# Accepted!

Paper accepted in IEICE.

# More to Go! Design of Short-Length Message Fountain Code for Erasure Channel Transmission.

# Hook up with Money

I went to a grant proposal defence few days ago. They only asked me two questions. 1. Is your work pattern-able? 2. Is your work commercialize-able. They want money-outcome now.

# Solving Math’s Assignment with Sage Someone sent me this question:

Solve for the currents in the circult of Figure 2, if E(t)=5H(t-2) and the initial currents are zero. [Hint : Use Lapalce transform to solve this problem.]

So, to solve it, form mesh analysis of two loops. Then, convert them from time domain to complex domain with Laplace transform. Next, solve I1 and I2 with normal algebra. Then only inverse I1 and I2 back to time domain.

Of cause, if you familiar with Sage, you can solve it within 30min (or lesser?).

t = var('t')
s = var('s')
I1 = var('I1')
I2 = var('I2')

E(t) = 5*unit_step(t-2)

E(s) = E(t).laplace(t, s); E(s)
# >> 5*e^(-2*s)/s

equation = [
-E(s) + I1*20*s + 10*(I1-I2) == 0,
10*(I2-I1) + I2*30*s + I2*10 == 0 ]

solution = solve(equation, I1, I2); solution
# >> [[I1 == 1/2*(3*s + 2)*e^(-2*s)/(6*s^3 + 7*s^2 + s), I2 == 1/2*e^(-2*s)/(6*s^3 + 7*s^2 + s)]]

# Note that Sage cannot inverse-Laplace time-delay function. So, taking out e^(-2*s)
I1(s) = 1/2*(3*s + 2)/(6*s^3 + 7*s^2 + s)
I2(s) = 1/2/(6*s^3 + 7*s^2 + s)

i1_temp(t) = I1(s).inverse_laplace(s, t); i1_temp
# >>t |--> -1/10*e^(-t) - 9/10*e^(-1/6*t) + 1

i2_temp(t) = I2(s).inverse_laplace(s, t); i2_temp
# >> t |--> 1/10*e^(-t) - 3/5*e^(-1/6*t) + 1/2

# Referring to Table. For G(s)= e^(as)F(s), the inverse is g(t) = f(t-a).
u(t) = unit_step(t)
i1(t) = u(t-2) * ( -1/10*e^(-(t-2)) - 9/10*e^(-1/6*(t-2)) + 1 ) # Answer for i1
i2(t) = u(t-2) * ( 1/10*e^(-(t-2)) - 3/5*e^(-1/6*(t-2)) + 1/2 ) # Answer for i2

p1 = plot(i1(t), 0, 10, color='blue', legend_label='i1(t)')
p2 = plot(i2(t), 0, 10, color='red', legend_label='i2(t)')
show(p1 + p2)


And, the final answers are: $i_1(t) =u(t)\left( \frac{-1}{10}e^{-(t-2)} -\frac{9}{10}e^{-(t-2)/6} +1 \right)$ $i_2(t) =u(t)\left( \frac{1}{10}e^{-(t-2)} -\frac{3}{5}e^{-(t-2)/6} +\frac{1}{2} \right)$ Current I1 and I2 vs time.

# Bye SAINT2012

I happened to be the last batch of speakers in SAINT2012 conference. Next year it will be merged with COMPSAC as “New COMPSAC”, which will be held in Kyoto in 2013. The name of “SAINT” will not be used anymore.