# In beginner analog electronics: in a Bode plot

Discussion in 'Homework Help' started by lse123, Jan 22, 2008.

1. ### lse123 Thread Starter Active Member

Oct 17, 2006
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In beginner analog electronics: in a Bode plot how ind for a particular frequency[eg f=789.987Hz] the AdB where we have logarithic scale AdB -->dB

What prerequisite courses (and textbooks~) has the "Principles of Electronic Communication Systems - Louis E. Frenzel" ?

2. ### Papabravo Expert

Feb 24, 2006
10,148
1,791
I don't understand your question. Can you restate it with a greater level of clarity?

3. ### lse123 Thread Starter Active Member

Oct 17, 2006
87
0
in a Bode plot how I find for a particular frequency[eg f=789.987Hz] , the AdB, where we have logarithic scale: AdB -->dB & f-->Hz decades ?

4. ### beenthere Retired Moderator

Apr 20, 2004
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I believe the horizontal scale plots the frequency. The plot should indicate a start and stop frequency, with the test frequencies linearly distributed between those extremes. I'm not certain the scaling is enough for 3 decimal place accuracy (but 790 Hz should be pretty close).

5. ### lse123 Thread Starter Active Member

Oct 17, 2006
87
0
yes but if f frequency for which find AdB is between two points indicating two decades how to calculate exact value for AdB ?

6. ### Papabravo Expert

Feb 24, 2006
10,148
1,791
Two methods come to mind. If you have the graph of the Bode Plot then you draw a vertical line at the frequency of interest and look at the value of AdB where the vertical line intersects the Bode Plot.

If you have a closed form expression for the trnasfer function then you can simply plug in the frequency and evaluate the magnitude.

These two things seem obvious to me so maybe there is still an aspect to your question that I don't understand.

If we go back to first principles then the first thing to remember is that there are multiple ways of working with frequency.
1. Frequency is measured in cycles per second or Hertz (Hz.)
2. Frequency is measured in radians per second which is 2*pi*f
3. Frequency is normalized with respect to another frequency.
An example of number 3 would be f/fo, where fo is some constant frequency of interest. f < fo means the ratio is less than 1 and f > fo means the ratio is greater than 1 and f = fo means the ratio is exactly 1.

The vertical axis in a Bode Plot of Voltage Gain is:
Code ( (Unknown Language)):
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2. 20 * log (Vo/Vi)
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4. where
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6. Vo is the output voltage of the filter or other system
7. Vi is the input voltage
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So if you have expressions for Vi and Vo that you can evaluate at a frequency of interest you are home free.

If all you have is the plot in dead tree form then you'll just have to be happy with an approximation. If your question was how do I find a particular frequency on a logarithmic scale then you just take the logarithm of the frequency. Throw away the characteristic and use the mantissa as the fraction of the interval from the start of the decade to the end of the decade.

Code ( (Unknown Language)):
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2. [SIZE=4][SIZE=2]log (789.987) = 2.89762[/SIZE][/SIZE]
3. [SIZE=4][SIZE=2]characteristic = 2 so pick the decade between 100 and 1000[/SIZE][/SIZE]
4. [SIZE=4][SIZE=2]mantissa = .89762 [SIZE=4][SIZE=2]≈ .90[/SIZE][/SIZE][/SIZE][/SIZE]
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So on a linear scale you go 0.9*length of the decade and that is where you draw the vertical line.