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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 🙏
Causal systems are right sided, hence their z-transform would contain terms like 1 + 2z^(-1) + 3z^(-2) + ... etc Hence, to converge a this series, clearly |z| > 1. If |z| < 1, then z^(a large negative number) would tend to infinity and hence H(z) would not converge Hence, this could only be achieved
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Very Good explanation sir 👍
Your positive comments motivates me.
Teachers like me just wants positive comment from student.
Love from you guys means a lot to me.
My goal is to create largest community of engineers in entire globe. Please help me by sharing this playlist with your friends.
Sir please tell some complex problems in this four methods
Thank you for the clear explanation.
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 🙏
thank you so much sir❤️...you are life saver❤️🎈☘️
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 🙏
Thank you sir
Your positive comments motivates me, Thanks and welcome 🙏
Sir which company Pen do u use for writing..?
Mitshibushi ballpoint pen
Sir z-1 (5/5z-1) y question bta dijiy please sir
Sir |z|>1 casual kse hoga pls explain
Causal systems are right sided, hence their z-transform would contain terms like 1 + 2z^(-1) + 3z^(-2) + ... etc
Hence, to converge a this series, clearly |z| > 1.
If |z| < 1, then z^(a large negative number) would tend to infinity and hence H(z) would not converge
Hence, this could only be achieved