Combination of 2 formulas for a graph

In summary, you are trying to plot the following equation:4 + 2 cos (2pi 10 x 10^3t) cos (2Pi 125 10^3t)
  • #1
leejohnson222
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6
Homework Statement
i have 2 formulas and need to combine them to create a new wav form, not exactly sure how to this
Relevant Equations
[4 + 2 cos (2pi 10 x 10^3t) cos (2Pi 125 10^3t)]
i need to combine the two to make a new 3rd one
I think the t element is the X axis scale ?
not quite sure how to do this, suggestions please ?
 
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  • #2
What are the "two formulas" ? Also, use LaTeX, please so other people can read it.
 
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  • #3
Can you state the problem as was given to you? This is already a waveform. What kind of "new" wave form are you looking for? Yes, the "t element" is the X axis scale.
 
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  • #4
ok sorry, i dont have laTex i will try to install and use it

𝑉𝑐 = 4cos (2𝜋125x10^3t)

𝑉𝑎 = 2cos (2𝜋10𝑥10^3t)

then i need to combine these 2 for a new output,
i think i have manged to get the 2 formulas in graph but i need the 3rd one to compare too,
 
  • #5
leejohnson222 said:
ok sorry, i dont have laTex i will try to install and use it

𝑉𝑐 = 4cos (2𝜋125x10^3t)

𝑉𝑎 = 2cos (2𝜋10𝑥10^3t)

then i need to combine these 2 for a new output,
i think i have manged to get the 2 formulas in graph but i need the 3rd one to compare too,
There is no end to the ways you could combine them:
add them, 4cos (2𝜋125x10^3t)+2cos (2𝜋10𝑥10^3t)
multiply them: 4cos (2𝜋125x10^3t)x2cos (2𝜋10𝑥10^3t)
nest them: 4cos (2𝜋125x10^3x2cos (2𝜋10𝑥10^3t))
any combination of the above…

There must be some other constraint to make this a worthwhile question.
As already requested, please state the whole problem as presented to you.
 
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  • #6
Combine how? Sum? Product?

It may be convenient to set [itex]x = 2\pi(5 \times 10^{-3})t[/itex] so that [itex]V_c = 4 \cos(25 x)[/itex] and [itex]V_a = 2 \cos(2 x)[/itex].
 
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  • #7
leejohnson222 said:
ok sorry, i dont have laTex i will try to install and use it
You don't need to install LaTeX. Just click the link "LaTeX Guide" on the lower left to learn how to embed it in your posts.
 
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  • #8
ok so i need to plot

1680704640698.png

as a combination of the 2 other outputs
 
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  • #9
leejohnson222 said:
ok so i need to plot

View attachment 324525
as a combination of the 2 other outputs
We cannot help if you do not tell us exactly what you have been tasked to do. So far, you might as well ask how to combine an apple and an orchestra.
 
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  • #10
leejohnson222 said:
ok so i need to plot

View attachment 324525
as a combination of the 2 other outputs

Is your problem to express that as a sum of cosines? If so, use the identity [tex]
\cos A \cos B \equiv \tfrac12\left(\cos(A - B) + \cos(A + B)\right).[/tex]
 
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  • #11
Hi @leejohnson222. There are plenty members ready and willing to help you! But the task is not clear. Here are a few guesses (which could be completely wrong). Maybe you can confirm if they are correct or not…

1. The question is about modulation. If so, is it AM (amplitude modulation) or FM (frequency modulation)?

2. ##V_c## represents the voltage waveform of the carrier signal (amplitude 4V, frequency 125kHz) and ##V_{\alpha}## represents that of the modulating signal (amplitude 2V, frequency 10kHz).

3. You want a voltage-time graph for the modulated carrier.
 
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  • #12
appologies,
so the question gives me 2 formulas for wav forms,
i have to plot both
then i need to multiply the two signals together to compare the 2 signals with the 3rd
make general observations of the 3 wav signals

This is an AM freq signal after i plot all 3 i need to focus on the 3rd combination wav and show how the signal comprises frequencies, 115khz, 125khz & 135khz

i have attempted this in desmos, but i dont know how to alter the view so the waves are more easy to view
i understand i need to have the time axis in micro seconds

https://www.desmos.com/calculator/sljnofb1yv
 
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  • #13
1. In posts #1 and #8 you referred to:
“[4 + 2 cos (2pi 10 x 10^3t) cos (2Pi 125 10^3t)]”

This is not the same as the product ##f(t)g(t)## where
##f(t) = 4 \cos(2 \pi \times 125 \times 10^3 t)## and
##g(t) = 2 \cos(2 \pi \times 10 \times 10^3 t) ##

I presume your original; formula is wrong and that ##f(t)g(t)## is the one you actually need.

2. In Desmos, you have entered ##f(x)## and ##g(x)## (where ##x## represents time) incorrectly. Note that ##10^{3x}## is not the same as ##10^3x##. For example:
##10^{3 \times 2} = 1,000,000##
##10^3 \times 2 = 2000##

3. Since the frequency is high, you need to change the time-scale so you don’t get a load of waves compressed so much they look like a single block of colour on the screen.

For example see https://www.desmos.com/calculator/xzzr1vf3k0. I used the ‘spanner’ tool to set the time-range (x-axis range) to be 0 to 0.0001s. Note the scale-marking are in seconds. You need to experiment with setting the time-range for yourself.

4. If you want the scale-markings in μs, you are going to have to introduce a factor ##10^{-6}## into your definitions of ##f(x)## and ##g(x)##.

5. If you have 3 overlapping graphs, you can toggle each one on/off using its coloured button on the left of the screen. So you can display the 3rd graph by itself if required.
 
  • #14
yes you are correct there never was F or G = in the original question, i was told (by a desmos user) that if you change the formula to this its easier to multiply the 2 wavs so that is what i did, but looks like its confused things somewhat.
yes from the formula t= Miliseconds i have substituted t for x
How do i best sort this out now ?
i guess i could just have the 2 wav forms and attempt to write out the multiplied one?
 
  • #15
leejohnson222 said:
yes you are correct there never was F or G = in the original question, i was told (by a desmos user) that if you change the formula to this its easier to multiply the 2 wavs so that is what i did, but looks like its confused things somewhat.
yes from the formula t= Miliseconds i have substituted t for x
How do i best sort this out now ?
i guess i could just have the 2 wav forms and attempt to write out the multiplied one?
You've lost me. Please answer the following 2 questions.

Q1. State the exact problem, word-for-word, as given to you (not your interpretation of it) including any formulae.

Q2. With SI units, what are the units of ##f## and ##t## in an expression of the form ##\cos (2 \pi f t)##?
 
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  • #16
You work for a company that produces radio transmitters and part of that circuit uses a ring modulator to multiply a high frequency sinusoidal carrier signal, with a much lower frequency audio signal to transmit over longer distances using AM

The ring modular circuit works by multiplying together the 125khz signal V1
And the10khz signal v 2 to produce the transmitted modulated signal output Vo.
The two signals and the output of the modulator are as below

V1 = 4cos(2pi 125x 103t)

V2 = 2cos (2pi 10x 103t)

V0 = [4 + 2cos(2pi10x103t)]cos (2Pi 125x103t)Using a graph software plot V1, V2, and Vo
Ensure the scale is carefully chosen to show all 3 wave forms

What observations do you notice

Looking at the equation for Vo and using trigonometrical identities,
Show that the output from the modulator comprises frequency components at 115khz,
125khz, 135khz

Plot your alternative mathematical expansion to demonstrate the the two versions fo the output signal, Vo from part 1 and part 2 are identical

I believe frequency is in Khz and T=mili
I hope that makes sense now, sorry for the confusion.
 
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  • #17
leejohnson222 said:
using trigonometrical identities,
… namely, the one given to you in post #10.
Your expression takes the slightly more complicated form ##(c+\cos(\omega t))\cos(\phi t)##, but you can still use the identity to expand that into a sum of cosines.
 
  • #18
leejohnson222 said:
The two signals and the output of the modulator are as below

V1 = 4cos(2pi 125x 103t)

V2 = 2cos (2pi 10x 103t)

V0 = [4 + 2cos(2pi10x103t)]cos (2Pi 125x103t)
Ok. These are the 3 graphs you need. Do not use 'f(x)g(x)' - you do not use the product of V1 and V2.

If using SI units, expressions such as ##\cos(2 \pi f t)## have ##f## in hertz and ##t## in seconds.

Post the link to your graphs so we can take a look.
 
  • #21
leejohnson222 said:
You need to add the ‘combined function’; this is simply the equation given for V0:
“V0 = [4 + 2cos(2pi10x103t)]cos (2Pi 125x103t) “
This is the correct formula for the 125kHz carrier amplitude-modulated by the 10kHz signal.

However, when I tried entering it, Desmos struggled and I couldn’t get it to work correctly. Maybe that’s what happened to you. A ‘work-around’ is needed.

Look at this carefully: https://www.desmos.com/calculator/sgu6tew3uj

Can you enter V2 and V0 in terms of f(t) and g(t) and show us your results?

Notes:

You can use ‘t’ rather than ‘x’ as the independent variable. See above link.

When entering “V2”, simply type ‘V’ followed by ‘2’ and Desmos will make ‘2’ the subscript to give you ##V_2##.

125kHz (for example) is the same as ##125 \times10^3## Hz.
Use ‘*’ for multilocation and ‘^’ to raise to a power.
So entering 125*10^3 into Desmos will give ##125·10^3## in the formula.

If you want/need to know where the formula
“V0 = [4 + 2cos(2pi10x103t)]cos (2Pi 125x103t)“
comes from, try here for example: https://electronicscoach.com/amplitude-modulation.html
 
  • #22
thanks for the help on this, its an added layer learning desmos on top of the theory
also that explaination is quite useful and it mirrors the music software i work with within synthesis design
i dont think i have this correct at all but
https://www.desmos.com/calculator/ejvuwcfxbg
 
  • #23
leejohnson222 said:
thanks for the help on this, its an added layer learning desmos on top of the theory
also that explaination is quite useful and it mirrors the music software i work with within synthesis design
i dont think i have this correct at all but
https://www.desmos.com/calculator/ejvuwcfxbg
Your Desmos formulae for ##V_1## and ##V_2## are correct. But not for ##V_0##.

You have used ##V_0 = V_1 \times V_2##, i.e. you have just multiplied the two signals. This is wrong.

The correct formula is given in the question as:
##V_0 = (4 + \text{something}) \times \text {some other thing}##.

Hints: you must have ‘4+’ somewhere in your equation.
See if you can see what the 'somethings' are - there is nothing difficult.
 
  • #24
ok i though you had to multiply, the original formula says Vo = (4+2cos) * the previous 2 wave files,(formulas)
problem is i have tried to type this into desmos and it doesnt produce any wav, i am clearly missing something here, i thought i could substitute the entire 2 formulas for (f) (g) that didnt work in desmos either?

https://www.desmos.com/calculator/5tlcgd31vq
 
  • #25
leejohnson222 said:
ok i though you had to multiply, the original formula says Vo = (4+2cos) * the previous 2 wave files,(formulas)
problem is i have tried to type this into desmos and it doesnt produce any wav, i am clearly missing something here, i thought i could substitute the entire 2 formulas for (f) (g) that didnt work in desmos either?

https://www.desmos.com/calculator/5tlcgd31vq
Nope!

In case it’s causing any confusion, note that your Post #1 equation for V0 has the right-square-bracket in the wrong place (corrected in later posts).
So
[4 + 2 cos (2pi 10 x 10^3t) cos (2pi 125 10^3t)]
should be
[4 + 2 cos (2pi 10 x 10^3t)] cos (2pi 125 10^3t)

Look at the structure carefully:
‘4’ is the amplitude of the carrier;
‘2 cos (2pi 10 x 10^3t)’ is the signal.
The above two items get added together first. Then their sum is multiplied by ’cos (2pi 125 10^3t)’ (which is the cosine part of the carrier).

##f(t) = \cos (2 \pi *125*10^3~ t)##
##g(t) = \cos (2 \pi *10*10^3 ~t)##
##V_1(t) = 4f(t)##
##V_2 (t)= 2g(t)##
##V_0(t) = ????##

Minor edit.
 
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  • #26
i know its right in front of my face but i cant grab it,
to me 4 + v2 * g(t) i had a few goes at it, but i gotta get up for work early, so will come back to this in a day or so
https://www.desmos.com/calculator/izg3kuu8s3
 
  • #27
leejohnson222 said:
i know its right in front of my face but i cant grab it,
to me 4 + v2 * g(t) i had a few goes at it, but i gotta get up for work early, so will come back to this in a day or so
https://www.desmos.com/calculator/izg3kuu8s3
There are 2 mistakes in "4 + v2 * g(t)"

1. You must add ##4## and ##V_2(t)## first, before the multiplication. You make sure it's done first by enclosing it in brackets:
##(4 + V_2(t))##

Think about the difference between:
1 + 2 * 3 = 7 and
(1 + 2) * 3 = 9

2. If you check carefully, you should see that you then need to multiply by ##f(t)##, not by ##g(t)##, giving:
##(4 + V_2(t)) f(t)##

Or you could use ##(4 + 2g(t)) f(t)##.
 
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  • #28
that is so frustrating as i think it typed something very close to that in desmos and then thought it was wrong, sometimes coming back after a break has a very positive effect.
I was getting f(t) and g(t) mixed up.
so does this look correct to you ? https://www.desmos.com/calculator/upxjlnnwbm
 
  • #29
leejohnson222 said:
so does this look correct to you ? https://www.desmos.com/calculator/upxjlnnwbm
Your 5th equation
##v_{0}\left(t\right)=\left[4f\left(+2g\left(t\right)\right)\right]\cdot\cos\left(g\left(t\right)\right)\ ##
and your 7th equation
##v_{0}\left(t\right)=\left[V_{1}\left(t\right)\ +2\right]\ \cdot g\left(t\right)##
are both wrong

Your 6th equation
##v_{0}\left(t\right)=\left(4+2g\left(t\right)\right)\cdot f\left(t\right)##
is correct.

Please don’t take offence but it looks like you are struggling with basic arithmetic/algebra. For example, if you are told that x = (a+b)c, which of the following is/are correct?
1) x = a +bc
2) x = ac + b
3) x = (a+c)b
4) x = ac + bc
5) x = (1+a)bc
6) x = abc

If you can’t do that easily, you might want to consider doing some revision!
 
  • #30
I would say number 3 and 4 are correct
 
  • #31
leejohnson222 said:
I would say number 3 and 4 are correct
No.

We are told x = (a+b)c.

We can choose any values we want for a, b and c. For example, suppose a= 1, b =2 and c = 5.

We can put the chosen values in so that we know what x is for our chosen values of a, b and c;
x = (a+b)c = (1+2)*5 = 3*5 = 15

Choice 3) is x = (a+c)b. This gives x = (1+5)*2 = 6*2 = 12. That’s not equal to 15! So choice 3) is wrong.

There is only one correct answer, choice 4), x = ac + bc = 1*5 + 2*5 = 5+10 = 15.

This is very basic maths. If you are on some sort of course which requires maths, you need to do some revision/catching-up. There’s plenty of online stuff available, for example BBC Bitesize Maths. Good luck.
 
  • #32
i appreaciate your assistance, yes i am on a course after a 15 year break from maths, and i know some of the basic rules are letting me down, with brackets etc, i actually did exactly this by substituing for numbers and went to work out all of the ones your wrote up, problem is its hard after a 10 hour day and sorting the kids out, anyway no excuse i need to improve here to progress. thanks again
 

1. What is a combination of 2 formulas for a graph?

A combination of 2 formulas for a graph is a mathematical technique used to create a new formula by combining two existing formulas. This allows for the representation of complex relationships between variables in a graph.

2. How is a combination of 2 formulas for a graph different from a single formula?

A single formula represents a direct relationship between two variables, whereas a combination of 2 formulas allows for the representation of more complex relationships, such as non-linear or exponential relationships.

3. What are the benefits of using a combination of 2 formulas for a graph?

Using a combination of 2 formulas for a graph allows for a more accurate representation of complex relationships between variables. It also allows for better analysis and prediction of data.

4. How do you determine which formulas to combine for a graph?

The choice of which formulas to combine depends on the relationship between the variables being represented. It is important to consider the type of relationship (linear, non-linear, exponential) and the data being analyzed.

5. Are there any limitations to using a combination of 2 formulas for a graph?

While a combination of 2 formulas can provide a more accurate representation of relationships between variables, it may not always be possible to find two formulas that accurately represent the data. It is also important to ensure that the combined formula is mathematically valid and does not produce unrealistic results.

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