A 10W signal is received at 5W. This is a ____________dB loss.

A 10W signal is received at 5W. This is a ____________dB loss.

• A. 1
• B. 3
• C. 5
• D. 10

Answer: (B)

To determine the decibel loss when a signal power changes, we can use the formula for power in decibels, which is given by: \[ dB = 10 \times \log_{10}\left(\frac{P_{received}}{P_{transmitted}}\right) \] In this scenario, the transmitted power is 10W and the received power is 5W. Plugging these values into the formula gives: \[ dB = 10 \times \log_{10}\left(\frac{5W}{10W}\right) \] \[ dB = 10 \times \log_{10}(0.5) \] Now, the logarithm of 0.5 can be calculated. Knowing that \(\log_{10}(0.5) \approx -0.301\): \[ dB = 10 \times (-0.301) \] \[ dB = -3.01 \] Rounding this to the nearest whole number gives a loss of approximately -3 dB. Since we are looking for the absolute value of the loss, it translates to a loss of 3 dB. Thus, the statement pertaining to a 3 dB loss is accurate,

To determine the decibel loss when a signal power changes, we can use the formula for power in decibels, which is given by:

[ dB = 10 \times \log_{10}\left(\frac{P_{received}}{P_{transmitted}}\right) ]

In this scenario, the transmitted power is 10W and the received power is 5W. Plugging these values into the formula gives:

[ dB = 10 \times \log_{10}\left(\frac{5W}{10W}\right) ]

[ dB = 10 \times \log_{10}(0.5) ]

Now, the logarithm of 0.5 can be calculated. Knowing that (\log_{10}(0.5) \approx -0.301):

[ dB = 10 \times (-0.301) ]

[ dB = -3.01 ]

Rounding this to the nearest whole number gives a loss of approximately -3 dB. Since we are looking for the absolute value of the loss, it translates to a loss of 3 dB.

Thus, the statement pertaining to a 3 dB loss is accurate,