document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=81388>Question 81388</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am taking an online electronics course and I am stumped on decibel/logarithm section relating to amplifiers.  My problems deal in power or voltage gain, and levels.  Here are 2 examples that I cannot solve. (1) A power amplifier has a gain of 26dB, and an input of 1 volt.  What is the voltage level? (2) An amplifier has a voltage gain of 60dB, the input is 10 microvolts, what is the output?  I have a couple formulas, and I know the answers, but I can't figure out how to get the answers from the formulas. The answer to problem one is 20, and the answer to problem 2 is 10. Here are the formulas:\r<BR>\n" );
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document.write( "Gain= signal out/signal in\r<BR>\n" );
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document.write( "dB= 10 x log(base 10) (W out/ W in)<BR>\n" );
document.write( "dB= 20 x log(base 10) (V out/ V in)\r<BR>\n" );
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document.write( "Output voltage= input level x gain ratio\r<BR>\n" );
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document.write( "I need a very detailed expression as to how to solve these problems.  I have looked in two textbooks for further explanation, but I cannot find the solutions.  \r<BR>\n" );
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document.write( "Thank you so much if you can help!!!! <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #58348 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=58348\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve these two problems you need only the voltage amplification equation:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dB+=+20%2ALog+%28V%5B2%5D%2FV%5B1%5D%29 ALIGN=MIDDLE  ALT=\"dB+=+20%2ALog+%28V%5B2%5D%2FV%5B1%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "and a knowledge of a few rules of Logarithms.<BR>\n" );
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document.write( "The dB equation defines decibels.  It uses the base 10 Logarithms. In this equation, <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V%5B2%5D ALIGN=MIDDLE  ALT=\"V%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the output voltage, and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V%5B1%5D ALIGN=MIDDLE  ALT=\"V%5B1%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the input voltage.  <BR>\n" );
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document.write( "Here are a couple of rules of Logarithms that you should be familiar with. Assume that <BR>\n" );
document.write( "the base of the Logarithm is b:<BR>\n" );
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document.write( "Rule I. …. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log+%28x%5En%29+=+n+%2A+Log+%28x%29 ALIGN=MIDDLE  ALT=\"Log+%28x%5En%29+=+n+%2A+Log+%28x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Rule II. …. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log+%28x%2Fy%29+=+Log%28x%29 ALIGN=MIDDLE  ALT=\"Log+%28x%2Fy%29+=+Log%28x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > – <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log%28y%29 ALIGN=MIDDLE  ALT=\"Log%28y%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Rule III. …. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log+%28x%2Ay%29+=+Log+%28x%29 ALIGN=MIDDLE  ALT=\"Log+%28x%2Ay%29+=+Log+%28x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > + <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log+%28y%29 ALIGN=MIDDLE  ALT=\"Log+%28y%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Rule IV. …. Log base b of x = y is equivalent to saying <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5Ey+=+x ALIGN=MIDDLE  ALT=\"b%5Ey+=+x\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Note that in decibel calculations b (the base) is 10.  This translates Rule IV to:<BR>\n" );
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document.write( "Log (x) = y is equivalent to  saying <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%5Ey+=+x ALIGN=MIDDLE  ALT=\"10%5Ey+=+x\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
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document.write( "If you need a further explanation of these rules, you can get it from an Algebra II text, <BR>\n" );
document.write( "from elsewhere on this site, or by doing a Google on Logarithms.<BR>\n" );
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document.write( "Time to work your first problem by using the voltage form of the decibel equation.  <BR>\n" );
document.write( "Given the amplifier has a power gain of 26 dB, and an input voltage of 1 volt.  Substitute <BR>\n" );
document.write( "those two values into the dB equation (26 for dB and 1 for the input voltage) and you get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=26+=+20%2ALog%28V%5B2%5D%2F1%29 ALIGN=MIDDLE  ALT=\"26+=+20%2ALog%28V%5B2%5D%2F1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "The plan is to work down to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V%5B2%5D ALIGN=MIDDLE  ALT=\"V%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so start by dividing both sides of this equation by <BR>\n" );
document.write( "20 to get rid of the 20 on the right side.  At the same time note that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V%5B2%5D%2F1 ALIGN=MIDDLE  ALT=\"V%5B2%5D%2F1\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "is just <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V%5B2%5D ALIGN=MIDDLE  ALT=\"V%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >. These simplifications lead to the equation becoming:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=26%2F20+=+Log+%28V%5B2%5D%29 ALIGN=MIDDLE  ALT=\"26%2F20+=+Log+%28V%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "And after dividing 26 by 20 on the left side the equation reduces to:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.3+=+Log+%28V%5B2%5D%29 ALIGN=MIDDLE  ALT=\"1.3+=+Log+%28V%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Time to apply Rule IV.  When you do you get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%5E1.3+=+V%5B2%5D ALIGN=MIDDLE  ALT=\"10%5E1.3+=+V%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Calculator time.  Your calculator probably works this way: enter 1.3 then press the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%5Ex ALIGN=MIDDLE  ALT=\"10%5Ex\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "key and you find the result to be 19.95262315 volts.  Close enough for electrical work.  <BR>\n" );
document.write( "Call it 20 volts.<BR>\n" );
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document.write( "Now to your second problem.  If the answer is to be 10, then you probably meant that the <BR>\n" );
document.write( "input voltage was 10 millivolts  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%2A10%5E-3+ ALIGN=MIDDLE  ALT=\"10%2A10%5E-3+\"   DISABLED_style= \"max-width:90%; height:auto;\" > instead of 10 microvolts <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%2A10%5E-6 ALIGN=MIDDLE  ALT=\"10%2A10%5E-6\"   DISABLED_style= \"max-width:90%; height:auto;\" >. <BR>\n" );
document.write( "So let’s use 10 millivolts.<BR>\n" );
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document.write( "For the given gain of 60 dB and an input of  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%2A10%5E-3 ALIGN=MIDDLE  ALT=\"10%2A10%5E-3\"   DISABLED_style= \"max-width:90%; height:auto;\" > volts the equation becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=60+=+20%2ALog+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29 ALIGN=MIDDLE  ALT=\"60+=+20%2ALog+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Just as was done in the last problem, divide both sides by 20 to reduce the problem to:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=60%2F20+=+Log+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29 ALIGN=MIDDLE  ALT=\"60%2F20+=+Log+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "The left side divides out to a quotient of 3 and the equation becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3+=+Log+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29 ALIGN=MIDDLE  ALT=\"3+=+Log+%28V%5B2%5D%2F%2810%2A10%5E-3%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "On the right side apply Rule II (division) to split the term into two separate logarithms:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3+=+Log+%28V%5B2%5D%29 ALIGN=MIDDLE  ALT=\"3+=+Log+%28V%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > – <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log+%2810%2A10%5E-3%29 ALIGN=MIDDLE  ALT=\"Log+%2810%2A10%5E-3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Notice that 10*10^-3 = 10^(-2).  Substitute this into the right hand logarithm and get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3+=+Log%28V%5B2%5D%29 ALIGN=MIDDLE  ALT=\"3+=+Log%28V%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > – <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log+%2810%5E%28-2%29%29 ALIGN=MIDDLE  ALT=\"Log+%2810%5E%28-2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Apply Rule I (exponent rule) to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log%2810%5E-2%29 ALIGN=MIDDLE  ALT=\"Log%2810%5E-2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > and get - <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2ALog+%2810%29 ALIGN=MIDDLE  ALT=\"2%2ALog+%2810%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  But by Rule IV <BR>\n" );
document.write( "you can see that Log base 10 of 10 = 1. (Think <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%5Ey+=+10 ALIGN=MIDDLE  ALT=\"10%5Ey+=+10\"   DISABLED_style= \"max-width:90%; height:auto;\" > which means y must be 1.) <BR>\n" );
document.write( "So - <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2A+Log%2810%29 ALIGN=MIDDLE  ALT=\"2%2A+Log%2810%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > equals -2*1 equals -2.  Substitute this result for – <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Log%2810%5E-2%29 ALIGN=MIDDLE  ALT=\"Log%2810%5E-2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "and the equation becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3+=+Log%28V%5B2%5D%29 ALIGN=MIDDLE  ALT=\"3+=+Log%28V%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > – <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-2%29 ALIGN=MIDDLE  ALT=\"%28-2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "And this simplifies to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3+=+Log%28V%5B2%5D%29+%2B+2 ALIGN=MIDDLE  ALT=\"3+=+Log%28V%5B2%5D%29+%2B+2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Subtract 2 from both sides and the equation becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1+=+Log%28V%5B2%5D%29 ALIGN=MIDDLE  ALT=\"1+=+Log%28V%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Apply Rule IV to this equation and you get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V%5B2%5D+=+10%5E1+=+10 ALIGN=MIDDLE  ALT=\"V%5B2%5D+=+10%5E1+=+10\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "This agrees with the answer you had for this problem, but recall that 10 millivolts <BR>\n" );
document.write( "was used in place of 10 microvolts to get the answer.<BR>\n" );
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document.write( "Hope this helps you get a handle on decibel equations.<BR>\n" );
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