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137338 here is the picture. When I use integral in Matlab, it gives error and when I use tapz, the result is wrong
The only thing that can saturate a transformer is voltage. More specifically it's flux respectively V*s integral which directly relates to mT core induction. But for a given core and number of turns, flux has also a fixed relation to magnetizing current. In so far it's quite right to look at instantaneous inductor current to deci
Hello I need to stack two helical antennas in cst because antennas are in uhf range these are wide around 2 wavelength and I used integral equation solver Now in order to stack two of these antennas how to set the excitation port correctly to get a single farfield? file is attached Thanks
The transformer is specified with a V*t integral of 250 ?Vs, so it should be able the transmit similar pulses without saturation. But the circuit matters - it must allow reset of transformer flux over a cycle, or in other words zero voltage average across the winding - it must drive the primary with sufficient low impedance to provide the mag
Hello, I'm designing a 21 turns 435 MHz helix antenna (L=2 wavelength) and simulate with transient analysis: with 4,9 M mesh cells it takes around > 28 hours and 1K staircase cells. Maximum gain simulated is at 400 MHz VS 435 MHz calculated from updated research With integral Solver and 34 steps/wavelength now the maximum gain si around 440 MHz
As for a free-run oscillator, the jitter accumulates over time. The wavescope calculates jitter in a certain time windows. My question is if I want to know the rms jitter of a oscillator in a time range of ,maybe 4us , then what the is the frequency range for the integral of phase noise in reponse. (mainly, the lower limit of integral
Momentum of ADS is a spectrum domain moment method. This uses green function with horizontal homogeneous and vertical inhomogeneous. So it is called as 2.5D EM solver. Then it formulates integral equation to be solved. When we use green function of full inhomogeneous, it is called as 3D even for moment method. However we can't get green functi
If I have a general integral equation ... Why do you call this an integral equation? See more on integral equations at The integral on the RHS is a definite integral. It has a pole (rather poles) at +/- w. Similar equations you will see in absorption and disper
took the integral of the instantaneous volt.current products from the start to the finish of a single recharge interval, then divided this by the recharge interval time (900ms) to get the average power over the recharge interval. The scope does this for you. We then divided this power reading by the recharge interval time in order to get the
I would try to emulate how semiconductor fab folks and modeling folks test it - apply a voltage ramp, acquire the current as you go, and do the math (integral of I being Q, Q being C*V, at any given V your C is I/(dV/dt). Pretty easy to set up if you have a Calculator in your waveform viewer with a deriv() function. Or you could dump raw dat
One possibility is to look among ads in the Business-Promotions-Advertisements section. Resourcefulness is called for. How much of the assembly needs to be replaced? How much can be left as is? Is the shaft encoder an integral part of the drilling equipment, or is it an add-on electronic component? Where does the signal from the encoder go? Do
Considering the complete elliptic integral of first kind K(m) I have had as a result, using both scilab or wolframalpha: K(0.5) = 1.8540747 But you have to consider that someone consider the argument of the function as square of two that means calculate K(m^2) instead of K(m). In your case K(0.5^2) = K(0.25) = 1.6857504
hey there, Can anyone help me writing a matlab program that computes the attached expressions? assume J(r') and r' have a lot of samples(lets say 100) and both contain 3 numbers (100 rows 3 columns) . I just dunno how to compute the 3D integral basically.... thank you!
I am trying to simulate a 2x2 array of helical antenna excited by 4 lumped ports 131916 However, when I try to simulate after defining two symmetry planes, I get the error: Some conductors could not be considered for port definition. Missing port information. For some discrete faceports the mesh information is missin
Hi Gib, The power flow should be the integral of the Poynting vector through the specified plane. To do this in CST MWS, you should first define an analytical face (the face is just for analysis, and does not interfere with the model) at the plane you are interested in. The power flow monitor should be activated prior to running the simulation. W
I presume you answered the question yourself with the simulation circuit. You can also calculate the output by referring to integral calculus rules in a math handbook. u(t) = ∫sin ωt dt = ? Consider that term "c" (initial value) in the equation is undetermined. Furthermore OP offset voltage causes the integrator output to slowl
hi antenna experts. i designed a Reflectarray unit cell with proper phase shift in CST. i used unit cells in array and fed it by a horn antenna. the solution type is integral equation. now i want to know how can i gain the phase of each element(phase of fields or reflection phase ) after feeding by horn. thank you all.
Follow the IEEE Standard 1241 about ADC measurement methodology. Apply sinusoidal wave to the input and sampling it with numbers of samples≥2^N and make FFT of collected samples to get SINAD. Next calculate ENOB from formula ENOB=(SINAD-1.76)/6.02 Of course the sampling frequency cannot be an integral multiplicity of input signal frequency to
Pulses are of narrow width and large amplitude. Just a pulse of voltage OR current has zero energy (theoretically speaking). If your pulses have definite width and separation, they become periodic. You can certainly measure integral(Vdt) but that will not give you the energy. But a simpler way (because you know their width) is just to count and not
Hi guys, I was reading in a circuit theory book, and after the definition of the Laplace transform it says how important is that the integral converges therefore the function f(t) defined for t > 0 and is zero for t <= 0. Then it continuous and I quote the book here, "Accordingly, for most of the functions encountered in electrical engineer