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shaun1702
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#1
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.
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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#2
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 1002974![Name: 02EB8A42-A49A-49EE-8609-D6164FC63423.jpeg
Views: 190
Size: 58.0 KB]()
Attachment 1002972
Attachment 1002974
Last edited by EmberPlayer; 1 year ago
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shaun1702
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#3
(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 1002974![Name: 02EB8A42-A49A-49EE-8609-D6164FC63423.jpeg
Views: 190
Size: 58.0 KB]()
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 1002974
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EmberPlayer
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#4
(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.
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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shaun1702
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#5
(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'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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shaun1702
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#6
(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.
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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shaun1702
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#7
(Original post by EmberPlayer)
What software are you using?
What software are you using?
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EmberPlayer
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#8
(Original post by shaun1702)
i have attached a screenshot of the software
i have attached a screenshot of the software
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MCCORMHEAD
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#9
Ive done the working outs but the figure im getting is 0.29 on the graph. When I worked it out I got 2.0833x10^-7 im confused shouldnt they be the same result.
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Anthc123
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#10
(Original post by MCCORMHEAD)
Ive done the working outs but the figure im getting is 0.29 on the graph. When I worked it out I got 2.0833x10^-7 im confused shouldnt they be the same result.
Ive done the working outs but the figure im getting is 0.29 on the graph. When I worked it out I got 2.0833x10^-7 im confused shouldnt they be the same result.
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