unknown transfer function R(s)->G(s)->C(s). if output signal is r(t)=2e^-3t, what was the transfer function?
143954
I'm not understanding the question.  R(s) is the input in the frequency domain, G(s) is the transfer function in the frequency domain, and C(s) is the output?  r(t)=2e^-3t is the output in the time domain?  What is the input in the time domain?  \n\nInput in the time domain convolved with the transfer function in the time domain gives the output in the time domain.  Alternatively, input in the frequency domain times the transfer function in the frequency domain gives the output in the frequency domain. Take the Fourier or LaPlace Transform of that to get output in the time domain--I think that is what you are asking.\n\nPer Table 5.9 in Kreysig, the LaPlace Transform of 1/(s-a) <=> e^at.  Using this transform and assuming that your input is an impulse, making the output equal to the transfer function, the transfer function would be 2/(s+3).\n\nIf you add some detail, I will check back on this question tomorrow.
680385
the maximum power (transfer) theorem states that to obtain maximum power from a source with a fixed internal resistance, the resistance of the load must be made the same as that of the source.\n\nas far as what equation?  hmmm...\n\nP = k T Bw, where P is the power generated internally in a Resistor, T is the temperature and Bw is the badwidth of the measuring system or amplifier. If the Resistor (a component) is modelled by a Thevenin equivalente circuit, i.e. a noise voltage generator with Vth and Rth (voltage source and its impedance) feeding a matched load impedance R = Rth, results in Maximum power transfer, giving: kTBw = Vth^2/(4Rth). From which the Nyquist equation results as Vth^2 = 4 k T Bw R.
958486
Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1
469104
The algebraic sum of the currents entering a node is equal to zero. Where current entering is positive and exiting is negative (or vice versa). In other words, what comes in must come out.\nNote: this is only true for DC signals.
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
1128734
I'm going to make a wild guess that you are trying to check an answer you got with the answer in the back of the book. I'm also going to assume that you were supposed to take a derivative of\n(x^3/(a-x))^(1/2).\nIn that case, you got the wrong answer because you didn't do the chain rule. You still need to multiply what you have by the derivative of the thing inside, i.e. the derivative of x^3/(a-x).\n\nIf I'm making a bad assumption and this didn't come from a differentiation problem, please ignore my answer.
708667
I think you mean\n\nS = A - RT/(1-R)\ntake A  to the left and negate\n\nA- S = RT/(1-R)\nmutiltply by (1-R) on both sides\n\n(A-S)(1-R) = RT\nor (A-S) -R(A-S) = RT\ntake R term to the right\n\nA- S = R(T+A-S)\n\nso R = (A-S)/(T+A-S)
486296
s(n) =  [1-r^n]*a/(1-r) + [(1-n)r^n]*d/(1-r) + [1-r^(n-1)]*dr/(1-r)^2\nfor n > 0\n\ns(infinity) = (a + r(d-a))/(1-r)^2\nfor |r| < 1\n\nSee simulation\nhttp://www.geocities.com/derekcowley/public/agp.xls\n\nproof:\n\ncompute the difference equation:\ny(n) = [a + (n - 1)d]*r^(n-1) = a*r^(n-1) + (n-1)d*r^(n-1)\ny(n+1) = [a+nd]*r^n = a*r^n + nd*r^n = y(n)*r  + d*r^n\ny(n+1) = y(n)*r  + d*r^n\n\ngenerate the LTI discrete time system state space:\n\nx2(n) = r^(n-1)\nx2(n+1) = r*x2(n)\nx1(n+1) = r*x1(n) + d*x2(n) \nx1(1) = a\nx2(1) = r\n\nX(n+1) = A*X(n) + B*delta(n)\ny(n+1) = C*X(n)\n\nwhere A is a 2x2 matrix [r,d;0,r]\nB is column vector [a;r]\nC is a row vector [1,0]\ndelta(n) = 1 for n=0, 0 otherwise\n\nCompute the Z-transform:\n\nH(z) = C*[(zI-A)^(-1)]*B = a/(z-r) + dr/(z-r)^2\n\ncompute running sum:\n\nAccumulator U(z) = z/(z-1)\nS(z) = H(z)*U(z) = az/(z-1)(z-r) + drz/(z-1)(z-r)^2\n\nCompute partial sum (partial fraction expansion):\n\nS(z) = e/(z-1) + f/(z-r) + g/(z-r)^2\n\ne = a/(1-r) + dr/(1-r)^2\nf = -ar/(1-r) - dr/(1-r)^2\ng = -dr^2/(1-r)\n\nInverse Z-transform\n\ns(n) = e  + f*r^(n-1) + g*(n-1)r^(n-2) [n>0]\n= see top line above\n\nTo see a spectral analysis of the sequence based on pole/zero locations see\nhttp://www.geocities.com/derekcowley/public/agp2.xls\nYou will need the "Analysis ToolPak" add-in in Excel to properly see this file.
929743
∫t²cos(1 - t³) dt\n\nI note that d(1 - t)/dt = -3t² \nThis is set for substitution of 1 - t³\n\nLet u = 1 - t Thus du = -3t²dt ie  -1/3 . du = t²dt\n\nSo ∫t²cos(1 - t³) dt\n= ∫cos(1 - t³) t²dt\n= ∫cosu (-1/3) . du\n= -1/3 sinu + c\n= -1/3 sin (1 - t³) + c
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
889055
Power:\nP=dE/dt = dW/dt\n\nForce:\nf=ma\n\nEnergy:\ne=mc^2
878725
antiderivative of:\n20-12cos2pi(x-0.1)\nis:\n\n20x - 12 sin( 2pi(x-0.1)) / 2pi +C\n=20x - 6 sin( 2pi(x-0.1)) / pi +C\nto double check just take the derivative of the function and see that you get:  20-12cos2pi(x-0.1) '
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
414104
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
436468
More info is needed. \n\nWhat is the pass band frequency, what is the spectral content of the input signal. Ideally how steep do you want it?