Sir thank you so much for lecture network ke mghe kvl kcl apply nhi krna ata tha apke lecture se hi maine sikha hai aap bhut easy method se sb smghate h
@karthikhkamath2 жыл бұрын
wait, why are the subscribes and views soo less for this good teacher !!! 😭😭😭
@KKSirEduHub2 жыл бұрын
GOD knows...help growing the channel ..promote it
@vaibhav639752 жыл бұрын
I didn't understand a single word in my class but after these lectures I will really perform well in my internals as well as my semester exams . 😄😄 thank you sir , for these lectures .
@KKSirEduHub2 жыл бұрын
👍stay connected and kindly promote the channel
@vivek35753 жыл бұрын
Sir app ka teaching ka tarika bahut achha h Or bhi video upload kr diziye sir SS ki......🙏
@KKSirEduHub Жыл бұрын
yes sure..stay connected
@VishalSharma-ri3ih3 жыл бұрын
How we check stable or unstable ? Is this done on a particular value of x(t) and Is that value given in the questions ? Please explain more sir ....
@KKSirEduHub Жыл бұрын
system stability is checked on the basis of ling term response of a system under few test signals
@63_tanishkgupta55 Жыл бұрын
Very nice..💐
@KKSirEduHub Жыл бұрын
thanks
@hk-rv7yb3 жыл бұрын
Very nice sir bibo ka example jyaad Hina cahiya tha
@KKSirEduHub Жыл бұрын
yes…separate lectures rakhenge us par
@himanshiaggarwal35043 жыл бұрын
Great work doing sir
@KKSirEduHub Жыл бұрын
thanks
@Abhi227243 жыл бұрын
Keep uploading sir👍
@KKSirEduHub Жыл бұрын
sure
@shwetasingh4703 жыл бұрын
y(t) =x(t+10) +x^2(t) will be dynamic, non linear, non causal, stable and time invariant in nature.
@satwikpal91613 жыл бұрын
how stable?
@AvinashSingh-rh3pk2 жыл бұрын
its an unstable one because of the x^2(t) the graph of x^2 (t) is un bounded
@anshagarwal5999 Жыл бұрын
@@AvinashSingh-rh3pk good
@whisperthestory10 ай бұрын
@@AvinashSingh-rh3pk It will stable only as if we give u(t) as input so u(t+10) is finite and similarly u^2(t) which is square of unit step signal is always finite in amplitude so both are finite and stable hence this given system is stable