Recent content by reece

thanks I have them pointing to the first state.

I am designing a Up/Down Gray code counter that has two redundant don't care conditions. The problem is I have no idea how to ensure the counter escapes from the redundant states. How do I work out where the redundant states should point too?

In trig FS, you evaluate each component a0, an, bn. Well the exponential replaces that I think. im guessing you use Euler's Rules (i think) to evaluate it which says sin theta = 1 / 2i [ e ^ i*theta - e ^ -i*theta] cos theta = 1 /2 [ e ^ i*theta + e ^ -i*theta] Hopefully that answered...

woops my bad, those t's arent spose to be in there. But the rest is right. And yes it is what you typed out. Im leading to think I shouldn't get the same answer as what I got using the Trig-Complex Relationship equation. Which is why I am getting something different. The exponential part is...

Homework Statement Solve \frac{1}{1} \int^{0}_{-1} -t e^{-j2\pi*nt}dtHomework Equations So I use integration by parts u = -t and dv = e^{-j2\pi*nt} , du= -1 and v = \frac{1}{-j2\pi*n}e^{-j2\pi*nt}The Attempt at a Solution after integration by parts I get: \frac{e^-j2\pi*nt}{j2pi*n} +...

Hi, I can solve homogeneous laplace fine, but with RHS i get stuck half way through. Q: y' +3y = 8e^{t} y(0) = 2 Working as if it was homogeneous.. sY(s) - 2 + 3Y(s) = 8 . \frac{1}{s-1} Y(s) (s+3) - 2 = 8 . \frac{1}{s-1} I think the next step is Y(s) = \frac{2}{s+3} +...

Hi, I can solve homogeneous laplace fine, but with RHS i get stuck half way through. Q: y' +3y = 8e^{t} y(0) = 2 Working as if it was homogeneous.. sY(s) - 2 + 3Y(s) = 8 . \frac{1}{s-1} Y(s) (s+3) - 2 = 8 . \frac{1}{s-1} I think the next step is Y(s) = \frac{2}{s+3} +...

Basically I don't know how F(s) can be split up to below. F(s) = \frac{1}{s^{2}(s-2)} = \frac{1}{4} ( - \frac{1}{s} - 2 \frac{1}{s^{2}} + \frac{1}{s-2} ) I thought it would be 1/s^2 - 1 / s-2 How does this work? Please explain. thanks