Found this diagram somewhere on the Internet. As it is said on the top, it is permitted to repost here. Very true and interesting!
Monday, February 25, 2013
Microsoft Empire
Found this diagram somewhere on the Internet. As it is said on the top, it is permitted to repost here. Very true and interesting!
Monday, February 18, 2013
Wednesday, February 13, 2013
New workbench for SMT works
My workbench (modification is still ongoing...waiting for the SMT solder tool to come. My new multi-meter is not yet shown there).
Oscilloscope: 100 MHz Tektronix 2232 (spec) (bought it from eBay)
Digital Analyzer (bottom left): Tektronix 1241 (bought it from eBay)
Multi-meter: UT61C (with RS232C connection), and two others (older ones, not seen)
SMT Soldering station: X-Tronic 9020 XTS (not-seen). Bought it from Amazon
Solder: Weller-SPG40 (on the bench) and SPG-100 (not seen as it is behind the lamp)
Power-supply: CircuitSpecialists CS11802X 0-18 V, 0-2A variable power supply
Vise: PANAVise
Small breadboard: TI msp430 LauchPad connected with my seven-segment circuit on breadboard
Board in the middle: PIC887A development board, measuring temperature and displaying it on the 16x2 LCD as well as sending the temperature to RS232.
Big breadboard: the old prototype of the same function as the PIC protoboard above.
Wishlist:
Digital Storage Oscilloscope with 100 MHz bandwidth
Higher frequency measurement for multi-tester (the UT61C freq. range is only 0-10 MHz, while UT61E has 0-220 MHz range)
Microscope 25-95x (this one looks cool for my bench, except I may not have space left)
Monday, January 28, 2013
Beckman 792 *** 661-1-R1K
It's a resistor array, each contains 1k resistor. The connection is:
Pin 16: common
Pin 1-15: individual resistor, each 1 kilo ohm.
Pin 16: common
Pin 1-15: individual resistor, each 1 kilo ohm.
Saturday, January 19, 2013
Roku 2 XS Internals
List of the chips inside:
- BCM 59002 From Broadcom: Mobile Power Management Unit
- SMC LAN 9512: Hi Speed USB 2.0 Hub and Ethernet controller
- AKM 4430 ET: 3.3volt 2 bit Stereo Audio digital audio converter
- BCM 4336 (inside the metal shield): Wireless LAN controller/transceiver
- BCM 20702 (inside the metal shield): Bluetooth controller
- TI 2051B: Power distribution switch
- Samsung K4P2G324EC: LPDDR2 RAM 2-Gigabit (512 MByte)
- Hynix HY27UF082G2B NAND Flash (2Gbit - 256MB)
The CPU inside the SoC chip (BCM-4336), according to Broadcom website, is based on ARM Cortex-M3 architecture. The chip integrates 802.11 a/b/g/n Wifi transceiver.
Another interesting site, but for Roku 2 XS: http://www.techrepublic.com/photos/cracking-open-the-roku-2-xs/6274959?seq=20&tag=thumbnail-view-selector;get-photo-roto
Friday, January 18, 2013
A Google interview
Question:
There are some data represented by(x,y,z). Now we want to find the Kth least data. We say (x1, y1, z1) > (x2, y2, z2) when value(x1, y1, z1) > value(x2, y2, z2) where value(x,y,z) = (2^x)*(3^y)*(5^z).
Now we cannot get it by calculating value(x,y,z) or through other indirect calculations as lg(value(x,y,z)). How to solve it?
Answer:
Say we have a set, H, which has these 3-tuple data.
Value(a) = 2x3y5z
if Value(H[m]) > Value(H[n]):
H[m] > H(n)
Test:
Assume h1 = (x1,y1,z1) = (0,0,0)
Value(h1) = 20 * 30 * 50 = 1*1*1 = 1
(x2,y2,z2) = (-1,-1,-1)
Value(h2) = 2-1* 3-1* 5-1 = 0.5 * 0.33 * 0.2, a value greater than 0 but less than 1
From the above, Value(h1) > Value(h2), so according to the statement above h1 should be > h2, or (0,0,0) > (-1,-1,-1).
One quick guess is that if z1 > z2, the Value(x1,y1,z1) would be greater than Value(x2,y2,z2), for z=[some number to infinite], because the 5^z would be more significant than the contribution of 2^x*3^y alone.
There are some data represented by(x,y,z). Now we want to find the Kth least data. We say (x1, y1, z1) > (x2, y2, z2) when value(x1, y1, z1) > value(x2, y2, z2) where value(x,y,z) = (2^x)*(3^y)*(5^z).
Now we cannot get it by calculating value(x,y,z) or through other indirect calculations as lg(value(x,y,z)). How to solve it?
Answer:
Say we have a set, H, which has these 3-tuple data.
Value(a) = 2x3y5z
if Value(H[m]) > Value(H[n]):
H[m] > H(n)
Test:
Assume h1 = (x1,y1,z1) = (0,0,0)
Value(h1) = 20 * 30 * 50 = 1*1*1 = 1
(x2,y2,z2) = (-1,-1,-1)
Value(h2) = 2-1* 3-1* 5-1 = 0.5 * 0.33 * 0.2, a value greater than 0 but less than 1
From the above, Value(h1) > Value(h2), so according to the statement above h1 should be > h2, or (0,0,0) > (-1,-1,-1).
One quick guess is that if z1 > z2, the Value(x1,y1,z1) would be greater than Value(x2,y2,z2), for z=[some number to infinite], because the 5^z would be more significant than the contribution of 2^x*3^y alone.
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