Motivation For Research

•December 9, 2008 • 1 Comment
Tokyo tower

Tokyo tower

Japan is not only famous of its manga and anime, but also of the advance of technology. One of the reason of this advance is because of so many researches conducted in universities and companies. Recently I just found out about the reason why students and lecturers in Japan willingly conduct so many researches. Not only to add more entries on their curriculum vitae, but also because they want to go abroad. As you may know, there are so many international conferences about technology. And since it’s international scale, most of the time it’s held outside Japan. By doing research, they can submit papers, and if the papers passed the review, they will have the chance to go abroad. They also don’t have to worry about the money, because the university will pay for them, everything. For university, money is not a problem. The more papers produced by their students/lecturers, the more famous the university will be.

Researching is like racing here. In a meeting few days ago, my sensei listed the international conferences that will be held next year like PIMRC in Tokyo, VTC in Alaska, and APWCS in Seoul. He encouraged everyone to submit a paper to any of the conferences. Not only the master student or PhD students, but also the bachelor students as well. Even one of my friend who is still in bachelor degree is ready to submit paper to APWCS cause he wants to go to Seoul.

I believe this is one of motivation that universities in Indonesia should have to advance in technology. Especially to ITB, since I heard ITB is planning to be a research based institute. Hopefully more good quality papers are produced from ITB.

Probability of Error for BPSK over Rayleigh channel with Maximal Ratio Combiner

•November 13, 2008 • 1 Comment

Aside from deriving BER for BPSK over rayleigh channel, I also got another assignment to derive the equation of probability of error (BER) for BPSK over rayleigh channel with maximal ratio combiner (mrc). After searching here and there, I found the solution in Digital Communications by Proakis page 825. But just like the previous assignment, Proakis didn’t derive the equation in detail. Even Mr. Google couldn’t help me this time 😦 So I tried to solve it myself, but I’m stuck. So I posted here my unfinished work, maybe you can help me figuring it out.

For a certain value of time, the bit error rate probability is


where \gamma_b=\frac{E_b}{N_o}\sum_{k=1}^L {h_k}^2 = \sum_{k=1}^L \gamma_k . \gamma_k is the instantaneous SNR on the kth channel.

When k=1 , \gamma_b\equiv\gamma_1 and

P(\gamma_1)=\frac{1}{\bar \gamma_c} e^{-\frac{\gamma_1}{\gamma_c}}

where \bar \gamma_c=\frac{E_b}{N_0}E({h_k}^2) (average SNR per channel). So the characteristic function is

\psi_{\gamma_1}(jv) = E(e^{jv\gamma_1})
=\int_{-\infty}^{\infty} e^{jv\gamma_1}\frac{1}{\bar \gamma_c}e^{-\frac{\gamma_1}{\bar \gamma_c}}d\gamma_1
=\frac{1}{\bar \gamma_c}\int_{-\infty}^{\infty}e^{jv\gamma_1}\frac{1}{\bar \gamma_c}e^{-\frac{\gamma_1}{\bar \gamma_c}}d\gamma_1
=\left. -\frac{1}{\bar \gamma_c}\frac{\bar \gamma_c}{1-jv\bar \gamma_c}e^{-\gamma_1\left(\frac{1}{\bar \gamma_c}-jv \right) }\right|_{-\infty}^{\infty}
=\frac{1}{1-jv\bar \gamma_c}

Since the fading on K channels is mutually statistically independent
\psi_{\gamma_b}(jv)=\frac{1}{(1-jv\bar \gamma_c)^K}

Using inverse fourier transform,

p(\gamma_b) = \frac{1}{2\pi}\int_{-\infty}^\infty \psi_\gamma(jv)e^{-jv\gamma_b}

=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{(1-jv\bar \gamma_c)}e^{-jv\gamma_b}

=\frac{1}{{\bar \gamma_c}^K \Gamma(K)}{\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}

=\frac{1}{(K-1)!{\bar \gamma_c}^K}{\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}

So, the Bit error rate (error probability)

P_2=\int_{-\infty}^\infty P_2(\gamma_b)p(\gamma_b)d\gamma_b

=\int_{-\infty}^\infty Q(\sqrt{2\gamma_b})p(\gamma_b)d\gamma_b

To solve equation above, use integration by parts.
By definition, \int f(x)g(x)dx = F(x)g(x)-\int F(x)g'(x)
which is another way of rewriting the usual \int udv=uv-\int vdu

Let f(\gamma_b)=p(\gamma_b) .
And by definition

\int x^ne^{ax}dx=\frac{e^{ax}}{a^{n+1}}\left[ (ax)^n-n(ax)^{n-1}+n(n-1)(ax)^{n-2}-\dots+(-1)^nn!\right]

So we have

F(\gamma_b) = \int f(\gamma_b)d\gamma_b = \int \frac{1}{(K-1)!{\bar \gamma_c}^K}{\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}

=\frac{1}{(K-1)!{\bar \gamma_c}^K} \int {\gamma_b}^{K-1}e^{-\frac{\gamma_b}{\bar \gamma_c}}

=\frac{e^{-\frac{\gamma_b}{\bar \gamma_c}}}{(-1)^K(K-1)!}\left[ \left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-1}-(K-1)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-2}+\right.

\left.(K-1)(K-2)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-3}-\dots+(-1)^{K-1}(K-1)!\right]

Let g(\gamma_b)=Q(\sqrt{2\gamma_b}) also

\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)dt=f(v(x))v'(x)-f(u(x))u'(x)


g'(\gamma_b)=\frac{d}{d\gamma_b}Q(\sqrt{2\gamma_b})=\left( -\frac{1}{\sqrt{2\pi}}e^{-\gamma_b}\right)\left( \frac{1}{\sqrt{2}}\gamma_b^{-\frac{1}{2}}\right)

Probability of error becomes

P_2=\int_0^\infty Q(\sqrt{2\gamma_b})p(\gamma_b)d\gamma_b=\left. F(\gamma_b)g(\gamma_b)\right|_0^\infty-\int_0^\infty F(\gamma_b)g'(\gamma_b)

=\frac{1}{2}-\frac{1}{(-1)^K(K-1)!}\frac{1}{2\sqrt{\pi}}\int_0^\infty \gamma_b^{-\frac{1}{2}}e^{-\gamma_b\left( 1+\frac{1}{\bar \gamma_c} \right)}\left[ \left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-1}-(K-1)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-2}\right.

\left.+(K-1)(K-2)\left(-\frac{\gamma_b}{\bar \gamma_c} \right)^{K-3}-\dots+(-1)^{K-1}(K-1)!\right]d\gamma_b

And somehow the equation becomes

P_2=\left[ \frac{1}{2}(1-\mu)\right]^K\sum_{k=0}^K \binom{K-1+k}{k}\left[ \frac{1}{2}(1+\mu)\right]^k

\mu=\sqrt{\frac{\bar \gamma_c}{1+\bar \gamma_c}}

So that’s it. Anybody can help me on the last part of the derivation? You can download the pdf version of that equation here.

PSP and Linux

•November 12, 2008 • 2 Comments

For Linux(nix) users, creating psp compatible files is so easy. Use the following.

How to convert videos to psp compatible format:
ffmpeg -i InputVideo -f psp -r 29.97 -b 768k -ar 24000 -ab 64k -s 320x240 outputvideo.mp4

How to convert pdf to image(s):
convert nani.pdf nani.jpg
or you can use ghostscript instead using this script:
if [ $# -ne 2 ];then
echo “Usage: $0 input.pdf outfile”
gs \
-sDEVICE=jpeg \
-r150 \
-dTextAlphaBits=4 \
-dGraphicsAlphaBits=4 \
-dMaxStripSize=8192 \
-sOutputFile=${INPUT}_%d.jpg \

To be able to execute convert and ffmpeg, you need to install them first of course.

BER for BPSK in Rayleigh Channel

•November 6, 2008 • 4 Comments

Last week I got a homework from my professor to derive the equation of bit error rate (BER) of binary phase shift keying (BPSK) in Rayleigh channel. As usual, mr. google was my best friend, then I found a nice explanation about BER for BPSK in Rayleigh channel. Unfortunately, the author also hasn’t figured out how to derive the equation yet. But thank God and my good friend who is studying KTH University now. He showed the way to derive it. Here’s the solution in pdf if you need it. Hope it will be useful for anyone who got the same homework as I did 😀

National Novel Writing Month

•November 6, 2008 • 2 Comments

When I opened my dashboard today, I accidentally saw this link to a blog about National Novel Writing Month in the What’s hot of wordpress news. It’s about writing a 50000 words novel along November and submit it at the end of November. No money prize, only a web badge and certificate. But most valuable prize is, aside from the manuscript itself as stated in the FAQ of NaNoWriMo, the pride and satisfaction of being able to write and finish a novel of your own which will be read by many people I assume.

I’ve been wanting to write some novel, and it seemed to be a good chance for me to start. But unfortunately, no idea yet. Furthermore, since the event started since November 1, I wonder if I can still join. But it seems that what matter is to finish 50000-words novel by the end of this month, I think it’s still possible to join. What do you think?

HP new Mini 1000 and MIE Linux: I want this!

•October 30, 2008 • Leave a Comment
HP Mini 1000

HP Mini 1000

I was thinking of buying a netbook recently, well not now or the near future cause no money yet :p But after I saw this one, I really like it! just look at it, isn’t it beautiful? And the MIE version (Mobile Internet Experience) which is based on Ubuntu, and which I will buy of course, is much cheaper than the MS one! I just found it here, that the reason why the windows version is more expensive is not only because windows is not free OS, but also because in MIE version, there’s no Windows button on the netbook. The Windows key on the keyboard is replaced by an HP key on the MIE models and pressing this key takes you straight to the MIE home key. And i don’t know if it’s true or not, but on that forum, that guy said that the Dell, HP, etc have to pay to MS for that button! Ckck…

Anyway, you can find the spec here, and I hope it’s released here in Japan soon in January, with the same price of course :p

Songbird: Autodetect Library

•October 28, 2008 • Leave a Comment

Yeah, for the first time, Songbird’s not responding, as expected. But it was only that one time 😀 Don’t know what happened, I just listened to the shoutcast as usual (I really like listening to 181.FM Classic 60’s and 70’s 😀 ) then suddenly it’s not responding, maybe because the internet connection suddenly went down or i don’t know.

Anyway, it would be nice if Songbird has the feature to auto detect the library, so that when there’s a new song(s), it will be automatically added to the library. Or is it there already? It’s annoying to add manually every time I have new songs.