Ap log jitna easy krte ho exam mai itna easy nhi ata thora hard koro tbhi doubt clear hoga

Thank you❤

Galat solve kiya hai rotation of plot -270 hoga aur apne 180 leke solve kiya hai

Thank you so much sir ..😊

Time waste

Bc sbb bhula diya khudd ko ata nhi dussro ko smjane luge ho

Nice👌👌

Your multiplication method is wrong Kya confidence se wrong sikarahahe

You are the bes of other ❤❤❤

H Next time use English sir😢

Thanks sir for your contrubtion.

wtf ! i dint even understand ... very low quality explanation

Super bro🎉

Damn I was thinking same

Multiple s*4

Multiple s*4

thank youu brooo for the shortcut method

Equation. Bhi likho sir ji ☠️

Thanks

Thanks

sir u did partitioning wrong in the beginning itself.. in the beginning anyhow u do its 0.2

best video!

Sir kaya bolu bass wa 😢❤❤❤❤❤

3:22 bhai subtraction galat h 6s/4 aana tha

Kal exam haii thank you so much yrr itna acche se smja ne ke liye🫂

vaah bhaiya aapke paas toh mera vaala pen hai

Sir, how to find gain margin and phase margin for this kind of problem.

Wrong approach don't follow this ❌ 1. Source Entropy ( 𝐻 H): The entropy of a source is calculated using the formula: 𝐻 = − ∑ 𝑖 = 1 𝑛 𝑝 𝑖 log ⁡ 2 ( 𝑝 𝑖 ) H=− i=1 ∑ n p i log 2 (p i ) where 𝑝 𝑖 p i is the probability of the 𝑖 𝑡 ℎ i th symbol. Given Probabilities: 𝑝 1 = 1 2 , 𝑝 2 = 1 4 , 𝑝 3 = 1 8 , 𝑝 4 = 1 16 , 𝑝 5 = 1 16 p 1 = 2 1 ,p 2 = 4 1 ,p 3 = 8 1 ,p 4 = 16 1 ,p 5 = 16 1 Now calculate 𝐻 H: 𝐻 = − ( 𝑝 1 log ⁡ 2 ( 𝑝 1 ) + 𝑝 2 log ⁡ 2 ( 𝑝 2 ) + 𝑝 3 log ⁡ 2 ( 𝑝 3 ) + 𝑝 4 log ⁡ 2 ( 𝑝 4 ) + 𝑝 5 log ⁡ 2 ( 𝑝 5 ) ) H=−(p 1 log 2 (p 1 )+p 2 log 2 (p 2 )+p 3 log 2 (p 3 )+p 4 log 2 (p 4 )+p 5 log 2 (p 5 )) Substitute the values: 𝐻 = − ( 1 2 log ⁡ 2 1 2 + 1 4 log ⁡ 2 1 4 + 1 8 log ⁡ 2 1 8 + 1 16 log ⁡ 2 1 16 + 1 16 log ⁡ 2 1 16 ) H=−( 2 1 log 2 2 1 + 4 1 log 2 4 1 + 8 1 log 2 8 1 + 16 1 log 2 16 1 + 16 1 log 2 16 1 ) Calculate each term: 1 2 log ⁡ 2 1 2 = 1 2 × ( − 1 ) = − 0.5 2 1 log 2 2 1 = 2 1 ×(−1)=−0.5 1 4 log ⁡ 2 1 4 = 1 4 × ( − 2 ) = − 0.5 4 1 log 2 4 1 = 4 1 ×(−2)=−0.5 1 8 log ⁡ 2 1 8 = 1 8 × ( − 3 ) = − 0.375 8 1 log 2 8 1 = 8 1 ×(−3)=−0.375 1 16 log ⁡ 2 1 16 = 1 16 × ( − 4 ) = − 0.25 16 1 log 2 16 1 = 16 1 ×(−4)=−0.25 (repeated twice for 𝑝 4 p 4 and 𝑝 5 p 5 ). Sum these values: 𝐻 = − ( − 0.5 − 0.5 − 0.375 − 0.25 − 0.25 ) = 1.875 bits/symbol H=−(−0.5−0.5−0.375−0.25−0.25)=1.875 bits/symbol 2. Information Rate ( 𝑅 R): The information rate is given by: 𝑅 = 𝐻 × symbols emitted per second R=H×symbols emitted per second The source emits 1 symbol every millisecond, so it emits 1000 symbols per second. 𝑅 = 𝐻 × 1000 = 1.875 × 1000 = 1875 bits/second R=H×1000=1.875×1000=1875 bits/second

Thak u bro

Tree matrix se match ho raha kya dekho pehle😂

How to mark j 1/2 and j -1/2

Bhai ye har bar aap Wpc nikalne k liye kabhi =180 se equatte karte ho kisi problem me -180 se main samjha nahi ye logic?😊

PELHE KHUD SIKH LE

Second year electrical engineering ECA subject Mumbai University

Pehle camera auto focus bnd kro...beech beech mai blur hota hai amd thats irritating

Very easy explanation. Thank you

Best ho bhai tum

apka answer wrong hai relative ka

Not understand Sorry sir but you can improve little bit 😃

Stop smoking prof

thank a lot

Thxx❤

Thank you bhaiya ☺️✨

🙌🙏

Bhaiya aap ki wajah se exam mein pass ho gayi

Thanks for your knowledge

Chup chap😂😂😂

Thanks🙏

IL = Sqrt(3) * Iph is OK, but Iph = Sqrt(3) * IL is wrong. It should be Iph = IL / Sqrt(3)

Thanku sir 👍