Calculating/understanding exponential/anti-log curves increase/decay

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I'm teaching myself about current sources in analog circuits - particularly the linear to anti-log converter - and I have a couple of questions about understanding/calculating exponential curves but I'm struggling with the maths theory.

I'm trying to plot an ideal exponential curve so I can match the performance of my circuit to it. The x axis shows Voltage from 0 to 2.50 and the y axis shows Gain (G) from 0 to 10. I would like to plot 6 points at 0,0.5,1,1.5,2 and 2.5 (Volts).

My target start and end points are: 0 Volts = 0 Gain and 2.5 Volts = 10 Gain

My first question is - how do I calculate down for each of the remaining 4 points from 10 to 0 in an exponential manner?

My second question - related to the first (I think) is how do I work out the exponent of the sequence below?:

1V = 100Hz
2V = 200Hz
3V = 400Hz
4V = 800Hz
5V - 1600Hz

Whatever the exponent(?) of the above sequence is - is the same exponent that should be used in plotting the sequence down from 10 in my first question.

I hope that makes sense, my electronics knowledge is self-taught and practical, this is one of those time where I need to calculate and predict a result rather than just trying to poke round until I find it.

Thanks in advance.
 


I have the answer for first question.

You are taking the Gain on Y-axis. First, convert that values in decibles by using 20*log(gain).

Then the plot become as exponential.
 

subbuindia said:
I have the answer for first question.
You are taking the Gain on Y-axis. First, convert that values in decibles by using 20*log(gain).
Then the plot become as exponential.
Click to expand...

Thanks for the reply.

Sorry if I have misunderstood your reply - but this equation: 20*log(gain) - looks like the one to convert Peak to Peak Voltage to dB?

My requirement is to calculate values for the 4 points between 0 Volts and 2.5 Volts - so I need to fill in the blanks here for an exponential curve...

Volts/Gain
0.0:0.0
0.5: ?
1.0: ?
1.5: ?
2.0: ?
2.5:10.00

I hope that makes more sense - again, sorry if I have misunderstood your answer.

Thanks.
 


Thank you for giving another thought.

Volts/Gain
0.0:0.0
0.5:1
1.0:2.5
1.5:4.5
2.0:7
2.5:10.00

I take it as: y=x^2 + x + (x/2)

If you find any mistake, please forgive that.

Thanks.
 

I hope this answer the question.

Code: 
F = ( 100 ) * ( 2 ^ ( V-1 ) )

+++
 

Hi

1) An exponential curve does not pass by (x,y)=(0,0), so what you want doesn't have an exact solution.
Nevertheless, some approximatins are possible, like a first linear chord starting at (0,0) that joins an exponential characteristic at which it is tangent.

Are you looking for exponential gain or for exponential input-output characteristic?


2) Voltage increases 1V each time frequency doubles. This means that voltage is proportional to the log of frequency, like voltage = A * log2(frequency / B) when A and B are proper constants.
Solving for A and B using 2 points gives this result:

voltage = 1V * log2(frequency / 50Hz)

frequency = 50Hz * 2^(voltage / 1V)


Regards

Z
 
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    juz_ad

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Hi zorro, thanks for the reply...


Good question. The honest answer is I'm not sure what the difference would be between the two? If I put a linear Voltage in e.g. 1, 2, 3, 4, 5 Volts and get an exponential(?) increase in Gain (in Volts) at the output e.g. 1, 2, 4, 8, 16 Volts - then does that satisfy both (a) Gain and (b) Input/Output?

Sorry if I'm missing your point - this level of maths does challenge me.


OK - that makes more sense. Thanks.

So maybe what I'm looking for is a log/antilog relationship between Voltage and Gain (the same way that you have demonstrated that the relationship between Voltage and Frequency is a log relationship) rather than a 'true' exponential relationship?

Anything you or anyone else could follow this up with would be very helpful.
 

zorro said:
frequency = 50Hz * 2^(voltage / 1V)
Click to expand...

Can I just clarify - in this equation 50Hz * 2^(voltage / 1V)

50Hz is always the base/root/lowest frequency and 'voltage' is the extra voltage I am adding to the existing voltage - i.e. if an input of 1 Volt gives me an output of 50Hz, then to calculate the output of each extra volt added to 1 Volt would be:

(2 Volts) 50Hz * 2^(1 / 1V) = 100Hz
(3 Volts) 50Hz * 2^(2 / 1V) = 200Hz
(4 Volts) 50Hz * 2^(3 / 1V) = 400Hz
(5 Volts) 50Hz * 2^(4 / 1V) = 800Hz

...and to work backwards, volt by volt, from 800Hz would be:

(4 Volts) 50Hz * 2^(-1 / 1V) = 400Hz
etc.

That looks like exactly what I need - just want to make sure I've understood it.

Thanks.
 

If I got the meaning of your table, these are some corrections in red:

(1 Volts) 50Hz * 2^(1 / 1V) = 100Hz
(2 Volts) 50Hz * 2^(2 / 1V) = 200Hz
(3 Volts) 50Hz * 2^(3 / 1V) = 400Hz
(4 Volts) 50Hz * 2^(4 / 1V) = 800Hz

...and to work backwards, volt by volt, from 800Hz would be:

(3 Volts) 800Hz * 2^(-1 / 1V) = 400Hz
etc.

Regards

Z
 

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