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

[em2682]

create an equatio for the graph using the y = ae^(bx)cos(cx) for the following graph
View attachment 8085

 

[MarkFL]

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

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

 

[HOI]

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.

 

[MarkFL]

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?

 

[em2682]

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 - - -

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 - - -

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

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

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

 

[MarkFL]

Well, that's a tiny graph, which isn't very easy to read.

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

Let's say:

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

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

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

(1) and (3) imply:

$$a=\frac{35}{2},\,d=\frac{3}{7}$$

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

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

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

And so we have:

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

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

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

[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]

 

[em2682]

Well, that's a tiny graph, which isn't very easy to read.

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

Let's say:

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

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

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

(1) and (3) imply:

$$a=\frac{35}{2},\,d=\frac{3}{7}$$

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

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

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

And so we have:

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

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

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

View attachment 8093
Does this help or change the values?

 

[I like Serena]

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.

 

[em2682]

Well, that's a tiny graph, which isn't very easy to read.

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

Let's say:

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

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

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

(1) and (3) imply:

$$a=\frac{35}{2},\,d=\frac{3}{7}$$

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

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

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

And so we have:

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

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

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

thank you so much! what were some importnt assumptions made to determine the equation?

 

[MarkFL]

thank you so much! what were some importnt assumptions made to determine the equation?

All the assumptions were posted.