Electronics Forum  Help Search Members Calendar 
Welcome Guest ( Log In  Register )  Resend Validation Email 
StevensElectronicAccount 
Posted: October 19, 2011 04:26 pm

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
If one takes the equation v(t+h)=v(t), one can use a taylor series on the left side. Doing the first few terms one gets v(t) + v'(t)h + v''(t)(h^2)/2 = v(t). And then one gets v'(t)h + v''(t)(h^2)/2 = 0. If one has an amplifier and uses feedback to make this condition true for the amplifier's output will the amplifier have an oscillating output or will it have a constant output when the amplifier is fed a constant voltage? In a real life circuit how many terms of the taylor series would be used?

Sch3mat1c 
Posted: October 19, 2011 09:00 pm

Forum Addict ++ Group: Moderators Posts: 20,521 Member No.: 73 Joined: July 24, 2002 
No, that's not quite right, you need:
v(t) + v'(t) (t+h) + v''(t) (t+h)^2 / 2 + ... If you approximate with the first three terms, you get: v(t+h) = v(t) ~= v(t) + v'(t) (t+h) + v''(t) (t+h)^2 / 2 If you cross out the v(t)'s, you get: 0 = v'(t) (t+h) + v''(t) (t+h)^2 / 2 We can factor out one (t+h), of course. Integrating will provide the solution to this differential equation, v(t) = c1  c2 / (t+h) where c1 and c2 are any real numbers  the solution is valid for a space of equations (including constant and reciprocally decaying features), which ones apply to a particular problem depend on initial conditions (i.e., the value of v(0) and v'(0), or at any other time points really). Now remember, the truncated Taylor series only approximates in the vicinity of t+h and is not true, by any means, for all t, h or v. For example, the Taylor expansion of 1/(x+a) is a "good try", but apparently diverges much quicker than other series, like e^x or x^(3/2) do for their corresponding series. In particular, reciprocal functions tend to get *worse* as the order of approximation goes up. The above solution may be applicable for small ranges of t, h and v, but it will not describe any useful portion of the function. In fact, although it might be tempting to suppose that, if h is small, it might still apply in the vicinity of 0<t<h, but this is not the case, because at t=h, the reciprocal term becomes undefined. Tim  Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.

StevensElectronicAccount 
Posted: October 19, 2011 09:45 pm

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
My line of reasoning is
One has f(x) is equal to the summation of the terms f^k (a) * (x  a)^k / k! f(a + h) is therefore equal to the summation of the terms f^k (a) * (a + h  a ) ^ k / k! The variable a cancels out and substituting x for a one gets f(x + h) is equal to the summation of f^k (a) * h ^ k / k! I don't think it is v(t) + v'(t) (t+h) + v''(t) (t+h)^2 / 2 + ... This seems untrue because v(t+0) does not equal v(t) + v'(t)t + v''(t)t^2/2 + ... unless v'(t)t + v''(t)t^2/2 + ... is equal to zero whereas my sequence reduces to v(t) if h is zero. Anyways I made a stupid error. I meant to ask how it'd work for the approximation v(t+h) = v(t) + v'(t)h + v''(t)h^2 / 2 + v'''(t)h^3 / 6 This would result in a sinusoidal instead of an exponential. Solving the question I didn't mean to ask is easy. 0 = v'(t)h + v''(t)h^2/2 We let p = v'(t) 0 = p(t) + p'(t)h/2 This is just an exponential and is simple. p(t)^(1) * p'(t)h/2 = 1 (h/2)*ln(p(t)) = t + c p(t) = k * e ^((2/h)*t) If one integrates p(t) to get v(t) that is just an exponential as well. I meant to ask the question for 0 = v'(t) + v''(t) h /2 + v'''(t) h ^ 2 / 6 
millwood 
Posted: October 26, 2011 01:04 am


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
tough to know exactly what you are trying to say.
one family of solutions for that would be v(t) = k, where k is a constant with regards to t. 

millwood 
Posted: October 26, 2011 01:12 am

Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
v(t+h) = v(t) can be rewritten as
v(t) = v(th), that is, the signal at t is the same as the signal h period ago, aka there is a delay. in continuous time domain, there is a transmission line with delay. in the discrete domain, it is like an array with a length of h. by itself, it do not oscillate  the device simply outputs what it received h time units ago. aka there is no feedback. now, if you were to feed it back, with a gain factor of beta ... 
millwood 
Posted: October 26, 2011 11:55 am


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
just to continue:
remember that v''(t) = dv'(t) / dt. so you can rewrite the above as dv'(t) / v'(t) = k*dt intergrate that, you get ln(v'(t)) = k1*t + k2, or v'(t) = exp (k1*t+k2). v'(t) = dv(t) / dt. so dv(t) = exp(k1*t+k2)*dt. integrate that, you get v(t) = k*exp (k1*t+k2). k/k1/k2 are simply some constants and in this series not necessarily the same from equation to equation. you can solve for them via boundary conditions. if the original equation has the t  term on the right (an incorrect tylor expansion for example), the solution is quite similar, except that you will now need to integrate over that tterm. and for an equation involving the 3rd order derivative, the process is the same. 

:: support us ::