shaun1702
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I'm doing HNC/D Mech Engineering. i was looking for help on this question i have answered but am sure is incorrect.

One of your commonly-used laboratory instantaneous test signal voltages (vs) is
described by the equation…

Vs = 6sin(2pift - (pi/4))

where f = 1MHz and t represents time.
Make time (t) the subject of this formula, and hence determine the first point in time when
the instantaneous signal voltage has a magnitude of +3V.

I've rearranged the formula thus t = sin(3/6)+(pi/4) / 2pi(1.10^6) = 2.013 x 10^-7

Any advise would be greatly appreciated.
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EmberPlayer
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Hi, you forgot to take the arcsin of both sides while rearranging to make time t the subject. From the plot, the X refers to the first point in time when Vs = 3V
Attachment 1002972
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Last edited by EmberPlayer; 2 weeks ago
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shaun1702
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(Original post by EmberPlayer)
Hi, you forgot to take the arcsin of both sides while rearranging to make time t the subject. From the plot, the X refers to the first point in time when Vs = 3V
Attachment 1002972
Attachment 1002974Name:  02EB8A42-A49A-49EE-8609-D6164FC63423.jpeg
Views: 4
Size:  58.0 KB
i understand this now thanks for that, the second part of the question asks me to make a graph like yours but im not sure if i am inputting it right into the software.
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EmberPlayer
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(Original post by shaun1702)
i understand this now thanks for that, the second part of the question asks me to make a graph like yours but im not sure if i am inputting it right into the software.
What software are you using?
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shaun1702
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(Original post by shaun1702)
I'm doing HNC/D Mech Engineering. i was looking for help on this question i have answered but am sure is incorrect.

One of your commonly-used laboratory instantaneous test signal voltages (vs) is
described by the equation…

Vs = 6sin(2pift - (pi/4))

where f = 1MHz and t represents time.
Make time (t) the subject of this formula, and hence determine the first point in time when
the instantaneous signal voltage has a magnitude of +3V.

I've rearranged the formula thus t = sin(3/6)+(pi/4) / 2pi(1.10^6) = 2.013 x 10^-7

Any advise would be greatly appreciated.
i have done this bit now, but i don't know how to make a visual ie put the information into a the graph software
.
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shaun1702
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(Original post by shaun1702)
i understand this now thanks for that, the second part of the question asks me to make a graph like yours but im not sure if i am inputting it right into the software.
its call graph
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shaun1702
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(Original post by EmberPlayer)
What software are you using?
i have attached a screenshot of the software
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EmberPlayer
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(Original post by shaun1702)
i have attached a screenshot of the software
I’m not personally familiar with that particular software, however I’d imagine all you need to do is change the function f(x) to equal 6*sin((2*pi*1*(10^6)*x) - (pi/4))
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