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Hi all, First post... be gentle, lol. Old dog trying to learn new tricks....
I'm doing HNC/D Mech Engineering and I've come across this one that has me struggling a little.
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 = asin(3/6)+(pi/4) / 2pi = 0.6486 (assuming Vs is the +3V. I'm not convinced about this, though)
Am I along the right lines? Any assistance would be greatly appreciated.
I'm doing HNC/D Mech Engineering and I've come across this one that has me struggling a little.
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 = asin(3/6)+(pi/4) / 2pi = 0.6486 (assuming Vs is the +3V. I'm not convinced about this, though)
Am I along the right lines? Any assistance would be greatly appreciated.
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#2
(Original post by t90ark)
Hi all, First post... be gentle, lol. Old dog trying to learn new tricks....
I'm doing HNC/D Mech Engineering and I've come across this one that has me struggling a little.
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 = asin(3/6)+(pi/4) / 2pi = 0.6486 (assuming Vs is the +3V. I'm not convinced about this, though)
Am I along the right lines? Any assistance would be greatly appreciated.
Hi all, First post... be gentle, lol. Old dog trying to learn new tricks....
I'm doing HNC/D Mech Engineering and I've come across this one that has me struggling a little.
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 = asin(3/6)+(pi/4) / 2pi = 0.6486 (assuming Vs is the +3V. I'm not convinced about this, though)
Am I along the right lines? Any assistance would be greatly appreciated.
Frequency scales time, in this case a lot :-).
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Ah, that's another area I'm unsure about; should f be 1 or 1e+6 as its MHz?
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#4
(Original post by t90ark)
Ah, that's another area I'm unsure about; should f be 1 or 1e+6 as its MHz?
Ah, that's another area I'm unsure about; should f be 1 or 1e+6 as its MHz?
1MHz is a million cycles per second.
Last edited by mqb2766; 1 month ago
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#6
(Original post by t90ark)
Hi all, First post... be gentle, lol. Old dog trying to learn new tricks....
I'm doing HNC/D Mech Engineering and I've come across this one that has me struggling a little.
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 = asin(3/6)+(pi/4) / 2pi = 0.6486 (assuming Vs is the +3V. I'm not convinced about this, though)
Am I along the right lines? Any assistance would be greatly appreciated.
Hi all, First post... be gentle, lol. Old dog trying to learn new tricks....
I'm doing HNC/D Mech Engineering and I've come across this one that has me struggling a little.
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 = asin(3/6)+(pi/4) / 2pi = 0.6486 (assuming Vs is the +3V. I'm not convinced about this, though)
Am I along the right lines? Any assistance would be greatly appreciated.
(asin(1/2)+pi/4)/(2*pi) = 0.208333...
Check your calc (and include f).
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(Original post by mqb2766)
Just checked your numbers. Ignoring f, I got
(asin(1/2)+pi/4)/(2*pi) = 0.208333...
Check your calc (and include f).
Just checked your numbers. Ignoring f, I got
(asin(1/2)+pi/4)/(2*pi) = 0.208333...
Check your calc (and include f).
2.08333...10-7 with f?
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#8
(Original post by t90ark)
yes, same now. Must have had parenthesis wrong, or been in degrees or something. Thanks for the spot.
2.08333...10-7 with f?
yes, same now. Must have had parenthesis wrong, or been in degrees or something. Thanks for the spot.
2.08333...10-7 with f?
I was going to mention the lack of brackets on both the numerator and denominator in the OP, but decided not to be picky :-). Putting them in is a good habit to get into and checking rad mode .
Last edited by mqb2766; 1 month ago
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(Original post by mqb2766)
Yes, that's correct.
I was going to mention the lack of brackets on both the numerator and denominator in the OP, but decided not to be picky :-). Putting them in is a good habit to get into.
Yes, that's correct.
I was going to mention the lack of brackets on both the numerator and denominator in the OP, but decided not to be picky :-). Putting them in is a good habit to get into.
Just got to suss the graph now.
Much appreciated.
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#10
Hi, im struggling on the same question but HNC elec, any joy with the graph yet? its all ive got left to do before submitting but cant get my head around it.
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#11
(Original post by sheardy1)
Hi, im struggling on the same question but HNC elec, any joy with the graph yet? its all ive got left to do before submitting but cant get my head around it.
Hi, im struggling on the same question but HNC elec, any joy with the graph yet? its all ive got left to do before submitting but cant get my head around it.
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#12
(Original post by mqb2766)
What is the question/your problem?
What is the question/your problem?
I have the answer to the question itself but now it is asking for a graph drawn using suitable software to show at least 2 cycles of this signal. Im attempting to use desmos but I cant work out what to input to get the sine wave correct.
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#13
(Original post by sheardy1)
One of your commonly-used laboratory instantaneous test signal voltages (vs) is described by the equation… Vs = 6sin (2pi f t - 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 have the answer to the question itself but now it is asking for a graph drawn using suitable software to show at least 2 cycles of this signal. Im attempting to use desmos but I cant work out what to input to get the sine wave correct.
One of your commonly-used laboratory instantaneous test signal voltages (vs) is described by the equation… Vs = 6sin (2pi f t - 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 have the answer to the question itself but now it is asking for a graph drawn using suitable software to show at least 2 cycles of this signal. Im attempting to use desmos but I cant work out what to input to get the sine wave correct.
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(Original post by sheardy1)
Hi, im struggling on the same question but HNC elec, any joy with the graph yet? its all ive got left to do before submitting but cant get my head around it.
Hi, im struggling on the same question but HNC elec, any joy with the graph yet? its all ive got left to do before submitting but cant get my head around it.
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#16
(Original post by shaun1702)
where does the a come from before the sine?
where does the a come from before the sine?
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#17
(Original post by uberteknik)
It's the scaling factor for the amplitude of the waveform. i.e. it scales the maximum sin value for the parameter in question, voltage or current.
It's the scaling factor for the amplitude of the waveform. i.e. it scales the maximum sin value for the parameter in question, voltage or current.
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#18
(Original post by shaun1702)
How would i find the value of the amplitude ?
How would i find the value of the amplitude ?
The maximum amplitude is therefore 6 volts in the context of the question. -pi/4 is a phase shift causing the output signal to lag the reference signal by -45o
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#19
(Original post by shaun1702)
where does the a come from before the sin?
where does the a come from before the sin?
If you're refering to post #1, asin is the inverse sine function, also known as arcsin, or
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#20
(Original post by ghostwalker)
If you're refering to post #1, asin is the inverse sine function, also known as arcsin, or
If you're refering to post #1, asin is the inverse sine function, also known as arcsin, or
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