Question

The output of a UHF modulator is -12 dBm. What will be the output power when a 20 dB amplifier is placed at the output of this modulator. (Hint: dBm or dBmw (decibel-milliwatts) is a unit of level used to indicate that a power level is expressed in decibels (dB) with reference to one milliwatt (mW).

Answer 1

When a 20 dB amplifier is placed at the output of the modulator, the output power will be approximately +13.0 dBm.

Step-by-step explanation:

To calculate the output power when a 20 dB amplifier is placed at the output of the modulator, you need to add the gain of the amplifier to the initial power level. Here's how you can do it step by step:

Convert the initial power level from dBm to milliwatts (mW):

The formula to convert dBm to milliwatts is: P(mW) = 10^(dBm/10)

where P(mW) is the power in milliwatts and dBm is the power level in decibels relative to one milliwatt.

Given the initial power is -12 dBm, let's calculate it in milliwatts:

P(mW) = 10^(-12/10) ≈ 0.251 mW (rounded to three decimal places).

Calculate the power after amplification:

The amplifier provides a 20 dB gain, which means the output power will be 20 dB higher than the input power.

P_out(mW) = P_in(mW) * 10^(Gain/10)

where P_out(mW) is the output power in milliwatts, P_in(mW) is the input power in milliwatts, and Gain is the amplifier gain in dB.

P_out(mW) = 0.251 mW * 10^(20/10) ≈ 0.251 mW * 100 ≈ 25.1 mW (rounded to one decimal place).

Convert the output power back to dBm:

To express the output power in dBm, you need to use the reverse formula:

P(dBm) = 10 * log10(P(mW))

where P(dBm) is the power in dBm and P(mW) is the power in milliwatts.

P(dBm) = 10 * log10(25.1 mW) ≈ 13.0 dBm (rounded to one decimal place).

So, when a 20 dB amplifier is placed at the output of the modulator, the output power will be approximately +13.0 dBm.