i have plotted 3 graphs for an AM modulation signal fine, but the second half of the equation wants me to expand v0:[4+2cos(2pi10*10^3t)cos(2pi125*10^3t).

i know the second part of the question wants me to use 1/2 cos (A+B)+cos(A-B) but im struggling to expand it.

any tips ??

i know the second part of the question wants me to use 1/2 cos (A+B)+cos(A-B) but im struggling to expand it.

any tips ??

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Original post by sly studies

i have plotted 3 graphs for an AM modulation signal fine, but the second half of the equation wants me to expand v0:[4+2cos(2pi10*10^3t)cos(2pi125*10^3t).

i know the second part of the question wants me to use 1/2 cos (A+B)+cos(A-B) but im struggling to expand it.

any tips ??

i know the second part of the question wants me to use 1/2 cos (A+B)+cos(A-B) but im struggling to expand it.

any tips ??

Can you post an image of the exact question? You seem to have lots of things multiplied together inside the cos functions but I'm not convinced that's how they should be!

Original post by davros

Can you post an image of the exact question? You seem to have lots of things multiplied together inside the cos functions but I'm not convinced that's how they should be!

for some reason its not letting me attach my screen shot, what format do images need to be in?

the cos function is for the modulation so its cos(2pi 10*10^3t) the 10*10^3t is for 10 khz frequency/ its the same for the 125*10^3t thats 125khz frequency

Original post by sly studies

for some reason its not letting me attach my screen shot, what format do images need to be in?

the cos function is for the modulation so its cos(2pi 10*10^3t) the 10*10^3t is for 10 khz frequency/ its the same for the 125*10^3t thats 125khz frequency

the cos function is for the modulation so its cos(2pi 10*10^3t) the 10*10^3t is for 10 khz frequency/ its the same for the 125*10^3t thats 125khz frequency

I think standard jpegs or pngs should work. So you're saying the 1st term is literally

$\displaystyle \cos(2\pi \times 10 \times 10^3 \times t)$ ?

Original post by davros

I think standard jpegs or pngs should work. So you're saying the 1st term is literally

$\displaystyle \cos(2\pi \times 10 \times 10^3 \times t)$ ?

$\displaystyle \cos(2\pi \times 10 \times 10^3 \times t)$ ?

its not working i've tried with a few images i'll try from my phone in a moment.

the only multiplication sign on the actual question is between the 10 and the 10^3.

to plot the graph of the modulation i replaced t with (x)

the long equation vo: from the original post i have plotted and looks correct

I am also struggling with this question also. There doesn't seem to be any question in the workbook which has a relevance to this one? I have spoken with one of the tutor's today and he pointed me in the direction of section 2.3.3.

Original post by lukefreddie123!

I am also struggling with this question also. There doesn't seem to be any question in the workbook which has a relevance to this one? I have spoken with one of the tutor's today and he pointed me in the direction of section 2.3.3.

2.3.3 is for after the expansion I believe

I think it's 4* cos(2π125*10^t)

Then 2cos(2π10*10t)cos(2π125*10^3t)

Once expanded use

1/2 cos (a+b)+cos(a-b) but I'm struggling with the expansion. I may be wrong but the question says expand then uses the appropriate trig identity

Original post by sly studies

2.3.3 is for after the expansion I believe

I think it's 4* cos(2π125*10^t)

Then 2cos(2π10*10t)cos(2π125*10^3t)

Once expanded use

1/2 cos (a+b)+cos(a-b) but I'm struggling with the expansion. I may be wrong but the question says expand then uses the appropriate trig identity

I think it's 4* cos(2π125*10^t)

Then 2cos(2π10*10t)cos(2π125*10^3t)

Once expanded use

1/2 cos (a+b)+cos(a-b) but I'm struggling with the expansion. I may be wrong but the question says expand then uses the appropriate trig identity

You really need to post a picture of the question. Click on the camera in the post / reply toolbar to upload an image or upload it to another image sharing site/social media and link it.

I will upload a link to the image this afternoon. As I cannot load a picture either

Original post by mqb2766

You really need to post a picture of the question. Click on the camera in the post / reply toolbar to upload an image or upload it to another image sharing site/social media and link it.

https ://ibb.co/xjHwKXP

Original post by lukefreddie123!

https://ibb.co/xjHwKXP

As per the earlier post, the OP has the right identity, for the term

2cos(2π125*10^3 t) cos(2π10*10^3 t)

So

cos(a)cos(b) = (cos (a+b)+cos(a-b)) / 2

so replace a and b with the appropriate expressions and add and subtract and you get (two of) the desired frequencies given in the question

(edited 1 year ago)

Hi Mqb, Thanks for your response. Yes i agree it's Cos(a)Cos(b) - (cos(a+b) +cos(a-b)/2. When adding cos (a) to cos(b) do you only add the magnitude of cos (2+1) of do you add the entire equation(2cos(2π125*10^3t) + (cos(2π10*10^3t).?

If the latter is correct would Cos (a)+Cos(b) = 3cos(4π135*10^6t^2) be correct?

If the latter is correct would Cos (a)+Cos(b) = 3cos(4π135*10^6t^2) be correct?

Original post by mqb2766

a = 2π125*10^3 t

b = 2π10*10^3 t

So add and subtract them and use them in the identity

cos(a)cos(b) = (cos (a+b)+cos(a-b)) / 2

Youre adding and subtracting the arguments of cos.

b = 2π10*10^3 t

So add and subtract them and use them in the identity

cos(a)cos(b) = (cos (a+b)+cos(a-b)) / 2

Youre adding and subtracting the arguments of cos.

What happens to the 4 before the 2 I assumed one side stayed 2 cos whole the other side became 4 cos I will read your response in more detail after work thanks for the reply

(edited 1 year ago)

Original post by sly studies

What happens to the 4 before the 2 I assumed one side stayed 2 cos whole the other side became 4 cos I will read your response in more detail after work thanks for the reply

That is just the identity part for

cos(a)cos(b) = ...

in the original question it is scaled by ... and there is another term added to it. I guess thats your job to make it work for your assignment.

Original post by mqb2766

That is just the identity part for

cos(a)cos(b) = ...

in the original question it is scaled by ... and there is another term added to it. I guess thats your job to make it work for your assignment.

cos(a)cos(b) = ...

in the original question it is scaled by ... and there is another term added to it. I guess thats your job to make it work for your assignment.

Do you know any good websites for researching this topic for some reson in still struggling I think I'm missing something basic

Original post by sly studies

Do you know any good websites for researching this topic for some reson in still struggling I think I'm missing something basic

To simplify it you could consider

(4 + 2 cos(1))*cos(3)

What would you get using the above identity? Then just put the more complex expresson in place of the cos() arguments 1 and 3.

Most of the questions are about using trig identities to prove others. Rather than just using them as youre doing here.

(edited 1 year ago)

are you still stuck on this i got a distinction so if you need help i am more then willing

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