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**For a communication link, the received power level is measured as 5 dBm. What does this mean?**

Here, we have a positive ‘dBm’ and hence we can say that the received power level is greater than the reference power level which is 1 mW.

$$5dBm=10log_{10}(\frac{P_{received}}{1mW})$$

$$log_{10}(\frac{P_{received}}{1mW})=\frac{5}{10}=0.5$$

$$(\frac{P_{received}}{1mW})=10^{0.5}=3.1622$$

$$P_{received}=3.1622mW$$

We can infer that the received power level is 3.1622 times greater than the reference power level. However, this doesn’t mean that the received power level is greater than the input or transmitted power level. In fact, the received power level can never be greater than the transmitter power. In practice, the received power level in mobile communications is several orders below the transmitted power level, and hence the received power level is often measured in dBm, dBµW or even dBnW at times. Similarly, 10 dBµV indicates how high the magnitude of received voltage when compared with the reference of 1 microvolt.

**Let us consider a communication link intended for long-distance communications. Let the initial transmission power be 5 kW. Since this link would support communications to long distances, the transmission power should be quite high to combat the effects of fading and that the received power level is above the noise floor. What will happen if we increase the transmission power by 3 dB and 10 dB?**

$$3dB=10log_{10}(\frac{P_{t\:new}}{5kW})$$

$$log_{10}(\frac{P_{t\:new}}{5kW})=0.3$$

$$(\frac{P_{t\:new}}{5kW})=10^{0.3}=2;\:P_{t\:new}=10kW$$

Thus,

$$5kW+3dB\equiv10kW$$

This can also be written as

$$37dBW+3dB\equiv40dBW$$

The reference power is taken as 1 W.

$$37dBW=10log_{10}(\frac{5kW}{1W})=10log_{10}5000$$

Therefore, we can observe that a 3 dB increment in the transmission power yields double the initially available transmission power.

Similarly, for a 10 dB increase, we have a ten-fold increase in transmit power.

**Find ‘X’ that satisfies the above equation.**

$$5kW+X(dB)=10kW$$

**Solution **− Here, the input power is 5 kW. It is increased by certain ‘dB’ to make it 10 kW and we have to find out the ‘X’. Let us use the formula − We first take the ratio of the two quantities i.e., power ratio

$$10log_{10}(\frac{10kW}{5kW})=X(dB)\sim\:3dB$$

**What value of ‘R’ satisfies the equation below?**

$$10dB\mu\:V-R(dB)=1dB\mu\:V$$

**Solution **− We can directly say that R equals 9 dB as we know that we can perform subtraction between the quantities dBµV and dB. Note that the output voltage is less than that of input and this might possibly be due to attenuation offered by the transmission medium.

**A transmission medium offers 10 dB attenuation. If the input power is 40 W, what will be the output power in dBm?**

**Solution **− Input power is 40 W. Attenuation is 10 dB. Now, we frame the equation −

$$40W-10dB=?$$

Since it is asked to express the output power in dBm, let us convert 40 W to dBm and proceed with the calculation.

$$40W\equiv10log_{10}(\frac{40W}{1mW})=46.02dBm$$

$$46.02dBm-10dB=36.02dBm$$

Let us convert 36.02 dBm back to linear scale. We get 4 W. Thus, we can observe that the output power is one-tenth of the input power. We can this observe that a 10 dB attenuation in input power gives us an output power that is one-tenth of the input power.

$$36.02dBm=10log_{10}(\frac{P_{out,mW}}{1mW})$$

$$log_{10}(\frac{P_{out,mW}}{1mW})=3.602$$

$$({P_{out,mW}})=10^{3.602}\approx4000mW=4W$$

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