Multiply the entire equation with z^4 and then rearrange from highest power. For example z^4x5=5z^4 will come first and the last one will be z^4x(1/z^4)=1 which is placed at the end
1.put 1/z in place of s in the given equation. 2.now multiply the new equation formed by z power 4. 3.rearrange it. (1st column of each row should be non- zero, if any one of the rows has zero in the 1st column then only this method works)
Thank you very much. These videos are very much helpful and wonderful explanation mam. Please continue to help us. I'm doing feedback control systems subject in Italy and this video is helping me a lot!
I'm writing a test on Tuesday on Stability System "Routh Hurwitz Criterion" I'm sure after watching these videos, I'm going to score 100%. many thanks ma'am.
If you could not grasp the idea of cross multiplying with 1/z here's what you can do simply take the LCM of ie 1/z^4 in this case and chage the partial fraction into a single fraction term .
@Devalla Rakesh what is the issue buddy if some is trying to learn some thing he doesn't know ... it is not a crime if he doesn't know Some thing at any point if he is studying for gate or any other competitive exam
Sign change occurs for 2 times in the equation ...which means that there are 2 points on the right side of s-plane ..these points are poles and they are responsible for unstable nature of system
there are two sign changes on the first column of the routh array , indicating that there are two poles at the right side of s-plane. Thus, the system is unstable because there is pole at the right side of s-plane
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1)..1/z⁴ + 1/z³ +2(1/Z²) +2(1/Z)+5=0
2)..1/Z⁴ +1/Z³ +2/Z² + 2/Z +5/1=0
3)..take common denominator z⁴
4)..(1+z+2z²+2z³+5z⁴)/z⁴=0
5)..5z⁴+2z³+2z²+1=0
Thanks now I understood
Thank you so much...now I understand 😌😌
Tqs😊😊😊😊😊
life saver dude !!!! thanks
Thanku 😊 😊 😊 😊 😊 😊
Multiply the entire equation with z^4 and then rearrange from highest power. For example z^4x5=5z^4 will come first and the last one will be z^4x(1/z^4)=1 which is placed at the end
Thanks!!
Thanks mate!! I was bit confused in that!😂😂
Ty so much
really thanks for you >>
thank you
Dear One, your teaching style is very good. Your video helped me a lot. I thank you very much
1.put 1/z in place of s in the given equation.
2.now multiply the new equation formed by z power 4.
3.rearrange it.
(1st column of each row should be non- zero, if any one of the rows has zero in the 1st column then only this method works)
Name of the video should be "Special Case of Routh Array Case I"
Ri88
Yes
Yea
ya
Now I feel that I am able to clear my sessional exam easily . I saw first teacher teaching in the way which is easier to understand.
Thanks a lot mam
How was your Sessional exams
@@sachinchoudhary5024 I don't remember now 😅.
@@himanshutanwar4179 😲
I have used your videos for 1 year now, the explanations are easy to follow 🙏. Be blessed ma’am
All of control system vedios are very nice mam....and they are very helpful to us...😘
Loved the simplicity in your teaching
Thank You So much madam for this amazing Playlist of Control System
Thank you very much. These videos are very much helpful and wonderful explanation mam. Please continue to help us. I'm doing feedback control systems subject in Italy and this video is helping me a lot!
Who tf asked
I'm writing a test on Tuesday on Stability System "Routh Hurwitz Criterion" I'm sure after watching these videos, I'm going to score 100%. many thanks ma'am.
DID YOU SCORE?
@@RajinderSingh-qm6zb 😂😂
If you could not grasp the idea of cross multiplying with 1/z here's what you can do simply take the LCM of ie 1/z^4 in this case and chage the partial fraction into a single fraction term .
Excellent teaching..very easy to understand..thanks a lot mam
Cross multiplication:
Take LCM of (1/z)^4 that means multiply and divide whole eqn with (1/z)^4 and solve it
The way of you explaining is very clear mam....it's very easy to understand
Thanks For my exam preparation 😃
To know how she cross multiplied just common (1/z)4 from whole equation than u take that z4 to other side it become 0×z4 i think now u guys understood
Multiplying the whole equation with Z4 .....u will get the answer....ryt.....I tried wat u said but didn't get answers
@Devalla Rakesh some people have issues understanding hence they are here, get a life and stop demotivating everyone
Thanks Now i get it
@Devalla Rakesh what is the issue buddy if some is trying to learn some thing he doesn't know ... it is not a crime if he doesn't know Some thing at any point if he is studying for gate or any other competitive exam
@@keerthanaanandan6062 just follow what I said above when solving the problem
Nice classes maam
But u should not skip those multiplication steps
It was really confusing
Can we take auxiliary equation when we get zero and then continue the process?
5:03 cross Multiple I'm Not understood plz tell the Formula..
(1/z)^4 wale puri equation ko z^4 se multiply kro or solve kro jo answer aye ushe phr re-arrange kr lo bas jawab a jye ga
Mujhe bhi samaj nhi aa raha
Koi samjao
@@Mubashirhassan-akb hamesha highest exponent power wala se hi multiply krna hai kya
@@Aditikumari16 yes
Should I write this in diploma msbte paper
Thanks to Tutorials India😊 and big thankyou to mam 😊🙏
I easily understood mam I was so much confused .
Thank u so much ❣️
Your lecturer is excellent mam
How cross multiplication is done
Ma’am after substituting 1/z if we again get 0 what should we do?
plzzz upload video for bode plots and polar plots problems and graphs
Why mam u have taken 1/z any reason plz mention
how to convert cross multiply one more time repeat for cross multiply
Better go to before class
In characteristic equation there is no constant value
Then what value should take
0 or 1
Reply
by seeing comments here, I realized why engineers don't get job easily
To know how she cross multiplied :
multiply by z^4 on both sides to get the new equation
Thanks bro
(2/z)^2 should be equals to 4z^2 right? here she only raise the z to the 2nd power not the whole term why?
because we are replacing ''S'' with 1/Z . hope u get the point .
Very good explanation
mam .... wonderful explanation mam....it is very helpful to us...thank u so much mam for making this vedios
Mam but in Routh array there is not rule to substitute in s=1/z
madam aapne title me mistake kar dia hai myself coming from village area aap kya fail kara ke hi maanogi....have a nice day.
If we have an odd power sequence such as s^5 + 4s^3 + 2s + 1. How do we solve
Mam can you show one example for higher to find stability R.H criteria
Very good explanation thanks
Thankiyou miss
Mam u are good lecturerer .....nice video👍👍👌👌
Thank you ma'am🥰
Thankyou so much 😊 mam....
how z2 last number is zero saying there is no classes,i didnt understand that part
Cross multiplication degara neat ga explain cheyandi mam ala vachindhi ani
Thanks madam , excellent explaination
Perfect teaching
If first column all value in -ve then system be
Stable or unstable
Plz reply anyone
Unstable
Thankyou sooo muuchhhhh ❤❤❤❤❤❤❤🎉🎉🎉🎉🎉
How to do cross multiplication of (1/z) 5
The title may be wrong. This is Case 1
How you cross multiplied don't understand....?🙄
take L.C.M of z^4 in denumerator and send to right side ... do with hands u will get the result .
kindly correct your title of video this is the first case and you write 2 case in video title
Anyone can help me what is conclusion is the system stable or marginally stable ?
superb mam
Why replace with 1/z?
Because of the zero term in the first column of the Routh array
Thank you soo much madam
Thank you teacher 😍
Naku ardam kaledu
Hi mam i month cs exam ide so please question paper solutions and shree video
thank you mam
how you do cross multiple
Samw question
multiple by z"4
I multiply by Z(to the power of 4) or Z4
@@karthi6843 .....thanks man
Deko (1/z)^4 wale puri equation ko z^4 se multiply kro or phr re-arrange kr lo bas ans a jye ga
Mam
5z^4+2z3.....pls tell me in details
Multiple all the values by z^4 and simplify
at 11:42, what did u say about the poles here?
Sign change occurs for 2 times in the equation ...which means that there are 2 points on the right side of s-plane ..these points are poles and they are responsible for unstable nature of system
You meant to say Epsilon !
How u cross multiplied
Mam 5th module ka problems videos upload kariye.....mam
gd gd gd video....watch from bd...eee
How she cross multiplied? please anyone show the system.
It's lcm process
Thank you
Evarina detail ga explain cheyandi
I did not get the two poles thing can you clear that plz
there are two sign changes on the first column of the routh array , indicating that there are two poles at the right side of s-plane. Thus, the system is unstable because there is pole at the right side of s-plane
Very difficult bana diya
Thanks
Mam pls do problems on rootlocus mam
Brain, she got it!
On the basis of two cases the two answers are different 🤔
How second part of the questions become 2Z^3
Multiply the LHS and RHS with Z^4.
Yeah cross multiplex kaise kiya bilkul nhi samjh aay ??
Aman Shrivastava bhai simple LCM liya h
Aman Shrivastava merko vi ...ganda padatii h ye mam.
@@simranjeetsingh2405 kaise bhai
(1/z)^4 wale puri equation ko z^4 se multiply kro or phr re-arrange kr lo bas ans a jye ga
The first method doesnt work all the time😌 but the second method does
Shukriya
Are there Havard students using these Tutorials to study?
I'm just concerned 😂
Please do speed up n stop repeating too much
this is case 1 ..please change the video name and also change another video name to 2.
Cross multiple kis kis se bhai .....
if is 0 always is system unstable ?
جمعه ت دئ هاوكئشا ره ئسي لئكداني (z^4) كه ن
Jedi power is too mainstream. Try Jed power.
1/z ka cross thikse bolke sikhate yaar usme kuchh pata hi nhi chala
2Z^3 kese aya
Bolo
multiply all equation by Z^4
how to drive cross multipication ,
plz describe.
@@puleekkar thnxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx nigga
I got it
Day by day RUclips changes into ads tube
Y is all ur video not clear
Or else just multiply both sides with z^4
this one is difficult.
I solve it in a easy way.
How u have done send me please 8812094922
👋👋🤌
Anyone can help me what is conclusion is the system stable or marginally stable ?
sorry teacher i and friends غيبتك