Problems on Root locus| Part-13 Stability analysis

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  • čas přidán 8. 09. 2024
  • In this video we have solved problems on root locus method

Komentáře • 114

  • @nazninsathi5332
    @nazninsathi5332 Před rokem +6

    The teacher is so excellent with his ausum tutorial vedio so helpful❤

  • @me-jz7uv
    @me-jz7uv Před rokem +11

    Thank you sir.. from Nigeria.. having a paper in the afternoon🙏

  • @gayathriar942
    @gayathriar942 Před 4 lety +9

    You make the concepts simple,clear and easy to understand."Thank you so much sir!!!"👍🏻

    • @anshulbajpei935
      @anshulbajpei935 Před 9 měsíci

      Aapki engineering complete ho gai

    • @gayathriar942
      @gayathriar942 Před 9 měsíci +1

      @@anshulbajpei935 yes

    • @anshulbajpei935
      @anshulbajpei935 Před 9 měsíci

      ​@@gayathriar942very nice but aap konse state se ho me Ahmedabad se hu

    • @gayathriar942
      @gayathriar942 Před 9 měsíci

      @@anshulbajpei935 Thank you , I'm from Karnataka

    • @anshulbajpei935
      @anshulbajpei935 Před 8 měsíci

      ​@@gayathriar942 So you must not know Hindi to speak.?

  • @kulfranc9726
    @kulfranc9726 Před 11 měsíci +2

    Am greatly humbled by d lecture

  • @mariamkhanum8043
    @mariamkhanum8043 Před 4 lety +5

    Easily understandable 👍

  • @kamaliyaraj2880
    @kamaliyaraj2880 Před rokem +2

    Interesting and easy to understand ❤

  • @DGXDX_
    @DGXDX_ Před 3 měsíci

    Thank you sir
    Nice explanation you made it so easy to understand step by step 👍

  • @mahanteshhalingali-mf9oq
    @mahanteshhalingali-mf9oq Před 2 měsíci +1

    Amazing explaination sir

  • @Hari-lx4yd
    @Hari-lx4yd Před 8 dny +2

    From mars, thanks sir

  • @arunak1716
    @arunak1716 Před 4 lety +1

    Very much helpful content sir......

  • @swapniljadhav9494
    @swapniljadhav9494 Před rokem +2

    great Explanation

  • @mtanusri6154
    @mtanusri6154 Před 4 lety +5

    tq sir its really helpful 👍

  • @chandanpateln.s8390
    @chandanpateln.s8390 Před 4 lety +2

    Good content and explanation sir 👍

  • @syedabrar1724
    @syedabrar1724 Před 2 lety +6

    EASY METHOD For intersection with imaginary axis
    Just substitute s=jw in the characteristic equation and equate the real and imaginary part to zero. And find the value of w, which is the intersection of root locus with both +ve and - ve imaginary axis.

  • @MR_UNKNOWN709
    @MR_UNKNOWN709 Před 6 měsíci +1

    Lot of help Sir 🎉

  • @aliyasiddiqua4861
    @aliyasiddiqua4861 Před 4 lety +1

    Explanation is very nice

  • @mdashraf9409
    @mdashraf9409 Před 9 měsíci +1

    nice explanation

  • @lakshmankumar-vz6ci
    @lakshmankumar-vz6ci Před 3 lety +1

    wonderfull

  • @omkaromsachi8052
    @omkaromsachi8052 Před 4 lety +1

    Your allway super sir

  • @arunmk21
    @arunmk21 Před 9 měsíci +1

    Thank you sir

  • @karthikms5407
    @karthikms5407 Před 2 lety +1

    Tqew Sir. For urs wonderful explanation ♥

  • @kennethabraham9997
    @kennethabraham9997 Před 4 lety +3

    Very helpful , thanks sir!

  • @syedirfan5560
    @syedirfan5560 Před 4 lety +1

    Good explanation 👍

  • @deeptidase363
    @deeptidase363 Před rokem +1

    Thank you so much sir🙏so calmly and nicely you have taught

  • @keshavkumar5774
    @keshavkumar5774 Před 2 lety +2

    Thank u very much sir .....

  • @pallavipallu727
    @pallavipallu727 Před 2 lety +1

    Thankyou sir because detailed explanation

  • @eceelectronicsandcommunica3986

    Nice

  • @hawaphiri
    @hawaphiri Před 10 měsíci

    Loving the video

  • @johnbritto4450
    @johnbritto4450 Před 4 lety +1

    Sir tqsm

  • @sumaiyafathima2107
    @sumaiyafathima2107 Před 4 lety +1

    Very helpful. Tq sir👍

  • @justinvarghese52
    @justinvarghese52 Před měsícem

    Hi sir great explanation, for the s=-4.7, the value of k is -230.723, I think might be calculations might be wrong..

    • @as-tutorials5537
      @as-tutorials5537  Před měsícem +1

      Hi Justin the calculation are correct. Please check by substituting s= -4.7 in K equation. k=-s^3-9s^2-18s
      k=(-(-4.7)^3) -(9*(-4.7)^2)-(18*(-4.7))
      Answer is k= -10.39

  • @indumathim7612
    @indumathim7612 Před 4 lety +3

    Thanks for explaining clearly sir!!!☺️

  • @gamingwithirfan3459
    @gamingwithirfan3459 Před rokem +1

    Superb ❤️

  • @DivyaS-du5wl
    @DivyaS-du5wl Před 4 lety +1

    Thank you sir 👍

  • @bushrahasan6500
    @bushrahasan6500 Před 4 lety +1

    👌

  • @adeebaroohi9745
    @adeebaroohi9745 Před 4 lety +1

    Understood sir 👍

  • @anamika7581
    @anamika7581 Před 4 lety +1

    Thanx sir

  • @vijaysinhjadeja5361
    @vijaysinhjadeja5361 Před 2 lety +1

    Thank you very much sir

  • @divyas3067
    @divyas3067 Před 4 lety +1

    Thanku sir

  • @user-lc8eo6rg4p
    @user-lc8eo6rg4p Před 3 měsíci +1

    Nice explanation sir, but in root locus how to get 's' values i didn't understand

  • @monikasr8999
    @monikasr8999 Před 2 lety +1

    Thank you so much sir 🙂 it's very helpful sir

  • @allwynpaul8061
    @allwynpaul8061 Před 4 lety +1

    Well explained sir:)

  • @gayathrikrishnan3566
    @gayathrikrishnan3566 Před 4 lety +1

    Thank you so much sir 👍

  • @syedshabazpasha2997
    @syedshabazpasha2997 Před 4 lety +1

    Thanks 🌝 sir

  • @MsanthoshkumarHKEC
    @MsanthoshkumarHKEC Před 4 lety +1

    Thank you sir,very helpful 🙂

  • @guruxxx
    @guruxxx Před 4 lety +1

    Thank you sir!!

  • @Not_in_use-455
    @Not_in_use-455 Před 4 lety +1

    Tqsm sir 👍🏻

  • @249srihari4
    @249srihari4 Před 8 měsíci +1

    thank u sir

  • @hemaojha2067
    @hemaojha2067 Před 4 lety +1

    Thank You Sir•

  • @TGnong
    @TGnong Před 2 lety +4

    Sir after dk/ds how to find value of s im stuck here... Please help me... S=1.21 how???

    • @as-tutorials5537
      @as-tutorials5537  Před 2 lety +9

      −3s^2−18s−18=0
      For this equation: a=-3, b=-18, c=-18
      −3s^2−18S−18=0
      Step 1: Use quadratic formula with a=-3, b=-18, c=-18.
      S=(−b ± (√b^2−4.a.c))/2.a
      Substitute a,b and c value in the above formula
      S=(−(−18)±√(−18)^2−4(−3)(−18))/2(−3)
      S=(18±√108)/−6
      S=−3−√3 and s=−3+√3
      S=-4.7 and s=-1.26

    • @TGnong
      @TGnong Před 2 lety +1

      Thankyou so much sir

  • @Suyash_Rasal
    @Suyash_Rasal Před 9 měsíci +1

    Thanku sir ❤

  • @erumbegum1107
    @erumbegum1107 Před 4 lety +1

    Understandable.Thank you sir!!

  • @dipteshsaha5288
    @dipteshsaha5288 Před 4 lety

    Thank u sir for your wonderful explain

  • @deepakravidas1983
    @deepakravidas1983 Před 3 měsíci +1

    Thank you sir Tomorrow is my exam

  • @fathenawaz2488
    @fathenawaz2488 Před 4 lety +1

    Thank you

  • @afaqakram330
    @afaqakram330 Před 4 lety

    Thanks.......

  • @afzalahmed.m5677
    @afzalahmed.m5677 Před 4 lety +1

    Thank you sir taking the effort n explaining the concept very clearly

  • @aishuaishu4386
    @aishuaishu4386 Před 8 měsíci +1

    but sir is k value 10.39 how sir😢

  • @praveenbasavaraj1408
    @praveenbasavaraj1408 Před 2 lety +2

    Sir your taking s=-1.26 not understand Plz exam Ripley

    • @as-tutorials5537
      @as-tutorials5537  Před 2 lety +2

      −3s^2−18s−18=0
      For this equation: a=-3, b=-18, c=-18
      −3s^2−18S−18=0
      Step 1: Use quadratic formula with a=-3, b=-18, c=-18.
      S=(−b ± (√b^2−4.a.c))/2.a
      Substitute a,b and c value in the above formula
      S=(−(−18)±√(−18)^2−4(−3)(−18))/2(−3)
      S=(18±√108)/−6
      S=−3−√3 and s=−3+√3
      S=-4.7 and s=-1.26

  • @anasarshad3453
    @anasarshad3453 Před rokem +1

    Love from Pakistan 🎉🎉🇵🇰🇵🇰🇵🇰

  • @gunjans6311
    @gunjans6311 Před 4 lety

    Thank you sir...

  • @manishplayz6687
    @manishplayz6687 Před měsícem +1

    How did u get the -1.26 anwser as a valid s ?

    • @as-tutorials5537
      @as-tutorials5537  Před měsícem

      Two ways to check the s valid or not
      1. Root Locus Check: If the value of 's' is on the root locus plath it’s valid.
      2. 'K ' Equation :substitute 's ' value in 'K' equation. If 'K' is positive, 's' is valid; if K is negative 's' isn’t valid.

  • @tanzimkakon8839
    @tanzimkakon8839 Před 2 lety +2

    checking valid break-away and break-in point is interesting

  • @alihaider1563
    @alihaider1563 Před 8 měsíci

    How do we find out the values for K for which the closed loop poles are real?

    • @carultch
      @carultch Před 6 měsíci

      Those are called break-in and break-away points. Given a transfer function of N(s)/D(s), the breakpoints will be found when the following equation holds true:
      N(s)*D'(s) - N'(s)*D(s)=0
      We can do this for a system with 1 zero and 2 poles as an example. (s+6)/[(s + 5)*(s + 2)]. It is very difficult to do this analytically for higher order systems, as it means solving higher order polynomials, that may be impossible to solve by hand.
      N(s) = s + 6
      N'(s) = 1
      D(s) = (s + 5)*(s + 2)
      D'(s) = 2*s + 7
      Construct equation to equate to zero:
      (s + 6)*(2*s + 7) - 1*(s + 5)*(s + 2) = 0
      Solve for s:
      s = -8, the break-in point
      s = -4, the break-away point
      Given a closed loop made from K and G(s), the closed loop transfer function will be K*G(s)/(1 + K*G(s)). The denominator of this will equal zero at the two closed-loop poles.
      1 + K*G(s) = 0
      When G(s) = N(s)/D(s):
      1 + K*N(s)/D(s) = 0
      Solve for K:
      K = -D(s)/N(s)
      Thus, for our example:
      K = -(s + 5)*(s + 2)/(s + 6)
      Plug in breakpoints for s:
      at s = -4:
      K = -(-4 + 5)*(-4 + 2)/(-4 + 6) = 1
      at s = -8:
      K = -(-8 + 5)*(-8 + 2)/(-8 + 6)
      K = 9
      So at 1 and 9, we have poles that transition between being real, and being a complex conjugate pair. When K=9, the poles are both real.

  • @murali2004-n8k
    @murali2004-n8k Před 8 dny

    How to get s value

  • @harsham9107
    @harsham9107 Před 4 lety

    👍👍👍👍👍

  • @harsham9107
    @harsham9107 Před 4 lety

    ☺☺👍

  • @Vasu0221
    @Vasu0221 Před rokem

    Sir I didn't understand how s=-1.26& s= -4.7

    • @as-tutorials5537
      @as-tutorials5537  Před rokem +3

      −3s^2−18s−18=0
      For this equation: a=-3, b=-18, c=-18
      −3s^2−18S−18=0
      Step 1: Use quadratic formula with a=-3, b=-18, c=-18.
      S=(−b ± (√b^2−4.a.c))/2.a
      Substitute a,b and c value in the above formula
      S=(−(−18)±√(−18)^2−4(−3)(−18))/2(−3)
      S=(18±√108)/−6
      S=−3−√3 and s=−3+√3
      S=-4.7 and s=-1.26

  • @kishoreburle727
    @kishoreburle727 Před rokem

    How to calucat breaking point

    • @as-tutorials5537
      @as-tutorials5537  Před rokem

      Go through this link. czcams.com/video/_GnUs2wwZ6k/video.html
      czcams.com/video/Q1aNPwwbr8c/video.html
      In this problem I have explained to calculate breakaway point

  • @siddheshwagh2300
    @siddheshwagh2300 Před rokem

    if there is no s in the denominator how to solve

    • @siddheshwagh2300
      @siddheshwagh2300 Před rokem

      mrans s/(s+3)(s+6)

    • @as-tutorials5537
      @as-tutorials5537  Před rokem

      No 's' in the denominator means no Poles at the origin. For s/(s+3)(s+6) in this numerator one zero that is s=0 and denominator 2 Poles that is s=-3 and s=-6. Therefore one pole s=-3 reaches to zero and another pole s=-6 move towards infinite path at an angle 180°

  • @New_movie007
    @New_movie007 Před rokem

    T che valu chukle

  • @ayeeshakhan5191
    @ayeeshakhan5191 Před 4 lety +1

    Thank you sir

  • @ayeshamuskan9739
    @ayeshamuskan9739 Před 4 lety +1

    Thank u so much sir..

  • @bhavanak4033
    @bhavanak4033 Před 4 lety +1

    👌

  • @ankitkumar-ne7fr
    @ankitkumar-ne7fr Před 4 lety +1

    Thank you sir 👌

  • @chetans4036
    @chetans4036 Před 4 lety +1

    Thank u sir ☺️

  • @resithasarvana5367
    @resithasarvana5367 Před 3 lety +1

    Thank you sirr

  • @maazahmed9787
    @maazahmed9787 Před 4 lety +1

    Thank u sir

  • @MariamIqra-HKEC
    @MariamIqra-HKEC Před 4 lety +1

    Thank you sir

  • @lopnasarrinmondal9483
    @lopnasarrinmondal9483 Před 4 lety +1

    Thank u sir

  • @nishantgaikwad4876
    @nishantgaikwad4876 Před 2 lety +2

    Thank You sir

  • @faisalzaffari5943
    @faisalzaffari5943 Před 4 lety +1

    Thank you sir

  • @shivanik2359
    @shivanik2359 Před 4 lety +1

    Thank u sir

  • @afrozkhan5878
    @afrozkhan5878 Před 4 lety +1

    Thank u sir 👍

  • @arusafathima3148
    @arusafathima3148 Před 4 lety +1

    Thank you sir👍

  • @861tummalapallibindhuhemad5

    Thank you sir

  • @aneesarahamath8307
    @aneesarahamath8307 Před 4 lety +1

    Thnks sir

  • @kavana6474
    @kavana6474 Před 4 lety +1

    Thank u sir