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mart [117]
2 years ago
14

A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the atte

nuation in decibels?
Computers and Technology
1 answer:
11Alexandr11 [23.1K]2 years ago
7 0

Answer:

Attenuation = 0.458\ db

Explanation:

Given

Power at point A = 100W

Power at point B = 90W

Required

Determine the attenuation in decibels

Attenuation is calculated using the following formula

Attenuation = 10Log_{10}\frac{P_s}{P_d}

Where P_s = Power\ Input and P_d  = Power\ output

P_s = 100W

P_d = 90W

Substitute these values in the given formula

Attenuation = 10Log_{10}\frac{P_s}{P_d}

Attenuation = 10Log_{10}\frac{100}{90}

Attenuation = 10 * 0.04575749056

Attenuation = 0.4575749056

Attenuation = 0.458\ db <em>(Approximated)</em>

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When this program is compiled and executed on an x86-64 Linux system, it prints the string 0x48\n and terminates normally, even
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Answer:

I get 0x55 and this the linking address of the main function.

use this function to see changes:

/* bar6.c */

#include <stdio.h>

char main1;

void p2()

{

printf("0x%X\n", main1);

}

Output is probably 0x0

you can use your original bar6.c with updaated foo.c

char main;

int main() // error because main is already declared

{

  p2();

   //printf("Main address is 0x%x\n",main);

  return 0;

}

Will give u an error

again

int main()

{

  char ch = main;

  p2(); //some value

  printf("Main address is 0x%x\n",main); //some 8 digit number not what printed in p2()

  printf("Char value is 0x%x\n",ch); //last two digit of previous line output

  return 0;

}

So the pain in P2() gets the linking address of the main function and it is different from address of the function main.

Now char main (uninitialized) in another compilation unit fools the compiler by memory-mapping a function pointer on a char directly, without any conversion: that's undefined behavior. Try char main=12; you'll get a multiply defined symbol main...

Explanation:

5 0
2 years ago
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