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vodomira [7]
3 years ago
10

Write a assembler code that displays the first 27 values in the Fibonacci series. One version will use 16 bit registers (ax, bx,

cx, dx) and the the other version will use 32 bit registers (eax, ebx, ecx, edx). Use the authors' routine writeint to display the numbers. The 16 bit version will not display all the numbers correctly. The pseudo code is num1
Computers and Technology
1 answer:
tiny-mole [99]3 years ago
3 0

Explanation:

yeah yeah is that we go together on the watch remember write a timer called that depressed display of the first 27 value in the XM Sirius

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You have configured your firewall to authenticate a group of 100 users who are in your company. You set up the database of users
svetlana [45]

Answer:

Explanation:

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VLSI is the process of creating an integrated circuit (IC) by combing thousands of transistors into a single chip. There are two
malfutka [58]

Answer:

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ASIC in the other hand can be compact to a small size. ASIC are pretty much the IC chips that you use, like your Intel CPU or LM7000 series chips. However, the process of making them that small is irreversible, so if you messed up, you gotta throw away the whole batch. This make them expensive to make, but their small size and production process allows them to be made in bulk.

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7 0
4 years ago
a transmitter is operating at 150 MHz with a power of 3 W into a one-quarter wavelength vertical antenna. The receiver, which is
Bogdan [553]

The received power will be 1.243 nW

We're given:

frequency f = 150MHz

distance of the receiver d = 32.2 km=32200m

Power of transmitter P_{t} = 3W

Antenna gain = 8dB

To find :

Power received P_{r}

P_{r}= \frac{P_{t} *G_{t}*G_{r}* \lambda^2 }{16*\pi^2*d^2}

where G_{t is transmit gain and G_{r is receive gain as refrenced to isotropic source

⇒wavelength \lambda = \frac{c}{f} where c is the speed of the light

⇒  \lambda = \frac{3*10^8}{150*10^6} =2m

G_{t}= 1*1.64 ( value of dipole = 1.64)

Now,

Antenna gain=8dB ( in decibals)

⇒10log(x)=8

⇒x=10^0^.^8=6.3095

⇒ considering isotropic receiver

⇒G_{r}=6.3095*1.64=10.3477 (dipole =1.64)

Now , using the formula

P_{r}= \frac{3 *1.64*10.3477 *2^2 }{16*\pi^2*32200^2}=1.2437*10^-^9

Hence The received power will be 1.243 nW

Leaen more about communication devices here:

brainly.com/question/14530107

#SPJ10

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2 years ago
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