The hands-on approach is one of the best ways of learning MATLAB. Since the example requires curve fitting, the MATLAB function polyfit will be covered. Curve fitting, which is an important consideration in engineering Numerical methods are generally used to obtain the best fit to a given set of data. The Ezyfit toolbox for Matlab enables you to perform simple curve fitting of one-dimensional data using arbitrary (non linear) fitting functions. EVERMOTION ARCHINTERIORS VOL 22 CGPERSIA TORRENT To properly quite CyberDuck, enable the. Very few of how are currently may be. A workaround is installed means that. Please note, the part on how Viewer :.
If you want to always have the Ezyfit menu in your figures, type 'efmenu install'. This will create or update your 'startup. Note: If you upgrade Matlab and you want to use your previous Ezyfit installation, you just have to follow the steps Frederic Moisy EzyFit 2. Retrieved June 20, Learn About Live Editor. Select a Web Site. Choose a web site to get translated content where available and see local events and offers.
Based on your location, we recommend that you select:. Select the China site in Chinese or English for best site performance. Other MathWorks country sites are not optimized for visits from your location. Toggle Main Navigation. Search MathWorks. Close Mobile Search. Trial software. You are now following this Submission You will see updates in your followed content feed You may receive emails, depending on your communication preferences. A free curve fitting toolbox for Matlab.
View Version History. The impedances are in ohms. The basic structure of a three-phase system consists of a three-phase voltage source connected to a three-phase load through transformers and transmission lines. The three-phase voltage source can be wye- or delta-connected. Also the three-phase load can be delta- or wye-connected. Figure 6. The method of symmetrical components can be used to ana- lyze unbalanced three-phase systems.
This is illustrated by the following ex- ample. Its complex frequency representation is also shown. From equation 6. The gen- eral form of polyval is polyval p, x 6. It is repeated here. As the resistance is decreased from 10, to Ohms, the bandwidth of the frequency response decreases and the quality factor of the circuit increases. Johnson, J. Compare your result with that obtained in part a. Plot the polynomial over the appropriate interval to verify the roots location.
The describing equations for the various two-port network represen- tations are given. Also, I 2 and V2 are output current and voltage, respectively. It is assumed that the linear two-port circuit contains no independent sources of energy and that the circuit is initially at rest no stored energy.
Furthermore, any controlled sources within the lin- ear two-port circuit cannot depend on variables that are outside the circuit. The following exam- ple shows a technique for finding the z-parameters of a simple circuit. Example 7. The following two exam- ples show how to obtain the y-parameters of simple circuits.
Find its y- parameters. The h-parameters of a bipolar junction transistor are determined in the following example. The negative of I 2 is used to allow the current to enter the load at the receiving end. Examples 7. These are shown in Figure 7. Figure 7. Z1 Y2 Figure 7. The resistance values are in Ohms. From Example 7. A termi- nated two-port network is shown in Figure 7. Z L is the load impedance. V2 b Obtain the expression for. Topics covered are Fou- rier series expansion, Fourier transform, discrete Fourier transform, and fast Fourier transform.
The term in 2 Equation 8. Equation 8. Figure 8. The coeffi- cient cn is related to the coefficients a n and bn of Equations 8. It provides information on the amplitude spectral compo- nents of g t. Example 8. If g t is continuous and non- periodic, then G f will be continuous and periodic.
The periodicity of the time-domain signal forces the spectrum to be dis- crete. It is also the total number frequency sequence values in G[ k ]. T is the time interval between two consecutive samples of the input sequence g[ n]. F is the frequency interval between two consecutive samples of the output sequence G[ k ]. This means that T should be less than the reciprocal of 2 f H , where f H is the highest significant frequency component in the continuous time signal g t from which the sequence g[ n] was obtained.
Several fast DFT algorithms require N to be an integer power of 2. A discrete-time function will have a periodic spectrum. In DFT, both the time function and frequency functions are periodic. In general, if the time-sequence is real-valued, then the DFT will have real components which are even and imaginary components that are odd. Simi- larly, for an imaginary valued time sequence, the DFT values will have an odd real component and an even imaginary component.
The FFT can be used to a obtain the power spectrum of a signal, b do digi- tal filtering, and c obtain the correlation between two signals. The vector x is truncated or zeros are added to N, if necessary. The sampling interval is ts. Its default value is 1. The spectra are plotted versus the digital frequency F. Solution a From Equation 8. With the sampling interval being 0. The duration of g t is 0.
The am- plitude of the noise and the sinusoidal signal can be changed to observe their effects on the spectrum. Math Works Inc. Using the FFT algorithm, generate and plot the frequency content of g t. Assume a sampling rate of Hz. Find the power spectrum. Diode circuit analysis techniques will be discussed. The electronic symbol of a diode is shown in Figure 9.
Ideally, the diode conducts current in one direction. The cur- rent versus voltage characteristics of an ideal diode are shown in Figure 9. The characteristic is divided into three regions: forward-biased, reversed- biased, and the breakdown. If we assume that the voltage across the diode is greater than 0.
The following example illustrates how to find n and I S from an experimental data. Example 9. Figure 9. The thermal voltage is directly propor- tional to temperature. This is expressed in Equation 9. The reverse satura- tion current I S increases approximately 7. T1 and T2 are two different temperatures. Assuming that the emission constant of the diode is 1. We want to determine the diode current I D and the diode volt- age VD. There are several approaches for solving I D and VD.
In one approach, Equations 9. This is illustrated by the following example. Assume a temperature of 25 oC. Then, from Equation 9. Using Equation 9. The iteration technique is particularly facilitated by using computers. It consists of an alternat- ing current ac source, a diode and a resistor. The battery charging circuit, explored in the following example, consists of a source connected to a battery through a resistor and a diode.
Use MATLAB a to sketch the input voltage, b to plot the current flowing through the diode, c to calculate the conduction angle of the diode, and d calculate the peak current. Assume that the diode is ideal. The output of the half-wave rectifier circuit of Figure 9. The smoothing circuit is shown in Figure 9. When the amplitude of the source voltage VS is greater than the output volt- age, the diode conducts and the capacitor is charged. When the source voltage becomes less than the output voltage, the diode is cut-off and the capacitor discharges with the time constant CR.
The output voltage and the diode cur- rent waveforms are shown in Figure 9. Therefore, the output waveform of Figure 9. When v S t is negative, diode D1 is cut-off but diode D2 conducts. The current flowing through the load R enters it through node A. The current entering the load resistance R enters it through node A. The output voltage of a full-wave rectifier circuit can be smoothed by connect- ing a capacitor across the load. The resulting circuit is shown in Figure 9. The output voltage and the current waveforms for the full-wave rectifier with RC filter are shown in Figure 9.
The capacitor in Figure 9. Solution Peak-to-peak ripple voltage and dc output voltage can be calculated using Equations 9. I ZM is the maximum current that can flow through the zener without being destroyed. A zener diode shunt voltage regulator circuit is shown in Fig- ure 9. Con- versely, if R is constant and VS decreases, the current flowing through the zener will decrease since the breakdown voltage is nearly constant; the output voltage will remain almost constant with changes in the source voltage VS.
Now assuming the source voltage is held constant and the load resistance is decreased, then the current I L will increase and I Z will decrease. Con- versely, ifVS is held constant and the load resistance increases, the current through the load resistance I L will decrease and the zener current I Z will increase.
In the design of zener voltage regulator circuits, it is important that the zener diode remains in the breakdown region irrespective of the changes in the load or the source voltage. From condition 1 and Equation 9. I Z ,max 9. Solution Using Thevenin Theorem, Figure 9. In addition, when the source voltage is 35 V, the output voltage is The zener breakdown characteristics and the loadlines are shown in Figure 9.
Lexton, R. Shah, M. Angelo, Jr. Sedra, A. Beards, P. Savant, Jr. Ferris, C. Ghausi, M. Warner Jr. Assume a temperature of 25 oC, emission coef- ficient, n , of 1. Both intrinsic and extrinsic semicon- ductors are discussed.
The characteristics of depletion and diffusion capaci- tance are explored through the use of example problems solved with MATLAB. The effect of doping concentration on the breakdown voltage of pn junctions is examined. Electrons surround the nucleus in specific orbits. The electrons are negatively charged and the nucleus is positively charged.
If an electron absorbs energy in the form of a photon , it moves to orbits further from the nucleus. An electron transition from a higher energy orbit to a lower energy orbit emits a photon for a direct band gap semiconductor. The energy levels of the outer electrons form energy bands. In insulators, the lower energy band valence band is completely filled and the next energy band conduction band is completely empty. The valence and conduction bands are separated by a forbidden energy gap.
In semicon- ductors the forbidden gap is less than 1. Some semiconductor materials are silicon Si , germanium Ge , and gallium arsenide GaAs. Figure Silicon has four valence electrons and its atoms are bound to- gether by covalent bonds. At absolute zero temperature the valence band is completely filled with electrons and no current flow can take place.
As the temperature of a silicon crystal is raised, there is increased probability of breaking covalent bonds and freeing electrons. The vacancies left by the freed electrons are holes. The process of creating free electron-hole pairs is called ionization.
The free electrons move in the conduction band. Since electron mobility is about three times that of hole mobility in silicon, the electron current is considerably more than the hole current. The following ex- ample illustrates the dependence of electron concentration on temperature. Solution From Equation The width of energy gap with temperature is given as . An n-type semiconductor is formed by doping the silicon crystal with elements of group V of the periodic table antimony, arse- nic, and phosphorus.
The impurity atom is called a donor. The majority car- riers are electrons and the minority carriers are holes. A p-type semiconductor is formed by doping the silicon crystal with elements of group III of the peri- odic table aluminum, boron, gallium, and indium.
The impurity atoms are called acceptor atoms. The majority carriers are holes and minority carriers are electrons. The law of mass action enables us to calculate the majority and minority car- rier density in an extrinsic semiconductor material. In an n-type semiconductor, the donor concentration is greater than the intrin- sic electron concentration, i.
Example It is used to describe the energy level of the electronic state at which an electron has the probability of 0. Equation In addition, the Fermi energy can be thought of as the average energy of mobile carriers in a semiconductor mate- rial. In an n-type semiconductor, there is a shift of the Fermi level towards the edge of the conduction band. The upward shift is dependent on how much the doped electron density has exceeded the intrinsic value. Drift current is caused by the application of an elec- tric field, whereas diffusion current is obtained when there is a net flow of car- riers from a region of high concentration to a region of low concentration.
This is shown in Figure Practical pn junctions are formed by diffusing into an n-type semiconductor a p-type impurity atom, or vice versa. Because the p-type semiconductor has many free holes and the n-type semiconductor has many free electrons, there is a strong tendency for the holes to diffuse from the p-type to the n-type semi- conductors.
Similarly, electrons diffuse from the n-type to the p-type material. When holes cross the junction into the n-type material, they recombine with the free electrons in the n-type. Similarly, when electrons cross the junction into the p-type region, they recombine with free holes.
In the junction a transition region or depletion region is created. In the depletion region, the free holes and electrons are many magnitudes lower than holes in p-type material and electrons in the n-type material. As electrons and holes recombine in the transition region, the region near the junc- tion within the n-type semiconductor is left with a net positive charge. The re- gion near the junction within the p-type material will be left with a net negative charge. This is illustrated in Figure Because of the positive and negative fixed ions at the transition region, an elec- tric field is established across the junction.
The electric field creates a poten- tial difference across the junction, the potential barrier. The potential barrier pre- vents the flow of majority carriers across the junction under equilibrium condi- tions. That is, from Figure Typically, VC is from 0.
For germanium, VC is ap- proximately 0. When a positive voltage VS is applied to the p-side of the junction and n-side is grounded, holes are pushed from the p-type material into the transition re- gion. In addition, electrons are attracted to transition region.
The depletion region decreases, and the effective contact potential is reduced. This allows majority carriers to flow through the depletion region. The depletion region increases and it become more difficult for the majority carriers to flow across the junction. The current flow is mainly due to the flow of minority carriers.
Using Equations The following example shows how I S is affected by tem- perature. During device fabrication, a p-n junction can be formed using process such as ion-implantation diffusion or epitaxy. The dop- ing profile at the junction can take several shapes. Two popular doping pro- files are abrupt step junction and linearly graded junction. In the abrupt junction, the doping of the depletion region on either side of the metallurgical junction is a constant.
This gives rise to constant charge densi- ties on either side of the junction. If the doping density on one side of the metallurgical junction is greater than that on the other side i. This condition is termed the one-sided step junction approximation. This is the practical model for shallow junctions formed by a heavily doped diffusion into a lightly doped region of opposite polarity . In a linearly graded junction, the ionized doping charge density varies linearly across the depletion region.
The charge density passes through zero at the metallurgical junction. It can be obtained from Equations Equations The positive voltage, VC , is the contact potential of the pn junction. However, when the pn junction becomes slightly forward biased, the capacitance increases rapidly. It is also used to plot the depletion ca- pacitance.
The holes are momentarily stored in the n-type material before they recombine with the majority carriers electrons in the n-type material. Similarly, electrons are injected into and temporarily stored in the p-type material. The electrons then recombine with the majority carriers holes in the p-type material. The diffusion capacitance is usually larger than the depletion capacitance [1, 6]. Typical values of Cd ranges from 80 to pF.
A small signal model of the diode is shown in Figure Cd rd Rs Cj Figure RS is the semiconductor bulk and contact resistance. The model of the diode is shown in Figure Cj Rs Rd Figure The diffusion capacitance is zero. The resistance Rd is reverse resistance of the pn junction normally in the mega-ohms range.
At a critical field, E crit , the accelerated carriers in the depletion region have sufficient energy to create new electron-hole pairs as they collide with other atoms. The secondary electrons can in turn create more carriers in the depletion region. This is termed the avalanche breakdown process. For silicon with an impurity concentration of cm-3, the critical electric field is about 2.
This high electric field is able to strip electrons away from the outer orbit of the silicon atoms, thus cre- ating hole-electron pairs in the depletion region. This mechanism of break- down is called zener breakdown. This breakdown mechanism does not involve any multiplication effect. Normally, when the breakdown voltage is less than 6V, the mechanism is zener breakdown process.
For breakdown voltages be- yond 6V, the mechanism is generally an avalanche breakdown process. For an abrupt junction, where one side is heavily doped, the electrical proper- ties of the junction are determined by the lightly doped side. The following example shows the effect of doping concentration on breakdown voltage.
Solution Using Equation Impurity Concentration' axis [1. Singh, J. Jacoboni, C. Mousty, F. Caughey, D. IEEE, Vol. Hodges, D. Neudeck, G. II, Addison-Wesley, Beadle, W. McFarlane, G. Sze, S. Plot the difference between of ni for Equations Assume donor concentrations from to Use impurity gradient values from to It can be used to perform the basic mathematical operations: addition, subtrac- tion, multiplication, and division.
They can also be used to do integration and differentiation. There are several electronic circuits that use an op amp as an integral element. Some of these circuits are amplifiers, filters, oscillators, and flip-flops. In this chapter, the basic properties of op amps will be discussed. The non-ideal characteristics of the op amp will be illustrated, whenever possi- ble, with example problems solved using MATLAB. Its symbol is shown in Figure It is a differ- ence amplifier, with output equal to the amplified difference of the two inputs.
It also has a very large input resistance to ohms. The out- put resistance might be in the range of 50 to ohms. The offset voltage is small but finite and the frequency response will deviate considerably from the infinite frequency response.
The common-mode rejection ratio is not infinite but finite. Table This condi- tion is termed the concept of the virtual short circuit. In addition, because of the large input resistance of the op amp, the latter is assumed to take no cur- rent for most calculations. Thus, Equation R2 R1 Vin Vo Figure With the assumptions of very large open-loop gain and high input resistance, the closed-loop gain of the inverting amplifier depends on the external com- ponents R1 , R2 , and is independent of the open-loop gain.
The integrating time con- stant is CR1. It behaves as a lowpass filter, passing low frequencies and at- tenuating high frequencies. However, at dc the capacitor becomes open cir- cuited and there is no longer a negative feedback from the output to the input.
The output voltage then saturates. To provide finite closed-loop gain at dc, a resistance R2 is connected in parallel with the capacitor. The circuit is shown in Figure The resistance R2 is chosen such that R2 is greater than R. From Equation This circuit is shown in Figure For Figure The input impedance of the amplifier Z IN approaches infinity, since the current that flows into the posi- tive input of the op-amp is almost zero.
R2 R1 Vo Vin Figure In addition, from Equation Plot the closed-loop gain as the open-loop gain increases from to The pole of the voltage amplifier and level shifter is KHz and has a gain of The pole of the output stage is KHz and the gain is 0. Sketch the magnitude response of the operational amplifier open-loop gain. This causes the op amp to have a single pole lowpass response. The process of making one pole dominant in the open-loop gain characteristics is called frequency compensation, and the latter is done to ensure the stability of the op amp.
For an op amp connected in an inverting configuration Figure Slew rate is important when an output signal must follow a large input signal that is rapidly changing. If the slew rate is lower than the rate of change of the input signal, then the output voltage will be distorted. The output voltage will become triangular, and attenuated. However, if the slew rate is higher than the rate of change of the input signal, no distortion occurs and input and output of the op amp circuit will have similar wave shapes.
As mentioned in the Section In addition, the op amp has a limited output current capability, due to the saturation of the input stage. The latter is the frequency at which a sinusoidal rated output signal begins to show distortion due to slew rate limiting.
The fol- lowing example illustrates the relationship between the rated output voltage and the full-power bandwidth. CMRR decreases as frequency increases. For an inverting amplifier as shown in Figure Thus, the common-mode input voltage is approximately zero and Equation A method normally used to take into account the effect of finite CMRR in cal- culating the closed-loop gain is as follows: The contribution of the output voltage due to the common-mode input is AcmVi ,cm.
This output voltage con- tribution can be obtained if a differential input signal, Verror , is applied to the input of an op amp with zero common-mode gain. Schilling, D. Wait, J. Irvine, R. The resistance values are in kilohms. A square wave of zero dc voltage and a peak voltage of 1 V and a frequency of KHz is connected to the input of the unity gain follower.
Assume the following values of the open-loop 3 5 7 9 gain: 10 , 10 , 10 and The operation of the BJT depends on the flow of both majority and minority carriers. There are two types of BJT: npn and pnp transistors. The electronic symbols of the two types of transistors are shown in Figure The model is shown in Figure In the case of a pnp transistor, the directions of the diodes in Figure In addition, the voltage polarities of Equations The four regions of operations are forward-active, reverse-active, saturation and cut-off.
Forward-Active Region The forward-active region corresponds to forward biasing the emitter-base junction and reverse biasing the base-collector junction. It is the normal operational region of transistors employed for amplifications.
The cut-off region corresponds to reverse biasing the base-emitter and base- collector junctions. The collector and base currents are very small compared to those that flow when transistors are in the active-forward and saturation regions.
Solution From Equations Assume a temperature of oK. The variation on V BE with temperature is similar to the changes of the pn junction diode voltage with temperature. The collector-to-base leakage current, I CBO , approximately doubles every 10o temperature rise. As discussed in Section 9. The change in collector current can be obtained using partial derivatives.
The derivation is assisted by referring to Figure Calculate the collector current at 25 oC and plot the change in collector current for temperatures between 25 and oC. At each temperature, the stability factors are calculated using Equations The change in I C for each temperature is calculated using Equation It is uneconomical to fabricate IC resistors since they take a disproportionately large area on an IC chip.
In addition, it is almost impossible to fabricate IC inductors. Biasing of ICs is done using mostly transistors that are connected to create constant current sources. Examples of integrated circuit biasing schemes are discussed in this section. The current mirror consists of two matched transistors Q1 and Q2 with their bases and emitters connected.
The transistor Q1 is connected as a diode by shorting the base to its collector. In the latter mode of transistor operation, the device Q2 behaves as a current source. To obtain an expression for the output current, we assume that all three transistors are identical. Solution We use Equation Similarly, we use Equation
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