CONCEPTS AND APPLICATIONS OF

LOGARITHMIC UNITSBY- B. B. KHANNA

The muscles of the iris can contract or dilate the pupils to adjust the amount of light coming into our eyes. In an analogous way, the middle ear has a mechanism which can adjust the intensity of sound waves striking our eardrums. This adjustment enables us to discriminate very small changes in the intensity of quiet sounds, but to be much less sensitive to volume changes in louder noises. This means that the human ear can safely hear a huge range of very soft to very loud sounds.

Our ears can compress sound waves

If you hear a sound of a certain loudness, and then are asked to choose a sound that is twice as loud as the first sound, the sound you choose will in fact be about ten times the intensity of the first sound. For this reason, a logarithmic scale, one that goes up by powers of ten, is used to measure the loudness of a sound. The exponent of a number (here we use only 10) is its logarithm. Example of a base 10 logarithm:

10 x 10 x 10 x 10 = 10,000 = 104

log10 10,000 = log 10,000 = 4

Logarithms and the decibel scale

Decibel (dB), is a specific function that operates on a unitless parameter

Why unitless parameter? Ans. Many values are unitless, such as ratios and

coefficients

What is a decibel?

dB= 10 log10 (x)where x is unitless

For example, amplifier gain is unitless value!e.g. amplifier gain is the ratio of the output power to the input power:

So, Gain in dB= 10 log10 G= G (dB)

The unit used to compare the intensity of sounds was originally the Bel (in commemoration of the work of Alexander Graham Bell), which was the logarithm of the intensity ratio 10:1. This unit was considered too large to be useful, so a unit one tenth the size of a Bel, the ‘decibel’ (dB), was adopted.

Bels and Decibels

Value of dB indicates:

1) +ve dB = P2 > P1

2) -ve dB= P2 < P1

3) 0 dB= P2 = P1

Thus, dB is used to indicate gain or loss in a system like amplifier or attenuator,

respectively

The power dissipated in a circuit is given by

The value in dB can be calculated as

Voltage or current ratios

10 log10 (10) = 10 dB

so, amplifier with gain G=10 will be said to have a gain of 10 dB

G= 10n G (dB) = 10n

10 log10 (2n) = n 10 log10 (2) ≈ 3n

Standard dB values

STANDARD REFERENCE LEVELS

USED INBROADCASTING

i.e., power is normalized w.r.t 1 W of power used when amount of power involved is

high

e.g. if P= 100 W Then, P(dBw) = +20 dBw Or P= 1mW Then P (dBw)= -30 dBw

dBw

EIRP= Effective Isotropic Radiated PowerLet a 10 kW power of a transmitter is

connected to a cable of loss 3 dB and an antenna of gain 10 dB. The power output from the antenna to space is calculated as:

Input power to cable: 10 kWLoss in cable: 3 dBOutput from the cable: 5 kWGain of antenna =10 dBOutput from antenna or EIRP = 50 kW

Example: EIRP of transmitter used in satellite communication is expressed in dBw

i.e., power in dB w.r.t 1 W of power which is 0.774 volt across 600 ohms

e.g. if P= 100 Watts P (dBm) = +50 dBm If P= 1 mW P(dBm) = 0 dBm

dBm

0.7746 volts is reference level (need not be measured across 600 ohms)

In AIR and DD : used in Meltron/ Keltron Audio Consoles

+ 8 dBu = 1.946 volt or 8 dB above reference level

The output of monitoring amplifier= 8 watts = +39 dBu

dBu

When reference level is taken as 1 µV

dBµv

When reference level is 1 µv/m Unit is used in field strength measurements

dBµv/m

When 1 kW is used as reference level Used in high power calculation To convert power in watts to dBk,

dBk= 10 (log P - 3) To convert from dBk to watts,

P(watts)= antilog (dBk/10 + 3)

dBk

The VU is the conventional unit for measurement of speech level. A VU can be related to a dBm only with a sinusoidal tone (a simple tone of one frequency) between 35 Hz and 10,000 Hz. The following relationship will be helpful:

Power level in dBm = VU − 1.4 dB (for complex audio signals).

One might ask: If the level reading on a broadcaster’s program channel is −11 VU, what would the equivalent be in dBm? Reading in VU − 1.4 dB = reading in dBm. Thus the answer is −11 VU − 1.4 dB = −12.4 dBm.

Volume Unit (VU)

10log10 [x y ] = 10log10 [x ]+10log10 [y ] 10log10 [x/y] =10log10 [x ] - 10log10 [y ]

Pout = G Pin

Multiplicative products and decibels

EXAMPLES

Find gain of amplifier which delivers output power of 5 watts to its load when given an input power of 5 milliwatts

Question 1

Solution 1

The power supplied to a sound programme line is 6 mW. The power at the receiving end is 6 mW.

What is line attenuation? Log 2= 0.3010 Log 6= 0.7781

Question 2

Solution 2

The input and output impedance of an amplifier are equal. A tone signal of 0.1 V produces an output voltage of 50 V. What is gain of amplifier in dB

Log 5= 0.698

Question 3

Solution 3

In TVRO, the signal received at the parabolic dish antenna is about -120 dBm. Find out the equivalent power.

Question 4

Solution 4

Power P1= 6 dBm is combined with power P2= 10 dBm

What is resulting total power?

QUESTION 5

P1= 6 dBm

=4 mW P2= 10 dBm

= 10 mW PT= P1 + P2

= 4 mW + 10 mW = 14 mW

PT(dBm)= 11.46 dBm

Solution 5