Region of Convergence for the z-Transform

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z-transforms of signals in general do not exist over the entire z-plane. The infinite series defining the z-transform only converges for a subset of values of z, termed the region of convergence. Multiple examples of deriving z-tranforms.

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24 сен 2024

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Комментарии : 59

@allsignalprocessing 11 лет назад

I will try to clarify my use of terms: 1) A causal signal has x[n] = 0 for n

@felixg4785 8 лет назад

Why can't lectures try to be like you? Thanks, you are awesome.

@allsignalprocessing 11 лет назад

We say this signal is non causal by viewing it as if it were the impulse response of an LTI system. The impulse response of a causal system must be zero for n

@allsignalprocessing 11 лет назад

The grt than1/4 applies to the first term alone in g[n] while less than 1/2 applies to the second term in g[n]. Hence for the z-transform of g[n] to converge, we can only include values of z for which BOTH terms in g[n] converge. This leads to the overall ROC being the INTERSECTION of grt than1/4 and less than 1/2, which results in the ring. I think it is easier to figure this out by looking at the individual terms. With a sum of signals the ROC will be the intersection of individ ROCs.

@maristelasantos8714 9 лет назад

You have such a great talent for teaching. Thank you veeery muuchhh for saving my final exam!

@morto00x 8 лет назад

By far the best ROC explanation I've found. Thanks

@augustoferreira3720 7 лет назад

in case 3 the plot is wrong. For y[n]=-α^n u[-n -1] , with 0 < α < 1 the sequence should grow unbounded as n decreases.

@Mani_KrSingh 6 лет назад

Yes you are true. The sequence will grow unbounded. For the decaying sequence as shown in Figure, it should have been a>1

@joshualiu8551 5 лет назад

Thanks, this comment should be pushed front

@mehrabzamanian581 2 года назад

I think it should -a^-n instead of -a^n

@jiwann4213 11 лет назад

you should have clarified on how to choose region of convergence in examples rather than in general. I am not clear with the explanation. At 11:46 how and why are they different? : |z|>1/4 but |z|

@KlaskeyProductions 9 лет назад

I could almost feel my hair blowing as your explanation at 13:15 flew right over my head lol.

@ShrenikJain 7 лет назад

Klae G's haha 🤣🤣🤣

@adityavinayak3989 5 лет назад

@@ShrenikJain i saw your videos. well, some people upload for sharing knowledge. and some like u open an online so-called last hours tuition school to make money lol

@bugzo0 11 лет назад

wait, u wrote my signals book ?

@creatiph 10 лет назад

Beginner here, what's u[n] ? It's in the X[n] equation at the start but it's there anymore when we switch to the X(z) form. Thanks

@nazcov 11 лет назад

Not that i'm an expert either but it's not non causal, its anti causal. A causal system can be past and present values, where a noncausal contains future values. I think the anticausal system is a subcategory for only past values, referring to what mr. Van Ven commented at the top. Feel free to correct me if im wrong.

@r410a8 3 месяца назад

What is the formula of u[n] in 3:26 such the a variable's value determes the value of x[n] such a way in each case,i don't understand. And why in 4:36 when you aply the z transform formula on to x[n] you write Sum n-oo to +oo a^n*z^-n instead of Sum -oo to +oo a^n*u[n]*z^-n suposed the x[n] is a^n*u[n] not only a^n ? I don't understand this neither.

@springbokmarine 8 лет назад

Thank you so so much for posting these videos!

@realislamicguidance2375 3 года назад

Suppose a signal is growing exponentially. We take its Z transform & find its ROC. So what are we supposed to do practically? Are we supposed to manipulate our O/P by multiplying any signal from ROC such that our O/P converges & system is stable? And what if our signal is exponentially decaying so whats the purpose of finding ROC in this case (except for determing stability/casuality)?

@abedsoubra2453 7 лет назад

how do you know if a series is converging or not ?

@zoala001 9 лет назад

I kind of got the idea of ROC but I have problem of knowing the casual and causal functions. I mean I can't really say if a series function is greater or smaller than alpha.

@JainamTalsania 8 лет назад

Your videos are awesome, keep doing great work.

@mohamedquabbou5253 7 лет назад

thank you for the for this vid you saved me from failing my semestre

@FatimaHassan-fn1vk 6 лет назад

Why do we assume the radius to be alpha?

@nomadic_rider42 10 лет назад

Great explanation, thanks a lot for the effort.

@uniqueaakash14 7 лет назад

you are a life saver

@oath0703 10 лет назад

Hi Sir, in the last example, when you rewrite, is it x[n] = u[n] +delta[n] - (-3/4)^n*u[-n-1] ? isn't it x[n] = u[n] - delta[n] + (-3/4)^n*u[-n-1] ? From your notation, two ways(before rewrite, after rewrite) of expression x[-1] will be x[n] = (-3/4)^-1 , x[n] = - (-3/4)^-1 I'm a bit confused on this stuff. I really do love your videos. Thank you so much

@allsignalprocessing 10 лет назад

Note that at 13:15 the expression is x[n] = u[n] +delta[n] - - (-3/4)^n*u[-n-1], that is there are two negative signs in between the delta[n] and (-3/4)^n terms and thus it is effectively a +. In your equation above you missed the second minus sign. You also have a - in front of delta[n], and that should be +. We can check as follows: x[-1] = - - (-3/4)^(-1) = (-3/4)^(-1) which agrees before and after the rewrite step. Also, x[0] = u[0] +(-3/4)^0 u[0] = 2 (before the rewrite step) and x[0] = u[0] + delta[0] = 2 (after the rewrite step). I think the algebra is ok as written, but that is a good question - you are paying very close attention.

@oath0703 10 лет назад

Barry Van Veen Ohh,, I see now. It wasn't tricky. I was just missing one negative sign. Thank you so much Sir. I wish more students and people can find this channel and benefit from this great lectures. I found a month before the final which is tomorrow while have been suffering for couple months from ...well. Appreciate your great effort.

@seimela 9 лет назад

Barry Van Veen thank you for explaining ......got confused also

@joshuasalvador8785 9 лет назад

How do I know if its case 2 or 3 when I'm doing the inverse z transform table look up?

@augustoferreira3720 7 лет назад

You must check the ROC.... if |z|>|α|, it is a right sided sequence α^n u[n]. if |z|

@danel5726 9 лет назад

Hi! This should be obvious, but in Case 2, why is x[n]=0 for negative values of n? Is u[n] a predefined function? Thank you!

@ashishgupta9332 8 лет назад

yes

@SamriddhaDhali 9 лет назад

THANK YOU SIR. best VIDEO FOR UNDERSTANDING ROC

@kousik_das 7 лет назад

Thank you sir, very nicely explained

@priteshparikh9324 9 лет назад

at 14:19 i did not get the roc condition. shouldnt it be z> -3/4 our only condition for infinite geometric series is that it should exist ...i.e denominator should not become zero or negative?

@ShrenikJain 7 лет назад

pritesh parikh see it's |z| and not just z. so you are right z >-3/4 so..... |z|