder if there are routines also for the efficient quadrature of integrands which cannot be analytically decomposed as $exp(i \omega t) A(t)+B(t)$ but which nevertheless feature a rapidly oscillatory behavior (you can assume fixed frequency) . I.e. they behave like $exp(i \omega t) A(t)+B(t)$ with some unknown $A$ and $B$. It appears to me that I could in principle do better than using a conventional quadrature for non-oscillating integrands as I could try to find $A$ and $B$ using smooth (and successively refining) fits and then run a quadrature for oscillating integrals in order to obtain a numerical estimate for the integral. I wonder if there are any known efficient methods for such problems.

• I don't have an answer, but I suggest you have a look at the publications of Daan Huybrechts on oscillatory integrals. I think he or his group at KULeuven are still quite active in the field. – GertVdE Feb 27 '14 at 9:27
• Your reference has already been helpful. Browsing it I found this math.ethz.ch/~hiptmair/Seminars/OSCINT/OLV06b.pdf which deals with multivariate quadrature of oscillatory integrands. It reminds me to mention that my original integral is similar, a two dimensional one on a certain domain. To perform the first integral which is strictly oscillatory (i.e.) I can use Levin quadrature. The remaining integral is of the form I describe above. So references which deal with 2-dim integrals are also useful. However my integrand can feature integrable singularities. – highsciguy Feb 27 '14 at 18:16
• Some basic literature would be good which explains the terms used in the paper, like 'non-resonance condition' etc. – highsciguy Feb 27 '14 at 18:18


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# Can you use Field Oriented Control with BLDC and if so what are the disadvantages?

From what I understand Field Oriented Control is meant to be used with PMSM as PMSM has sinusoidal back EMF due to distributed stator windings.

On the other hand BLDC's have concentrated stator windings resulting in a trapezoidal back EMF.

Six Step commutation uses strictly two activated stator windings, 1 deactivated winding (lets assume 3 stator 1 pole pair motor). Whereas FOC uses variable current through all the coils to create a stator magnetic field that is always orthogonal to the rotor.

Does the concentrated structure of the BLDC stator windings make FOC performance worse, and if so why and how does it make it worse?

Imagine you have a DC motor. $$$$\tau=ki$$$$ Now apply a rectified 3 phase sine wave current to the motor to get the average torque. $$$$\tau_{avg}=\frac{3\sqrt{3}}{2\pi}ki_{peak}\\$$$$ Now imagine that there will be some min and max current applied from the rectified sine wave current. This will generate the torque ripple. https://www.electronics-tutorials.ws/power/three-phase-rectification.html.


text:
Doctoral Dissertations

Title

Numerical and theoretical investigation of the variational formulation of a water wave problem

January 1997

Mathematics

Ph.D.

Abstract

A function $\eta$ is a wave shape if it, along with its streamfunction, satisfies Bernoulli's condition. Thus far only constant, or flat water, shapes are guaranteed to exist. Intuition suggests that there should be non-constant shapes. Here we provide and investigate a variational interpretation of this problem. We numerically study the water wave problem at various vorticities and pressure values. The approach proves well-suited for the implementation of numerical Mountain Pass techniques developed by Choi and McKenna and, using such techniques, we locate numerous non-constant solutions. ^

COinS


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 1 0 0 0.000%

## 문제

There are a lot of mountain ranges in the Neverlands. Every mountain range consists of a number of valleys and summits. The slope between two consecutive summit and valley is always either 1 or -1, and all the summits and the valleys have integer heights. A bowling ball is swinging on a part of a mountain range consisting of n valleys and n-1 summits. The ball is always touching down the surface of the mountain range (it does not jump). Mountains before the first and after the last valleys are too high such that the ball can never exit the mountain range. At time t0 the ball is located in the valley number s moving to the upper-right direction and it has an initial kinetic energy K0. The following figure shows a mountain range with 4 valleys and 3 summits, and the ball located in the 2nd valley (enumerated from left to right).

By the simple physics we know that at any time t the ball has a gravitational potential energy Pt=mgh and also a kinetic energy Kt=(1/2)mv2, where m is the mass of the ball, g is the constant of the Earth gravity (here equals to 10), and h and v are the height and the velocity of the ball at time t. By the transformation of energy from potential to kinetic or vice versa, the total energy of the ball Pt+Kt is fixed during its movements, unless it falls into a valley: at the ith valley (from left), ci units of the kinetic energy is lost due to friction or it stops if its kinetic energy is below ci but consider no friction in other locations. Note that the ball loses cs unit of energy when it leaves the starting valley at time t0. You can assume the ball diameter is equal to 0 and its mass is equal to 1. Your task is to find the valley or summit at which the ball will stop.

## 입력

There are multiple test cases in the input. The first line of each test case contains three space-separated positive integers n, and s and K0(1 ≤ n ≤ 3000, 1 ≤ s ≤ n, 1 ≤ K0 ≤ 1015). Each of the following n lines contains two integers hi and ci, the height and friction of the ith valley. The jth line of the next n-1 lines contains Hj, the height of The jth summit from the left (0 ≤ hi,ci,Hj ≤ 109). It is guaranteed that at least one of cis is greater than zero. The input terminates with “0 0 0 0” which should not be processed.

## 출력

For each test case, output a line conforming one of the following formats depending on whether the ball stops at either a valley or a summit.

• If the ball stops at valley number k, output “Valley: k” (omit the quotes.)
• If the ball stops at Summit number k, output “Summit: k (omit the quotes.)

## 예제 입력 1

4 2 17
1 1
2 1
1 1
1 2
3
3
2
1 1 1000000000000000
1 1
0 0 0 0

## 예제 출력 1

Summit: 2
Valley: 1


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# Linear regression without scikit-learn#

In this notebook, we introduce linear regression. Before presenting the available scikit-learn classes, we will provide some insights with a simple example. We will use a dataset that contains measurements taken on penguins.

Note

If you want a deeper overview regarding this dataset, you can refer to the Appendix - Datasets description section at the end of this MOOC.

import pandas as pd

Flipper Length (mm) Body Mass (g)
0 181.0 3750.0
1 186.0 3800.0
2 195.0 3250.0
3 193.0 3450.0
4 190.0 3650.0

We will formulate the following problem: using the flipper length of a penguin, we would like to infer its mass.

import seaborn as sns

feature_name = "Flipper Length (mm)"
target_name = "Body Mass (g)"
data, target = penguins[[feature_name]], penguins[target_name]

ax = sns.scatterplot(data=penguins, x=feature_name, y=target_name,
color="black", alpha=0.5)
ax.set_title("Body Mass as a function of the Flipper Length")

Text(0.5, 1.0, 'Body Mass as a function of the Flipper Length')

Tip

The function scatterplot from seaborn take as input the full dataframe and the parameter x and y allows to specify the name of the columns to be plotted. Note that this function returns a matplotlib axis (named ax in the example above) that can be further used to add elements on the same matplotlib axis (such as a title).

In this problem, penguin mass is our target. It is a continuous variable that roughly varies between 2700 g and 6300 g. Thus, this is a regression problem (in contrast to classification). We also see that there is almost a linear relationship between the body mass of the penguin and its flipper length. The longer the flipper, the heavier the penguin.

Thus, we could come up with a simple formula, where given a flipper length we could compute the body mass of a penguin using a linear relationship of the form y = a * x + b where a and b are the 2 parameters of our model.

def linear_model_flipper_mass(flipper_length, weight_flipper_length,
intercept_body_mass):
"""Linear model of the form y = a * x + b"""
body_mass = weight_flipper_length * flipper_length + intercept_body_mass
return body_mass

Using the model we defined above, we can check the body mass values predicted for a range of flipper lengths. We will set weight_flipper_length to be 45 and intercept_body_mass to be -5000.

import numpy as np

weight_flipper_length = 45
intercept_body_mass = -5000

flipper_length_range = np.linspace(data.min(), data.max(), num=300)
predicted_body_mass = linear_model_flipper_mass(
flipper_length_range, weight_flipper_length, intercept_body_mass)

We can now plot all samples and the linear model prediction.

label = "{0:.2f} (g / mm) * flipper length + {1:.2f} (g)"

ax = sns.scatterplot(data=penguins, x=feature_name, y=target_name,
color="black", alpha=0.5)
ax.plot(flipper_length_range, predicted_body_mass)
_ = ax.set_title(label.format(weight_flipper_length, intercept_body_mass))

The variable weight_flipper_length is a weight applied to the feature flipper_length in order to make the inference. When this coefficient is positive, it means that penguins with longer flipper lengths will have larger body masses. If the coefficient is negative, it means that penguins with shorter flipper lengths have larger body masses. Graphically, this coefficient is represented by the slope of the curve in the plot. Below we show what the curve would look like when the weight_flipper_length coefficient is negative.

weight_flipper_length = -40
intercept_body_mass = 13000

predicted_body_mass = linear_model_flipper_mass(
flipper_length_range, weight_flipper_length, intercept_body_mass)

We can now plot all samples and the linear model prediction.

ax = sns.scatterplot(data=penguins, x=feature_name, y=target_name,
color="black", alpha=0.5)
ax.plot(flipper_length_range, predicted_body_mass)
_ = ax.set_title(label.format(weight_flipper_length, intercept_body_mass))

In our case, this coefficient has a meaningful unit: g/mm. For instance, a coefficient of 40 g/mm, means that for each additional millimeter in flipper length, the body weight predicted will increase by 40 g.

body_mass_180 = linear_model_flipper_mass(
flipper_length=180, weight_flipper_length=40, intercept_body_mass=0)
body_mass_181 = linear_model_flipper_mass(
flipper_length=181, weight_flipper_length=40, intercept_body_mass=0)

print(f"The body mass for a flipper length of 180 mm "
f"is {body_mass_180} g and {body_mass_181} g "
f"for a flipper length of 181 mm")

The body mass for a flipper length of 180 mm is 7200 g and 7240 g for a flipper length of 181 mm

We can also see that we have a parameter intercept_body_mass in our model. This parameter corresponds to the value on the y-axis if flipper_length=0 (which in our case is only a mathematical consideration, as in our data, the value of flipper_length only goes from 170mm to 230mm). This y-value when x=0 is called the y-intercept. If intercept_body_mass is 0, the curve will pass through the origin:

weight_flipper_length = 25
intercept_body_mass = 0

# redefined the flipper length to start at 0 to plot the intercept value
flipper_length_range = np.linspace(0, data.max(), num=300)
predicted_body_mass = linear_model_flipper_mass(
flipper_length_range, weight_flipper_length, intercept_body_mass)

ax = sns.scatterplot(data=penguins, x=feature_name, y=target_name,
color="black", alpha=0.5)
ax.plot(flipper_length_range, predicted_body_mass)
_ = ax.set_title(label.format(weight_flipper_length, intercept_body_mass))

Otherwise, it will pass through the intercept_body_mass value:

weight_flipper_length = 45
intercept_body_mass = -5000

predicted_body_mass = linear_model_flipper_mass(
flipper_length_range, weight_flipper_length, intercept_body_mass)

ax = sns.scatterplot(data=penguins, x=feature_name, y=target_name,
color="black", alpha=0.5)
ax.plot(flipper_length_range, predicted_body_mass)
_ = ax.set_title(label.format(weight_flipper_length, intercept_body_mass))

In this notebook, we have seen the parametrization of a linear regression model and more precisely meaning of the terms weights and intercepts.


text:
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If xy is not equal to 0, then which of the following is the equivalent

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If xy is not equal to 0, then which of the following is the equivalent [#permalink]

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05 Sep 2017, 00:46
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If xy is not equal to 0, then which of the following, is the equivalent of $$\frac{(x^2y)^3+x^5y^3}{(\frac{x}{y})^{-1}}$$?

A. $$x^6y^2(x + 1)$$

B. $$x^4y^4(x + 1)$$

C. $$2x^4y^4$$

D. $$2x^5y^3$$

E. $$2x^6y^2$$

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Re: If xy is not equal to 0, then which of the following is the equivalent [#permalink]

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05 Sep 2017, 01:46
Bunuel wrote:
If xy is not equal to 0, then which of the following, is the equivalent of $$\frac{(x^2y)^3+x^5y^3}{(\frac{x}{y})^{-1}}$$?

A. $$x^6y^2(x + 1)$$

B. $$x^4y^4(x + 1)$$

C. $$2x^4y^4$$

D. $$2x^5y^3$$

E. $$2x^6y^2$$

X^5y^3 + x^5y^3 / (y/x)

= x^5y^2(x+1) *x/y
= x^6y^2(x+1)

A should be correct.

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Re: If xy is not equal to 0, then which of the following is the equivalent [#permalink]

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05 Sep 2017, 08:05
Bunuel wrote:
If xy is not equal to 0, then which of the following, is the equivalent of $$\frac{(x^2y)^3+x^5y^3}{(\frac{x}{y})^{-1}}$$?

A. $$x^6y^2(x + 1)$$

B. $$x^4y^4(x + 1)$$

C. $$2x^4y^4$$

D. $$2x^5y^3$$

E. $$2x^6y^2$$

$$\frac{(x^2y)^3+x^5y^3}{(\frac{x}{y})^{-1}}$$

$$\frac{x^6y^3+x^5y^3}{(\frac{y}{x})}$$

$$\frac{x^5y^3(x +1)}{(\frac{y}{x})}$$

$$(x*y^{-1})x^5y^3(x +1)$$

$$x^6y^2(x +1)$$

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If xy is not equal to 0, then which of the following is the equivalent [#permalink]

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06 Sep 2017, 13:42
Bunuel wrote:
If xy is not equal to 0, then which of the following, is the equivalent of $$\frac{(x^2y)^3+x^5y^3}{(\frac{x}{y})^{-1}}$$?

A. $$x^6y^2(x + 1)$$

B. $$x^4y^4(x + 1)$$

C. $$2x^4y^4$$

D. $$2x^5y^3$$

E. $$2x^6y^2$$

It took me a long time to figure out that the denominator is $$(\frac{x}{y})^{-1}$$, not $$(\frac{x}{y}) - 1$$

$$\frac{(x^2y)^3+x^5y^3}{(\frac{x}{y})^{-1}}$$ =

$$\frac{x^6y^3+x^5y^3}{(\frac{y}{x})}$$ =

$$(\frac{x}{y}) * (x^5y^3)(x + 1)$$ =

$$(x^1y^{-1}) * (x^5y^3)(x + 1)$$=

$$x^{(1+5)}y^{(-1+3)} * (x + 1)$$

$$x^6y^2(x + 1)$$

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Re: If xy is not equal to 0, then which of the following is the equivalent [#permalink]

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07 Sep 2017, 16:41
Bunuel wrote:
If xy is not equal to 0, then which of the following, is the equivalent of $$\frac{(x^2y)^3+x^5y^3}{(\frac{x}{y})^{-1}}$$?

A. $$x^6y^2(x + 1)$$

B. $$x^4y^4(x + 1)$$

C. $$2x^4y^4$$

D. $$2x^5y^3$$

E. $$2x^6y^2$$

Let’s first simplify the numerator of the given expression:

(x^6)(y^3) + (x^5)(y^3)

(x^5)(y^3)(x + 1)

Simplifying the denominator, we have: (x/y)^-1 = y/x

Now we have:

(x^5)(y^3)(x + 1)/(y/x) = (x^5)(y^3)(x + 1)*(x/y) = (x^6)(y^2)(x + 1)

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Re: If xy is not equal to 0, then which of the following is the equivalent   [#permalink] 07 Sep 2017, 16:41
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text:
## List of models supported by timm

timm supports a wide variety of pretrained and non-pretrained models for number of Image based tasks.

To get a complete list of models, use the list_models function from timm as below. The list_models function returns a list of models ordered alphabetically that are supported by timm. We just look at the top-5 models below.

import timm

timm.list_models()[:5]

['adv_inception_v3',
'cspdarknet53',
'cspdarknet53_iabn',
'cspresnet50',
'cspresnet50d']

In general, you always want to use factory functions inside timm. Particularly, you want to use create_model function from timm to create any model. It is possible to create any of the models listed in timm.list_models() using the create_model function. There are also some wonderful extra features that we will look at later. But, let's see a quick example.

import random
import torch

random_model_to_create = random.choice(timm.list_models())
random_model_to_create

'resnet50d'
model = timm.create_model(random_model_to_create)
x     = torch.randn(1, 3, 224, 224)
model(x).shape

torch.Size([1, 1000])

In the example above, we randomly select a model name in timm.list_models(), create it and pass some dummy input data through the model to get some output. In general, you never want to create random models like this, and it's only an example to showcase that all models in timm.list_models() are supported by timm.create_model() function. It's really that easy to create a model using timm.

## Does timm have pretrained weights for these models?

Of course! timm wants to make it super easy for researchers and practioners to experiment and supports a whole lot of models with pretrained weights. These pretrained weights are either:

1. Directly used from their original sources
2. Ported by Ross from their original implementation in a different framework (e.g. Tensorflow models)
3. Trained from scratch using the included training script (train.py). The exact commands with hyperparameters to train these individual models are mentioned under Training Scripts.

To list all the models that have pretrained weights, timm provides a convenience parameter pretrained that could be passed in list_models function as below. We only list the top-5 returned models.

timm.list_models(pretrained=True)[:5]

['adv_inception_v3',
'cspdarknet53',
'cspresnet50',
'cspresnext50',
'densenet121']

## My dataset doesn't consist of 3-channel images - what now?

As you might already know, ImageNet data consists of 3-chanenl RGB images. Therefore, to be able to use pretrained weights in most libraries, the model expects a 3-channel input image.

### torchvision raises Exception

import torchvision

m = torchvision.models.resnet34(pretrained=True)

# single-channel image (maybe x-ray)
x = torch.randn(1, 1, 224, 224)

# torchvision raises error
try: m(x).shape
except Exception as e: print(e)

Given groups=1, weight of size [64, 3, 7, 7], expected input[1, 1, 224, 224] to have 3 channels, but got 1 channels instead

As can be seen above, these pretrained weights from torchvision won't work with single channel input images. As a work around most practitioners convert their single channel input images to 3-channel images by copying the single channel pixels accross to create a 3-channel image.

Basically, torchvision above is complaining that it expects the input to have 3 channels, but got 1 channel instead.

# 25-channel image (maybe satellite image)
x = torch.randn(1, 25, 224, 224)

# torchvision raises error
try: m(x).shape
except Exception as e: print(e)

Given groups=1, weight of size [64, 3, 7, 7], expected input[1, 25, 224, 224] to have 3 channels, but got 25 channels instead

Again, torchvision raises an error and this time there is no workaround to get past this error apart from just not using pretrained weights and starting with randomly initialized weights.

### timm has a way to handle these exceptions

m = timm.create_model('resnet34', pretrained=True, in_chans=1)

# single channel image
x = torch.randn(1, 1, 224, 224)

m(x).shape

torch.Size([1, 1000])

We pass in a parameter in_chans to the timm.create_model function and this somehow just magically works! Let's see what happens with the 25-channel image?

m = timm.create_model('resnet34', pretrained=True, in_chans=25)

# 25-channel image
x = torch.randn(1, 25, 224, 224)

m(x).shape

torch.Size([1, 1000])

This works again! :)

### How is timm able to use pretrained weights and handle images that are not 3-channel RGB images?

timm does all this magic inside the load_pretrained function that get's called to load the pretrained weights of a model. Let's see how timm achieves loading of pretrained weights.

from timm.models.resnet import ResNet, BasicBlock, default_cfgs
from copy import deepcopy

Below, we create a simple resnet34 model that can take single channel images as input. We make this happen by passing in in_chans=1 to the ResNet constructor class when creating the model.

resnet34_default_cfg = default_cfgs['resnet34']
resnet34 = ResNet(BasicBlock, layers=[3, 4, 6, 3], in_chans=1)
resnet34.default_cfg = deepcopy(resnet34_default_cfg)

resnet34.conv1

Conv2d(1, 64, kernel_size=(7, 7), stride=(2, 2), padding=(3, 3), bias=False)
resnet34.conv1.weight.shape

torch.Size([64, 1, 7, 7])

As we can see from the first convolution of resnet34 above, the number of input channels is set to 1. And the conv1 weights are of shape [64, 1, 7, 7]. This means that the number of input channels is 1, output channels is 64 and kernel size is 7x7.

But what about the pretrained weights? Because ImageNet consists of 3-channel input images, the pretrained for this conv1 layer would be [64, 3, 7, 7].Let's confirm that below:

resnet34_default_cfg

{'url': 'https://github.com/rwightman/pytorch-image-models/releases/download/v0.1-weights/resnet34-43635321.pth',
'num_classes': 1000,
'input_size': (3, 224, 224),
'pool_size': (7, 7),
'crop_pct': 0.875,
'interpolation': 'bilinear',
'mean': (0.485, 0.456, 0.406),
'std': (0.229, 0.224, 0.225),
'first_conv': 'conv1',
'classifier': 'fc'}

Let's load the pretrained weights from the model and check the number of input channels that conv1 expects.

import torch

Great, so we have loaded the pretrained weights of resnet-34 from 'https://github.com/rwightman/pytorch-image-models/releases/download/v0.1-weights/resnet34-43635321.pth' URL, let's now check the shape of the weights for conv1 below:

state_dict['conv1.weight'].shape

torch.Size([64, 3, 7, 7])

So this layer expects the number of input channels to be 3!

### So how is timm able to load these weights?

Something very clever happens inside the load_pretrained function inside timm. Basically, there's two main cases to consider when the expected number of input channels is not equal to 3. Either the input channels are 1 or not. Let's what happens in either case.

When the number of input channels is not equal to 3, then timm updates the conv1.weight of the pretrained weights accordingly to be able to load the pretrained weights.

#### Case-1: When the number of input channels is 1

If the number of input channels is 1, timm simply sums the 3 channel weights into a single channel to update the shape of conv1.weight to be [64, 1, 7, 7]. This can be achieved like so:

conv1_weight = state_dict['conv1.weight']
conv1_weight.sum(dim=1, keepdim=True).shape

>> torch.Size([64, 1, 7, 7])

And thus by updating the shape of the first conv1 layer, we can now safely load these pretrained weights.

#### Case-2: When the number of input channels is not 1

In this case, we simply repeat the conv1_weight as many times as required and then select the required number of input channels weights.

As can be seen in the image above, let's say our input images have 8 channels. Therefore, number of input channels is equal to 8.

But, as we know our pretrained weights only have 3 channels. So how could we still make use of the pretrained weights?

Well, what happens in timm has been shown in the image above. We copy the weights 3 times such that now the total number of channels becomes 9 and then we select the first 8 channels as our weights for conv1 layer.

This is all done inside load_pretrained function like so:

conv1_name = cfg['first_conv']
conv1_weight = state_dict[conv1_name + '.weight']
conv1_type = conv1_weight.dtype
conv1_weight = conv1_weight.float()
repeat = int(math.ceil(in_chans / 3))
conv1_weight = conv1_weight.repeat(1, repeat, 1, 1)[:, :in_chans, :, :]
conv1_weight *= (3 / float(in_chans))
conv1_weight = conv1_weight.to(conv1_type)
state_dict[conv1_name + '.weight'] = conv1_weight

Thus, as can be seen above, we first repeat the conv1_weight and then select required number of in_chans from these copied weights.


text:
MTM - Maple Programming Help

Home : Support : Online Help : Connectivity : MTM Package : MTM/disp

MTM

 disp
 display objects

 Calling Sequence disp(expr)

Parameters

 expr - expression

Description

 • The disp command displays a Maple object via the print and/or printf commands.  Arrays with more than 2 dimensions are displayed in 2-dimensional slices.

Examples

 > $\mathrm{with}\left(\mathrm{MTM}\right):$
 > $\mathrm{disp}\left(\mathrm{Array}\left(1..3,1..2,1..2\right)\right)$
 (:,:,1) =
 (:,:,2) =
 $\left[\begin{array}{rr}{0}& {0}\\ {0}& {0}\\ {0}& {0}\end{array}\right]$
 $\left[\begin{array}{rr}{0}& {0}\\ {0}& {0}\\ {0}& {0}\end{array}\right]$ (1)


text:
## Calculus (3rd Edition)

We are given: $f(x)$ discontinuous in $x=c$ $g(x)$ discontinuous in $x=c$ We have to check in $f(x)+g(x)$ is also discontinuous in $c$. Consider the example: $f(x)=\begin{cases} x+1,\text{ for }x\leq 2\\ 2x+1,\text{ for }x>2 \end{cases}$ $g(x)=\begin{cases} -2x+2,\text{ for }x\leq 2\\ -x-2,\text{ for }x>2 \end{cases}$ Compute $f(x)+g(x)$: $f(x)+g(x)=\begin{cases} x+1-2x+2,\text{ for }x\leq 2\\ 2x+1-x-2,\text{ for }x>2 \end{cases}$ $f(x)+g(x)=\begin{cases} -x+3,\text{ for }x\leq 2\\ x-1,\text{ for }x>2 \end{cases}$ The functions $f$ and $g$ have a jump discontinuity in $x=2$. We check if $f+g$ is continuous in $x=2$: $\displaystyle\lim_{x\rightarrow 2^{-}} (f(x)+g(x))=\displaystyle\lim_{x\rightarrow 2^{-}} (-x+3)=1$ $\displaystyle\lim_{x\rightarrow 2^{+}} (f(x)+g(x))=\displaystyle\lim_{x\rightarrow 2^{-}} (x-1)=1$ $(f+g)(2)=-2+3=1$ As the left and right hand limits and the value of $(f_g)$ in $x=2$ are equal, the sum function is continuous in $x=2$. Therefore the statement is FALSE.


text:
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# If the area of the right triangle above is 18, what is the value of x?

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If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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01 Sep 2017, 07:39
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If the area of the right triangle above is 18, what is the value of x?

(A) 60
(B) 50
(C) 45
(D) 30
(E) cannot be determined

Attachment:

2017-09-01_1836.png [ 9.91 KiB | Viewed 1045 times ]

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Re: If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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01 Sep 2017, 07:49
Bunuel wrote:

If the area of the right triangle above is 18, what is the value of x?

(A) 60
(B) 50
(C) 45
(D) 30
(E) cannot be determined

Attachment:
2017-09-01_1836.png

Length of height $$= 6$$

Let the length of base be $$= a$$

Area $$= \frac{1}{2} * base *height = \frac{1}{2} * 6 * a = 18$$

$$3a = 18$$

$$a = \frac{18}{3} = 6$$

Therefore its isosceles right triangle.

$$x = 45$$

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If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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01 Sep 2017, 14:59
Similarly, the problem can be solved through trigonometry:
Given that tg(alpha)= (cathetus #1 / cathetus #2), and, given that the unknown cathetus of this triangle can be computed via the above suggested method, we are settled down to observe that (cathetus #1 / cathetus #2), in fact, obtains the value of 1. The direct implication of this derivation is - guess what? - the value of the angle of our concern - 45 degrees.

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Re: If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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07 Sep 2017, 13:59
In right angle triangle , perpendicular is 6 and Area of triangle is 18.

thus , 1/2 * B * H = 18
B = 6

As two sides of triangle are equal , it is known as right isosceles triangle.
Now, the value of x = 45

Option : C
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Re: If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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12 Sep 2017, 02:05
Hi, I appreciate your help with the following:

The question asks regarding the value of x when the triangle's area is above 18 (i.e., when x > 6 because when a question says Area > 18 this means that 1/2 x b x h > 18 = 1/2 x b x 6 > 18 = 1/2 b > 3 -> b > 6) therefore the area is 18 and not above. I frankly don't quite get the point here?!
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Posts: 47112
Re: If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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12 Sep 2017, 02:14
Rabz wrote:
Hi, I appreciate your help with the following:

The question asks regarding the value of x when the triangle's area is above 18 (i.e., when x > 6 because when a question says Area > 18 this means that 1/2 x b x h > 18 = 1/2 x b x 6 > 18 = 1/2 b > 3 -> b > 6) therefore the area is 18 and not above. I frankly don't quite get the point here?!

"Above" here does not mean "more than", it refers to the triangle which is above the text.
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Re: If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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12 Sep 2017, 03:29
Given a rt angle triangle : Area = 1/2 * 6 * base = 18 -----> Base = 6
tan x = 6/6 ----> x = tan-1(1) ----> x= 45
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Re: If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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15 Sep 2017, 10:13
Bunuel wrote:

If the area of the right triangle above is 18, what is the value of x?

(A) 60
(B) 50
(C) 45
(D) 30
(E) cannot be determined

Recall that the formula for the area of a triangle is A = bh/2. We can let b = the base of the triangle and create the following equation:

(6 x b)/2 = 18

6b = 36

b = 6

We see that the triangle is an isosceles right triangle. Since x is the measure of an acute angle of an isosceles right triangle, it must be 45.

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Re: If the area of the right triangle above is 18, what is the value of x?  [#permalink]

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15 Sep 2017, 10:58
1/2 Base * Height is the area of the triangle.

-- 6 (height) * 3 (half the base) = 18 so the base is 6

If two sides of a right triangle are equal, their angles must correspond with 45, 45, 90. Hence, the right answer is (C), 45 degrees.
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text:
Prove that $c_0$ is closed in $\ell^\infty$

Define $c_0$ to be the space of real sequences in $\ell^\infty$ that converge to $0$ and $\ell^\infty$ be the set of all bounded sequences. Show that $c_0$ is closed in $\ell^\infty$.

I am reading in the solution in here and I don't understand how we define $$x^{k}_n - \omega_n$$

Because $(x_n^k) = ( \{x^1_1,x_2^1,\dots \}, \{x_1^2, x_2^2,\dots \}, \dots)$ and $(\omega_n) = (\omega_1, \omega_2, \dots)$

The objects in $(x_n^k)$ are sequences (sets) and the objects in $\omega_n$ are real numbers, how can you subtract a number from a set?

No, $x^{(k)}$ is assumed to be a sequence of points of $c_0$. Here $k$ is the index. So $x^{(1)}$ is a sequence (namely a point in $c_0$), which has itself indices, which are denoted by $n$. So we write $x^{(1)} = (x^{(1)}_0, x^{(1)}_1,x^{(1)}_2,\ldots)$ .So every $x^{(k)}_n$ is indeed some real number: the $n$-th member of the sequence that is the point $x^{(k)}$ of $c_0$.
Note that there is a difference between writing $x^{(k)}$ (a single point, i.e. a single sequence), and $(x^{(k)})$, i.e. a sequence of points (sequences); the latter should really be written $(x^{(k)})_k$, so show the index set (the one that runs), especially when we make the indices of each sequence point explicit: $(x^{(k)}_n)_k$.
• As I said, this is true. So $x^{(k)}_n$ is a number. – Henno Brandsma Sep 20 '14 at 21:41
• Yes, I agree $x^{k}$ is a number. But we are dealing with the collection $(x^{k}_n)_{k \geq 1}$ which is a set of sequences. – Hawk Sep 20 '14 at 21:44
• $x^{(k)}$ is not a number, but a sequence (=point). The proof writes $|x^{(k)}_n - \omega_n|$, which is a simple absolute value of a difference between reals. Note that $k$ and $n$ are bound variables there! – Henno Brandsma Sep 20 '14 at 21:47
• I meant to say "$x_n^k$", I have a cold right now so I am making a bunch of typos. – Hawk Sep 20 '14 at 21:52


text:
# Math Help - Solving for interest rate

1. ## Solving for interest rate

At what rate of interest does it take six years of compounding monthly for an account to double in value?

How do you get from,
(1+r/12)^72=2 to

r= 12(2^(1/72)-1)

2. i really don't get what you needed.. but if you are just solving for $r$ in $\left({\frac{1+r}{12}}\right)^{72} = 2$, then that is just simple algebra(and i think, there was a mistake in your grouping..)..

first, get the 72nd root of both sides, you will have $\frac{1+r}{12} = 2^{\frac{1}{72}}$

then multiply 12 to both sides, $1+r = (12)(2^{\frac{1}{72}})$

lastly, subtract 1 to both sides, $r = (12)(2^{\frac{1}{72}}) - 1$

now, is that what you want?

3. ## Yes

Thank you very much


text:
# The Mathematical Principles of Natural Philosophy (1846)/BookI-III

SECTION III.

Of the motion of bodies in eccentric conic sections.

PROPOSITION XI. PROBLEM VI.

If a body revolves in an ellipsis; it is required to find the law of the centripetal force tending to the focus of the ellipsis.

Let S be the focus of the ellipsis. Draw SP cutting the diameter DK of the ellipsis in E, and the ordinate Qv in x; and complete the parallelogram QxPR. It is evident that EP is equal to the greater semi-axis AC: for drawing HI from the other focus H of the ellipsis parallel to EC, because CS, CH are equal, ES, EI will be also equal; so that EP is the half sum of PS, PI, that is (because of the parallels HI, PR, and the equal angles IPR, HPZ), of PS, PH, which taken together are equal to the whole axis 2AC. Draw QT perpendicular to SP, and putting L for the principal latus rectum of the ellipsis (or for ${\displaystyle \scriptstyle {\frac {2BC^{2}}{AC}}}$), we shall have L ${\displaystyle \scriptstyle \times }$ QR to L ${\displaystyle \scriptstyle \times }$ Pv as QR to Pv, that is, as PE or AC to PC; and L ${\displaystyle \scriptstyle \times }$ Pv to GvP as L to Gv; and GvP to Qv² as PC² to CD²; and by (Corol. 2, Lem. VII) the points Q and P coinciding, Qv² is to Qx² in the ratio of equality; and Qx² or Qv² is to QT² as EP² to PF², that is, as CA² to PF², or (by Lem. XII) as CD² to CB². And compounding all those ratios together, we shall have L ${\displaystyle \scriptstyle \times }$ QR to QT² as AC ${\displaystyle \scriptstyle \times }$ L ${\displaystyle \scriptstyle \times }$ PC² ${\displaystyle \scriptstyle \times }$ CD², or 2CB² ${\displaystyle \scriptstyle \times }$ PC² ${\displaystyle \scriptstyle \times }$ CD² to PC ${\displaystyle \scriptstyle \times }$ Gv ${\displaystyle \scriptstyle \times }$ CD² ${\displaystyle \scriptstyle \times }$ CB², or as 2PC to Gv. But the points Q and P coinciding, 2PC and Gv are equal. And therefore the quantities L ${\displaystyle \scriptstyle \times }$ QR and QT², proportional to these, will be also equal. Let those equals be drawn into ${\displaystyle \scriptstyle {\frac {SP^{2}}{QR}}}$, and L ${\displaystyle \scriptstyle \times }$ SP² will become equal to ${\displaystyle \scriptstyle {\frac {SP^{2}\times QT^{2}}{QR}}}$. And therefore (by Corol. 1 and 5, Prop. VI) the centripetal force is reciprocally as L ${\displaystyle \scriptstyle \times }$ SP², that is, reciprocally in the duplicate ratio of the distance SP.   Q.E.I.

The same otherwise.

Since the force tending to the centre of the ellipsis, by which the body P may revolve in that ellipsis, is (by Corol. 1, Prop. X.) as the distance CP of the body from the centre C of the ellipsis; let CE be drawn parallel to the tangent PR of the ellipsis; and the force by which the same body P may revolve about any other point's of the ellipsis, if CE and PS intersect in E, will be as ${\displaystyle \scriptstyle {\frac {PE^{3}}{SP^{2}}}}$ (by Cor. 3, Prop. VII.); that is, if the point S is the focus of the ellipsis, and therefore PE be given as SP² reciprocally.   Q.E.I.

With the same brevity with which we reduced the fifth Problem to the parabola, and hyperbola, we might do the like here: but because of the dignity of the Problem and its use in what follows. I shall confirm the other cases by particular demonstrations.

PROPOSITION XII. PROBLEM VII.

Suppose a body to move in an hyperbola; it is required to find the law of the centripetal force tending to the focus of that figure.

Let CA, CB be the semi-axes of the hyperbola; PG, KD other conjugate diameters; PF a perpendicular to the diameter KD; and Qv an ordinate to the diameter GP. Draw SP cutting the diameter DK in E, and the ordinate Qv in x, and complete the parallelogram QRPx. It is evident that EP is equal to the semi-transverse axis AC; for drawing HI, from the other focus H of the hyperbola, parallel to EC, because CS, CH are equal, ES, EI will be also equal; so that EP is the half difference of PS, PI; that is (because of the parallels IH, PR, and the equal angles IPR, HPZ), of PS, PH, the difference of which is equal to the whole axis 2AC. Draw QT perpendicular to SP; and putting L for the principal latus rectum of the hyperbola (that is, for ${\displaystyle \scriptstyle {\frac {2BC^{2}}{AC}}}$, we shall have L ${\displaystyle \scriptstyle \times }$ QR to L ${\displaystyle \scriptstyle \times }$ Pv as QR to Pv, or Px to Pv, that is (because of the similar triangles Pxv, PEC), as PE to PC, or AC to PC. And L ${\displaystyle \scriptstyle \times }$ Pv will be to Gv ${\displaystyle \scriptstyle \times }$ Pv as L to Gv; and (by the properties of the conic sections) the rectangle GvP is to Qv² as PC² to CD²; and by (Cor. 2, Lem. VII.), Qv² to Qx² the points Q and P coinciding, becomes a ratio of equality; and Qx² or Qv² is to QT² as EP² to PF², that is, as CA² to PF², or (by Lem. XII.) as CD² to CB²: and, compounding all those ratios together, we shall have L ${\displaystyle \scriptstyle \times }$ QR to QT² as AC ${\displaystyle \scriptstyle \times }$ L ${\displaystyle \scriptstyle \times }$ PC² ${\displaystyle \scriptstyle \times }$ CD², or 2CB² ${\displaystyle \scriptstyle \times }$ PC² ${\displaystyle \scriptstyle \times }$ CD² to PC ${\displaystyle \scriptstyle \times }$ Gv ${\displaystyle \scriptstyle \times }$ CD² ${\displaystyle \scriptstyle \times }$ CB², or as 2PC to Gv. But the points P and Q coinciding, 2PC and Gv are equal. And therefore the quantities L ${\displaystyle \scriptstyle \times }$ QR and QT², proportional to them, will be also equal. Let those equals be drawn into ${\displaystyle \scriptstyle {\frac {SP^{2}}{QR}}}$, and we shall have L ${\displaystyle \scriptstyle \times }$ SP² equal to ${\displaystyle \scriptstyle {\frac {SP^{2}\times QT^{2}}{QR}}}$. And therefore (by Cor. I and 5, Prop. VI.) the centripetal force is reciprocally as L ${\displaystyle \scriptstyle \times }$ SP², that is, reciprocally in the duplicate ratio of the distance SP.   Q.E.I.

The same otherwise.

Find out the force tending from the centre C of the hyperbola. This will be proportional to the distance CP. But from thence (by Cor. 3, Prop. VII.) the force tending to the focus S will be as ${\displaystyle \scriptstyle {\frac {PE^{3}}{SP^{2}}}}$, that is, because PE is given reciprocally as SP².   Q.E.I.

And the same way may it be demonstrated, that the body having its centripetal changed into a centrifugal force, will move in the conjugate hyperbola.

LEMMA XIII.
The latus rectum of a parabola belonging to any vertex is quadruple the distance of that vertex from the focus of the figure.

This is demonstrated by the writers on the conic sections.

LEMMA XIV.
The perpendicular, let fall from the focus of a parabola on its tangent, is a mean proportional between the distances of the focus from the point of contact, and from the principal vertex of the figure.

For, let AP be the parabola, S its focus, A its principal vertex, P the point of contact, PO an ordinate to the principal diameter, PM the tangent meeting the principal diameter in M, and SN the perpendicular from the focus on the tangent: join AN, and because of the equal lines MS and SP, MN and NP, MA and AO, the right lines AN, OP, will be parallel; and thence the triangle SAN will be right-angled at A, and similar to the equal triangles SNM, SNP; therefore PS is to SN as SN to SA.   Q.E.D.

Cor. 1. PS² is to SN² as PS to SA.

Cor. 2. And because SA is given, SN² will be as PS.

Cor. 3. And the concourse of any tangent PM, with the right line SN. drawn from the focus perpendicular on the tangent, falls in the right line AN that touches the parabola in the principal vertex.

PROPOSITION XIII. PROBLEM VIII.
If a body moves in the perimeter of a parabola; it is required to find the law of the centripetal force tending to the focus of that figure.

Retaining the construction of the preceding Lemma, let P be the body in the perimeter of the parabola; and from the place Q, into which it is next to succeed, draw QR parallel and QT perpendicular to SP, as also Qv parallel to the tangent, and meeting the diameter PG in v, and the distance SP in x. Now, because of the similar triangles Pxv, SPM, and of the equal sides SP, SM of the one, the sides Px or QR and Pv of the other will be also equal. But (by the conic sections) the square of the ordinate Qv is equal to the rectangle under the latus rectum and the segment Pv of the diameter; that is (by Lem. XIII.), to the rectangle 4PS ${\displaystyle \scriptstyle \times }$ Pv, or 4PS ${\displaystyle \scriptstyle \times }$ QR; and the points P and Q coinciding, the ratio of Qv to Qx (by Cor. 2, Lem. VII.,) becomes a ratio of equality. And therefore Qx², in this case, becomes equal to the rectangle 4PS ${\displaystyle \scriptstyle \times }$ QR. But (because of the similar triangles QxT, SPN), Qx² is to QT² as PS² to SN², that is (by Cor. 1, Lem. XIV.), as PS to SA; that is, as 4PS ${\displaystyle \scriptstyle \times }$ QR to 4SA ${\displaystyle \scriptstyle \times }$ QR, and therefore (by Prop. IX. Lib. V., Elem.) QT² and 4SA ${\displaystyle \scriptstyle \times }$ QR are equal. Multiply these equals by ${\displaystyle \scriptstyle {\frac {SP^{2}}{QR}}}$, and ${\displaystyle \scriptstyle {\frac {SP^{2}\times QT^{2}}{QR}}}$ will become equal to SP² ${\displaystyle \scriptstyle \times }$ 4SA: and therefore (by Cor. 1 and 5, Prop. VI.), the centripetal force is reciprocally as SP² ${\displaystyle \scriptstyle \times }$ 4SA; that is, because 4SA is given; reciprocally in the duplicate ratio of the distance SP.   Q.E.I.

Cor. 1. From the three last Propositions it follows, that if any body P goes from the place P with any velocity in the direction of any right line PR, and at the same time is urged by the action of a centripetal force that is reciprocally proportional to the square of the distance of the places from the centre, the body will move in one of the conic sections, having its focus in the centre of force; and the contrary. For the focus, the point of contact, and the position of the tangent, being given, a conic section may be described, which at that point shall have a given curvature. But the curvature is given from the centripetal force and velocity of the body being given; and two orbits, mutually touching one the other, cannot be described by the same centripetal force and the same velocity.

Cor. 2. If the velocity with which the body goes from its place P is such, that in any infinitely small moment of time the lineola PR may be thereby described; and the centripetal force such as in the same time to move the same body through the space QR; the body will move in one of the conic sections, whose principal latus rectum is the quantity ${\displaystyle \scriptstyle {\frac {QT^{2}}{QR}}}$ in its ultimate state, when the lineolae PR, QR are diminished in infinitum. In these Corollaries I consider the circle as an ellipsis; and I except the case where the body descends to the centre in a right line.

PROPOSITION XIV. THEOREM VI.
If several bodies revolve about one common centre, and the centripetal force is reciprocally in the duplicate ratio of the distance of places from the centre; I say, that the principal latera recta of their orbits are in the duplicate ratio of the areas, which the bodies by radii drawn to the centre describe in the same time.

For (by Cor. 2, Prop. XIII) the latus rectum L is equal to the quantity ${\displaystyle \scriptstyle \scriptstyle {\frac {QT^{2}}{QR}}}$ in its ultimate state when the points P and Q coincide. But the lineola QR in a given time is as the generating centripetal force; that is (by supposition), reciprocally as SP² . And therefore ${\displaystyle \scriptstyle \scriptstyle {\frac {QT^{2}}{QR}}}$ is as QT² ${\displaystyle \scriptstyle \scriptstyle \times }$ SP²; that is, the latus rectum L is in the duplicate ratio of the area QT ${\displaystyle \scriptstyle \scriptstyle \times }$ SP.   Q.E.D.

Cor. Hence the whole area of the ellipsis, and the rectangle under the axes, which is proportional to it, is in the ratio compounded of the subduplicate ratio of the latus rectum, and the ratio of the periodic time. For the whole area is as the area QT ${\displaystyle \scriptstyle \scriptstyle \times }$ SP, described in a given time, multiplied by the periodic time.

PROPOSITION XV. THEOREM VII.
The same things being supposed, I say, that the periodic times in ellipses are in the sesquiplicate ratio of their greater axes.

For the lesser axis is a mean proportional between the greater axis and the latus rectum; and, therefore, the rectangle under the axes is in the ratio compounded of the subduplicate ratio of the latus rectum and the sesquiplicate ratio of the greater axis. But this rectangle (by Cor. 3. Prop. XIV) is in a ratio compounded of the subduplicate ratio of the latus rectum, and the ratio of the periodic time. Subduct from both sides the subduplicate ratio of the latus rectum, and there will remain the sesquiplicate ratio of the greater axis, equal to the ratio of the periodic time.   Q.E.D.

Cor. Therefore the periodic times in ellipses are the same as in circles whose diameters are equal to the greater axes of the ellipses.

PROPOSITION XVI. THEOREM VIII.
The same things being supposed, and right lines being drawn to the bodies that shall touch the orbits, and perpendiculars being let fall on those tangents from the common focus; I say, that the velocities of the bodies are in a ratio compounded of the ratio of the perpendiculars inversely, and the subduplicate ratio of the principal latera recta directly.

From the focus S draw SY perpendicular to the tangent PR, and the velocity of the body P will be reciprocally in the subduplicate ratio of the quantity ${\displaystyle \scriptstyle \scriptstyle {\frac {SY^{2}}{L}}}$. For that velocity is as the infinitely small arc PQ described in a given moment of time, that is (by Lem. VII), as the tangent PR; that is (because of the proportionals PR to QT, and SP to SY), as ${\displaystyle \scriptstyle {\frac {SP\times QT}{SY}}}$; or as SY reciprocally, and SP ${\displaystyle \scriptstyle \times }$ QT directly; but SP ${\displaystyle \scriptstyle \times }$ QT is as the area described in the given time, that is (by Prop. XIV), in the subduplicate ratio of the latus rectum.   Q.E.D.

Cor. 1. The principal latera recta are in a ratio compounded of the duplicate ratio of the perpendiculars and the duplicate ratio of the velocities.

Cor. 2. The velocities of bodies, in their greatest and least distances from the common focus, are in the ratio compounded of the ratio of the distances inversely, and the subduplicate ratio of the principal latera recta directly. For those perpendiculars are now the distances.

Cor. 3. And therefore the velocity in a conic section, at its greatest or least distance from the focus, is to the velocity in a circle, at the same distance from the centre, in the subduplicate ratio of the principal latus rectum to the double of that distance.

Cor. 4. The velocities of the bodies revolving in ellipses, at their mean distances from the common focus, are the same as those of bodies revolving in circles, at the same distances; that is (by Cor. 6, Prop. IV), reciprocally in the subduplicate ratio of the distances. For the perpendiculars are now the lesser semi-axes, and these are as mean proportionals between the distances and the latera recta. Let this ratio inversely be compounded with the subduplicate ratio of the latera recta directly, and we shall have the subduplicate ratio of the distance inversely.

Cor. 5. In the same figure, or even in different figures, whose principal latera recta are equal, the velocity of a body is reciprocally as the perpendicular let fall from the focus on the tangent.

Cor. 6. In a parabola, the velocity is reciprocally in the subduplicate ratio of the distance of the body from the focus of the figure; it is more variable in the ellipsis, and less in the hyperbola, than according to this ratio. For (by Cor. 2, Lem. XIV) the perpendicular let fall from the focus on the tangent of a parabola is in the subduplicate ratio of the distance. In the hyperbola the perpendicular is less variable; in the ellipsis more.

Cor. 7. In a parabola, the velocity of a body at any distance from the focus is to the velocity of a body revolving in a circle, at the same distance from the centre, in the subduplicate ratio of the number 2 to 1; in the ellipsis it is less, and in the hyperbola greater, than according to this ratio, (by Cor. 2 of this Prop.) the velocity at the vertex of a parabola is in this ratio, and (by Cor. 6 of this Prop. and Prop. IV) the same proportion holds in all distances. And hence, also, in a parabola, the velocity is everywhere equal to the velocity of a body revolving in a circle at half the distance; in the ellipsis it is less, and in the hyperbola greater.

Cor. 8. The velocity of a body revolving in any conic section is to the velocity of a body revolving in a circle, at the distance of half the principal latus rectum of the section, as that distance to the perpendicular let fall from the focus on the tangent of the section. This appears from Cor. 5.

Cor. 9. Wherefore since (by Cor. 6, Prop. IV), the velocity of a body revolving in this circle is to the velocity of another body revolving in any other circle reciprocally in the subduplicate ratio of the distances; therefore, ex aequo, the velocity of a body revolving in a conic section will be to the velocity of a body revolving in a circle at the same distance as a mean proportional between that common distance, and half the principal latus rectum of the section, to the perpendicular let fall from the common focus upon the tangent of the section.

PROPOSITION XVII. PROBLEM IX.
Supposing the centripetal force to be reciprocally proportional to the squares of the distances of places from the centre, and that the absolute quantity of that force is known; it is required to determine the line which a body will describe that is let go from a given place with a given velocity in, the direction of a given right line.

Let the centripetal force tending to the point S be such as will make the body p revolve in any given orbit pq; and suppose the velocity of this body in the place p is known. Then from the place P suppose the body P to be let with a given velocity in the direction of the line PR; but by virtue of a centripetal force to be immediately turned aside from that right line into the conic section PQ. This, the right line PR will therefore touch in P. Suppose likewise that the right line pr touches the orbit pq in p, and if from S you suppose perpendiculars let fall on those tangents, the principal latus rectum of the conic section (by Cor. 1, Prop. XVI) will be to the principal latus rectum of that orbit in a ratio compounded of the duplicate ratio of the perpendiculars, and the duplicate ratio of the velocities; and is therefore given. Let this latus rectum be L; the focus S of the conic section is also given. Let the angle RPH be the complement of the angle RPS to two right; and the line PH, in which the other focus H is placed, is given by position. Let fall SK perpendicular on PH, and erect the conjugate semi-axis BC; this done, we shall have SP² - 2KPH + PH² = SH² = 4CH² = 4BH² - 4BC² = ${\displaystyle \scriptstyle {\overline {SP+PH^{2}}}}$ - L ${\displaystyle \scriptstyle \times {\overline {SP+PH}}}$ = SP² + 2SPH + PH² - L ${\displaystyle \scriptstyle \times {\overline {SP+PH}}}$. Add on both sides 2KPH - SP² - PH² + L ${\displaystyle \scriptstyle \times {\overline {SP+PH}}}$, and we shall have L ${\displaystyle \scriptstyle \times {\overline {SP+PH}}}$ = 2SPH + 2KPH, or SP + PH to PH, as 2SP + 2KP to L. Whence PH is given both in length and position. That is, if the velocity of the body in P is such that the latus rectum L is less than 2SP + 2KP, PH will lie on the same side of the tangent PR with the line SP; and therefore the figure will be an ellipsis, which from the given foci S, H, and the principal axis SP + PH, is given also. But if the velocity of the body is so great, that the latus rectum L becomes equal to 2SP + 2KP, the length PH will be infinite; and therefore, the figure will be a parabola, which has its axis SH parallel to the line PK, and is thence given. But if the body goes from its place P with a yet greater velocity, the length PH is to be taken on the other side the tangent; and so the tangent passing between the foci, the figure will be an hyperbola having its principal axis equal to the difference of the lines SP and PH, and thence is given. For if the body, in these cases, revolves in a conic section so found, it is demonstrated in Prop. XI, XII, and XIII, that the centripetal force will be reciprocally as the square of the distance of the body from the centre of force S; and therefore we have rightly determined the line PQ, which a body let go from a given place P with a given velocity, and in the direction of the right line PR given by position, would describe with such a force.   Q.E.F.

Cor. 1. Hence in every conic section, from the principal vertex D, the latus rectum L, and the focus S given, the other focus H is given, by taking DH to DS as the latus rectum to the difference between the latus rectum and 4DS. For the proportion, SP + PH to PH as 2SP + 2KP to L, becomes, in the case of this Corollary, DS + DH to DH as 4DS to L, and by division DS to DH as 4DS - L to L.

Cor. 2. Whence if the velocity of a body in the principal vertex D is given, the orbit may be readily found; to wit, by taking its latus rectum to twice the distance DS, in the duplicate ratio of this given velocity to the velocity of a body revolving in a circle at the distance DS (by Cor. 3, Prop. XVI.), and then taking DH to DS as the latus rectum to the difference between the latus rectum and 4DS.

Cor. 3. Hence also if a body move in any conic section, and is forced out of its orbit by any impulse, you may discover the orbit in which it will afterwards pursue its course. For by compounding the proper motion of the body with that motion, which the impulse alone would generate, you will have the motion with which the body will go off from a given place of impulse in the direction of a right line given in position.

Cor. 4. And if that body is continually disturbed by the action of some foreign force, we may nearly know its course, by collecting the changes which that force introduces in some points, and estimating the continual changes it will undergo in the intermediate places, from the analogy that appears in the progress of the series.

SCHOLIUM.

If a body P, by means of a centripetal force tending to any given point R, move in the perimeter of any given conic section whose centre is C; and the law of the centripetal force is required: draw CG parallel to the radius RP, and meeting the tangent PG of the orbit in G; and the force required (by Cor. 1, and Schol. Prop. X., and Cor. 3, Prop. VII.) will be as ${\displaystyle \scriptstyle {\frac {CG^{3}}{RP^{2}}}}$.


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Home > Quantization Error > Quantization Error Derivation

# Quantization Error Derivation

## Contents

Due to the sampling step of an ADC, these harmonics get folded to the Nyquist band, pushing the total noise power into the Nyquist band and with an approximately white spectrum The application of such compressors and expanders is also known as companding. Quantization error models In the typical case, the original signal is much larger than one least significant bit (LSB). In contrast, mid-tread quantizers do have a zero output level, and can reach arbitrarily low bit rates per sample for input distributions that are symmetric and taper off at higher magnitudes. http://vealcine.com/quantization-error/quantization-noise-model-quantization-error.php

doi:10.1109/JRPROC.1948.231941 ^ Seymour Stein and J. When the input data can be modeled as a random variable with a probability density function (pdf) that is smooth and symmetric around zero, mid-riser quantizers also always produce an output Examples of fields where this limitation applies include electronics (due to electrons), optics (due to photons), biology (due to DNA), physics (due to Planck limits) and chemistry (due to molecules). The potential signal-to-quantization-noise power ratio therefore changes by 4, or 10 ⋅ log 10  ( 4 )   =   6.02 {\displaystyle \scriptstyle 10\cdot \log _{10}(4)\ =\ 6.02} https://en.wikipedia.org/wiki/Quantization_(signal_processing)

## Quantization Noise Power Formula

Around the quantum limit, the distinction between analog and digital quantities vanishes.[citation needed] See also Analog-to-digital converter Beta encoder Data binning Discretization Discretization error Posterization Pulse code modulation Quantile Regression dilution The step size Δ = 2 X m a x M {\displaystyle \Delta ={\frac {2X_{max}}{M}}} and the signal to quantization noise ratio (SQNR) of the quantizer is S Q N R For example, for N {\displaystyle N} =8 bits, M {\displaystyle M} =256 levels and SQNR = 8*6 = 48dB; and for N {\displaystyle N} =16 bits, M {\displaystyle M} =65536 and

• Gray, Vector Quantization and Signal Compression, Springer, ISBN 978-0-7923-9181-4, 1991. ^ Hodgson, Jay (2010).
• Satish Kashyap 18.745 προβολές 35:35 Vector Quantization Part-1 - Διάρκεια: 8:22.
• Rounding and truncation are typical examples of quantization processes.
• doi:10.1109/MCOM.1977.1089500 ^ Rabbani, Majid; Joshi, Rajan L.; Jones, Paul W. (2009). "Section 1.2.3: Quantization, in Chapter 1: JPEG 2000 Core Coding System (Part 1)".

Please try the request again. is since Vfs = 2n q, then which simplifies to N.B. The difference between the original signal and the reconstructed signal is the quantization error and, in this simple quantization scheme, is a deterministic function of the input signal. Quantization Error In Analog To Digital Conversion Berklee Press. ^ William Fleetwood Sheppard, "On the Calculation of the Most Probable Values of Frequency Constants for data arranged according to Equidistant Divisions of a Scale", Proceedings of the London

Madhan Mohan 27.677 προβολές 4:03 Signal-to-Noise Ratio - Διάρκεια: 13:17. Quantization Noise In Pcm doi:10.1109/TIT.1984.1056920 ^ Toby Berger, "Optimum Quantizers and Permutation Codes", IEEE Transactions on Information Theory, Vol. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply.

If it is assumed that distortion is measured by mean squared error, the distortion D, is given by: D = E [ ( x − Q ( x ) ) 2

For the example uniform quantizer described above, the forward quantization stage can be expressed as k = ⌊ x Δ + 1 2 ⌋ {\displaystyle k=\left\lfloor {\frac {x}{\Delta }}+{\frac {1}{2}}\right\rfloor } Uniform Quantization The distinguishing characteristic of a mid-riser quantizer is that it has a classification threshold value that is exactly zero, and the distinguishing characteristic of a mid-tread quantizer is that is it The difference between the blue and red signals in the upper graph is the quantization error, which is "added" to the quantized signal and is the source of noise. Although r k {\displaystyle r_{k}} may depend on k {\displaystyle k} in general, and can be chosen to fulfill the optimality condition described below, it is often simply set to a

## Quantization Noise In Pcm

For signals whose amplitude is less than the FSR the Signal - to - Noise Ratio will be reduced. Recording and Producing in the Home Studio, p.38-9. Quantization Noise Power Formula The Art of Digital Audio 3rd Edition. Quantization Error Example It has been shown to be a valid model in cases of high resolution quantization (small Δ {\displaystyle \Delta } relative to the signal strength) with smooth probability density functions.[4][15] However,

Principles of Digital Audio 2nd Edition. http://vealcine.com/quantization-error/quantization-of-signals-quantization-error.php IT-44, No. 6, pp. 2325–2383, Oct. 1998. Madhan Mohan 15.150 προβολές 4:08 Digital Audio 102 - PCM, Bit-Rate, Quantisation, Dithering, Nyquists Sampling Theorum - PB15 - Διάρκεια: 6:06. When the input signal has a high amplitude and a wide frequency spectrum this is the case.[16] In this case a 16-bit ADC has a maximum signal-to-noise ratio of 98.09dB. How To Reduce Quantization Error

Rich Radke 8.100 προβολές 1:03:51 Quantization and Coding in A/D Conversion - Διάρκεια: 8:31. By using this site, you agree to the Terms of Use and Privacy Policy. The input and output sets involved in quantization can be defined in a rather general way. http://vealcine.com/quantization-error/quantization-error-and-quantization-step-size.php Your cache administrator is webmaster.

Lloyd, "Least Squares Quantization in PCM", IEEE Transactions on Information Theory, Vol. What Is Quantization The additive noise model for quantization error A common assumption for the analysis of quantization error is that it affects a signal processing system in a similar manner to that of For a fixed-length code using N {\displaystyle N} bits, M = 2 N {\displaystyle M=2^{N}} , resulting in S Q N R = 20 log 10  2 N = N

## Please try the request again.

However, in some quantizer designs, the concepts of granular error and overload error may not apply (e.g., for a quantizer with a limited range of input data or with a countably For some probabilistic source models, the best performance may be achieved when M {\displaystyle M} approaches infinity. Gray and David L. Quantization Step Size Formula doi:10.1109/TIT.1972.1054906 ^ Toby Berger, "Minimum Entropy Quantizers and Permutation Codes", IEEE Transactions on Information Theory, Vol.

The value of the measured signal is Vm = Vs - e, where Vm is the measured value, Vs is the actual value, and e is the error. In more elaborate quantization designs, both the forward and inverse quantization stages may be substantially more complex. The quantization error creates harmonics in the signal that extend well above the Nyquist frequency. this content Your cache administrator is webmaster.

doi:10.1109/TIT.2005.846397 ^ Pohlman, Ken C. (1989). The set of possible output values may be finite or countably infinite. Quantization replaces each real number with an approximation from a finite set of discrete values (levels), which is necessary for storage and processing by numerical methods. After defining these two performance metrics for the quantizer, a typical Rate–Distortion formulation for a quantizer design problem can be expressed in one of two ways: Given a maximum distortion constraint


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# Maths Functions

## What Are Functions in Math?

A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.

Representation-

Functions are generally represented as $f(x)$

Let , $f(x)= x^{3}$

It is said as f of x is equal to x cube.

Functions can also be represented by g(), t(),… etc.

 Lets Work Out- Example- Find the output of the function $g(t)= 6t^{2}+5$ at (i) t = 0 (ii) t = 2 Solution- The given function is $g(t)= 6t^{2}+5$ (i) At t = 0, $g(0)= 6(0)^{2}+5$     = 5 (ii) At t = 2, $g(2)= 6(2)^{2}+5$     = 29

Types of Functions-

One-to-One function – It signifies that one element of a set is related to one element of the other set.

Many-to-One-function- It signifies that many elements of a set is related to one element of the other set.

 Note : One-to-Many- A function cannot have a one to many relation ie. one element of a set cannot be related to more than one element of the other set.

Vertical Line Test-

Vertical line test is used to determine whether a curve is a  function or not.

If any curve cuts a vertical line at more than one points then the curve is not a function.

Types of Functions-

### Polynomial function-

A polynomial function can be expressed as :

$f(x)= a_{n}x^{n}+a_{n-1}x^{n-1}+…..+a_{1}x^{1}+a_{0}$

The highest power in the expression is known as the degree of the polynomial function. The different types of polynomial functions based on the degree are:

1. The polynomial function is called as Constant function if the degree is zero.
2. The polynomial function is called as Linear if the degree is one.
3. The polynomial function is Quadratic if the degree is two.
4. The polynomial function is Cubic if the degree is three.

### Constant Polynomial Functions

The polynomial of 0th degree where f(x) = f(0) = $a_{0}$=c. Regardless of the input, the output always result in constant value. The graph for this is a horizontal line.

### Linear Polynomial Functions

A linear polynomial function is a first degree polynomial where the input needs to be multiplied by m and added to c. It can be expressed by f(x) = mx + c.

For example, f(x) = 2x + 1 at x = 1

f(1) = 2.1 + 1 = 3

f(1) = 3

A Quadratic polynomial function is a second degree polynomial and it can be expressed as;

F(x) = ax 2 + bx + c, and a is not equal to zero.

Where a, b, c are constant and x is a variable.

Example, f(x) = 2x 2 + x – 1 at x = 2

If x = 2, f(2) = 2.2 2 + 2 – 1 = 9

### Cubic Polynomial Function

A cubic polynomial function is a polynomial of degree three and can be expressed as;

F(x) = ax 3 + bx 2 + cx + d and a is not equal to zero.

### Algebraic Functions

A function that consists of a finite number of terms involving powers and roots of independent variable x and fundamental operations such as addition, subtraction, multiplication, and division is known as an algebraic equation.

For Example-

$f(x)=5x^{3}-2x^{2}+3x+6$, $g(x)=\frac{\sqrt{3x+4}}{(x-1)^{2}}$..

Rational functions, irrational functions and Polynomials functions are examples of algebraic functions.


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# Partitioning the Dutch national flag

Yesterday’s post from Programming Praxis asks us to solve a problem known as the Dutch National Flag problem (attributed to Edsgar Dijkstra): sort an array of red, white and blue symbols so that all reds come together, followed by all whites, followed finally by all blues.

For once, the first algorithm to come to mind was actually the correct one. This doesn’t always happen quite so nicely, but it is pleasant when it does. The basic idea is similar to the in-place partition often used with quicksorts ((This is actually more of a partitioning problem than a sorting one)): Start with labels at the beginning and end of the array; advance the labels towards each other, swapping elements on the wrong side of the pivot.

In this case, there are three labels (one for each color), but that doesn’t actually change the algorithm much at all. Here are the basic steps:

• Start with three labels Lr = 0, Lw = 0, Lb = |data| - 1
• While Lw < Lb, check the element at Lw:
• Red: Swap Lr and Lr and increment Lr and Lw
• White: Increment Lw
• Blue: Swap Lw and Lb and decrement Lb

That’s all there is to it. Here’s how you might translate that into Racket:

(define (vector-swap! v i j)
(define temp (vector-ref v i))
(vector-set! v i (vector-ref v j))
(vector-set! v j temp))

(define (partition-flag! flag)
(let loop ([r 0] [w 0] [b (- (vector-length flag) 1)])
(cond
[(> w b) flag]
[(eq? (vector-ref flag w) 'red)
(vector-swap! flag r w)
(loop (+ r 1) (+ w 1) b)]
[(eq? (vector-ref flag w) 'white)
(loop r (+ w 1) b)]
[(eq? (vector-ref flag w) 'blue)
(vector-swap! flag w b)
(loop r w (- b 1))])))

Since the code is relative straight forward, I went ahead off the deep end and worked out a visualization that will run in your web browser. If everything works as it should, clicking ‘Run Demo’ below should show you exactly what happens when partitioning a 50-element vector.

### Demo

jQuery(function($) { // Context to draw with var canvas =$('#canvas')[0];
var context = canvas.getContext('2d');
var progress = $('#progress'); // Various constants var demo_size = 50; var demo_step = 250; var left_offset = canvas.width / 10; var top_offset = canvas.height / 6; var block_width = 4/5 * canvas.width / demo_size; var block_height = 1/3 * canvas.height; var label_width = 3; var label_height = block_height / 3; // Data for the simulation var styles = new Array('red', 'white', 'blue'); var labels = new Array(0, 0, 0); var data = new Array(); var tick = 0; var comparisons = 0; var swaps = 0; // Logging function function log(msg) { progress.html( 'Step ' + tick + ':
' + msg + '

' + 'So far:
' + comparisons + ' comparisons
' + swaps + ' swaps
' + demo_size + ' elements' ); } // Run the entire demo function showDemo() { // Generate a new set of data data = new Array(demo_size); for (var i = 0; i < data.length; i++) { data[i] = Math.floor(Math.random() * 3); } // Reset the arrays to left, left, and right labels = new Array(0, 0, data.length - 1); // Clear progress and raw everything the first time progress.text(''); draw(); // Run the update functions comparisons = 0; swaps = 0; tick = 0; function simStep() { tick += 1; if (step()) setTimeout(simStep, demo_step); else log("finished"); draw(); } setTimeout(simStep, demo_step); } // Swap two data points function swap(i, j) { swaps += 1; var temp = data[i]; data[i] = data[j]; data[j] = temp; } // Take a single step of the sorting function function step() { // Stop when the white and blue pointers cross if (labels[1] > labels[2]) { return false; } // On red: swap to red and increment red and white else if (data[labels[1]] == 0) { log('red, swap to red and increment red and white'); swap(labels[0], labels[1]); labels[0] += 1; labels[1] += 1; } // On white: increment white else if (data[labels[1]] == 1) { log('white, increment white'); labels[1] += 1; } // On blue: swap to blue and decrement blue else if (data[labels[1]] == 2) { log('blue, swap to blue and decrement blue'); swap(labels[1], labels[2]); labels[2] -= 1; } // Done, signify that we took a step comparisons += 1; return true; } // Draw the blocks and labels function draw() { // Clear the screen context.clearRect(0, 0, canvas.width, canvas.height); // Draw blocks context.lineWidth = 1; context.strokeStyle = 'black'; for (var i = 0; i < data.length; i++) { context.beginPath(); context.rect( left_offset + i * block_width, top_offset, block_width, block_height ); context.fillStyle = styles[data[i]]; context.fill(); context.stroke(); } // Draw labels var center_x = 0, center_y = 0; for (var i = 0; i < labels.length; i++) { center_x = left_offset + labels[i] * block_width + block_width / 3; center_y = 2 * top_offset + block_height + i * label_height; context.beginPath(); context.rect( center_x - label_width / 2, center_y - label_height / 2, label_width, label_height ); context.fillStyle = styles[i]; context.fill(); context.stroke(); } }$('#run').click(showDemo);
});


text:
# DirichletBC¶

class dolfin.fem.bcs.DirichletBC(*args, **kwargs)

Interface for setting (strong) Dirichlet boundary conditions. u = g on G, where u is the solution to be computed, g is a function and G is a sub domain of the mesh. A DirichletBC is specified by the function g, the function space (trial space) and boundary indicators on (a subset of) the mesh boundary. The boundary indicators may be specified in a number of different ways. The simplest approach is to specify the boundary by a SubDomain object, using the inside() function to specify on which facets the boundary conditions should be applied. The boundary facets will then be searched for and marked only on the first call to apply. This means that the mesh could be moved after the first apply and the boundary markers would still remain intact. Alternatively, the boundary may be specified by a MeshFunction over facets labeling all mesh facets together with a number that specifies which facets should be included in the boundary. The third option is to attach the boundary information to the mesh. This is handled automatically when exporting a mesh from for example VMTK. The ‘method’ variable may be used to specify the type of method used to identify degrees of freedom on the boundary. Available methods are: topological approach (default), geometric approach, and pointwise approach. The topological approach is faster, but will only identify degrees of freedom that are located on a facet that is entirely on the boundary. In particular, the topological approach will not identify degrees of freedom for discontinuous elements (which are all internal to the cell). A remedy for this is to use the geometric approach. In the geometric approach, each dof on each facet that matches the boundary condition will be checked. To apply pointwise boundary conditions e.g. pointloads, one will have to use the pointwise approach. The three possibilities are “topological”, “geometric” and “pointwise”. Note: when using “pointwise”, the boolean argument on_boundary in SubDomain::inside will always be false. The ‘check_midpoint’ variable can be used to decide whether or not the midpoint of each facet should be checked when a user-defined SubDomain is used to define the domain of the boundary condition. By default, midpoints are always checked. Note that this variable may be of importance close to corners, in which case it is sometimes important to check the midpoint to avoid including facets “on the diagonal close” to a corner. This variable is also of importance for curved boundaries (like on a sphere or cylinder), in which case it is important not to check the midpoint which will be located in the interior of a domain defined relative to a radius. Note that there may be caching employed in BC computation for performance reasons. In particular, applicable DOFs are cached by some methods on a first apply . This means that changing a supplied object (defining boundary subdomain) after first use may have no effect. But this is implementation and method specific.

Copy constructor. Either cached DOF data are copied.

Parameters: DirichletBC & bc (const) – (DirichletBC &) The object to be copied. [direction=in]


text:
# What formatting should I use on 'd' when writing an integral? [duplicate]

Possible Duplicate:
What’s the proper way to typeset a differential operator?

What formatting should I use on 'd' when writing an integral?

I normally use

$$\int_{-\infty}^{infty} e^{-x^2} \mathoperator{d}\!x = \pi$$

which produces

$\int_{-\infty}^{infty} e^{-x^2} dx = \pi$

Is there a better formatting for dx?

• Jan 2, 2013 at 3:42
• not really, perhaps \delta x, and if you are doing partial integration, then \partial x Jan 2, 2013 at 3:48
• Please let us know if the link provided by Scott resolves the issue :) Jan 2, 2013 at 4:28
• Yes, it resolves it. Jan 2, 2013 at 7:21
• Good to hear :) Jan 2, 2013 at 8:06


text:
# Every locally free sheaf is Cohen-Macaulay on complex variety with at canonical singualities?

Suppose $$X$$ is a normal complex space with at most singularities, can we say any locally free sheaf on it is Cohen-Macauly? Recall that a coherent sheaf $$\mathcal{F}$$ over $$X$$ is called (maximal) Cohen-Macaulay if $$depth_{\mathcal{O}_x}(\mathcal{F}_x) = \dim\mathcal{O}_x$$ for any $$x\in X$$, where $$\mathcal{O}_x$$ is the local ring of $$X$$ at $$x$$.

How badly the singularities can the complex space $$X$$ possess to remain the above property?

Thanks!

• Please confer the following MO question: mathoverflow.net/questions/69436/… Oct 19, 2022 at 11:26
• This holds if and only if $X$ is Cohen-Macaulay.
– abx
Oct 19, 2022 at 11:59
• Sorry, I hadn't paid attention to your title. If $X$ has canonical singularities, it is Cohen-Macaulay by Elkik's theorem.
– abx
Oct 19, 2022 at 12:19
• There is also a variant of canonical in the non-Q-Gorenstein case due to de Fernex-Hacon which doesn't imply Cohen-Macaulay if I recall correctly. Oct 19, 2022 at 18:46