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I have no issue finding a and b. However to find K, I would like to know if I'm taking the right approach. I found $$a = 1/24$$ and $$b = 4$$ So how I'm planning on calculating K is as followed: $$2k/a = 8$$ since $$a = 1/24$$ then $$k = 1/6$$ Am I right? if not, please explain why this approach is wrong. Thank you. enter image description here

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  • \$\begingroup\$ This question appears to be unrelated to electronics design \$\endgroup\$ – Chetan Bhargava Apr 30 '14 at 4:45
  • \$\begingroup\$ Where did you get the 8 for (2K/a) = 8 from? I'm pritty sure you shouldn't include the a. It's because your trasfer cunion should be normalized in the form of (s/a),(s/b). Then you take 20log(constant), which in your case 20log(2K) and determine where the graph starts from. \$\endgroup\$ – Adel Bibi Apr 30 '14 at 4:58
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    \$\begingroup\$ @ChetanBhargava drawing bode plots is quite a basic EE thing actually. \$\endgroup\$ – Vladimir Cravero Apr 30 '14 at 5:02
  • \$\begingroup\$ @VladimirCravero more towards mathematical side if you agree. \$\endgroup\$ – Chetan Bhargava Apr 30 '14 at 5:15
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    \$\begingroup\$ "By signing this document, you are pledging to adhere to the honor system stating that all work performed here is strictly your own effort", the irony is too much for me. \$\endgroup\$ – hesson Apr 30 '14 at 5:21
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Your calculations to find a and b looks good to me but I don't agree on K.

You are given only the "unity frequency" or better "unity pulsation" (hope that's correct in English). You should take your transfer function and approximate it appropriately, then fill in the informations you have, i.e. at pulsation 8 gain is 1. The TF you are looking for is the TF stripped of all its poles and zeroes except the "origin pole" (again, hope that's correct), i.e.: $$G(s)=\frac{2K}{s}$$ that's because you are assuming \$s\ll s_l\$ where \$s_l\$ is the lowest singularity. Solving the previou equation for k: $$K=\frac{s\cdot G(s)}{2}=\frac{8\cdot 1}{2} = 4$$ that's about 12dB.

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  • \$\begingroup\$ Sorry, I can't see any values on the y axis, how did you know that the gain is 1 at pulstation 8? \$\endgroup\$ – Adel Bibi Apr 30 '14 at 5:26
  • \$\begingroup\$ I am assuming that the pulsation axis is located at 0dB as it usually is, and as you can see (it took a while for me) the first slope is prolonged till the axis so the gain is one at pulsation 8 only for the approximate transfer function. \$\endgroup\$ – Vladimir Cravero Apr 30 '14 at 5:39
  • \$\begingroup\$ Sweet! I got another misconception now, why did you determine that the magnitude at s=8 is the same at the beginning of the graph. I mean why did you take s=8? I'm asking because I can see the 20bd/decade between s=4 and s=8, which is ((8-4)/10 *20) = 8 db increment. This can never equal to the -20db/decade for s=1 to s=2 which has a 2db increment. \$\endgroup\$ – Adel Bibi Apr 30 '14 at 5:47
  • \$\begingroup\$ That is not the same. Look closely at my calculations, my G(s) expression contains an s so it is pulsation dependant. And there's no beginning of the graph on a bode plot... \$\endgroup\$ – Vladimir Cravero Apr 30 '14 at 5:51
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    \$\begingroup\$ Sorry, was confused for some time. The "there's no beginning of the graph on a bode plot" got me on the track, I was completely missing that part. Thanks! \$\endgroup\$ – Adel Bibi Apr 30 '14 at 5:55

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