how to draw the Nyquist Plot for this system:\n((S+10)^2)/(s^3)
200361
One Way: let s=j*w, then substitute and obtain real and imaginary parts.  Then let w vary over range of interest, and plot real and imaginary parts in R and Theta (Polar).\nBut Nyquist plot is really concerned with stability of closed loops and not with just a frequency plot like you show. You can of course plot in gain magnitude or db.\nNow, your expression is an open loop gain,then we usually express it as 10*(1+s/10)/s^3. Using a Bode plot is more convenient.All we want to know for simple stability of the loop when we close it is: Is the phase shift greater than minus180 deg when we close the loop.We work in a log  base 10 scale for w, and gain in db.10/s^3 has gain=1 (0db) at w=2.155 and you have a lead at w=10, then draw thhe lines and see the result. To state Nyquist's criterion goes like this:The net encirclements of the point (-1.0)= the difference between the number of poles and zeros of the open loop transfer function in  the right half-plane.\nBode,Nichols, and Nyquist are famous names in the field fron the WWII era and later in to sampled data methods.\ny
433746
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
393074
I'm sorry, the question isn't written out clearly enough.\n\nIs it supposed to be\n\n5r^3st^2\n-----------\n-10r^6s^3t        ?\n\nThen to write it as a question in here, it's best to use brackets:\n\n(5r^3st^2)/(-10r^6s^3t) (Okay, having seen someone else's interpretation, then perhaps something more along the lines of \n(5r^3 s t^2)/(-10 r^6 s^3 t))\n\nIf my interpretation is correct, then let's look at this in a different way:\n\n5*r*r*r*s*t*t\n-----------------------------\n-10 *r*r*r*r*r*r*s*s*s*t\n\n5/-10 gives you -2 in your denominator, not the numerator. Eliminate the variables that are equivalent to 1. That means you'll remove 3 r's from the top along with 3r's from the bottom, 1 s from the top and 1 s from the bottom, 1 t from the top and the t from the bottom. This leaves you with :\n\nt\n-----------\n-2r^3s^2
343722
sorry i am no good at math, but this may help:\nhttp://www.fourmilab.ch/earthview/pacalc.html
137715
Found a good diagram here\nhttp://www.laynetworks.com/Block%20diagram%20of%20the%20Pentium.htm
655332
Hint: It sounds like a line. Look at it again and describe the points you plotted in relation to where x=2 and y= any real number (infinity).
80191
I haven't worked with this in a long time and I'm not sure this will work, but try starting w/ this:\n1) Define an equation that describes the " ∞ " with polar coordinates (something like r = C1 - C2*theta);\n2) If L is lengh, delta.L = delta.theta / (2 pi) * r\nand dL = r/(2 pi) dtheta  ;\n3) substitute for r and integrate from 0 to 2pi (or the necessary for one complete " ∞ ".\n \nProbably you'll find more than one mistakes here, but I think the general idea is correct.
500894
are you talking about a polynomial :\na + bx +cx^2 + dx^3 +...\n\nbecause those a,b,c,d... do not appear any where on the graph... you have to look at the intercepts and the extreama, and the inflection points to get any good idea about the shape of the graph
517694
Just add them together sample by sample. Squaring them is destroying the negative values. (Just think of a sine wave.) \n The distortion must be terrible!\n\n Note however if when you add them the sum exceeds a certain value you will get some overload. If the sample values are modest to start with this will not happen.
366056
This one's a little tricky.\n\nAllow me to use "t" instead of "delta"\n\n\nint ((cos(2t))^3)dt\n\nSince (cos(2t))^2 = 1 - (sin(2t))^2\n\nSubstitute.  this becomes:\n\nint ((1 - (sin(2t))^2) * cos(2t))dt\n\nSubstitute u = sin(2t).  Then du = cos(2t)dt.\nSo dt = du / cos(2t)\nThe key here is when you plug this in, the cos(2t) values cancel.\nSo we get:\n\nint (1 - u^2)du\n= u - (1/3)u^3 + C\n= sin(2t) - ((sin(2t))^3)/3 + C
528149
Correct formula should be as follows:\n\nE = 1/2*I*w^2
335710
View source of the that page.\nYou will see the formulas.
76377
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
293581
Hertzsprung-Russell  Diagram.  See the wikipedia article for more info.
506152
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
1649
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
364846
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck