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Logarithms

Module by: Scott Starks. E-mail the author

Summary: This module is part of a collection of modules intended for use by preengineering students enrolled in MATH 108 at the University of Texas at El Paso. This module addresses some applications of logarithms in several fields of engineering. Examples are presented.

Logarithms

Introduction

This module is intended to present some areas of engineering in which logarithms are used. By reading the material and solving the associated problems, you will learn about some important applications of logarithms in engineering.

Decibels

The decibel (dB) is a logarithmic unit that indicates the ratio of a physical quantity relative to a specified or implied reference level. The decibel is used for a wide variety of measurements in science and engineering, most prominently in acoustics, electronics, communications, radar, sonar and control systems.

Decibels are frequently used as a means to express the power ratio for physical systems. It is computed by multiplying the factor 10 by the base 10 logarithm of the ratio of the quantities under consideration. Equation (1) shows the computation that is used to express the ratio of two powers using decibels

L DB = 10 log 10 P 2 P 1 L DB = 10 log 10 P 2 P 1 size 12{L rSub { size 8{ ital "DB"} } ="10"`"log" rSub { size 8{"10"} } left ( { {P rSub { size 8{2} } } over {P rSub { size 8{1} } } } right )} {}
(1)

Gain of an Amplifier: We will begin our discussion of decibels with an application in the field of electronics. An amplifier is an electronic device that is capable of boosting the power present in an input signal to produce an output signal with more power. It can be thought of as a black box as shown in Figure 1.

Figure 1: Block diagram of an amplifier.
Figure 1 (graphics1.jpg)

In practical cases, the ratio of the power in the output signal to the power in the input signal is a positive quantity whose value is greater than unity. The decibel measurement of this ratio of power is often called the gain of the amplifier and is given as

Gain = 10 log 10 P output P input dB Gain = 10 log 10 P output P input dB size 12{ ital "Gain"="10"`"log" rSub { size 8{"10"} } left ( { {P rSub { size 8{ ital "output"} } } over {P rSub { size 8{ ital "input"} } } } right )~ ital "dB"} {}
(2)

Question: An electronic signal is passed through an amplifier. Suppose that the power present in the signal at the input to the amplifier is 10 W. The power present in the signal at the output of the amplifier is 20 W. Express the gain of the amplifier in decibels.

We can use equation (2) to easily express the gain of the amplifier in terms of decibels

Gain = 10 log 10 20 W 10 W = 10 log 10 ( 2 ) = 3 . 01 dB 3 dB Gain = 10 log 10 20 W 10 W = 10 log 10 ( 2 ) = 3 . 01 dB 3 dB size 12{ ital "Gain"="10"`"log" rSub { size 8{"10"} } left ( { {"20"`W} over {"10"`W} } right )="10"`"log" rSub { size 8{"10"} } \( 2 \) =3 "." "01"` ital "dB" approx 3` ital "dB"} {}
(3)

Signal to Noise Ratio

Electrical signals are often corrupted by a random phenomenon known as noise when they are transmitted from one point to another . Because it is impossible to know the exact value of the noise at any point in time, it is often becomes difficult to extract the orignal signal at the receiver without the application of some form of signal processing algorithm such as a filter. The situation is depicted in Figure 2.

Figure 2: Communication system with signal processing.
Figure 2 (graphics2.jpg)

A common figure of merit of communication systems is the signal-to-noise ratio. Communication systems that are characterized by high signal-to-noise ratios are in general superior to those that are characterized by low signal-to-noise ratios.

By definition the signal-to-noise ratio or SNR is given as the ratio of the power in a signal divided by the power in the noise that is responsible for corrupting the signal. The signal-to-noise ratio can be expressed in decibels as follows

SNR = 10 log 10 P signal P noise dB 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"} {}
(4)

where Psignal and Pnoise 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 SNRin and that at its output as SNRout. For a good designs, SNRout will be significantly higher that that of SNRin. 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 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"} {}
(5)

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 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"} {}
(6)

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

Figure 3: Switched RC circuit.
Figure 3 (graphics3.jpg)

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 (V0) is 10 V.

At t =0, the switch is closed. Because the voltage across the capacitor cannot change instantaneously, the value of the voltage across the resistor immediately after the closing of the swith (v(0+) will be 10 V. After the switch has been closed, current will flow throughout the circuit and the voltage across the resistor will diminish exponentially. Because the decay can be described by the use of an exponential function, this is an example of exponential decay.

We can write an expression for the transient response, that is, the voltage across the resistor for positive time

v ( t ) = 10 e t / RC v ( t ) = 10 e t / RC size 12{v \( t \) ="10"`e rSup { size 8{ - t/ ital "RC"} } } {}
(7)

Here, the units of v(t) are Volts.

The product of the resistance and the capacitance is called the time constant for the circuit and is often denoted as (τ). The time constant is measured in seconds. For this example, τ=(100×103)(8×106)=800×103=0.800sτ=(100×103)(8×106)=800×103=0.800s size 12{τ= \( "100" times "10" rSup { size 8{3} } \) ` \( 8 times "10" rSup { size 8{ - 6} } \) ="800" times "10" rSup { size 8{ - 3} } =0 "." "800"`s} {}

So the transient response of the circuit becomes

Let us now work a problem about this circuit that involves logarithms.

Question: At what instant of time will the transent response be equal to 5 Volts?

Solution: We have been asked to find the value of t that satisfies the following equation.

5 = 10 e t / 0 . 8 5 = 10 e t / 0 . 8 size 12{5="10"`e rSup { size 8{ - t/0 "." 8} } } {}
(8)

Let us divide each side of the equation by (10) and interchange the sides

e t / 0 . 8 = 0 . 5 e t / 0 . 8 = 0 . 5 size 12{e rSup { size 8{ - t/0 "." 8} } =0 "." 5} {}
(9)

We now take the natural logarithm of each side

t 0 . 8 = 0 . 693 t 0 . 8 = 0 . 693 size 12{ { { - t} over {0 "." 8} } = - 0 "." "693"} {}
(10)
t = ( 0 . 8 ) ( 0 . 693 ) = 0 . 555 t = ( 0 . 8 ) ( 0 . 693 ) = 0 . 555 size 12{t= \( 0 "." 8 \) ` \( 0 "." "693" \) =0 "." "555"} {}
(11)

So we conclude that at the time 0.555 seconds after the switch closes, the value of the voltage across the resistor will be 5.0 volts.

Compound Interest

It is always essential to consider the financial aspects of an engineering project. The field of industrial engineering provides us with the analytical tools necessary to weigh the merits of competing designs.

Most situations surrounding the financial aspects of an engineering project involve the determination of what is most economical in the long run. That is, engineers must be aware of the costs and benefits of a project over a considerable period of time. In situations as these, it is important to consider the time value of money. Because of the existence of interest, the current value of a dollar is worth more than the value of a dollar some time in the future.

Interest can be defined as money that is paid for the use of money that has been borrowed. The rate of interest is the ratio between the interest chargeable or payable at the end of a period of time. This period of time is typically yearly, quarterly or monthly. In this module, we will restrict our attention to interest that is paid yearly or per annum.

As an example, a sum of money is invested at an annual rate of interest of 4%. One year later, the interest that would be paid on the investment would be $40 (4% of $1,000). So after one year, the initial sum would grow to a value of $1,040 one year later.

Suppose that the $1,040 were invested for a second year at the end of the first year. At the end of the second year, the amount of interest that would be payable would be 4% of $1,040 or $41.60. The amount of interest earned in the second year exceeds the amount earned in the first year because of a phenomenon known as compound interest.

Interest calculations can be quantified by mathematical formulas. Suppose that we are concerned with making an investment of P dollars at an annual rate of interest of i for n years. Here P denotes the present value of the investment, n represents the number of years the money is to be invested, and i is the interest rate per annum. Let us denote by F the future value of the investment at the end of n years. The value of F can be calculated via the formula

F = P ( 1 + i ) n F = P ( 1 + i ) n size 12{F=P` \( 1+i \) rSup { size 8{n} } } {}
(12)

Let us illustrate the use of this formula by means of a problem.

Question: A couple has just had a baby. The couple wishes to make an investment in an account that will be used to fund college expenses. The couple visits an investment advisor and establishes a tax-free college savings account. By tax-free, we mean an account where income tax is not paid yearly. The couple is advised that this feature will enable the account to grow faster.

The couple deposits $25,000 in the account. The couple is told that the value of the account will appreciate at an annual rate of 8% . What will be the value of the account when the child turns 18 years of age?

Solution: We are asked to find the value of F. From the problem statement, P is $25,000, i is 0.08 and n is 18. We substitute into the interest formula to obtain the result

F = ( 25 , 000 ) ( 1 + 0 . 08 ) 18 = ( 25 , 000 ) ( 1 . 08 ) 18 = ( 25 , 000 ) ( 3 . 996 ) = 99 , 900 F = ( 25 , 000 ) ( 1 + 0 . 08 ) 18 = ( 25 , 000 ) ( 1 . 08 ) 18 = ( 25 , 000 ) ( 3 . 996 ) = 99 , 900 size 12{F= \( "25","000" \) ` \( 1+0 "." "08" \) rSup { size 8{"18"} } = \( "25","000" \) ` \( 1 "." "08" \) rSup { size 8{"18"} } = \( "25","000" \) ` \( 3 "." "996" \) ="99","900"} {}
(13)

We conclude that the account will be worth $99,900 at the end of 18 years.

Let us now a related problem that involves the use of logarithms.

Question: An individual inherits $10,000 from a relative. The individual wishes to invest the sum in a tax-free account. He/she plans to use the proceeds of this account as a down payment on a future purchase of a home. The annual interest rate of the account is 6%,

The individual anticipates that he/she will need at least $20,000 for the down payment. How long will it take for the value of the account to grow to $20,000?

Solution: Once again, we begin with the identification of the parameters of the problem. Here, P is 10,000, i is 0.06, and F is 20,000. We incorporate these values into the interest formula

20 , 000 = ( 10 , 000 ) ( 1 + 0 . 06 ) n 20 , 000 = ( 10 , 000 ) ( 1 + 0 . 06 ) n size 12{"20","000"= \( "10","000" \) ` \( 1+0 "." "06" \) rSup { size 8{n} } } {}
(14)

Let us divide each side by (10,000) and re-arrange terms

( 1 . 06 ) n = 2 ( 1 . 06 ) n = 2 size 12{ \( 1 "." "06" \) rSup { size 8{n} } =2} {}
(15)

Now let us take the logarithm of each side of the equation. We will use 1.06 as the base of the logarithm

n = log 1 . 06 ( 2 ) n = log 1 . 06 ( 2 ) size 12{n="log" rSub { size 8{1 "." "06"} } \( 2 \) } {}
(16)

The base 1.06 logarithm is related to the base 10 logarithm as follows

log 1 . 06 ( x ) = log 10 ( x ) log 10 ( 1 . 06 ) log 1 . 06 ( x ) = log 10 ( x ) log 10 ( 1 . 06 ) size 12{"log" rSub { size 8{1 "." "06"} } \( x \) = { {"log" rSub { size 8{"10"} } \( x \) } over {"log" rSub { size 8{"10"} } \( 1 "." "06" \) } } } {}
(17)

We will use this relationship to help us find the value for n

n = log 10 ( 2 ) log 10 ( 1 . 06 ) = 0 . 301 0 . 0253 = 11 . 90 n = log 10 ( 2 ) log 10 ( 1 . 06 ) = 0 . 301 0 . 0253 = 11 . 90 size 12{n= { {"log" rSub { size 8{"10"} } \( 2 \) } over {"log" rSub { size 8{"10"} } \( 1 "." "06" \) } } = { {0 "." "301"} over {0 "." "0253"} } ="11" "." "90"} {}
(18)

So the individual should plan on waiting 11.90 years for the account to grow to a value of $20,000.

Problems such as the previous one often make use of the relationship

log a ( x ) = log b ( x ) log b ( a ) log a ( x ) = log b ( x ) log b ( a ) size 12{"log" rSub { size 8{a} } \( x \) = { {"log" rSub { size 8{b} } \( x \) } over {"log" rSub { size 8{b} } \( a \) } } } {}
(19)

This relationship allows one to convert from the logarithm of any base to a logarithm of another base. Typically, scientific calculators are only able to compute base 10 and natural (base e) logarithms. One should become acquainted with the conversion of logarithms of other bases to base 10 or base e logarithms.

Exercises

  1. The power of the signal entering an amplifier is 15 mW. The power of the signal that leaves the amplifier is 25 W. Express the gain of the amplifier in decibels.
  2. Unlike an amplifier, some devices reduce the power of input signals. This process is called attenuation. Suppose that the power that enters an attenuator is 60 W and the power at the output is 0.9 W. Express the gain of the attenuator in decibels.
  3. Consider a signal processing scheme such as that shown in Figure 2. The power of the input signal before it passes through the noisy channel is 40 W. After passing through the noisy channel, the original signal is corrupted by noise. The noise component has a power of 10 W. What is the Signal to Noise ratio of the signal that emerges from the noisy channel?
  4. Consider the situation described in exercise 3. The noisy signal enters a signal processor. The signal processor diminishes the noise power of the signal by 5%, while diminishing the power in the noise component by 95%, What is the SNR of the output of the signal processor?
  5. Consider the circuit shown in Figure 3. Let us replace the resistor with another whose value is 200 kΩ. (a) What is the new value of the time constant of the circuit. (b) Find the value of the transient response when t = 0.8 seconds. (c) Determine the value of time at which the transient response decays to a value of 1 Volt.

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