I am designing a Up/Down Gray code counter that has two redundant dont 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 shouldnt get the same answer as what I got using the Trig-Complex Relationship equation. Which is why im getting something different.
The exponential part is...
Homework Statement
Solve \frac{1}{1} \int^{0}_{-1} -t e^{-j2\pi*nt}dt
Homework 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