Thank you sir for this much awaited tutorial ✌. Looking forward to EM simulation video
@technologiesdiscussion16763 күн бұрын
Thank you for your strong support too. Noted on the EM simulation video. :)
@Hari-uy3rm3 күн бұрын
What is the criteria to decide the width of the resonator like how you decided the width to be 1mm here. Should it be near 50 ohms?
@technologiesdiscussion16763 күн бұрын
@@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.
@yongkang-z8c5 күн бұрын
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! :)
@technologiesdiscussion16765 күн бұрын
I am happy is useful. I also from NTU, Singapore many many years ago. Hahaha.
@smollande9 күн бұрын
👍无得顶
@technologiesdiscussion16768 күн бұрын
You mean in Cantonese. :) Thank you so much.
@smollande8 күн бұрын
@@technologiesdiscussion1676 i learned a lot from the video. Thank you too. I asked a question in the first comment below ... can take a look?
@technologiesdiscussion16768 күн бұрын
KZbin never tell me, I have new message. Hahaha. I have answer that question. Pls let me know if you need more explaination. Thanks again.
@paran0id.10 күн бұрын
at 16:43 from y value containing s variable how to calculate impedance
@technologiesdiscussion167610 күн бұрын
Y is admittance. S is Siemen. So to get impedance will be Z = 1/Y. Or 1/ S. One siemens is equal to the reciprocal of one Ohm. :)
@srijandwivedi29410 күн бұрын
Thanks
@technologiesdiscussion167610 күн бұрын
A simple thanks means so much. Thank you so much for your support!!! :)
@srijandwivedi29410 күн бұрын
@@technologiesdiscussion1676 Literally it was 😁helpful good content
@technologiesdiscussion167610 күн бұрын
I am simply Glad that this is useful. You can help this Channel by Like & Subscribe. Thank you :)
@AlbertRei342415 күн бұрын
Any software that calculate s parameters? Free software?
@technologiesdiscussion167615 күн бұрын
Free, I think a bit difficult. I use Mathlab the last time. :)
@_boss_baby_17 күн бұрын
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.!
@technologiesdiscussion167617 күн бұрын
Nice :) Thank again for your kind comment. :)
@_boss_baby_17 күн бұрын
which all area includes ur teachings?
@technologiesdiscussion167617 күн бұрын
I am in EMC, Filter Design, RF Circuit, LPWAN...
@_boss_baby_17 күн бұрын
your videos are so easy to understand. Although i dont require this for now but still watching with intresst.!
@technologiesdiscussion167617 күн бұрын
Is always nice to receive positive feedback. Once again, thank you so much for your strong support!!!
@_boss_baby_17 күн бұрын
Great Explaination !!!
@technologiesdiscussion167617 күн бұрын
I am simply Glad that this is useful. You can help this Channel by Like & Subscribe. Thank you :)
@_boss_baby_17 күн бұрын
@@technologiesdiscussion1676 done sir!!
@technologiesdiscussion167617 күн бұрын
Thank you so much :)
@engineerimrana221327 күн бұрын
Excellent explanation 👌
@technologiesdiscussion167627 күн бұрын
Glad is useful. You can help this Channel by Like & Subscribe. Thank you :)
@jadyaacoub29 күн бұрын
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
@technologiesdiscussion167629 күн бұрын
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. :)
@huyo9407Ай бұрын
thank you so much
@technologiesdiscussion1676Ай бұрын
You're welcome! You can help this Channel by Like & Subscribe.
@mohamedgourche7936Ай бұрын
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
@technologiesdiscussion1676Ай бұрын
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!!!
@mohamedgourche7936Ай бұрын
Thank you so much sir
@technologiesdiscussion1676Ай бұрын
You are most welcome!!! :)
@mohamedgourche7936Ай бұрын
thank you sir
@technologiesdiscussion1676Ай бұрын
Thanks again for your support!!! :)
@dasaridheeraj22Ай бұрын
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
@technologiesdiscussion1676Ай бұрын
Ok. Will do it. :) But need to wait for some time. Thank you so much for your strong support!!! :)
@antonivanov3830Ай бұрын
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
@technologiesdiscussion1676Ай бұрын
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,
@antonivanov3830Ай бұрын
@@technologiesdiscussion1676 ok, thank you!)
@technologiesdiscussion1676Ай бұрын
NP :) You can help this Channel by Like & Subscribe. Thank you :)
@antonivanov3830Ай бұрын
Thank you for your videos! It is unclear, how did you get V+\I+ = -V-/I-, it looks you just considered only incident and only reflected waves
@technologiesdiscussion1676Ай бұрын
Thank you for your question. Z0 = V+/I+ = -V-/I- Z0 is the same for incident or reflected becos at the same pt, hence I have the equation on top.
@antonivanov3830Ай бұрын
@@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🤔
@technologiesdiscussion1676Ай бұрын
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 :)
@gurudutt2261Ай бұрын
Pls continue discussion on LISN. It was really a helpful video. 🙂🙂🙂🙂🙂
@technologiesdiscussion1676Ай бұрын
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.
@gurudutt2261Ай бұрын
@@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 🙏🏼
@technologiesdiscussion1676Ай бұрын
Next in line for EMC, is the common CE stds. Think from here, u guys will have the ideas how to do all these test. :)
@theoryandapplication7197Ай бұрын
Thank you so much boro you are excellent
@technologiesdiscussion1676Ай бұрын
You are most welcome. Thanks again for your strong support!!! :)
@theoryandapplication7197Ай бұрын
Thank you so much sir
@technologiesdiscussion1676Ай бұрын
Most welcome!!! :)
@theoryandapplication7197Ай бұрын
useful and perfect thank you for sharing it is useful
@technologiesdiscussion1676Ай бұрын
I hope u also like the 1st series discussion of LPWAN (Part 1).
@theoryandapplication7197Ай бұрын
thank you for sharing it is useful
@technologiesdiscussion1676Ай бұрын
Glad it was helpful! U saw this :) Thank you again for your strong support!!!
@theoryandapplication7197Ай бұрын
good luck and thank you for sharing
@technologiesdiscussion1676Ай бұрын
Thanks for watching!
@theoryandapplication7197Ай бұрын
thank you very much pro , actually it is useful video and learn form your esteem channel
@technologiesdiscussion1676Ай бұрын
I appreciate your strong support. Thank you again!!! :)
@theoryandapplication7197Ай бұрын
thank you for sharing it is useful
@technologiesdiscussion1676Ай бұрын
I am happy that this is useful. :) You can help this Channel by Like & Subscribe. Thanks, again :)
@theoryandapplication7197Ай бұрын
@@technologiesdiscussion1676 of course dear , and also going to share your video in my channel
@technologiesdiscussion1676Ай бұрын
Thank you so much :)
@zandanshahАй бұрын
Keep the good work going
@technologiesdiscussion1676Ай бұрын
Thank you so much!!! :) You can help this Channel by Like & Subscribe. Thanks, again :)
@ahmednor5806Ай бұрын
Once I see your notifications bell of your channel I l make a space time to watch and get the best info 🙏💐💐
@technologiesdiscussion1676Ай бұрын
I really appreciate your strong support!!! :) Thanks again :)
@akkarabc8339Ай бұрын
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,
@technologiesdiscussion1676Ай бұрын
I am happy to do that. :) But I may need time. Thks. Pls help to Like & Subscribe. Thanks :)
@ahmednor5806Ай бұрын
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 💐💐💪💪🥰
@technologiesdiscussion1676Ай бұрын
U make my day!!! Hahaha But I am so grateful for your strong support!!!
@AlbertRei3424Ай бұрын
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.
@technologiesdiscussion1676Ай бұрын
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.
@AlbertRei3424Ай бұрын
@@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
@technologiesdiscussion1676Ай бұрын
@@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. :)
@bernieblundell69982 ай бұрын
Great content many thanks...
@technologiesdiscussion16762 ай бұрын
My pleasure! Pls help by Like the Video and Subscribe to this Channel. Thank you again for your feedback!!! :) Cheer!!!
@mrfredvjy2 ай бұрын
Very helpfull to make pasif crossover audio
@technologiesdiscussion16762 ай бұрын
Thank you for your suggestion. Frankly, I do not know how to do it. :)
@ahmednor58062 ай бұрын
Thanks for your effort and informative topics 🙏🙏🙏🙏
@technologiesdiscussion16762 ай бұрын
It's my pleasure. Always happy if this can help. :)
@ahmednor58062 ай бұрын
🙏🙏🌹🌹
@technologiesdiscussion16762 ай бұрын
Thank you :) I am happy to have strong support from you :)
@ferdind12 ай бұрын
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?
@technologiesdiscussion16762 ай бұрын
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!!!
@ferdind12 ай бұрын
@@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?
@technologiesdiscussion16762 ай бұрын
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.
@islem76er2 ай бұрын
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
@technologiesdiscussion16762 ай бұрын
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 !!!
@seshansesha76452 ай бұрын
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 ?
@technologiesdiscussion16762 ай бұрын
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.
@thoalfeqar2 ай бұрын
Great explanation 🎉Thank you sir.
@technologiesdiscussion16762 ай бұрын
You are welcome. Pls help by like this video and subscribe to this channel.
@buraktalhasumer8712 ай бұрын
bro i didnt see the pip installation can u you help me about that?
@technologiesdiscussion16762 ай бұрын
Thank you for your comment. What is pip installation? I am sorry I dun understand. Thanks.
@spotify_ERROR4042 ай бұрын
thank you 3:51 what is a reflector antenna whos the most beautiful
@technologiesdiscussion16762 ай бұрын
Hahaha. Sometime is so challenging to make a 'Perfect' video. Thanks. :) Cheers!!!
@sruthisruthi34162 ай бұрын
Super
@technologiesdiscussion16762 ай бұрын
Thank you so much. Pls help by Like the Video and Subscribe to this Channel. Thank you again for your feedback!!! :)
@ahmednor58062 ай бұрын
The Best and Informative channel about technology and communication engineering --Thanks too much 🌹🌹🌹🌹
@technologiesdiscussion16762 ай бұрын
Thank you so much. :)
@kumara52882 ай бұрын
He just reading the presentation what written..not proper explanation… waste of time.. I am sorry to tell this
@technologiesdiscussion16762 ай бұрын
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.
@donepearce2 ай бұрын
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.
@technologiesdiscussion16762 ай бұрын
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
@user-vr7qp9vb9d2 ай бұрын
hi, so long to see you.!! welcome back
@technologiesdiscussion16762 ай бұрын
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!!! :)
@user-dw9tt1tv4kАй бұрын
Could you tell me about Weight measurement in cispr11 please?
@technologiesdiscussion1676Ай бұрын
In fact, I am about to do this soon. After already do up the slides but have not made video yet.
@pardeepmanro74712 ай бұрын
Great personality is giving lecture.
@technologiesdiscussion16762 ай бұрын
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!!!
@user-pd7ms2ml5o2 ай бұрын
Thank you for the detailed explanation, great series!
@technologiesdiscussion16762 ай бұрын
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!!!
@skykai86732 ай бұрын
Why C1g don't you consider? Why Is it not contributing electric coupling part?
@technologiesdiscussion16762 ай бұрын
The voltage at Vi is the same as voltage at C1G (parallel) so therefore I remove C1G.