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Look simply 0.19 lemda + 0.375 lemda = 0.565 lemda okay. But since 1 circle is 0.5 lemda okay u complete one circle. That means 0.5 lemda is done. Now remaining is 0.565 - 0.5 = 0.065 So from 0.0 move anticlockwise upto 0.065 lemda ✨
Your positive comments motivates me, my goal is to create largest community of engineers in entire globe, so please help me for that by sharing this lecture series(playlist) with your friends in social media (watsapp, telegram etc). Thanks and welcome 🙏
Your positive comments motivates me, my goal is to create largest community of engineers in entire globe, so please help me for that by sharing this lecture series(playlist) with your friends in social media (watsapp, telegram etc). Thanks and welcome 🙏
Professor, Thanks for teaching: I don't understand how you moved .375 wavelength to the new location -j0.424 from the starting point -j2.5? I know when using the Smith Chart, λ is equivalent to 720 degrees, by using this relationship: 0.375 λ (720 degrees / λ ) = 270 degrees. So from -j2.5 going counterclockwise 270 degrees we arrive at -j0.424. Is this correct? Lambda = λ = 720 degrees 0.19 λ + 0.375 λ = 0.565 λ 136.8 degrees + 270 degrees = 406.8 degrees Since 0.500 λ is One Full Circle, it must be subtracted. 0.565 λ - 0.500 λ = 0.065 λ, 406.8 degrees - 360 degrees = 46.8 degrees [46.8 degrees ( λ / 720 degrees)] = 0.065 λ Finally location start + movement anticlockwise / toward the load = end location -->> 0.19 λ + 0.375 λ = 0.065 λ
sir i have transmiter working on 5.8ghz ...i try to test with 50ohm load ...but my swr meter show 2:1 ...how do i know the impedance of my transmitter with smith chart ?
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I have a CAT in 20 minutes . Can't thank you enough buddy 🙏 Thank you Very important 😊
Best wishes 🙏
we mzee ,,mhindi anaokoa😅
Such a nice explanation ,I have studied this just 2 hrs before by watching your video.. It was great learning with you sir .Thank u sir
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I'm having a CAT in about 3 hours. I can't thank you enough
Best wishes
This is understanding Smith chart made simple 👌. Thanks Prof.
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great sir your lectures help me a lot
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Excelente videos and Explanations.
I have calculated everything in software ao everything is fine
What we can write the value of Vmin. For this question in examination please reply sir....
0.375 lambda u draw it from point approx 0.31 to 0.44 how does it make 0.375 lambda.... can u plz clear it
Look simply 0.19 lemda + 0.375 lemda = 0.565 lemda okay. But since 1 circle is 0.5 lemda okay u complete one circle. That means 0.5 lemda is done. Now remaining is 0.565 - 0.5 = 0.065
So from 0.0 move anticlockwise upto 0.065 lemda ✨
Thank you
Such a good explanation sir❤️
Your positive comments motivates me, my goal is to create largest community of engineers in entire globe, so please help me for that by sharing this lecture series(playlist) with your friends in social media (watsapp, telegram etc). Thanks and welcome 🙏
great sir
how to find distance vmin in this case
I not understand how to move 3/2 in smith chart to calculate the load impedance
Sir .I got doubts on load impedance
sir,what is the value of Vmin ?
Hello, Sir. I have a question. What's the value of Vmin and Vmax?
Yes sir tell me
@@kesarapupavitra8550 I ask, What's the value of Vmin and Vmax?
Thank you ❤️❤️
Your positive comments motivates me, my goal is to create largest community of engineers in entire globe, so please help me for that by sharing this lecture series(playlist) with your friends in social media (watsapp, telegram etc). Thanks and welcome 🙏
appreciated
Why it is moved 0.375Lamda in anticlockwise direction not in clockwise direction?
Clockwise means towards the generator or from the load.
So sir Maximum Impedance is always real?
Since it lies on line where imaginary part is zero
Nope
@@EngineeringFunda Then how to locate the Maximum Impedance?
Professor, Thanks for teaching: I don't understand how you moved .375 wavelength to the new location -j0.424 from the starting point -j2.5?
I know when using the Smith Chart, λ is equivalent to 720 degrees, by using this relationship: 0.375 λ (720 degrees / λ ) = 270 degrees. So from -j2.5 going counterclockwise 270 degrees we arrive at -j0.424. Is this correct?
Lambda = λ = 720 degrees
0.19 λ + 0.375 λ = 0.565 λ
136.8 degrees + 270 degrees = 406.8 degrees
Since 0.500 λ is One Full Circle, it must be subtracted.
0.565 λ - 0.500 λ = 0.065 λ,
406.8 degrees - 360 degrees = 46.8 degrees
[46.8 degrees ( λ / 720 degrees)] = 0.065 λ
Finally location start + movement anticlockwise / toward the load = end location -->> 0.19 λ + 0.375 λ = 0.065 λ
1st explanation is correct by you
can we also do 3*lambda/8 = 270 degree and then plot Zl
Nice
Is there anyway to cross check the answer?
Sir is the load impedance normalized?
On smithchart everything is normalized
@@EngineeringFunda thank you 😊 sir
@@EngineeringFunda sir now how should we find the actual load impedance from this normalized value
@@umaschannel3244 multiply Zo with it to get actual load Impedance
@@EngineeringFunda
So sir Maximum Impedance is always real?
Since it lies on line where imaginary part is zero
Assalam o alaikum Sir g,
How can we find point if lamda=0.375 ,clock and anti clack wise,I don't know how you draw this
Clock and anticlockwise is well defined and given on the scale of Smith chart.
Sir what is the value of lambda 3lambda by 8 i am not getting it
When load is short how can we draw on Smith chart please explain
How can we calculate input admittance?
sir i have transmiter working on 5.8ghz ...i try to test with 50ohm load ...but my swr meter show 2:1 ...how do i know the impedance of my transmitter with smith chart ?
From VSWR find reflection coefficient.
From reflection coefficient we can have Zl,
Reflection coefficient = (Zl - Zo)/ (Zl + Zo)
How to get the value of 0.375* lambda in degree?
Sir Vmax should be equal to (0.313 × lambda ), how is Vmax equal to infinity ♾️ ?
👏👏
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v max and vmin prb
And ur application not available pdfs