For calculating the omega w value, perform the following steps. Shift 90 degrees to the other side, so -tan^-1(w)-tan^-1(w/2) =270 degrees. Or, tan^-1(w)+tan^-1(w/2)= -270 degrees. Now, apply the formula of tan^-1(x)+tan^-1(y)= tan^-1{(x+y)/(1-xy)}. So, tan^-1(w)+tan^-1(w/2)= tan^-1{(w+w/2)/(1-w*w/2)} = -270 degrees. Apply tan to both sides. So (w+w/2)/(1-w*w/2) = tan(-270)= undefined. Undefined means that the denominator is zero. Therefore, 1-w*w/2= 0. Or, 1-w^2/2 = 0. Or, w^2/2= 1. Or, w^2= 2. Or, w= rt(2). Thus, the value of omega w is sq. root 2 or 1.414.
Omega value can be calculated just follow my steps 1.send the 90 to LHS [~(-180+90=-90] 2.get the equation as (-90)=-tan^(w) - tan^(w/2).... {~remove negitive sign both side} 3.apply tan on both sides Tan(90)=tan(tan^(w)+tan^(w/2)) 4.we have formula.. Tan(a+b) 5..1/0=tan(tan^(w)+tan^(W/2))/1+tan(tan^w)*Tan(tan^w/2)........{~WRITE TAN 90 =1/0} 6..1/0=(W+W/2)/1+W2/2......{HERE W2 IS W SQUARE } 7.cross multiply 8. U gets 2-w2=0.......{here w2 is w square} 9.hence u got w=root 2.... Try it and be care full at cross multiplication only this one step i didn't wrote completely... Cross multiplication will me like this ex. A/B=C/D... Cross multiplication we gets.... AD=BC...... I spended 1hr to get it. I hope with my suggestion u all will get it in 10m...
For those who have doubt in finding Omega Value move -90 to otherside of eqal use -tan^1(x)-tan^1(y) = tan^1((x+y)/1-xy), then take tan on both side...Find w
Bro first of all, tan-¹x + tan-¹y = tan-¹[(x+y)/(1-xy)] Secondly after taking the 90° on the other side it sums up to be 270 and tan 270 is not defined.
Saket Fule yes am finding the same 0.16 but when making magnitude isn't it supposed to be 1/(w^2......) Rather than 1/(w^1.....) As it is. Please keep in touch @lusangw3 Twitter
Madam I have followed your videos regularly. Super explanation. Madam I want how to find K value when phase and gain margin is given by using polar plot..
How u r able to take out the value of w=?!. The first most difficultly I face why she have written +180° in the negative real axis which should be -180° if I consider in the direction of Type2(T2) system which brings down the final equation of phase angle to -180°=-90°- tan^-1(w) -tan^-1(w/2)... Solving this equation by multiplying tan on both sides the value of w becomes infinite
First substitute fie as 180°than move the 180°on RHS side than move tan inverse on LHS side use the trigonometric formula of tan inverse x + tan inverse y =tan inverse (x+y)/1-xy than solve it
The procedure use for w is wrong think so...there is another procedure in which we have to convert the denominator of the transfer function into value having only real parts (i.e by cross multiplying) after that the we will be getting a imaginary parts to in the numerator which we have to make equal to zero (ex - num = 15w +j(5w²-10) , then we have to make (5w² -10 = 0) then we will get w = +-√2) and if do same in this example also we will get w=+-√2 ...but magnitude will be 0.166. hope this may help..
