Create an equation for the graph using the y = ae^(bx)cos(cx)

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In summary: We should be careful with that term, as one could also call it the measured values that a measuring device outputs.I'm still in doubt whether we should try to fit the data to the given formula, or to the formula that I made up, or to the formula that Country Boy made up, or whether we should just make up a formula ourselves that seems to fit the data.Notice that the graph does not show the actual (x,y) points. It is more like the graph of a continuous function that passes through the points. It looks like the graph of a function of the form y=a*e^(bx)*cos(cx)+d, but the values of a, b, c, and d are not clearly indicated by the
  • #1
em2682
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create an equatio for the graph using the y = ae^(bx)cos(cx) for the following graph
View attachment 8085
 

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  • #2
It appears to me that you need something of the form:

\(\displaystyle y=a\left(e^{bx}+d\right)\cos(cx)+f\)
 
  • #3
I am inclined to think that MarkFL's "d" could be 0 but "f" clearly can't be. Without a non-zero f, the first loop of the function would go from a down to -a but you want it to go from 140 to 100. The average of those two is 120 so you want it to go from 120+ 20 to 120- 20.
 
  • #4
Country Boy said:
I am inclined to think that MarkFL's "d" could be 0 but "f" clearly can't be. Without a non-zero f, the first loop of the function would go from a down to -a but you want it to go from 140 to 100. The average of those two is 120 so you want it to go from 120+ 20 to 120- 20.

If $d=0$ and presumably $b<0$, then what would prevent the amplitude from tending to zero as $x\to\infty$, instead of some value greater than zero?
 
  • #5
MarkFL said:
If $d=0$ and presumably $b<0$, then what would prevent the amplitude from tending to zero as $x\to\infty$, instead of some value greater than zero?

are your able to work at the actual equation and show working?

- - - Updated - - -

Country Boy said:
I am inclined to think that MarkFL's "d" could be 0 but "f" clearly can't be. Without a non-zero f, the first loop of the function would go from a down to -a but you want it to go from 140 to 100. The average of those two is 120 so you want it to go from 120+ 20 to 120- 20.

are your able to work at the actual equation and show working?

- - - Updated - - -

MarkFL said:
It appears to me that you need something of the form:

\(\displaystyle y=a\left(e^{bx}+d\right)\cos(cx)+f\)

are your able to work at the actual equation and show working?
 
  • #6
Well, that's a tiny graph, which isn't very easy to read.

I would begin with the amplitude \(\displaystyle A(x)=a\left(e^{bx}+d\right)\)

Let's say:

\(\displaystyle A(0)=a\left(1+d\right)=25\tag{1}\)

\(\displaystyle A(50)=a\left(e^{50b}+d\right)=15\tag{2}\)

\(\displaystyle A(\infty)=ad=\frac{15}{2}\tag{3}\)

(1) and (3) imply:

\(\displaystyle a=\frac{35}{2},\,d=\frac{3}{7}\)

And so, substituting these values into (2), we find:

\(\displaystyle \frac{35}{2}\left(e^{50b}+\frac{3}{7}\right)=15\)

\(\displaystyle b=\frac{1}{50}\ln\left(\frac{3}{7}\right)\)

And so we have:

\(\displaystyle A(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\)

The period appears to be about $10\text{ s}$ and the vertical shift is \(\displaystyle f=\frac{245}{2}\) and so we have:

\(\displaystyle y(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\cos\left(\frac{\pi}{5}x\right)+\frac{245}{2}\)

[DESMOS]advanced: {"version":5,"graph":{"squareAxes":false,"viewport":{"xmin":0,"ymin":80,"xmax":250,"ymax":160}},"expressions":{"list":[{"id":"1","type":"expression","latex":"y=\\frac{35}{2}\\left(\\left(\\frac{3}{7}\\right)^{\\frac{x}{50}}+\\frac{3}{7}\\right)\\cos\\left(\\frac{\\pi}{5}x\\right)+\\frac{245}{2}","color":"#c74440"}]}}[/DESMOS]
 
  • #7
MarkFL said:
Well, that's a tiny graph, which isn't very easy to read.

I would begin with the amplitude \(\displaystyle A(x)=a\left(e^{bx}+d\right)\)

Let's say:

\(\displaystyle A(0)=a\left(1+d\right)=25\tag{1}\)

\(\displaystyle A(50)=a\left(e^{50b}+d\right)=15\tag{2}\)

\(\displaystyle A(\infty)=ad=\frac{15}{2}\tag{3}\)

(1) and (3) imply:

\(\displaystyle a=\frac{35}{2},\,d=\frac{3}{7}\)

And so, substituting these values into (2), we find:

\(\displaystyle \frac{35}{2}\left(e^{50b}+\frac{3}{7}\right)=15\)

\(\displaystyle b=\frac{1}{50}\ln\left(\frac{3}{7}\right)\)

And so we have:

\(\displaystyle A(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\)

The period appears to be about $10\text{ s}$ and the vertical shift is \(\displaystyle f=\frac{245}{2}\) and so we have:

\(\displaystyle y(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\cos\left(\frac{\pi}{5}x\right)+\frac{245}{2}\)

View attachment 8093
Does this help or change the values?
 

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  • #8
Here's a slightly different approach.

It looks like the signal that may come from an oscilloscope.
For such a signal it is not unusual that an offset was added to the original signal to get it nicely on the screen.
That means that yes, it has an offset, but we don't care about the offset.

What we typically want to know, is:
  1. The initial amplitude $A_0$.
  2. The characteristic time $\tau$ that the amplitude decays with a factor $e^{-1}$.
  3. The angular frequency $\omega$.
The first top is at $t=0$, which is $150$. The second top is $145$, so it decays about $\frac {150-145}2 \approx 2.5$ in the first half cycle.
The first bottom is at $99$. Subtract $2.5$ (slightly round up) to compensate, and we get $96$.

So:
$$A_0 = \frac{150 - 96}{2} = 27 \tag 1$$
The offset is then at
$$\text{Height offset} = \frac{150+96}{2} = 123 \tag 2$$

At the characteristic time $\tau$, we will have a height of $123 \pm \frac{27}{e} = 123 \pm \frac{27}{2.71} = 123\pm 10$.
Read the graph to see where heights $113$ and $133$ are, which is a bit difficult. I'll estimate
$$\tau=100 \text{ s} \tag 3$$
since it does seem we have nice round numbers in this exercise. After all, $27$ divides very nicely by $e\approx 2.71$.

Finally, following top-to-top, we have $24$ cycles in $250 \text{ s}$.
(I recounted to check if I was 1 off, but apparently I was not.)
That gives us:
$$\omega = 2\pi f= 2\pi\cdot\frac{24}{250} = 0.60\text{ rad/s} \tag 4$$

So we get:
$$A_0\, e^{-t/\tau}\cos(\omega t) = 27\,e^{-t/100}\cos(0.60\,t) \tag 5$$
Matching it with the given formula $ae^{bx}\cos{cx}$ gives us:
$$a=27,\ b=-0.010\text{ s}^{-1},\ c=0.60\text{ rad/s} \tag 6$$

With TikZ/pgfplots and its default settings (apparently the original image was made the same way):
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[xlabel={Seconds},ylabel={Height},ytick distance=10,xmin=0,xmax=250,ymin=90,ymax=150]
\addplot[red, smooth, thick, samples=1000, domain=0:250] (x, {27 * exp(-0.010 * x) * cos(deg(0.60 * x)) + 123});
\end{axis}
\end{tikzpicture}
Here's the original image for comparison:
https://www.physicsforums.com/attachments/8085

As we can see, the given formula does not match as MarkFL and Country Boy already indicated.
It doesn't decay to zero amplitude, but it seems to decay to some fixed amplitude.
Conclusion: the given formula does not describe the measured signal.
 
  • #9
MarkFL said:
Well, that's a tiny graph, which isn't very easy to read.

I would begin with the amplitude \(\displaystyle A(x)=a\left(e^{bx}+d\right)\)

Let's say:

\(\displaystyle A(0)=a\left(1+d\right)=25\tag{1}\)

\(\displaystyle A(50)=a\left(e^{50b}+d\right)=15\tag{2}\)

\(\displaystyle A(\infty)=ad=\frac{15}{2}\tag{3}\)

(1) and (3) imply:

\(\displaystyle a=\frac{35}{2},\,d=\frac{3}{7}\)

And so, substituting these values into (2), we find:

\(\displaystyle \frac{35}{2}\left(e^{50b}+\frac{3}{7}\right)=15\)

\(\displaystyle b=\frac{1}{50}\ln\left(\frac{3}{7}\right)\)

And so we have:

\(\displaystyle A(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\)

The period appears to be about $10\text{ s}$ and the vertical shift is \(\displaystyle f=\frac{245}{2}\) and so we have:

\(\displaystyle y(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\cos\left(\frac{\pi}{5}x\right)+\frac{245}{2}\)
thank you so much! what were some importnt assumptions made to determine the equation?
 
  • #10
em2682 said:
thank you so much! what were some importnt assumptions made to determine the equation?

All the assumptions were posted.
 

FAQ: Create an equation for the graph using the y = ae^(bx)cos(cx)

1. What is the purpose of creating an equation for a graph using the y = ae^(bx)cos(cx) format?

The equation y = ae^(bx)cos(cx) is a mathematical expression that can be used to model various real-world phenomena, such as population growth, radioactive decay, and oscillating systems. By understanding the components of this equation, scientists can make predictions and analyze data related to these phenomena.

2. How do you determine the values of a, b, and c in the equation y = ae^(bx)cos(cx)?

The values of a, b, and c can be determined by analyzing the given graph. The value of a represents the amplitude of the graph, which is the maximum value of y. The values of b and c determine the frequency and period of the oscillations, respectively. They can be calculated by finding the horizontal and vertical stretches of the graph.

3. Can the y = ae^(bx)cos(cx) equation be used for all types of graphs?

No, this equation is specifically used for graphs that exhibit exponential growth or decay, combined with oscillations. It is not applicable for linear or polynomial graphs.

4. How does changing the values of a, b, and c affect the shape and position of the graph?

The value of a determines the amplitude of the graph, so changing it will affect the height of the oscillations. The values of b and c affect the frequency and period of the oscillations, respectively. Changing these values will result in a change in the number of oscillations per unit and the time it takes for one complete oscillation.

5. Can the y = ae^(bx)cos(cx) equation be used to make accurate predictions?

Yes, as long as the values of a, b, and c are properly determined and the graph accurately represents the phenomenon being modeled, the equation can be used to make accurate predictions about the behavior of the system. However, it is important to note that any predictions made using this equation are only as accurate as the data and assumptions used to determine its values.

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