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.

Scroll to see replies

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.

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.

Just about. Divide by 2*pi*f.

Frequency scales time, in this case a lot :-).

Original post by t90ark

Ah, that's another area I'm unsure about; should f be 1 or 1e+6 as its MHz?

The latter. Just think about the effect it has on the time scaling. 1Hz is one cycle per second.

1MHz is a million cycles per second.

(edited 3 years ago)

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.

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.

Just checked your numbers. Ignoring f, I got

(asin(1/2)+pi/4)/(2*pi) = 0.208333...

Check your calc (and include f).

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).

(asin(1/2)+pi/4)/(2*pi) = 0.208333...

Check your calc (and include 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?

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?

2.08333...10-7 with f?

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 and checking rad mode .

(edited 3 years ago)

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.

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.

oh please..... be picky. A mathematician I am not, but I'm trying, lol.

Just got to suss the graph now.

Much appreciated.

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.

What is the question/your problem?

Original post by mqb2766

What is the question/your problem?

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.

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.

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.

You could share the desmos graph. I guess you realise the time (x) axis range will be very small as f=10^6.

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.

I've done what I think is right, lol. I used a combination of excel and graph. Did the curve points in excel and then took them into graph.

Original post by shaun1702

where does the a come from before the sine?

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.

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.

How would i find the value of the amplitude ?

Original post by shaun1702

How would i find the value of the amplitude ?

The original question stated the equation as 6sin(2.pi.f.t - (pi/4)).

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 -45

Original post by shaun1702

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 $\sin^{-1}$

Original post by ghostwalker

If you're refering to post #1, asin is the inverse sine function, also known as arcsin, or $\sin^{-1}$

yes i was , thanks for that

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