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171 Threads found on edaboard.com: Right Hand
Hello,in the CST manual which is presented bellow,they dont go to the specifics of the meaning of "combining results",i khow that there should be a phase shift of 90 degreen between the components in order to see if its right hand or left hand polarization. but what should i do specificly in CST to determine this? Or if i Plot in (...)
Hi, at the left side of the attached image, there is a commercial flash circuit. It charges the electrolytic to HV using a rectifier diode. I was wondering why not use another diode, as shown in the right hand side schematic, to charge the electrolityc faster? (using the positive pulses out of the transformer as well) Do you see any problems with
First the element should be right-hand or left-hand circular polarization. Then design the feeding network for the array. You could follow the example in the antenna books or papers.
How do you expect someone to help you fix the problem if you have not posted any code, or even reported precisely what is or is not working ? If you on the other hand are requesting a ready-to-use code, this is not the right way. You should first try to turn ON at least one of the segments of the display, and only after it this step, in the case of
Horizontal and vertical polarization, yes. right hand circularly and left hand circularly polarization, probably not, since they are time related and a 3D pattern is just a single snapshot in time.
It's a boost converter. The two transistors are arranged to increase gain (similar to a darlington or sziklai pair). The capacitor charge and discharges, causing voltage variations at the left-hand transistor. It creates hysteretic action turning the right hand transistor On and Off. The joule thief is a cousin of this circuit. The joule (...)
It sounds as though you are to contact every internet search engine, and ask what they charge you for placing an ad on their webpage, but only when a user searches on words which indicate a possible interest in your products and services. I imagine your ad is supposed to look like a 'hit'. (I often see such ads at the right-hand side of the screen
Could my pulse watch be similar to what you're making? I press my finger on a metal plate, and my pulse rate appears onscreen. I have wondered, how can it detect my pulse? All I can figure, is that a tiny voltage differential is generated between my left hand and my right hand. And the watch contains a high-gain amplifier. I touch the leads (...)
Impedance matching means that one side impedance equals or joins another impedance with a power transfer without reflection. Your picture 1 shows a line with 50 Ohms facing two parallel 50-Ohm lines: this is a mismatch as the impedances on the left and to the right of the junction are not equal: one is 50 Ohms, the other 25 Ohms. Fig.2 shows 50 Oh
hi Do you mean like this.? If Yes, On LTS Menu bar Windows/Tile Vertically and then grab the bottom right hand corner of the displayed window/s with your cursor mouse and drag to the required size. To move a window around the LTS screen grab the top bar of the window/s and move the grabbed window. E
See attached right figure. This is an ideal_balun in "analogLib/ideal_balun" Cadence DFII. If we want to satisfy impedance matching at left side, R2 has to be equal to R1/2. R2=R1/2. On the hand, if we want to satisfy impedance matching at right side, R1 has to be equal to 4*R2. R2=R1/4. These impedance characteristics are (...)
looks like you are designing patch antenna. patch antennas has ground plane underneath and those are Left hand circular polarized antenna (LHCP). you peak gain will be right above you patch antenna! i dont know what AMC stands for but i assume another PCB board. having another metal underneath patch antenna should not effect the patch because it ha
Hi there! I just started working with HFSS a while ago, so im still not totally aware of all relevant settings in my simulations. The following problem is nagging me right now: I am designing patch antennas that should work for a frequency range from 76-81GHz and transmit into a WR12 Waveguide. I got a few promising designs out of it and
The expression you have given cannot be evaluated in matlab. The series must be finite. You need to figure out the physical significance of the terms on the right hand side. A well written script has plenty of comments and with some basic idea, you should be able to figure out the physical meaning. As written, it simply means the sum of the odd t
Circular polarization will have equal magnitudes of rEtheta and rEPhi. And how to know whether it is right or left hand circular?
download the attachment vi on the right hand side at given on above link
Hello! After simulate this simple amp (all properly biased), I simulated the graph on the right hand side: Looks like for the low freqs, it's got a PSR of 0dB (ac disturbance in supply, measuring ac at the output). Isn't that weird? Doesn't 0dBs mean that transistor 3 acts as a resistor? Can someone explain to me the theory behind that? Could
Welcome AGKY, The current value is not set in the source properties, but rather in the source excitations. right click on "Field Overlays" in the left hand menu and select 'Edit Sources..". You should be able to set the value there. Good Luck
Welcome jwh, The setup you have will work. In the left hand menu, right click on "Field Overlays" -> "Edit Sources". In the window that pops up, you can set the phase of each excitation.
Hi All, How to add signals from the Source Code Window to Wave Window? When I right click on the signals and choose 'Add' submenu, all the options are in grey color. Why? Thank you!