At a particular test point in a radio receiver, the signal power is 24 milliwatt and the noise is 6.7 microwatt. The interface impedance is 50 ohm. What is the Signal-to-Noise ratio in dB? (Round off your answer to the first decimal point)
Question
At a particular test point in a radio receiver, the signal power is 24 milliwatt and the noise is 6.7 microwatt. The interface impedance is 50 ohm. What is the Signal-to-Noise ratio in dB? (Round off your answer to the first decimal point)
Solution
To calculate the Signal-to-Noise ratio in dB, we first need to find the ratio of the signal power to the noise power. Then, we convert this ratio to decibels using the formula 10 * log10(S/N).
Step 1: Convert the powers to the same units. The signal power is given as 24 milliwatts and the noise power is given as 6.7 microwatts. To compare them directly, we need to convert them to the same units.
1 milliwatt = 1000 microwatts, so the signal power is 24 * 1000 = 24000 microwatts.
Step 2: Calculate the Signal-to-Noise ratio. The Signal-to-Noise ratio (S/N) is the ratio of the signal power to the noise power. So, S/N = 24000 microwatts / 6.7 microwatts = 3582.09.
Step 3: Convert the Signal-to-Noise ratio to decibels. To convert the Signal-to-Noise ratio to decibels, we use the formula 10 * log10(S/N). So, the Signal-to-Noise ratio in dB is 10 * log10(3582.09) = 35.5 dB (rounded to the first decimal point).
So, the Signal-to-Noise ratio at this test point in the radio receiver is 35.5 dB.
Similar Questions
Question 13 :If the received signal strength is -55 dBm and the noise floor is -95 dBm, then the signal-to-noise ratio will be ______ dB.This type of question contains radio buttons and checkboxes for selection of options. Use Tab for navigation and Enter or space to select the option.optionA75 optionB20 optionC40 optionD150
Suppose we have cable carrying a signal accompanied by noise, for which we know the SNR in dB. As a result of improvements in cable shielding and electronic design, we manage to reduce the noise power (in Watts) by 50%. By how much does the SNR change? (Give the number in dB only - do not include your calculation and do not include units.)
For an amplifier with an output signal voltage of 4V, an output noisevoltage 0.005 V and an input and output resistance of 50 , determine the signal-to-noise power ratio
Ex 5 : For a non-ideal amplifier with a following parameters, determinea. input S/N ratio (dB)b. output S/N ratio (dB)c. noise factor and noise figureInput signal power = 2 x 10 -10 WInput noise power = 2 x 10 -18 WPower gain = 1000000Internal noise N d = 6 x 10 -12 W
For an amplifier operating at 27 °C with.a bandwidth of 20KHz.determine a) thermal noise power in watts and dBm, b) Calculatenoise power in watts if the bandwidth is doubled, c) RMS noisevoltage for 50 ohm load resistance and 50 ohm internal resistance,d) noise factor for an equivalent noise temperature of 75 K and e)noise figure in dB.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.