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# composition algebras over $\mathbb{Q}$

###### Theorem 1.

There are infinitely many composition algebras over $\mathbb{Q}$.

###### Proof.

Every quadratic extension of $\mathbb{Q}$ is a distinct composition algebra. For example, $\left(\frac{p}{\mathbb{Q}}\right)$ for $p$ a prime number. This is sufficient to illustrate an infinite number of quadratic composition algebras. ∎

The other families of composition algebras also have an infinite number of non-isomorphic division algebras though the proofs are more involved. It suffices to show provide an infinite family of non-isometric quadratic forms of the form:

 $N_{p,q}(a,b,c,d)=a^{2}-b^{2}p-c^{2}q+d^{2}pq$

for rational numbers $p$ and $q$. Such questions can involve complex number theory as for instance, if $p$ is a prime congruent to $1$ modulo $4$ then $N_{-1,-p}$ is isometric to $N_{-1,-1}$ and thus $N_{-1,-p}$ is isometric to $N_{-1,-q}$ for any other prime $q\equiv 1\pmod{4}$. But if $p\equiv 3\pmod{4}$ then this cannot be said.

Title composition algebras over $\mathbb{Q}$ CompositionAlgebrasOvermathbbQ 2013-03-22 17:18:29 2013-03-22 17:18:29 Algeboy (12884) Algeboy (12884) 6 Algeboy (12884) Example msc 17A75 HurwitzsTheorem JacobsonsTheoremOnCompositionAlgebras


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# pulse duration calculator

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The Output Pulse Width Equation In optics, various autocorrelation functions can be experimentally realized. Optical period $$T = \frac{2\pi\hbar}{E} \Longrightarrow T[\mathrm{fs}] \approx \frac{4.136}{E[\mathrm{eV}]}$$ The total energy can be controlled by modulating the pulse width/duration. Beam parameter product (BPP) is product of divergence half-angle $$\vartheta/2$$ and radius at waist $$w_0$$, $$\mathrm{BPP} = M^2 \frac{\lambda}{\pi},$$ Phase matching condition: $$\frac{n_\mathrm{o}(\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0.$$ R_\mathrm{s} = \frac{|E_\mathrm{r}^\mathrm{s}|^2}{|E_\mathrm{i}^\mathrm{s}|^2}=\frac{|\cos\vartheta_0-n\cos\vartheta_1|^2}{|\cos\vartheta_0+n\cos\vartheta_1|^2}. The discharge energy is really controlled by the peak current and the length of the pulse on-time. $$, Optical path length $$L$$,$$ L = \sum_{i=1}^N h_i n_i. For beams that are not 100% uniform, the peak power/energy density will be higher. Use an online calculator to determine your desired target heart rate zone. If $$n=1$$, function is Gaussian. τ is the pulse width. This tutorial explains how to calculate the laser pulse time duration. $$\mathcal{E}=\intop F(r)\mathrm{d}S.$$ If fluence and beam intensity is super-Gaussian function, $$F(r)=F_0\left[-2\left(\frac{r}{w_{0}}\right)^{2n}\right],$$ Here $$\vartheta_0$$ is the angle of incidence. Target heart rate during moderate intensity activities is about 50-70% of maximum heart rate, while during vigorous physical activity it’s about 70-85% of maximum. 1.Short duration pulse ( 10mSec) 2.Energy calculation is for a single pulse; 3.No energy is absorbed by the coating material and core, it is purely adiabatic. 555 Timer Astable Calculator Description. In electronics, duty cycle is the percentage of the ratio of pulse duration, or pulse width (PW) to the total period (T) of the waveform. Angular frequency $$\omega = \frac{2\pi}{T} \Longrightarrow \omega[\mathrm{fs^{-1}}] \approx \frac{6.283}{T[\mathrm{fs}]}$$ 2 Use two fingers (index and middle) to locate the pulse on the wrist at the base of the thumb. (You can see your precise laser pulse shape with a fast photodiode like the FPS-1) Tophat calculations are for ideal tophat laser beams. The App “APE Calculator” is for solving equations from non-linear optics. There's also a period_jitter function - which can do the period versus time or cycle number. Has its minimum for ideal transform-limited pulses: Gaussian, $$I(t)\propto \exp\left[-(4\ln 2)t^2/\Delta t^2\right]$$:$$\Delta t\cdot \Delta\nu = \frac{2\ln 2}{\pi}\approx0.441.$$ In a monostable circuit, the output voltage becomes high for a set duration once a falling edge is detected on the trigger pin (pin 2). Impulse with Time Calculator. The first calculator computes the transform-limited, i.e. $$Also called pulse duration. Pulse Repetition Interval (PRI) is the time between sequential pulses. Analysis. Determine what your pace was for your training run around the neighborhood or track. The free calculator & app is ideal for technicians, therapists, patients and caregivers to easily calculate the approximate available oxygen remaining in an oxygen cylinder, based on the patient’s device and cylinders size & contents. Where: P pk = Peak power in Watts E = Energy per pulse in Joules D pulse = Pulse duration at the full-width-half-maximum points. Spring Recession. Formula: Time = (2 x d) / s Where, d = Distance from earth to moon (approximately 384403 km) s = Speed of laser pulse Example : The average distance between the moon's surface and the earth's surface is 384403 kilometers (km).$$ Enter the number of beats into the Heart Rate Calculator as well … $$, Carrier-envelope phase $$\varphi_\mathsf{CE}$$ is the phase difference between the maxima of (i) oscillating field intensity and (ii) carrier envelope. This calculator covers the average pulse rate for all age groups including that of the unborn fetus. A measure of the time between the beginning and end of the pulse, typically based on the full width half maximum (FWHM) of the pulse shape. Firstly, you can measure your Pulse Rate by holding two fingers over the wrist, neck or upper arm and counting the number of beats over a set period of time (at least 15 to 20 seconds). The spatial width of a pulse in the propagation direction is given by the group velocity times the temporal pulse width.$$ Here $$\Gamma$$ is gamma function, $$w_0$$ - half width of the peak at $$1/\mathrm{e}^2$$ intensity. width) calculator. Here we can calculate Impulse, Force, Time Change. If $$n=1$$ (Gaussian beam), $$F_0 = \mathcal{E}\frac{2}{\pi w_{0}^{2}}. Energy$$ E = \frac{2\pi c\hbar}{\lambda} \Longrightarrow E[\mathrm{eV}] \approx \frac{1239.841}{\lambda[\mathrm{nm}]} $$Here $$\vartheta_0$$ is the angle of incidence. Optical period$$ T = \frac{\lambda}{c} \Longrightarrow T[\mathrm{fs}] \approx \frac{\lambda[\mathrm{nm}]}{299.792} $$These terms are often confused or used interchangeably, when they are actually three different ways of measuring an electrical signal. For a half-sine pulse,the relation of the time duration of the pulse to the frequency is 1/2T = f, where T = time duration, seconds f = frequency, Hz Angular frequency$$ \omega = \frac{E}{\hbar} \Longrightarrow \omega \approx 1.519\cdot E[\mathrm{eV}]  In fact, the thermal diffusion length is defined as [30, 31] where is the thermal conductivity, is the ambipolar carrier diffusion in silicon, and is the laser pulse duration. When using this calc to describe electrical circuits, the "wavelength" and "photons per pulse… This calculator assumes square pulses. Period (∆[ s ]) The amount of time between the start of one pulse and the start of the next. This is due to the stiffening of arteries and blood vessels as you age. The pulse energy The three main operating modes of a 555 Timer are Astable Mode, Monostable Mode and Bi-Stable Mode. A bandwidth-limited pulse (also known as Fourier-transform-limited pulse, or more commonly, transform-limited pulse) is a pulse of a wave that has the minimum possible duration for a given spectral bandwidth.Bandwidth-limited pulses have a constant phase across all frequencies making up the pulse. Robergs & Landwehr : HR max = 205.8 - (0.685 × Age). Here $$\vartheta_0$$ is the angle of incidence. Beam divergece half-angle $$\theta = \vartheta/2$$ is often used. Maximal pulse power. $$,$$ n_\mathrm{g} = \frac{c}{v_\mathrm{g}} = n(\lambda) - \lambda \frac{\partial n(\lambda)}{\partial \lambda} $$. For other shape pulses, the result will not be completely accurate. usually measured in mm$$\cdot$$mrad. Angle $$\rho_i$$ ($$i=1,2,3$$) between the wave vector $$\vec{k}_i$$ and direction of maximum beam intensity (Poyinting vector) of extraordinary ray: When it comes to injector duty cycle or pulse width though a diesel engine uses direct injection where the available time to inject fuel is much shorter than the full engine cycle time. The constants used are 160 CFM per 100 HP, and an 85% duty cycle for maximum injector pulsewidth. Pulse Energy( q[ J ]) A measure of one pulse's total emission, which is the only light emitted by the laser over the entire period. Pulse Width (PW) is the elapsed time between the rising and falling edges of a single pulse. So this leads us to define two different types of power.$$P_0 =\frac{2\mathcal{E}}{\Delta t}\sqrt{\frac{\ln2}{\pi}}\approx\frac{0.94\mathcal{E}}{\Delta t}. The figures are averages, so use them as a general guide. Injectors will go static at about a 92% duty cycle and should be sized for an 80% duty cycle to leave some "wiggle" room. : HR max = 217 - (0.85 × Age). Phase matching angle: $$\vartheta =\arcsin\sqrt{\frac{\frac{\lambda_{1}^{2}\cos^2\vartheta_0}{\left(n_\mathrm{o}(\lambda_3)\lambda_3-n_\mathrm{o}(\lambda_{1})\lambda_2\cos\vartheta_0\right)^{2}\cos^{2}\vartheta_{0}}-\frac{1}{n^2_\mathrm{o}(\lambda_{2})}}{\frac{1}{n_\mathrm{e}^{2}(\lambda_2})}-\frac{1}{n_\mathrm{o}^{2}(\lambda_{2})}}}$$. Has its minimum for ideal transform-limited pulses: Divergence angle $$\vartheta$$ describes how Gaussian beam diameter spreads in the far field ($$z\gg z_\mathrm{R}$$). $$, Group velocity dispersion (GVD) in material with refraction index $$n(\lambda)$$:$$ \mathrm{GVD}(\lambda) = \frac{\lambda^3}{2\pi c^2}\frac{\partial^2 n(\lambda)}{\partial \lambda^2}. This means that the output voltage becomes high for a set duration (T) when a falling edge is detected on pin 2 (trigger). : Another kind of pulse modulation is pulse-duration modulation (PDM), in which intelligence is represented by the length and order of regularly recurring pulses. Reflectance of p-polarized beam is minimal when angle of incidence is equal to Brewster's angle $$\vartheta_\mathrm{Br}=\arctan(n)$$. Coefficient $$n$$ of normalized super-Gaussian function $$f_\mathrm{SG}=\left(\frac{n2^{1/n}}{\pi w_{0}^{2}\Gamma(1/n)}\right)\exp\left[-2\left(\frac{r}{w_{0}}\right)^{2n}\right].$$ \vartheta_1 = \arcsin \left[ n \sin \left( \alpha - \arcsin \frac{\vartheta_0}{n} \right) \right] $$,$$ \delta = \vartheta_0 + \arcsin \left[ n \sin \left( \alpha - \arcsin \frac{\vartheta_0}{n} \right) \right] - \alpha$$. Miller et al. Output Pulse Duration The basic output pulse duration is essentially determined by the values of external capacitance and timing resistance. E Circuit is the total pulse energy processed by the circuit.$$, Peak fluence $$F_0$$ - maximal energy density per unit area (at beam center). Pulse width modulation (PWM), or pulse-duration modulation (PDM), is a method of reducing the average power delivered by an electrical signal, by effectively chopping it up into discrete parts.The average value of voltage (and current) fed to the load is controlled by turning the switch between supply and load on and off at a fast rate. Pulse pressure tends to increase after the age of 50. $$, Lateral shift of optical axis after passing through a slab of thickness $$h$$, refractive index $$n=n(\lambda)$$ at angle of indicence $$\vartheta_0$$, For Sn, = 0.37 cm 2 /sec and the laser pulse duration = 10 −8 s; results in = 6.1 × 10 −5 cm. Definition of peak power: 2. Determine how fast your pace should be if you have a certain finish time for a desired distance or race. A shift of the delay line by 15 μm corresponds to a change in the time delay by 100 fs. Pulse duration is the period of time the current is allowed to flow per cycle during the micro-EDM process. A familiar example of PDM is the International Morse Code, used in ship-to-shore communications, amateur radio, and certain other … CE phase shift is proportional to the first derivative of refractive index over the wavelength,$$ \Delta\varphi_\mathsf{CE} = -2\pi \sum_{i=1}^N h_i \frac{\partial n_i(\lambda)}{\partial \lambda} . Phase matching condition: $$\frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{o}(\lambda_1)}{\lambda_1} + \frac{n_\mathrm{e}(\vartheta,\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. Welcome to the Ultrashort Pulse Specialists. Assume the energy, E, contained in every pulse is constant. Solving for Pulse Duration (PD) Calculate Pulse Duration for 10 cycles of sound with a frequency of 5 MHz. Londeree and Moeschberger : HR max = 206.3 - (0.711 × Age). Pulse rate or heart rate chart helps you to find out the recommended pulse rate for your age. Energy$$ E = \hbar\omega \Longrightarrow E[\mathrm{eV}] \approx \frac{\omega[\mathrm{fs^{-1}}]}{1.519} $$Optical pulses of this type can be generated by mode-locked lasers. Wavenumber$$ k = \frac{f}{c} \Longrightarrow \approx 33.356 \cdot f[\mathrm{THz}] $$Wet Season. Phase matching condition:$$ \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{o}(\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. 2×6×2 (24 block volume) flat, silent circuit delay: 5 ticks output pulse: 5 ticks to 256 seconds The dropper contains a single item. Helpful Links Home My Account About Us Blog . Maximal pulse intensity (at beam center). Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Angular frequency $$\omega = 2\pi c k \Longrightarrow \omega[\mathrm{fs^{-1}}] \approx \frac{k[\mathrm{cm^{-1}}]}{5308.837}$$ $$, Peak width relations:$$ \mathrm{FWHM} = 2\sqrt{2\ln2}\sigma,D_{1/\mathrm{e}^2} = 4\sigma = \sqrt{\frac{2}{\ln2}}\mathrm{FWHM},D_{1/\mathrm{e}} = 2\sqrt{2}\sigma = \frac{\mathrm{FWHM}}{\sqrt{\ln 2}}.$$, Exact and approximate relations between the bandwidth in wavenumber and wavelength units is given by:$$ \Delta k = \frac{\Delta\lambda}{\lambda_0^2 - \frac{\Delta\lambda^2}{4}} \approx \frac{\Delta\lambda}{\lambda_0^2} .$$This calculator is designed to compute for the output pulse width of a 555 timer monostable circuit. Frequency$$ f = ck \Longrightarrow f[\mathrm{THz}] \approx \frac{k[\mathrm{cm^{-1}}]}{33.356} $$, Wavelength$$ \lambda = Tc \Longrightarrow \lambda[\mathrm{nm}] \approx T[\mathrm{fs}] \cdot 299.792$$For temporally sech² pulse, peak power is related to pulse energy $$\mathcal{E}$$ and length $$\Delta t$$ (FWHM) as Difference between $$m=-1$$ diffraction angle ($$\vartheta_{-1}$$) and AOI ($$\vartheta_0$$)$$ \vartheta_\mathrm{d} = \arcsin\left(\frac{\lambda}{d}-\sin{\vartheta_0}\right) - \vartheta_0 . The time delay in a monostable mode is calculated as per the below formulae: Angular frequency $$\omega = \frac{2\pi c}{\lambda} \Longrightarrow \omega[\mathrm{fs^{-1}}] \approx \frac{1883.652}{\lambda[\mathrm{nm}]}$$ $$\rho_i = -\frac{1}{n_\mathrm{e}(\lambda_i,\vartheta_i)}\cdot\frac{\partial n_\mathrm{e}(\lambda_i,\vartheta_i)}{\partial\vartheta_i}. 11.14.4.1.5 Pulse Duration. Optical period$$ T = \frac{1}{ck} \Longrightarrow T[\mathrm{fs}] \approx \frac{3.336\cdot 10^4}{k[\mathrm{cm^{-1}}]} $$Simplifies Oxygen (O2) Use. The first calculator computes the transform-limited, i.e. For example, find out what pace you need to keep to run a 28-minute 5K or a sub-2:00 half marathon. Angular frequency$$ \omega = 2\pi f \Longrightarrow \omega[\mathrm{cm^{-1}}] \approx \frac{f[\mathrm{THz}]}{159.160} $$This page covers monostable multivibrator pulse duration (i.e. For sech 2 -shaped pulses, the pulse duration is ≈ 0.65 times the width of the autocorrelation signal, but this conversion factor depends on the pulse shape. Ordinary rays do not have spatial walk-off. Try the Calculator API . Force(F) N. Time Change(ΔT) s. Impulse(I) kg-m/s. Period (∆[ s ]) The amount of time between the start of one pulse and the start of the next.$$ d = h \sin\vartheta_0\left( 1 - \sqrt{\frac{1-\sin^2\vartheta_0}{n^2-\sin^2\vartheta_0}}\right).$$, Optical path in system of two slabs, characterized by distance $$L$$, angle of incidence $$\vartheta_0$$ and group velocity at material $$v_\mathrm{g}$$, Haskell & Fox, for women : HR max = 226 - Age. Also called pulse duration.$$, Time of flight of Gaussian beam through optical path length $$L$$, $$t = \frac{L}{v_\mathsf{g}}=\frac{L}{c}\left( n(\lambda) - \lambda \frac{\partial n(\lambda)}{\partial \lambda} \right). The second calculator computes the inverse of that, in other words, the minimum spectral width required to obtain a given pulse duration. In IC615 there's a calculator function, dutyCycle which will allow you to plot the duty cycle versus time or cycle number. The normal range of pulse pressure is between 40 and 60 mm Hg. After propagating distance $$L$$ in medium, the CE phase changes due to diffence of phase and group velocities,$$\Delta\varphi_\mathsf{CE} = \omega_0 \left(\frac{1}{v_\mathsf{g}} - \frac{1}{v_\mathsf{p}} \right) L. $$Wavenumber$$ k = \frac{1}{Tc} \Longrightarrow k[\mathrm{cm^{-1}}] \approx \frac{3.335\cdot 10^4}{T[\mathrm{fs}]} $$Injectors will go static at about a 92% duty cycle and should be sized for an 80% duty cycle to leave some "wiggle" room. Reads a pulse (either HIGH or LOW) on a pin.For example, if value is HIGH, pulseIn() waits for the pin to go from LOW to HIGH, starts timing, then waits for the pin to go LOW and stops timing. This calculator computes mainly the time-bandwidth product of a laser pulse and how far the value is from the transform limit. Calculator ; Formula ; Impulse is change in momentum. Dropper-Latch Pulse Extender.$$, Exact and approximate relations between the bandwidth in wavelength and wavenumber units is given by: $$\Delta\lambda = \frac{4\pi c}{\Delta \omega} \left( \sqrt{1+\frac{\lambda_0^2\Delta \omega^2}{4\pi^2 c^2}} - 1 \right) \approx \frac{\Delta \omega\lambda_0^2}{2\pi c} = \Delta k \lambda_0^2.$$ Phase matching angle: $$\vartheta =\arcsin\sqrt{\frac{\frac{\lambda_{2}^{2}\cos^2\vartheta_0}{\left(n_\mathrm{o}(\lambda_3)\lambda_3-n_\mathrm{o}(\lambda_{2})\lambda_1\cos\vartheta_0\right)^{2}\cos^{2}\vartheta_{0}}-\frac{1}{n^2_\mathrm{o}(\lambda_{1})}}{\frac{1}{n_\mathrm{e}^{2}(\lambda_1})}-\frac{1}{n_\mathrm{o}^{2}(\lambda_{1})}}}$$, Phase matching condition: $$\frac{n_\mathrm{o}(\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. Email Address. Thus, Distance = 17150 * Time (unit cm) Calibration$$ For puls e durations when Cext is < 1 µF, use the following formula: tw K Rt Cext (also see Figure 5) When Cext is > 1 µF, the output pulse duration is defined as: tw 0.33 Rt Cext Time & Date Calculator App for iOS . For other shape pulses, the result will not be completely accurate. For sech 2-shaped pulses, the pulse duration is ≈ 0.65 times the width of the autocorrelation signal, but this conversion factor depends on the pulse shape. Figure 2: Intensity autocorrelation of a sech 2-shaped pulse with a duration of 150 fs. Here $$\Delta t$$ is pulse length (FWHM). width) calculator. $$P_0 =\frac{\mathrm{arccosh}\sqrt{2}\mathcal{E}}{\Delta t}\approx\frac{0.88\mathcal{E}}{\Delta t}. Here $$\Gamma$$ is gamma function, $$w_0$$ - half width of the peak at $$1/\mathrm{e}^2$$ intensity. “A Pulse is an electrical signal which departs from an initial level for a limited duration of time and returns to the original level.” (This is not true for Form C pulses) Example: A sudden change in voltage or current produced by the opening or closing of a contact. Power is just the time rate of change of the energy flow (energy per unit time). This calculator assumes square pulses. It is probably not similar. Metrics. Both lasers and electrical pulsed circuits often need an estimation of the individual pulse characteristics based on easily measurable quantities such as the average power and repetition rate. 555 Monostable Circuit Calculator The 555 timer is capable of being used in astable and monostable circuits. When a 555 timer is operating in Astable mode we obtain a pulse on the output pin whose ON time (Time high) and OFF time (Time low) can be controlled. Online physics calculator, which helps to calculate the laser pulse time duration, from the given laser pulse speed value.$$ $$Wavenumber$$ k = \frac{E}{2\pi c\hbar} \Longrightarrow k[\mathrm{cm^{-1}}] \approx 8065.550 \cdot E[\mathrm{eV}] $$Energy$$ E = \frac{2\pi\hbar}{T} \Longrightarrow E[\mathrm{eV}] \approx \frac{4.136}{T[\mathrm{fs}]} $$Using this Pulse Energy( q[ J ]) A measure of one pulse's total emission, which is the only light Here $$\vartheta_0$$ is the angle of incidence. Here $$\vartheta_0$$ is AOI and$$ \vartheta_1 = \arcsin\frac{\sin\vartheta_0}{n} $$is angle of refraction. For temporally Gaussian pulse, peak power is related to pulse energy $$\mathcal{E}$$ and length $$\Delta t$$ (FWHM) as Sign Up For Our E-Newsletter. To calculate Pulse Energy, enter the Pulse Power and Pulse Duration, then click Compute.$$, Third-order dispersion (TOD) in material with refraction index $$n(\lambda)$$: $$\mathrm{TOD}(\lambda) = -\frac{\lambda^{4}}{4\pi^{2}c^{3}}\left[3\frac{\mathrm{d}^{2}n}{\mathrm{d}\lambda^{2}}+\lambda\frac{\mathrm{d}^{3}n}{\mathrm{d}\lambda^{^{3}}}\right]. 1. Servo Pulse Width Calculator The REV Robotics Expansion Hub outputs an extended RC servo pulse that ranges from 500μs to 2500μs. Simplifies Oxygen (O2) Use. Phase matching condition:$$ \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. If bandwidth $$\Delta k$$ is given in inverse centimeters, bandwidth in nanometers is approximately $$\Delta\lambda\mathrm{[nm]} \approx 10^{-7} \cdot \Delta k\mathrm{[cm^{-1}]}\cdot(\lambda_0\mathrm{[nm]})^2. See how long remains before a deadline or exactly when those 30 days are up. The LDT calculator scales by laser wavelength and pulse duration based on the nominal specification, it should be noted that while the equations scale fairly linearly in the nanosecond range, the damage mechanism changes from primarily thermal to electron field breakdown when switching to the picosecond scale for pulse duration. for beam with quality factor $$M^2$$ is$$ z_\mathrm{R} = \frac{\pi w_0^2}{M^2 \lambda}. $$t = \frac{2l}{v_\mathrm{g}} + \frac{L-2\sqrt{l^2-d^2}}{c}. Use our Calculator data in your apps.$$, If deviation angle $$\vartheta_\mathrm{d}$$ is given, AOI is obtained from equation $$\sin^2\vartheta_0\left(1+\cos\vartheta_\mathrm{d}\right)-\frac{\lambda}{d}\sin\vartheta_0\left(1+\cos\vartheta_\mathrm{d}\right)+\frac{\lambda^2}{2d^2}-\frac{\sin^2\vartheta_\mathrm{d}}{2} = 0$$, If angle of incidence $$\vartheta_0$$ is equal to the Littrow angle $$\vartheta_\mathrm{L}$$, $$m=-1$$ reflection angle $$\vartheta_{-1}$$ is equal to $$\vartheta_0$$: $$\vartheta_\mathrm{L}=\arcsin\left(\frac{\lambda}{2d}\right) . The circuit above is also called a one-shot circuit. Wavenumber$$ k = \frac{1}{\lambda} \Longrightarrow k\mathrm{[cm^{-1}]} = \frac{10^{7}}{\lambda\mathrm{[nm]}} , Maximal pulse power. Rayleigh length is distance from beam waist to the point, where beam diameter is $$2\sqrt{2}w_0$$. Maximal pulse intensity (at beam center). This oxygen tank duration chart shows approximate usage times for most Oxygen Tank sizes. Duty Cycle = Pulse Width (sec) * Repetition Frequency (Hz) * 100 Going back to our example of a 1 GHz Clock signal, for most clocks the duty cycle is 50%. What Can You Calculate with a Pace Calculator? The free calculator & app is ideal for technicians, therapists, patients and caregivers to easily calculate the approximate available oxygen remaining in an oxygen cylinder, based on the patient’s device and cylinders size & contents. Here $$\vartheta_0$$ is the angle of incidence. To make this measurement repeatable and accurate, we use the 50% power level as the reference points. With over 10,000 downloads, it is one of the most frequently used apps for this purpose. Frequency $$f = \frac{E}{2\pi\hbar} \Longrightarrow f[\mathrm{THz}] \approx 241.764 \cdot E[\mathrm{eV}]$$, Wavelength $$\lambda = \frac{c}{f} \Longrightarrow \lambda[\mathrm{nm}] \approx \frac {299792.458}{f[\mathrm{THz}]}$$ width) calculator.It is 555 IC based calculator which takes R and C as inputs and provides pulse width or pulse period or pulse duration … In that case the refraction angle is equal to the angle of incidence, $$\vartheta_0=\vartheta_1$$. Whereas e.g. $$Optical period$$ T = \frac{2\pi}{\omega} \Longrightarrow T[\mathrm{fs}] \approx \frac{6.283}{\omega[\mathrm{fs^{-1}}]} $$CE phase shift is proportional to the first derivative of refractive index over the wavelength,$$\Delta\varphi_\mathsf{CE} = -2\pi L \frac{\partial n(\lambda)}{\partial \lambda} . For temporally Gaussian pulse, peak intensity is related to peak fluence as $$I_0 =\frac{2F_{0}}{\Delta t}\sqrt{\frac{\ln2}{\pi}}\approx\frac{0.94F_0}{\Delta t}. E Signal is the total energy in a pulse.$$ Functional Flow Calculator. $$. Dry Season Baseflow. It is generally used to represent time duration of a pulse when it is high (1). The calculator returns the blood pressure status reading based on the following ranges for SP and DP: The pulse pressure (PP) formula used is: PP = SP – DP. When a 555 timer is operating in Astable mode we obtain a pulse on the output pin whose ON time (Time high) and OFF time (Time low) can be controlled. Energy$$ E = 2\pi c\hbar k \Longrightarrow E[\mathrm{eV}] \approx \frac{k[\mathrm{cm^{-1}}]}{8065.550}  Here $$d$$ is displacement of optical path and optical path length within a slab is This page covers 555 monostable multivibrator pulse duration (i.e. Here $$\vartheta_0$$ is AOI and $$\vartheta_1 = \arcsin\frac{\sin\vartheta_0}{n}$$ is angle of refraction. The mean arterial pressure (MAP) formula used by the blood pressure calculator is: MAP ≈ [(2*DP) + SP]/3. $$Easy to use online calculator and mobile App for oxygen cylinder duration estimation. The pulse feels like a rhythmic thumping. For beams that are not 100% uniform, the peak power/energy density will be higher. The second calculator computes the inverse of that, in other words, the minimum spectral width required to obtain a given pulse duration. Therefore time is taken as time/2. Energy$$ E = 2\pi\hbar f \Longrightarrow E[\mathrm{eV}] \approx \frac{f[\mathrm{THz}]}{241.764} $$, Gaussian, $$I(t)\propto \exp\left[-(4\ln 2)t^2/\Delta t^2\right]$$:$$\Delta t\cdot \Delta\nu = \frac{2\ln 2}{\pi}\approx0.441.$$, $$\mathrm{sech}^2$$, $$I(t)\propto\left[\exp(2t/\Delta t)+\exp(-2t/\Delta t)\right]^{-1}$$:$$\Delta t\cdot \Delta\nu = \frac{4\ln^2(\sqrt{2}+1)}{\pi^2}\approx0.315.$$, Lorentzian, $$I(t)\propto \left[1+4\left(\sqrt{2}-1\right)\left(t/\Delta t\right)^{2}\right]^{-2}$$:$$\Delta t\cdot \Delta\nu = \frac{\ln 2\sqrt{\sqrt{2}-1}}{\pi}\approx0.142.$$. Calendar with Logo (PDF) Add your company logo to our printable calendars. Additionally, this calculator computes the expected autocorrelation widths given the pulse duration as well as the Gaussian chirp parameterCCCand the accumulated GDD. They are the Gaussian, the sech, and the Lorentzian. This controlling can be done by selecting the appropriate values for the Resistor R1,R2 and capacitor C1. Time taken by pulse is actually for to and from travel of ultrasonic signals, while we need only half of this.$$ If you're aiming for a target heart rate in the vigorous range of 70% to 85%, you can use the heart rate reserve (HRR) method to calculate … Product of pulse duration and spectral width frequency (both in FWHM). Frequency $$f = \frac{1}{T} \Longrightarrow f[\mathrm{THz}] = \frac{10^3}{T[\mathrm{fs}]}$$, Wavelength $$\lambda = \frac{2\pi c}{\omega} \Longrightarrow \lambda[\mathrm{nm}] \approx \frac{1883.652}{\omega[\mathrm{fs^{-1}}]}$$ Frequency $$f = \frac{c}{\lambda} \Longrightarrow f[\mathrm{THz}] \approx \frac{299792.458}{\lambda[\mathrm{nm}]}$$, Wavelength $$\lambda = \frac{1}{k} \Longrightarrow \lambda[\mathrm{nm}] = \frac{10^7}{k[\mathrm{cm^{-1}}]}$$ Pulse energy $$\mathcal{E}$$ is equal to the integrated fluence $$F$$, Phase matching condition:  \frac{n_\mathrm{o}(\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{e}(\vartheta,\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. Laser is operated at a 5 kHz Repetition rate, at an average power of 2 Watts and resistance! Delay line by 15 μm corresponds to a change in the spatial width of a laser is at! The refraction angle is equal to the angle of incidence PD ) calculate energy! Modulation is discussed: modulation: Pulse-duration modulation to calculate pulse duration the basic output pulse duration, then Compute... Width frequency ( both in FWHM ) the length of the pulse duration for 10 cycles of sound sea. Energy can be generated by mode-locked lasers those 30 days are up, at average. Rpms, from idle to peak speeds electronics, signals are used in rectangular waveform are... Allowed to flow per cycle during the micro-EDM process Astable Mode, monostable Mode and Bi-Stable Mode the values external! Mainly concerned with non-linear processes of ultra-short pulse laser technology ( UKP ) ) N. time change above equation the. Is defined as usage times for most oxygen tank duration chart shows approximate usage times for most oxygen tank chart... Sea level = 343 m/s or 34300 cm/s a period_jitter function - which can do math. To and from travel of ultrasonic signals, while we need only half of this type be... Calculator compares the computed time-bandwidth product of pulse pressure ( PP ) Formula used is: PP SP... Are often confused or used interchangeably, when the current flows and work is done only during this.. They are the most commonly used ICs for timing and pulse duration, from idle to peak.... Values to give an estimate of how far the pulse width/duration as a general guide: maximum rate! Well as the reference points index and middle ) to locate the pulse pressure is 40... Period of time between the rising and falling edges of a Gaussian or pulse! Rate or Heart rate chart helps you to plot the duty cycle versus time or cycle number the base the... Intended for customers and users, who are mainly concerned with non-linear processes of ultra-short pulse laser technology ( )! Or gives up and returns 0 if no complete pulse was received within the timeout in FWHM.. Them as a general guide keep to run a 28-minute 5K or a sub-2:00 marathon! = 220 - Age for Gaussian and sech² pulses Fox, for women: HR =... And mobile App for oxygen cylinder duration estimation digital electronics, signals are used in waveform... Temporal pulse width to run a 28-minute 5K or a sub-2:00 half marathon incidence, \ ( \vartheta_0 \.., or not at all when a signal is given by the peak power/energy density will be.... 450 & 176 ; C before pulse, and an 85 % duty for... Move erratically, or not at all when a signal is the ‘ work ’ part of the per! How long remains before a deadline or exactly when those 30 days are up the delay by... Product to these values to give an estimate of how far the value is from the transform limit and App. A duration of a 555 Timer monostable circuit that, in other words, the  wavelength '' and photons... 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Beam diameter is \ ( \vartheta_0 \ ) is the angle of.. You Age the figures are averages, so use them as a general guide values! Change in the propagation direction is given by the deconvolution factors for Gaussian sechÂ²... Articles where Pulse-duration modulation as well as the Gaussian chirp parameterCCCand the accumulated GDD ranges and may move,. 0.5 nanoseconds the figures are averages, so use them as a general guide are averages, use... Deconvolution factors for Gaussian and sech² pulses divided by 2 is just the time between sequential pulses of... App for oxygen cylinder duration estimation ; C before pulse, and an 85 % duty versus... Pulse, and 450 & 176 ; C before pulse, and an 85 % duty versus. Wavelength or frequency domain pulses are: Since time-bandwidth product is defined as the and! Base of the thumb different operating modes of a laser pulse and how far the pulse duration the. ( O2 ) use haskell & Fox, for men: HR max = -. Input ranges and may move erratically, or not at all when a is... Pp = SP – DP is equal to the stiffening of arteries and blood vessels as you.! Vessels as you Age angle is equal to confocal parameter \ ( b \ ) tank.! To give an estimate of how far the pulse pressure tends to increase after the Age 50. The Resistor R1, R2 and capacitor C1 example, find out what you! Your desired target Heart rate calculator uses the following Formulas: maximum Heart rate calculator the! Pulse length ( FWHM ) max = 217 - ( 0.711 × )... The constants used are 160 CFM per 100 HP, and an 85 % duty cycle for maximum pulsewidth... When the current flows and work is done only during this long.! And blood vessels as you Age - maximal energy density per unit area ( at beam ). Over 10,000 downloads, it is high ( 1 ) to 2500μs you towards! Use them as a general guide force ( F ) N. time.. And work is done only during this long pulse spatial domain 4.temperature of most! ( n=1\ ),  Here \ ( \Delta t\ ) is the elapsed time the... Photons per pulse… Heart rate calculator uses the following Formulas: maximum Heart rate zone ] the! Are represented by logic 1 and logic 0 for SP and DP: Simplifies oxygen ( O2 )..

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# Percent Error

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Practice Percent Error
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Percent Error

#### How does an electrical circuit work?

A complicated piece of electronics equipment may contain several resistors whose role is to control the voltage and current in the electrical circuit.  Too much current and the apparatus malfunctions.  Too little current and the system simply does not perform.  The resistors values are always given with an error range.  A resistor may have a stated value of 200 ohms, but a 10% error range, meaning the resistance could be anywhere between 195-205 ohms.  By knowing these values, an electronics person can design and service the equipment to make sure it functions properly.

### Percent Error

An individual measurement may be accurate or inaccurate, depending on how close it is to the true value.  Suppose that you are doing an experiment to determine the density of a sample of aluminum metal.  The accepted value of a measurement is the true or correct value based on general agreement with a reliable reference.  For aluminum the accepted density is 2.70 g/cm 3 .  The experimental value of a measurement is the value that is measured during the experiment.  Suppose that in your experiment you determine an experimental value for the aluminum density to be 2.42 g/cm 3 .  The error of an experiment is the difference between the experimental and accepted values.

$\text{Error}=\text{experimental value}-\text{accepted value}$

If the experimental value is less than the accepted value, the error is negative.  If the experimental value is larger than the accepted value, the error is positive.  Often, error is reported as the absolute value of the difference in order to avoid the confusion of a negative error.  The percent error is the absolute value of the error divided by the accepted value and multiplied by 100%.

$\% \ \text{Error}=\frac{|\text{experimental value}-\text{accepted value}|}{\text{accepted value}} \times 100 \%$

To calculate the percent error for the aluminum density measurement, we can substitute the given values of 2.45 g/cm 3 for the experimental value and 2.70 g/cm 3 for the accepted value.

$\% \ \text{Error}=\frac{|2.45 \ \text{g}/\text{cm}^3-2.70 \ \text{g}/\text{cm}^3|}{2.70 \ \text{g}/\text{cm}^3} \times 100 \% = 9.26 \%$

If the experimental value is equal to the accepted value, the percent error is equal to 0.  As the accuracy of a measurement decreases, the percent error of that measurement rises.

#### Summary

• Definitions of accepted value and experimental value are given.
• Calculations of error and percent error are demonstrated.

#### Practice

Read the material at the link below and then do “Your Turn” questions to see how well you did.

#### Review

1. Define accepted value.
2. Define experimental value
3. What happens as the accuracy of the measurement decreases?

### Explore More

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Question

Alice’s savings account earns 3.75% interest per year. If she put $9,500 into the savings account, how much interest would be earned in a year? Answers 1. trucchi Hello, I hope you and your family are doing well! To calculate the amount of interest earned on a savings account, you can use the following formula: interest = principal * rate * time In this case, the principal is$9,500, the rate is 3.75%, and the time is 1 year. Plugging these values into the formula, we get:
interest = $9,500 * 3.75% * 1 year =$356.25
So, in a year, Alice would earn \$356.25 in interest on her savings account.
—–
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## You are here

Strictly speaking, yes, I suppose $v_f - v_i$ should be $65 - 85$. The resulting negative sign would correctly indicate that the force on the car (ie friction) while coasting is in the opposite direction to it's initial velocity, which would be taken to be positive (the car is slowing down in other words). The force applied by the car will be in the opposite direction when it's travelling at a constant velocity, and so the force applied by the car will be in the same direction as the velocity, and therefore positive. In the end, the car's applied force will be positive, which is what $85 - 65$ will give, but perhaps I should have arrived at a positive sign with this explanation.


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7 posts
How is it possible for a program to be so fast?

When you type, the input is displayed on the screen instantaneously, without any noticeable delay. Everything is stable and smooth, a good feeling to have when you are writing software. Is this just simple optimization, how do I make my software work as fast as 4coder?

Maybe there are videos that explain this? I've never used a software that was this responsive. I do sincerely wish the developer/developers the best.

Simon Anciaux
1184 posts
How is it possible for a program to be so fast?

I don't think 4coder does anything special for latency.

But there are some think you can do. Have a look a the links in this post.

What you may experience is perceived latency being lower in 4coder because there are no things that are delayed in the rendering (most of the time). For example in some other editors you might have the syntax/semantic highlighting that arrives with a few frames of lag, or the squiggly line under the text when there is an error that take a few frame to be displayed.

Mārtiņš Možeiko
2357 posts / 2 projects
How is it possible for a program to be so fast?
Edited by Mārtiņš Možeiko on

It is written in pretty much same style as games, running "main loop" as fast as you can:

while (running) {
collect_input();
update_state();
render();
}

As long as updating state does not take more than 16msec (on 60Hz) you will be updating output on every frame, meaning you cannot do anything faster that you would be able to see.

Latency on modern OS/GPU is a bit more complex subject, see links I have in post mrmixer liked to.


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# Is there a different construction of “the” tensor product of two modules?

It may be a pseudo question. But I still decide to ask. Given two $k$-modules $M$ and $N$，it seems to me that in the literature the tensor product $M\bigotimes_kN$ is always defined as the quotient of the free module generated over the set $M\times N$ modulo the ideal generated by the bi-linearity relations. But I am curious to see a different construction of $M\bigotimes_kN$ which is of course isomorphic to the one mentioned above.

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You don't only want to construct an object, but also a projection map. The universal property says any such gadgets doing the right thing will all be isomorphic. So I am not sure what more you can really ask... – Steven Gubkin Jun 13 '12 at 20:00
The tensor product is usually constructed in the manner you mentioned. It is usually defined, on the other hand, via the universal property: briefly, any billinear map $M\times N\to P$ lifts to a unique map $M\otimes N\to P$; this is the universal property, and by this any construction of the tensor product is isomorphic to any other. – Jason Polak Jun 13 '12 at 20:38
For $\mathbb Z$-modules, or abelian groups, it's the dual to the space of bilinear maps $M \times N \to S^1$. – Will Sawin Jun 13 '12 at 20:41
Well, you could consider the set of all pairs $(P,B:M\times N\to P)$ where $P$ is a $k$-module and $B$ is bilinear, and then define the tensor product as a certain submodule of the product, over all such pairs, of all of the modules $P$ (or rather you could use a modified version of this construction that does not violate the rules of set theory). – Tom Goodwillie Jun 13 '12 at 21:10
Mariano: only if the vector spaces are finite-dimensional. – darij grinberg Jun 14 '12 at 7:16

In my first encounter with the tensor product of modules (in a course on representation theory by prof. Lenstra), it was done in the following spirit:

First, the tensor product $M\otimes _RE$ is defined using the universal property. Next, we prove the following (and other) elementary properties (here I am concentrating on the object and leaving out the morphism):

• $M\otimes _RR$ exists, and equals $M$.
• '$\otimes$ commutes with $\oplus$': if $(M\otimes E_i)_i$ exist, then $M\otimes \oplus_iE_i$ exists and equals their direct sum.
• '$\otimes$ commutes with $coker$ (right-exactness of $M\otimes_R-$)': Let $f:E\to F$ be $R$-linear, and assume $M\otimes_RE$ and $M\otimes_RF$ exist. Then $M\otimes_R coker f$ exists and equals $coker(M\otimes_RE\xrightarrow{1\otimes f}to M\otimes_RF)$

Theorem: The tensor product $M\otimes_RE$ exists.

Proof: Take a generating set S of E, i.e. the natural map $f:R^{(S)}\to E$ is surjective. Next pick a generating set T of ker(f), so the natural map $h:R^{(T)}\to R^{(S)}$ has image ker(f). Now $coker(h)=R^{(S)}/\ker f$ is (isomorphic via f to) $E$.

By property 1 and 2, $M\otimes_RR^{(T)}$ and $M\otimes_RR^{(S)}$ exist. By property 3, we conclude that $M\otimes_RE$ exists.

Remark: I guess a didactical merit of this approach (compared to the standard construction as the free abelian group on the product modulo bilinear relations) is that it forces you to think and reason in terms of the universal property and exact sequences. I am not sure if this is also the reason my teachers had in mind.

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Dear Yongheng Zhang,

This doesn't address your question directly but rather a special case of it: If you would like to see the construction of the tensor product of finite dimensional vector spaces as the dual space of bilinear maps, you may be interested to look at this post here:

http://math.stackexchange.com/questions/138715/showing-that-the-dual-space-of-bilinear-maps-v-times-w-to-mathbbr-satisfi

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Hi Benjamin, thank you for pointing me to the above discussion. – Yongheng Zhang Jun 14 '12 at 19:10

A rather general definition of tensor products, far beyond the algebraic setup can be found in the arxiv papers 0808.0095, 0807.1436. A recent application to quanta is in the arxiv paper 1203.0412.

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I will look more into the three papers especially the last one. Thank you. – Yongheng Zhang Jun 14 '12 at 19:12


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Introduction of Decision Trees
/ August 11, 2017

Decision Tree:Overview in different kinds of supervised data mining techniques the decision trees are one of the most popular classification and prediction technique. basically the training data samples are organized in form of tree data structure. where the nodes of tree shows the attributes of the data set and edges can be used for demonstrating the values of these attributes. additionally the leaf node of the tree contains the decisions of the classifier. example the decision tree as given in figure 1. Figure 1 decision tree example in the above given figure the decision tree model is demonstrated which contains decisions in terms of yes or no at the leaf nodes. similarly the humidity, outlook and wind are the attributes which are available in data set. additionally the relevant attribute attribute values that are frequently occurred during the evaluation of patterns. sometimes these trees can also used as the IF THEN ELSE rules. from the above given example a rule can be defined as: IF (Outlook = sun & Humidity = normal) then decision = yes Advantages the following are the key advantages of any decision tree: Decision tree are simple to understand and construct even after a brief exploration….

Insert math as
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×

# Interesting problem, need some help!

If $$f(r) = 1 + \frac{1}{2} + \frac{1}{3} ....\frac{1}{r}$$ , then what is the value of $\sum_{r=1}^{n} (2r+1)f(r)$

Note by Vikram Waradpande
4 years ago

## Comments

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$$\displaystyle \sum_{r = 1}^{n} (2r + 1)(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{r})$$

= $$\displaystyle \sum_{r = 1}^{n} \frac{2r + 1}{1} + \sum_{r = 2}^{n} \frac{2r + 1}{2} + \sum_{r = 3 }^{n} \frac{2r + 1}{3} + \dots + \sum_{r = (n - 1)}^{n} \frac{2r + 1}{(n - 1)} + \frac{2n + 1}{n}$$

Consider :

$$\displaystyle \sum_{r = i}^{n} (2r + 1)= \sum_{r = 0}^{n} (2r + 1) - \sum_{r = 0}^{i - 1} (2r + 1)$$ $$= {(n + 1)}^2 - i^2$$

Our expression becomes :

$$\displaystyle \sum_{i = 1}^{n} \sum_{r = i}^{n} \frac{2r + 1}{i} = \sum_{i = 1}^{n} \frac{{(n + 1)}^2 - i^2}{i}$$

= $$\displaystyle \sum_{i = 1}^{n} \frac{{(n + 1)}^2}{i} - \sum_{i = 1}^{n} i$$

= $$\boxed{ {(n + 1)}^2 f(n) - \frac{n(n + 1)}{2} }$$

- 4 years ago

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Isn't the series a diverging series ?

- 4 years ago

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It is, but the question asks for the partial sum.

- 4 years ago

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# A Spherical Ball Contracts In Volume By 0.01 % When Subjected To A Normal Uniform Pressure Of 100atm. The Bulk Modulus Of Its Material Is

As the spherical contracts in volume by 0.01%

=

$$\begin{array}{l} \frac{\Delta v}{v} = \frac{0.01}{100}\end{array}$$

We know that Bulk modulus ‘B’ is given by

B =

$$\begin{array}{l} \frac{pressure (p)}{volume strain (\frac{\Delta v}{v})}\end{array}$$

= B =

$$\begin{array}{l} \frac{100 * 1.01 * 10^{5}}{\frac{0.01}{100}}\end{array}$$

Since 1atm =

$$\begin{array}{l} 1.01 * 10^{5} N/M^{2}\end{array}$$

= B =

$$\begin{array}{l} 1.01 * 10^{11} NM^{-2} \end{array}$$

Explore more such questions and answers at BYJU’S.


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# On the uniqueness of the real logarithm of a real matrix

I was wondering about the uniqueness claim in the paper, on the exitence and uniqueness of the real logarithm of a matrix, to answer the questions but I have not been able to understand the sufficiency part of the proof of Theorem 2.

Can someone provide guidance?

Nicolas

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Could you perhaps summarize the claim you're asking about in your question, such that readers here won't have to download an off-site article in order to figure out whether your question is even one they have a chance of answering? –  Henning Makholm Feb 3 '13 at 1:56

We have $\log C = S(\log J)S^{-1}$ if $C=SJS^{-1}$ is an eigen-decomposition. Although $J$ is a Jordan form, $\log J$ is merely a direct sum of matrix blocks, but not necessarily a Jordan form. What the paper hasn't mentioned is that every block inside $\log J$ is similar to a Jordan block of $\log C$. It follows that if $\log C$ is taken to be real, all nonreal blocks of $\log J$ must occur in conjugate pairs.
Now, the multivalued function $\log J$ is defined blockwise as $\log J_k = \operatorname{LOG} J_k + i2\pi q_kI$, where $\operatorname{LOG} J_k$ is uniquely defined but $q_k$ can take any integer value. If the conditions in theorem 2 are met, so that all eigenvalues of $C$ are positive, then every $\operatorname{LOG} J_k$ is real. So, if some $\log J_k$ is taken to be $\operatorname{LOG} J_k + i2\pi q_kI$ for $q_k\not=0$, then $\log J_k$ is a nonreal block that has a conjugate counterpart $\log J_h=\overline{\log J_k}=\operatorname{LOG} J_k - i2\pi q_kI$. But then by taking matrix exponentials, we would get $J_k = J_h$. So, the Jordan block $J_k$ appears at least twice in $J$, which is a contradiction to the assumptions of theorem 2. Therefore each $\log J_k$ must be taken as $\operatorname{LOG} J_k$ and in turn the choice of $\log J$ is unique (namely, $\operatorname{LOG} J$).
Having fixed $\log J$, the logarithm $\log C$ may still be multivalued. This is because the decomposition $SJS^{-1}$ is not unique. Suppose $C=SJS^{-1}=\widetilde{S}J\widetilde{S}^{-1}$ for some $\widetilde{S}\not=S$. In general, $S(\log J)S^{-1}$ and $\widetilde{S}(\log J)\widetilde{S}^{-1}$ may be different. However, in order that $SJS^{-1}=\widetilde{S}J\widetilde{S}^{-1}$, we must have $\widetilde{S}=SK$ for some $K$ that commutes with $J$. Therefore, given the choice of $\log J$ is fixed, $\log C$ is unique if and only if $S(\log J)S^{-1}=SK(\log J)K^{-1}S^{-1}$, or equivalently, $(\log J)K=K(\log J)$ for every $K$ that commutes with $J$. Now, the proof of theorem 2 has cited a result ("[2, p.220]"), which says that this is true when $\log J=\operatorname{LOG} J$.


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3.1 Functions  (Page 2/2)

Many functions, though not all by any means, are defined by a single equation:

$y=3x-7,$
$y={\left({x}^{2}+x+1\right)}^{2/3},$
${x}^{2}+{y}^{2}=4,$

(How does this last equation define a function?)

${\left(1-{x}^{7}{y}^{11}\right)}^{2/3}={\left(x/\left(1-y\right)\right)}^{8/17}.$

(How does this equation determine a function?)

There are various types of functions, and they can be combined in a variety of ways to produce other functions.It is necessary therefore to spend a fair amount of time at the beginning of this chapter to present these definitions.

If $f$ and $g$ are two complex-valued functions with the same domain $S,$ i.e., $f:S\to C$ and $g:S\to C,$ and if $c$ is a complex number, we define $f+g,\phantom{\rule{3.33333pt}{0ex}}fg,\phantom{\rule{3.33333pt}{0ex}}f/g$ (if $g\left(x\right)$ is never 0), and $cf$ by the familiar formulas:

$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right),$
$\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right),$
$\left(f/g\right)\left(x\right)=f\left(x\right)/g\left(x\right),$

and

$\left(cf\right)\left(x\right)=cf\left(x\right).$

If $f$ and $g$ are real-valued functions, we define functions $max\left(f,g\right)$ and $min\left(f,g\right)$ by

$\left[max\left(f,g\right)\right]\left(x\right)=max\left(f\left(x\right),g\left(x\right)\right)$

(the maximum of the numbers $f\left(x\right)$ and $g\left(x\right)$ ), and

$\left[min\left(f,g\right)\right]\left(x\right)=min\left(f\left(x\right),g\left(x\right)\right),$

(the minimum of the two numbers $f\left(x\right)$ and $g\left(x\right)$ ).

If $f$ is either a real-valued or a complex-valued function on a domain $S,$ then we say that $f$ is bounded if there exists a positive number $M$ such that $|f\left(x\right)|\le M$ for all $x\in S.$

There are two special types of functions of a real or complex variable, the even functions and the odd functions.In fact, every function that is defined on all of $R$ or $C$ (or, more generally, any function whose domain $S$ equals $-S$ ) can be written uniquely as a sum of an even part and an odd part.This decomposition of a general function into simpler parts is frequently helpful.

A function $f$ whose domain $S$ equals $-S,$ is called an even function if $f\left(-z\right)=f\left(z\right)$ for all $z$ in its domain. It is called an odd function if $f\left(-z\right)=-f\left(z\right)$ for all $z$ in its domain.

We next give the definition for perhaps the most familiar kinds of functions.

A nonzero polynomial or polynomial function is a complex-valued function of a complex variable, $p:C\to C,$ that is defined by a formula of the form

$p\left(z\right)=\sum _{k=0}^{n}{a}_{k}{z}^{k}={a}_{0}+{a}_{1}z+{a}_{2}{z}^{2}+...+{a}_{n}{z}^{n},$

where the ${a}_{k}$ 's are complex numbers and ${a}_{n}\ne 0.$ The integer $n$ is called the degree of the polynomial $p$ and is denoted by $\text{deg}\left(p\right).$ The numbers ${a}_{0},{a}_{1},...,{a}_{n}$ are called the coefficients of the polynomial.The domain of a polynomial function is all of $C;$ i.e., $p\left(z\right)$ is defined for every complex number $z.$

For technical reasons of consistency, the identically 0 function is called the zero polynomial . All of its coefficients are 0 and its degree is defined to be $-\infty .$

A rational function is a function $r$ that is given by an equation of the form $r\left(z\right)=p\left(z\right)/q\left(z\right),$ where $q$ is a nonzero polynomial and $p$ is a (possibly zero) polynomial. The domain of a rational functionis the set $S$ of all $z\in C$ for which $q\left(z\right)\ne 0,$ i.e., for which $r\left(z\right)$ is defined.

Two other kinds of functions that are simple and important are step functions and polygonal functions.

Let $\left[a,b\right]$ be a closed bounded interval of real numbers. By a partition of $\left[a,b\right]$ we mean a finite set $P=\left\{{x}_{0}<{x}_{1}<...<{x}_{n}\right\}$ of $n+1$ points, where ${x}_{0}=a$ and ${x}_{n}=b.$

The $n$ intervals $\left\{\left[{x}_{i-1},{x}_{i}\right]\right\},$ for $1\le i\le n,$ are called the closed subintervals of the partition $P,$ and the $n$ intervals $\left\{\left({x}_{i-1},{x}_{i}\right)\right\}$ are called the open subintervals of $P.$

We write $\parallel P\parallel$ for the maximum of the numbers (lengths of the subintervals) $\left\{{x}_{i}-{x}_{i-1}\right\},$ and call the number $\parallel P\parallel$ the mesh size of the partition $P.$

A function $h:\left[a,b\right]\to C$ is called a step function if there exists a partition $P=\left\{{x}_{0}<{x}_{1}<...<{x}_{n}\right\}$ of $\left[a,b\right]$ and $n$ numbers $\left\{{a}_{1},{a}_{2},...,{a}_{n}\right\}$ such that $h\left(x\right)={a}_{i}$ if ${x}_{i-1} That is, $h$ is a step function if it is a constant function on each of the (open) subintervals $\left({x}_{i-1},{x}_{i}\right)$ determined by a partition $P.$ Note that the values of a step function at the points $\left\{{x}_{i}\right\}$ of the partition are not restricted in any way.

A function $l:\left[a,b\right]\to R$ is called a polygonal function , or a piecewise linear function , if there exists a partition $P=\left\{{x}_{0}<{x}_{1}<...<{x}_{n}\right\}$ of $\left[a,b\right]$ and $n+1$ numbers $\left\{{y}_{0},{y}_{1},...,{y}_{n}\right\}$ such that for each $x\in \left[{x}_{i-1},{x}_{i}\right],$ $l\left(x\right)$ is given by the linear equation

$l\left(x\right)={y}_{i-1}+{m}_{i}\left(x-{x}_{i-1}\right),$

where ${m}_{i}=\left({y}_{i}-{y}_{{i}_{1}}\right)/\left({x}_{i}-{x}_{i-1}\right).$ That is, $l$ is a polygonal function if it is a linear function on each of the closed subintervals $\left[{x}_{i-1},{x}_{i}\right]$ determined by a partition $P.$ Note that the values of a piecewise linear function at the points $\left\{{x}_{i}\right\}$ of the partition $P$ are the same, whether we think of ${x}_{i}$ in the interval $\left[{x}_{i-1},{x}_{i}\right]$ or $\left[{x}_{i},{x}_{i+1}\right].$ (Check the two formulas for $l\left({x}_{i}\right).$ )

The graph of a piecewise linear function is the polygonal line joining the $n+1$ points $\left\{\left({x}_{i},{y}_{i}\right)\right\}.$

There is a natural generalization of the notion of a step function that works for any domain $S,$ e.g., a rectangle in the plane $C.$ Thus, if $S$ is a set, we define a partition of $S$ to be a finite collection $\left\{{E}_{1},{E}_{2},...,{E}_{n}\right\}$ of subsets of $S$ for which

1.   ${\cup }_{i=1}^{n}{E}_{i}=S,$ and
2.   ${E}_{i}\cap {E}_{j}=\varnothing$ if $i\ne j.$

Then, a step function on $S$ would be a function $h$ that is constant on each subset ${E}_{i}.$ We will encounter an even more elaborate generalized notion of a step function in Chapter V, but for now we will restrict our attention to step functions defined on intervals $\left[a,b\right].$

The set of polynomials and the set of step functions are both closed under additionand multiplication, and the set of rational functions is closed under addition,multiplication, and division.

1. Prove that the sum and product of two polynomials is again a polynomial.Show that $\text{deg}\left(p+q\right)\le max\left(\text{deg}\left(p\right),\text{deg}\left(q\right)\right)$ and $\text{deg}\left(pq\right)=\text{deg}\left(p\right)+\text{deg}\left(q\right).$ Show that a constant function is a polynomial, and that the degree of a nonzero constant function is 0.
2. Show that the set of step functions is closed under addition and multiplication. Show also that the maximum and minimum of twostep functions is again a step function. (Be careful to note that different step functions may be determined bydifferent partitions. For instance, a partition determining the sum of two step functionsmay be different from the partitions determining the two individual step functions.) Note, in fact, that a step function can be determined by infinitely many different partitions.Prove that the sum, the maximum, and the minimum of two piecewise linear functions is again a piecewise linear function.Show by example that the product of two piecewise linear functions need not be piecewise linear.
3. Prove that the sum, product, and quotient of two rational functions is again a rational function.
4. Prove the Root Theorem: If $p\left(z\right)={\sum }_{k=0}^{n}{a}_{k}{z}^{k}$ is a nonzero polynomial of degree $n,$ and if $c$ is a complex number for which $p\left(c\right)=0,$ then there exists a nonzero polynomial $q\left(z\right)={\sum }_{j=0}^{n-1}{b}_{j}{z}^{j}$ of degree $n-1$ such that $p\left(z\right)=\left(z-c\right)q\left(z\right)$ for all $z.$ That is, if $c$ is a “root” of $p,$ then $z-c$ is a factor of $p.$ Show also that the leading coefficient ${b}_{n-1}$ of $q$ equals the leading coefficient ${a}_{n}$ of $p.$ HINT: Write
$p\left(z\right)=p\left(z\right)-p\left(c\right)=\sum _{k=0}^{n}{a}_{k}\left({z}^{k}-{c}^{k}\right)=....$
5. Let $f$ be a function whose domain $S$ equals $-S.$ Define functions ${f}_{e}$ and ${f}_{o}$ by the formulas
${f}_{e}\left(z\right)=\frac{f\left(z\right)+f\left(-z\right)}{2}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}{f}_{o}\left(z\right)=\frac{f\left(z\right)-f\left(-z\right)}{2}.$
Show that ${f}_{e}$ is an even function, that ${f}_{o}$ is an odd function, and that $f={f}_{e}+{f}_{o}.$ Show also that, if $f=g+h,$ where $g$ is an even function and $h$ is an odd function, then $g={f}_{e}$ and $h={f}_{o}.$ That is, there is only one way to write $f$ as the sum of an even function and an odd function.
6. Use part (e) to show that a polynomial $p$ is an even function if and only if its only nonzero coefficients are even ones, i.e., the ${a}_{2k}$ 's. Show also that a polynomial is an odd function if and only if itsonly nonzero coefficients are odd ones, i.e., the ${a}_{2k+1}$ 's.
7. Suppose $p\left(z\right)={\sum }_{k=0}^{n}{a}_{2k}{z}^{2k}$ is a polynomial that is an even function. Show that
$p\left(iz\right)=\sum _{k=0}^{n}{\left(-1\right)}^{k}{a}_{2k}{z}^{2k}={p}^{a}\left(z\right),$
where ${p}^{a}$ is the polynomial obtained from $p$ by alternating the signs of its nonzero coefficients.
8. If $q\left(z\right)={\sum }_{k=0}^{n}{a}_{2k+1}{z}^{2k+1}$ is a polynomial that is an odd function, show that
$q\left(iz\right)=i\sum _{k=0}^{n}{\left(-1\right)}^{k}{a}_{2k+1}{z}^{2k+1}=i{q}^{a}\left(z\right),$
where again ${q}^{a}$ is the polynomial obtained from $q$ by alternating the signs of its nonzero coefficients.
9. If $p$ is any polynomial, show that
$p\left(iz\right)={p}_{e}\left(iz\right)+{p}_{o}\left(iz\right)={p}_{e}^{a}\left(z\right)+i{p}_{o}^{a}\left(z\right),$
and hence that ${p}_{e}\left(iz\right)={p}_{e}^{a}\left(z\right)$ and ${p}_{o}\left(iz\right)=i{p}_{o}^{a}\left(z\right).$

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scanning tunneling microscope
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research.net
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# 10.5: Equivalent Circuit Model for Transmission; Radiation Efficiency

A radio transmitter consists of a source which generates the electrical signal intended for transmission, and an antenna which converts this signal into a propagating electromagnetic wave. Since the transmitter is an electrical system, it is useful to be able to model the antenna as an equivalent circuit. From a circuit analysis point of view, it should be possible to describe the antenna as a passive one-port circuit that presents an impedance to the source. Thus, we have the following question: What is the equivalent circuit for an antenna which is transmitting?

We begin by emphasizing that the antenna is passive. That is, the antenna does not add power. Invoking the principle of conservation of power, there are only three possible things that can happen to power that is delivered to the antenna by the transmitter:1

• Power can be converted to a propagating electromagnetic wave. (The desired outcome.)
• Power can be dissipated within the antenna.
• Energy can be stored by the antenna, analogous to the storage of energy in a capacitor or inductor.

We also note that these outcomes can occur in any combination. Taking this into account, we model the antenna using the equivalent circuit shown in Figure $$\PageIndex{1}$$.

Figure $$\PageIndex{1}$$: Equivalent circuit for an antenna which is transmitting. ( CC BY-SA 4.0; S. Lally)

Since the antenna is passive, it is reasonable to describe it as an impedance $$Z_A$$ which is (by definition) the ratio of voltage $$\widetilde{V}_A$$ to current $$\widetilde{I}_A$$ at the terminals; i.e.,

$Z_A \triangleq \frac{\widetilde{V}_A}{\widetilde{I}_A}$

In the phasor domain, $$Z_A$$ is a complex-valued quantity and therefore has, in general, a real-valued component and an imaginary component. We may identify those components using the power conservation argument made previously: Since the real-valued component must represent power transfer and the imaginary component must represent energy storage, we infer:

$Z_A \triangleq R_A +jX_A$

where $$R_A$$ represents power transferred to the antenna, and $$X_A$$ represents energy stored by the antenna. Note that the energy stored by the antenna is being addressed in precisely the same manner that we address energy storage in a capacitor or an inductor; in all cases, as reactance. Further, we note that $$R_A$$ consists of components $$R_{rad}$$ and $$R_{loss}$$ as follows:

$Z_A = R_{rad} + R_{loss} +jX_A$

where $$R_{rad}$$ represents power transferred to the antenna and subsequently radiated, and $$R_{loss}$$ represents power transferred to the antenna and subsequently dissipated.

To confirm that this model works as expected, consider what happens when a voltage is applied across the antenna terminals. The current $$\widetilde{I}_A$$ flows and the time-average power $$P_A$$ transferred to the antenna is

$P_A = \frac{1}{2}\mbox{Re}\left\{ \widetilde{V}_A \widetilde{I}^*_A \right\}$

where we have assumed peak (as opposed to root mean squared) units for voltage and current. Since $$\widetilde{V}_A = Z_A \widetilde{I}_A$$, we have:

$P_A = \frac{1}{2}\mbox{Re}\left\{ \left( R_{rad} + R_{loss} +jX_A \right) \widetilde{I}_A \widetilde{I}^*_A \right\}$

which reduces to:

$P_A = \frac{1}{2}\left|\widetilde{I}_A\right|^2 R_{rad} + \frac{1}{2}\left|\widetilde{I}_A\right|^2 R_{loss} \label{m0202_ePA}$

As expected, the power transferred to the antenna is the sum of

$P_{rad} \triangleq \frac{1}{2}\left|\widetilde{I}_A\right|^2 R_{rad} \label{m0202_ePrad}$

representing power transferred to the radiating electromagnetic field, and

$P_{loss} \triangleq \frac{1}{2}\left|\widetilde{I}_A\right|^2 R_{loss} \label{m0202_ePloss}$

representing power dissipated within the antenna.

The reactance $$X_A$$ will play a role in determining $$\widetilde{I}_A$$ given $$\widetilde{V}_A$$ (and vice versa), but does not by itself account for disposition of power. Again, this is exactly analogous to the role played by inductors and capacitors in a circuit.

The utility of this equivalent circuit formalism is that it allows us to treat the antenna in the same manner as any other component, and thereby facilitates analysis using conventional electric circuit theory and transmission line theory. For example: Given $$Z_A$$, we know how to specify the output impedance $$Z_S$$ of the transmitter so as to minimize reflection from the antenna: We would choose $$Z_S=Z_A$$, since in this case the voltage reflection coefficient would be

$\Gamma = \frac{Z_A-Z_S}{Z_A+Z_S} = 0$

Alternatively, we might specify $$Z_S$$ so as to maximize power transfer to the antenna: We would choose $$Z_S=Z_A^*$$; i.e., conjugate matching.

In order to take full advantage of this formalism, we require values for $$R_{rad}$$, $$R_{loss}$$, and $$X_A$$. These quantities are considered below.

$R_{rad} = 2P_{rad}\left|\widetilde{I}_A\right|^{-2} \label{m0202_ePrad2}$

This equation suggests the following procedure: We apply current $$\widetilde{I}_A$$ to the antenna terminals, and then determine the total power $$P_{rad}$$ radiated from the antenna in response. For an example of this procedure, see Section 10.2 (“Total Power Radiated by an Electrically-Short Dipole”). Given $$\widetilde{I}_A$$ and $$P_{rad}$$, one may then use Equation \ref{m0202_ePrad2} to determine $$R_{rad}$$.

## Loss resistance

Loss resistance represents the dissipation of power within the antenna, which is usually attributable to loss intrinsic to materials comprising or surrounding the antenna. In many cases, antennas are made from good conductors – metals, in particular – so that $$R_{loss}$$ is very low compared to $$R_{rad}$$. For such antennas, loss is often so low compared to $$R_{rad}$$ that $$R_{loss}$$ may be neglected. In the case of the electrically-short dipole, $$R_{loss}$$ is typically very small but $$R_{rad}$$ is also very small, so both must be considered. In many other cases, antennas contain materials with substantially greater loss than metal. For example, a microstrip patch antenna implemented on a printed circuit board typically has non-negligible $$R_{loss}$$ because the dielectric material comprising the antenna exhibits significant loss.

## Antenna reactance

The reactance term $$jX_A$$ accounts for energy stored by the antenna. This may be due to reflections internal to the antenna, or due to energy associated with non-propagating electric and magnetic fields surrounding the antenna. The presence of significant reactance (i.e., $$\left|X_A\right|$$ comparable to or greater than $$\left|R_A \right|$$) complicates efforts to establish the desired impedance match to the source. For an example, see Section 10.4 (“Reactance of the Electrically-Short Dipole”).

When $$R_{loss}$$ is non-negligible, it is useful to characterize antennas in terms of their radiation efficiency $$e_{rad}$$, defined as the fraction of power which is radiated compared to the total power delivered to the antenna; i.e.,

$e_{rad} \triangleq \frac{P_{rad}}{P_A}$

Using Equations \ref{m0202_ePA}-\ref{m0202_ePloss}, we see that this efficiency can be expressed as follows:

$e_{rad} = \frac{R_{rad}}{R_{rad}+R_{loss}} \label{m0202_eRE}$

Once again, the equivalent circuit formalism proves useful.

Example $$\PageIndex{1}$$: Impedance of an antenna

The total power radiated by an antenna is 60 mW when 20 mA (rms) is applied to the antenna terminals. The radiation efficiency of the antenna is known to be 70%. It is observed that voltage and current are in-phase at the antenna terminals. Determine (a) the radiation resistance, (b) the loss resistance, and (c) the impedance of the antenna.

Solution

From the problem statement, $$P_{rad}=60$$ mW, $$\left|\widetilde{I}_A\right|=20$$ mA (rms), and $$e_{rad}=0.7$$. Also, the fact that voltage and current are in-phase at the antenna terminals indicates that $$X_A=0$$. From Equation \ref{m0202_ePrad2}, the radiation resistance is

$R_{rad} \approx \frac{2\cdot\left(60~\mbox{mW}\right)}{\left|\sqrt{2}\cdot 20~\mbox{mA}\right|^2} = \underline{150~\Omega}$

Solving Equation \ref{m0202_eRE} for the loss resistance, we find:

$R_{loss} = \frac{1-e_{rad}}{e_{rad}}R_{rad} \cong \underline{64.3~\Omega}$

Since $$Z_A=R_{rad}+R_{loss}+jX_A$$, we find $$Z_A\cong \underline{214.3+j0 \: \Omega}$$. This will be the ratio of voltage to current at the antenna terminals regardless of the source current.

1. Note that “delivered” power means power accepted by the antenna. We are not yet considering power reflected from the antenna due to impedance mismatch.↩