Recent content by exidez

This is not a question from any textbook or assignment. It is a concept question in order to get a task done. I need to size a motor but just need to confirm my suspicion. I am sizing a motor needed to open/close a roller door that curls around a drum. Not one of those modern ones. I need to...

ok, that makes a lot more sense. I guess I was fixed on the idea that it was one dimensional that I missed all the references to 2D. zero padding is just for computer efficiency is that correct? or is it more to it than that?

Spatial Frequency Transform by doing FFT twice ?? This is for my own knowledge in relations to acoustics. I am trying to determine the location of sound using an array of sensors. Very similar to what they are doing in the webpage given. Im looking at the material at...

this comes strait from a textbook: http://higheredbcs.wiley.com/legacy/college/nise/0471794759/appendices/app_i.pdf I am looking at how they obtained (I.24) from (I.25) on page 3 and 4. Firstly we have: e^{-\textbf{A}t}x(t)-x(0)=\int{e^{-\textbf{A}t}\textbf{Bu}(\tau)d\tau} Then this...

ahhhhh yes. Then i get a term involving \ddot{x}, take it over the other side, factor the common \ddot{x} out and take what is in the bracket on the other side. I got the correct answer now! Also For the \ddot{\theta}. Thank a lot for you help. It is greatly appreciated! I had learned and...

The Two equations: V-mg=m\frac{d^{2}}{dt^{2}}(Lcos(\theta)) I\ddot{\theta}+c\dot{\theta}=VLsin(\theta)-HLcos(\theta) Derevation: \ddot{\theta}=\frac{1}{I}(VLsin(\theta)-HLcos(\theta)-c\dot{\theta}) \ddot{\theta}=\frac{1}{I}(VLsin(\theta)-HLcos(\theta)-c\dot{\theta})...

The two equations: F-H=M\ddot{x}+k\dot{x}) H=m\frac{d^{2}}{dt^{2}}(x-Lsin(\theta)) Derivation: \ddot{x}=\frac{1}{M}(F-H-k\dot{x}) \ddot{x}=\frac{1}{M}(F-m\frac{d^{2}}{dt^{2}}(x-Lsin(\theta))-k\dot{x}) \ddot{x}=\frac{1}{M}(F-m\frac{d}{dt}(\frac{d}{dt}(x+Lsin(\theta)))-k\dot{x})...

it would be faster if i just took a photo of my working I changed the variable of M to Mc to ensure i wasnt getting mixed up with m. They are two different masses. A little messy but you can see how i got 1/I and 1/M...

Thanks tiny-tim, I understand it correctly now and i have the matching answer but with one tiny problems.. If you look at equation 5 and 6they have a multiplyer at the front being 1/(I+L^2m) and 1/(M+m) respectively however, when in my answers they are simply 1/I and 1/M respectively as that...

Ok so now I am lost again.. I tried... If i just work out one, i can get the other but i still can't get the one.. \frac{d^2}{dt^2}(x+Lsin(\theta)) First working out the first derivative \frac{d}{dt}(x+Lsin(\theta)) = \frac{d}{d\theta}\frac{d\theta}{dt}(x+Lsin(\theta))...

ahhh of course they are! Why do i always forget the fundamentals!

In the attached image, how are equations 1, 2, 3 and 4 used to come to the final equation of 5 and 6? I am suspecting it has something to do with the derivative with respect to t, but I don't know how they remove it to get the final solution. Am I missing something incredibly simple that is not...

ok, i think i got it! i go the other way and make n = -1 I have never seen this but just to clear this up: if \frac{d}{ds}}F(S) is the derivative of F(S) then \frac{d^{-1}}{ds^{-1}}F(S) is the same as the integration of F(S) right?

Homework Statement the inverse laplace transform of ln\frac{s+2}{s-5} using the inverse Laplace transform of the derivative Homework EquationsL^{-1}{\frac{d^{n}}{ds^{n}}F(S)} = (-1)^{n}t^{n}f(t) The Attempt at a Solution the integral of ln\frac{s+2}{s-5} I worked to be (s+2)ln(s+2)-(s+2)...

Homework Statement \displaystyle\int\int\sqrt{4-x^2-y^2} dA R{(x,y)|x^2+y^2\leq4 .. 0\leq x} The Attempt at a Solution So far i have: \displaystyle\int^{\pi}_{0}\int^{r}_{0}\sqrt{4-r^2} rdrd\theta Solving i get...