# HNC engineering help

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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.
0
1 year ago
#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
Last edited by EmberPlayer; 1 year ago
0
#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
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.
0
1 year ago
#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.
What software are you using?
0
#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 have done this bit now, but i don't know how to make a visual ie put the information into a the graph software
.
0
#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.
its call graph
0
#7
(Original post by EmberPlayer)
What software are you using?
i have attached a screenshot of the software
0
1 year ago
#8
(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))
0
8 months ago
#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.
0
4 months ago
#10
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.
hi, did you manage to get it in the end? im stuck on the same question now, ive been on this for days trying to work out what to input into Graph
0
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