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KU EECS 622 - dB, dBm, dBw

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2/15/2005 dB.doc 1/9 Jim Stiles The Univ. of Kansas Dept. of EECS For example, amplifier gain is a unitless value! E.G., amplifier gain is the ratio of the output power to the input power: outinPGP= ()10 Gain in dB 10 logGGdB∴=  dB, dBm, dBw Decibel (dB), is a specific function that operates on a unitless parameter: Q: A unitless parameter! What good is that ! ? A: Many values are unitless, such as ratios and coefficients. 10dB 10 log (x) where x is unitless!2/15/2005 dB.doc 2/9 Jim Stiles The Univ. of Kansas Dept. of EECS Q: Wait a minute! I’ve seen statements such as: Of course, Power is not a unitless parameter!?! A: True! But look at how power is expressed; not in dB, but in dBm or dBw. Q: What the heck does dBm or dBw refer to ?? A: It’s sort of a trick ! Say we have some power P. Now say we divide this value P by one 1 Watt. The result is a unitless value that expresses the value of P in relation to 1.0 Watt of power. For example, if 2500PmW= , then 125PW .=. This simply means that power P is 2.5 times larger than one Watt! Since the value 1PW is unitless, we can express this value in decibels! …. the output power is 5 dBw …. or …. the input power is 17 dBm ….2/15/2005 dB.doc 3/9 Jim Stiles The Univ. of Kansas Dept. of EECS Specifically, we define this operation as: For example, 100P= Watts can alternatively be expressed as ()20PdBw dBw=+ . Likewise, 1PmW= can be expressed as ()30PdBw dBw=− . Q: OK, so what does dBm mean? A: This notation simply means that we have normalized some power P to one Milliwatt (i.e., 1PmW)—as opposed to one Watt. Therefore: For example, 100P= Watts can alternatively be expressed as ()50PdBm dBm=+ . Likewise, 1PmW= can be expressed as ()0PdBm dBm= . Make sure you are very careful when doing math with decibels! ()10PPdBw 10 log1 W⎛⎞⎜⎟⎝⎠ ()10PPdBm 10 log1 mW⎛⎞⎜⎟⎝⎠2/15/2005 dB.doc 4/9 Jim Stiles The Univ. of Kansas Dept. of EECS Standard dB Values Note that ()1010 10 10log dB= Therefore an amplifier with a gain G = 10 is likewise said to have a gain of 10 dB. Now consider an amplifier with a gain of 20 dB…… A: NO! Do not make this mistake! Recall from your knowledge of logarithms that: []10 1010log 10 10log 10 10 nnn⎡⎤==⎣⎦ Q: Yes, yes, I know. A 20 dB amplifier has gain G=20, a 30 dB amp has G=30, and so forth. Please speed this lecture up and quit wasting my valuable time making such obvious statements!2/15/2005 dB.doc 5/9 Jim Stiles The Univ. of Kansas Dept. of EECS Therefore, if we express gain as 10nG=, we conclude: ()10 10nGGdBn=↔ = In other words, G =100 = 102 (n =2) is expressed as 20 dB, while 30 dB (n =3) indicates G = 1000 = 103. Likewise 100 mW is denoted as 20 dBm, and 1000 Watts is denoted as 30 dBW. Note also that 0.001 mW = 10-3 mW is denoted as –30 dBm. Another important relationship to keep in mind when using decibels is []1010log 2 3.0≈ . This means that: []10 1010log 2 10log 2 3 nnn⎡⎤=⎣⎦ Therefore, if we express gain as 2nG=, we conclude: ()2 3nGGdBn=↔ As a result, a 15 dB (n =5) gain amplifier has G = 25 = 32. Similarly, 1/8 = 2-3 mW (n =-3) is denoted as –9 dBm.2/15/2005 dB.doc 6/9 Jim Stiles The Univ. of Kansas Dept. of EECS Multiplicative Products and Decibels Other logarithmic relationship that we will find useful are: [][][]10 10 1010log 10log 10logxy x y=+ and its close cousin: [][]10 10 1010log 10log 10logxxyy⎡⎤= −⎢⎥⎢⎥⎣⎦ Thus, the relationship Pout = G Pin is written in decibels as: []10 1010 10 101110log 10log1110log()() ()10log 10log11out ininoutinoutinouout outtPGPGPPmW mWGPPmW mWPPGmWPdBm GdB PdBmmW==⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤⎡⎤=+⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦=+⎣⎦ It is evident that “deebees” are not a unit! The units of the result can be found by multiplying the units of each term in a summation of decibel values.2/15/2005 dB.doc 7/9 Jim Stiles The Univ. of Kansas Dept. of EECS For example, say some power 16PdBm= is combined with power 210PdBm= . What is the resulting total power 12TPPP=+ ? A: NO! Never do this either! Logarithms are very helpful in expressing products or ratios of parameters, but they are not much help when our math involves sums and differences! []1010log ????xy+= So, if you wish to add P1 =6 dBm of power to P2 =10 dBm of power, you must first explicitly express power in Watts: P1 =10 dBm = 10 mW and P2 =6 dBm = 4 mW Q:This result really is obvious—of course the total power is: ()()( )1261016TP dBm P dBm P dBmdBm dBmdBm=+=+=2/15/2005 dB.doc 8/9 Jim Stiles The Univ. of Kansas Dept. of EECS Thus, the total power PT is: 1240 10014 0TPPP.mW .mW.mW=+=+= Now, we can express this total power in dBm, where we find: ()1014 010 11 4610T.mWPdBm log . dBm.mW⎛⎞==⎜⎟⎝⎠ The result is not 16.0 dBm !. We can mathematically add 6 dBm and 10 dBm, but we must understand what result means (nothing useful!). 10 10210224106 10 10log 10log1140 10log116 relative to 1 mWmW mWdBm dBmmW mWmWmWdB⎡⎤⎡⎤+= +⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤⎢⎥=⎢⎥⎣⎦= Thus, mathematically speaking, 6 dBm + 10 dBm implies a multiplication of power, resulting in a value with units of Watts squared !2/15/2005 dB.doc 9/9 Jim Stiles The Univ. of Kansas Dept. of EECS A few more tidbits about decibels: 1. 1 0 0.dB↔ 2. 0 0.dB↔−∞ 3. 57nndB↔  (can you show why?) I wish I had a nickel for every time my software has crashed-oh wait, I


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