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This article presents some of the numerical problems on SNR.

**At the transmitter, the signal power is 23 mW. The input SNR is 40 dB. The channel offers 3 dB attenuation to the signal and the output noise is thrice the input noise level. Determine the SNR at the output.**

**Soln **− $SNR_{i/p}=\frac{S_{i/p}}{N_{i/p}}$

**Calculation Of Output Power Level**

An attenuation of 3 dB equals halving the input transmission power. If the ratio of two quantities on the linear scale is 1/2, it translates to -3 dB on the dB scale which is indicated as attenuation. So, the output signal power is 23mW/2 =11.5 mW.

**Calculation Of Input Noise Level**

The input SNR is 40 dB. This means that the input power level is 10000 times greater than the input noise level.

$$\frac{10000}{1}=10000;\:10log_{10}(\frac{10000}{1})=40dB$$

$$\frac{23mW}{N_{i/p}}=10000$$

The input noise level is 2.3 μW.

In the question, it is given that the output noise is thrice the input noise. Thus, the output

noise level is 2.3μW x 3 = 6.9 μW.

**Calculation Of Output SNR**

$$SNR_{o/p}=\frac{S_{o/p}}{N_{o/p}}$$

Output signal power = 11.5 mW

Output noise power = 6.9 μW.

The ratio of the output signal power to the output noise power gives the output SNR at the receiver.

$$SNR_{o/p}=\frac{11.5mW}{6.9\mu\:W}=1666.67$$

$$SNR_{o/p}=10log_{10}1666.67=32.22dB$$

**Inference**

The input SNR is 40 dB while the output SNR is 32.22 dB. Due to the channel noise, the output SNR has decreased by 8 dB. However, the signal power is still large enough than the noise power to have a faithful detection and decoding at the receiver.

**The initial SNR measured at the transmitter was 20 dB. In order to combat the channel conditions, the signal power was doubled prior to transmission. What is the new SNR at the transmitter?**

**Soln **− Initial SNR = 20 dB. Let SP denote the initial signal power and S_{p}’ denote the new signal power such that S_{p}’ = 2 S_{P}. Let N_{p} denote the noise power. Let us first find convert the initial SNR to absolute value.

$$SNR=10\:log_{10}(\frac{S_{p}}{N_{p}})$$

$$20dB=10\:log_{10}(\frac{S_{p}}{N_{p}})$$

$$log_{10}(\frac{S_{p}}{N_{p}})=2;\:(\frac{S_{p}}{N_{p}})=10^{2}=100$$

$$(\frac{S_{p}}{N_{p}})=100\Rightarrow\:S_{p}=100N_{p}$$

Let SNR’ denote the new SNR. We know that S_{p}’ = 2 S_{P}

$$SNR'=10\:log_{10}(\frac{S_{p'}}{N_{p}})=10\:log_{10}(\frac{2S_{p}}{N_{p}})$$

$$Since,S_{p}=100N_{p}$$

$$SNR'=10\:log_{10}(\frac{200N_{p}}{N_{p}})=10\:log_{10}200\sim\:23dB$$

Therefore, the new SNR is 23 dB.

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