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SNR = 10 log 10 P signal P noise dB size 12{ ital "SNR"="10"`"log" rSub { size 8{"10"} } left ( { {P rSub { size 8{ ital "signal"} } } over {P rSub { size 8{ ital "noise"} } } } right )~ ital "dB"} {}

where P signal and P noise represent the values for the signal power and noise power respectively.

At the receiver, the original signal arrives with an added noise component. Communication engineers use signal processing algorithms in the form of filters to minimize the power of the noise that is present in the received noisy signal while seeking to retain as much of the power of the original signal as possible. Such algorithms are based on the premise that although random, noise may well fall into different regions of angular frequency in the electromagnetic spectrum than the signal itself. Thus, frequency selective filters can be utilized to increase the signal-to-noise ratio.

Suppose that we represent the signal-to-noise ratio at the input to the signal processing algorithm as SNR in and that at its output as SNR out . For a good designs, SNR out will be significantly higher that that of SNR in . This increase in signal-to-noise ratio results in superior performance and more reliable communication of information.

Question: A signal with power 30 W is sent through a noisy communication channel. The noise introduced by the communication channel has an associated power of 5 W. What is the signal-to-noise ratio of the noisy signal that is received from the communication channel in decibels?

Solution: We can find the SNR by substituting the appropriate values for the parameters into equation (4)

SNR = 10 log 10 P signal P noise = 10 log 10 30 W 5 W = 10 log 10 ( 6 ) = 7 . 78 dB size 12{ ital "SNR"="10"`"log" rSub { size 8{"10"} } left ( { {P rSub { size 8{ ital "signal"} } } over {P rSub { size 8{ ital "noise"} } } } right )="10"`"log" rSub { size 8{"10"} } left ( { {"30"~W} over {5~W} } right )="10"`"log" rSub { size 8{"10"} } \( 6 \) =7 "." "78"~ ital "dB"} {}

Question: Consider the noisy signal described in Example 2. Suppose that a signal processing filter is applied to it. The filter reduces the noise power by 95% while it reduces the signal power by only 10%. What is the signal-to-noise ratio of the filtered output?

Solution: The amount of signal power present in the filtered output is 30 W – 0.1 (30 W) or 27 W. The amount of noise power present in the filtered output is 5 W – 0.95 (5 W) or 0.25 W.

The signal-to-noise ratio of the filtered output is

SNR = 10 log 10 27 W 0 . 25 W = 10 log 10 ( 108 ) = 10 2 . 03 = 20 . 33 dB size 12{ ital "SNR"="10"`"log" rSub { size 8{"10"} } left ( { {"27"~W} over {0 "." "25"~W} } right )="10"`"log" rSub { size 8{"10"} } \( "108" \) ="10" cdot 2 "." "03"="20" "." "33"~ ital "dB"} {}

We see that the primary result obtained by the application of the signal processing filter is to raise the signal to noise ratio from 7.78 dB to 20.33 dB . This represents an increase in SNR of 20.33 – 7.78 dB = 12.55 dB . Along with this increase in SNR, one would observe an increase in reliable communication of information.

Transient response (exponential decay)

The transient response of a circuit is response of a circuit that is expressed as a function of time. Determining the transient response of a circuits is important in that knowledge of the transient response reveals how a circuit will behave whenever perturbations arise, such as the opening or closing of a switch.

Let us examine the RC circuit present in the figure below

Switched RC circuit.

This circuit is called an RC circuit in that it contains a resistor and a capacitor.

The switch is open for values of time ( t ) less than 0. Thus for negative values of time there is no current flow. Because there is no current flow, the voltage across the resistor ( v ( t )) is zero. The capacitor is initially charged. The initial voltage across the capacitor (V 0 ) is 10 V.

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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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