@@Hari-uy3rm Actually dun have. When u use hairpin u want to have a compact size. Therefore, the width needs to be small. So just choose 1mm and 2mm for the gap.
Thank you Prof, these videos that you made really helped me in my EEE studies in NTU! Hope you can continue to make these videos to pass on your valuable knowledge to the next generation! :)
finally completed entire series in One sitting. very intresting i had this topic in my Masters level but had lots of doubts but i think now its clear.!
the high Freq equation is a little bit misleading because, there wont be a frequency that high, practically infinity, so low frequency equation actually is the one that should be used in real life, adding to that a DC source does not have EM waves radiating
Thank you for your comment. You are right, there is no infinity but large enough for us to consider it to be a very big value. In engineering, we need to study by assume certain conditions like zero or infinity in order to understand the key concept. Eg, we cant have infinity Ohm, but large enough to limit the current flow. :)
I love your excellent way of explaining, I am an EMC engineer working in Germany, I am honored to meet you sir. If you have an interest to work with us here in Germany welcome
Thank you so much for your offer. For me, Germany always has the best engineer. How can I call myself expert when I am there. Hahaha. I have high respect for Engineer from German. Salute!!!
Dear sir, In continuation to this series can you please explain the following topics: 1. Time domain solution for transmission lines 2. Bounce diagram 3. High speed digital interconnection and signal integrity
07:12, where does it come from, if Z0>ZL then Г=(Z0-ZL)/(Z0+ZL) such case wasn't considered in Part 7 video I think the issue is that the numerator of Г should be taken by modulo thank you
Thank you for your question. I just want to simplify the Eq of Г. You can remain at this form: Г=(Z0-ZL)/(Z0+ZL) However, I simply the Eq so that u can cancel the term Z0 or ZL when needed. These simplify eqs are needed later on in order to make the eqn management. Thank you. :) I hope I ask your question,
@@technologiesdiscussion1676 ah, I needed to watch Part 7 to answer to my question:) but here in this video you sum up incident and reflected voltages and currents, which confused me🤔
Thank you for your support!!! :) You really need to see the complete series to have a complete understand. :) You can help this Channel by Like & Subscribe. Thanks, again :)
Thank you for your support!!! I really appreciate!!! I will do more detailed discussion on LISN but maybe will be later. Now, I want to discuss on the test and measurement procedure for Conducted Emission.
@@technologiesdiscussion1676 Thankyou for quick response sir. I have a request that for CE please give a detailed discussion as per CISPR 25. I’m pretty sure more than 90% audience who are watching your channel is from automobile industry and hence for us CISPR 25 is very important for us. I’m very indebted to you for your content 🙏🏼
Thanks a lot for sharing much knowledge in simple explanation. I watched your FM series that improved my understanding for those and today I found the antenna series. Can you simply explain some antenna's parameters such as Effective aperture , Electrical length of Antenna and please describe the meaning of Field Strength unit (dBuV/m) Thank you very much,
You have the talent to simplify the most complex topics in short videos..thanks for your help and effort keek going someone at middle east waits your lectures 💐💐💪💪🥰
I also read that a common mode current is a current flowing without any opposite return current nearby. That definition implies that 1 single current flowing in 1 single wire without any return nearby is a common mode current Do you agree with that definition? we usually say that common mode currents are 2 or more current flowing in the same direction,but the other definition makes more sense to me.
I have to disagree with you. 😊 All current flow out, need to return. Maybe indirectly, via gnd plane, chasis… without u knowing. I guess this is the key principle. 😊 Base on your description, I have to say is single mode rather than common mode. I hope I have answered your question. Thanks.
@@technologiesdiscussion1676 Hi, I do know that current flows in loop. My point here is that sometimes, the return current of a noise can be very far from it and hence do not cancel the radiation, So maybe that current can be called Common Mode
@@AlbertRei3424 Thank you again. I am sorry that I misunderstood your question. I dun think we called this common mode current. The current may cancel or not cancel the radiation. I can explain the differential and common mode concept but need to wait a while. Thanks. :)
So, for an ungrounded shield, won't the incoming H field induce eddy currents in the shield which in turn creates a field that opposes and mitigates the incoming field, thus partly shielding the inner conductor?
Thank you for your question. Yes, you are right. :) An ungrounded shield can still provide partial shielding due to the formation of eddy currents which create opposing magnetic fields, thereby reducing the effect of the incoming magnetic field on the inner conductor. I mention partial because other issues may surface. For Eg, E-field may become an issue. You can consider seeing my EMC consideration discussion from EMC Part 19 to EMC part 31. With these, you have a full understand on the shielding. :) Thank you so much for your support too!!!
@@technologiesdiscussion1676 So in the video you mention that the ungrounded shield doesnt affect the magnetic field, is this wrong, or is this because of the geometry of the shield? Or related to frequency, or nearfield far field? Also in the shielding videos with reflection, absorbtion etc, you say that most of the e field is reflected at the first boundry, but this doesnt seem to rely on whether the shield is grounded or not. Why is this? I thought grounding was determinental to the shielding of e field? Is the situation different when it comes to far fields?
I guess I have to make this clear. It is always impt to gnd the shield. Imagine this, if the shield is gnd, then the eddy effect will be shunt to gnd and so called "removed". The shield without gnd in fact can be a patch antenna and may create another issues.
Thank you so much bro , i have a resit exam tomorrow and i literally had no clue what analogic communication (frequency modulation) really is . Shout out to you sir <3
NP :) Think this video was done quickly to let my students understand it. Hahaha, never can I imagine another bro can benefit from it. Thank you so much for your support and all the best for your exam. Pls help to Like this video and consider to Subscribe. Thanks, bro !!!
If there is a 10 cm long pcb trace between source and load, where should we place the L and C impedance matching components? Both L and C should be close to load ?
10 cm can be long or short but mainly depend on freq. You are right, an impedance matching network should ideally be placed as close to the load as possible. The reasons as following: Minimizing Reflections: Placing the matching network close to the load minimizes the reflections that occur at the load interface, which can be critical in high-frequency applications where signal integrity is important. Effective Impedance Transformation: When the network is placed near the load, it more effectively transforms the load impedance to match the source impedance, thereby ensuring maximum power transfer. Simplified Design: By placing the matching network close to the load, you can simplify the design and analysis of the network, as the impedance seen by the source is the transformed impedance rather than a combination of transmission line effects and mismatched load impedance. Minimizing Transmission Line Effects: If the matching network were placed close to the source, any transmission line between the matching network and the load could introduce additional impedance variations, phase shifts, and potential mismatches, complicating the overall network design.
Thank you for your frank comment. I accept it 😊. I'd like to provide an explanation. In my earliest videos, I had few words on the slides, requiring me to explain more in detail. At that time, I included subtitles, so my viewers understood my explanations. However, to save time, I decided to drop the subtitles. With fewer words on the slides, I still explained in detail. Now, without subtitles, some of my viewers cannot fully understand what I'm trying to explain. More than five viewers have shared this feedback with me. They suggested that I write on the slides what I want to explain so they can fully understand it. As a result, you can see that my style has changed. Now, I include most of the words on the slides, and you might say that I read from them 😊. However, all the words on the slides are my own work. I realize I may have difficulty meeting the needs of all my viewers. I would appreciate any feedback to help me improve my work. Appreciate the rest of my viewers comments, pls. 😊 Thank you.
The discussion is not ended until you build the filter and show its performance. After that you can start a new discussion about why the performance was wrong.
Your expertise in this area is evident. You are absolutely right! After fabrication, it is exciting to obtain test and measurement results, but later you may find many mismatches and not know how to proceed. In fact, when I served as a reviewer for IEEE, I questioned some works on this topic. I know a few tricks to correlate test and measurement results with simulated results, but this is a completely new discussion, and I am afraid I can't discuss it fully here
Hahaha. I guess u mean EMC video right? :) I am not familiar with video edit and therefore need so long time to do one. Now, I am better and hopefully dun need so long to do the next one. Like to thank you for your strong support!!! :)
I am simply happy that these are useful. :) You can help this Channel by Like this Video, Subscribe to this Channel & Turn on your Notification bell. Thanks ahead!!!
Thank you for your support!!! :) Really appreciate that. You can help this Channel by Like this Video, Subscribe to this Channel & Turn on your Notification bell. Thanks ahead!!!
