Time Reversal Property of Z Transform

[@TutorialsPoint_]

To Get Certification, Click Here: bit.ly/3oCIvLA Use coupon "KZbin12" to get ‘’FLAT 12%’’ OFF at Checkout.

[@robbiebeenham5980]

It shouldn't matter much because in essence it is still the same sum, but surely the limits change signs. n=infinity to -infinity.

[@comedyspot8938]

Limits are wrong it should be reversed

[@jagadishware922]

Exactly

[@aidanabregov1412]

I'm assuming that the -inf to inf bounds aren't switched because you're still covering the bounds from -inf to inf, but then doesn't that only apply to EVEN functions where: x[n] = x[-n] ?

[@madusudan2474]

Thanks a lot for explain the example....

[@mr_ig_king_2002]

Mam aap pahle engineering funda se sikh ke aaye please 🙏

[@chamber946]

There's catch, This is summation, not integral. Summation just sums up the values, no matter lower to upper or upper to lower. therefore it doesn't matter

[@omkargurav9022]

explained very good

[@hasanalikhan3543]

And how about 10^n u(-n-5) how to calculate it's z transform plz help

[@aidanabregov1412]

10 is your a^n and your step function is u[ --( n + 5 ) ], hope this helps you with the first step (look up the z-trans of u[n] to a fraction and check what time reversal and delay (+5) does to the step function)

[@norahali5507]

good job......

[@madusudan2474]

Super mam

[@sudeepKumariit]

In each lecture, you are doing some mistakes.. plz avoid..

[@hasanalikhan3543]

2^n u(-n) to udhar hi reh gaya

[@abdullahx4908]

NO Not another algorithm

[@a.k.dwivedi71]

Hlo mam how we can write (1/2)^-n as 2^n plz rply

[@manjulas1861]

(1/2)^-n=(2^-1)^-n We know that (a^m)^n=a^mn Hence it is 2^n