I am struggling with a question on my electrical engineering course. The question requires me to 'draw at least two cycles of this signal and annotate the drawing so that your non-technical colleagues may understand the relevant information it contains. It asks me to use the graphing calculator Desmos | Graphing Calculator.

the question is as follows...

one of your labratory instantaneous test signal voltages (Vs) is described by the equation....

Vs = 8sin (2 π f t - ( π / 4)

where f=500kHz and t represents time.

Make t the subject and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +5v.

I have ended up with...

t = (sin^-1(5/8)+ π/4) / 2 π f

f = 5 x 10^5

I then get 4.649005 x10^-7

I am now unsure how to input this in to the graphing calculator?!

Any help would be much appreciated.

Thanks

the question is as follows...

one of your labratory instantaneous test signal voltages (Vs) is described by the equation....

Vs = 8sin (2 π f t - ( π / 4)

where f=500kHz and t represents time.

Make t the subject and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +5v.

I have ended up with...

t = (sin^-1(5/8)+ π/4) / 2 π f

f = 5 x 10^5

I then get 4.649005 x10^-7

I am now unsure how to input this in to the graphing calculator?!

Any help would be much appreciated.

Thanks

Scroll to see replies

Original post by richjpark

I am struggling with a question on my electrical engineering course. The question requires me to 'draw at least two cycles of this signal and annotate the drawing so that your non-technical colleagues may understand the relevant information it contains. It asks me to use the graphing calculator Desmos | Graphing Calculator.

the question is as follows...

one of your labratory instantaneous test signal voltages (Vs) is described by the equation....

Vs = 8sin (2 π f t - ( π / 4)

where f=500kHz and t represents time.

Make t the subject and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +5v.

I have ended up with...

t = (sin^-1(5/8)+ π/4) / 2 π f

f = 5 x 10^5

I then get 4.649005 x10^-7

I am now unsure how to input this in to the graphing calculator?!

Any help would be much appreciated.

Thanks

the question is as follows...

one of your labratory instantaneous test signal voltages (Vs) is described by the equation....

Vs = 8sin (2 π f t - ( π / 4)

where f=500kHz and t represents time.

Make t the subject and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +5v.

I have ended up with...

t = (sin^-1(5/8)+ π/4) / 2 π f

f = 5 x 10^5

I then get 4.649005 x10^-7

I am now unsure how to input this in to the graphing calculator?!

Any help would be much appreciated.

Thanks

It would want you to plot the t-Vs curve for a couple of cycles and maybe include the Vs=5 line so you can see where it intersects with the curve. So the x axis is time and the y axis is voltage. Note that the frequency is large so you may need to zoom in quite a bit to see just two cycles. Have a go and post your attempt if still stuck.

Original post by richjpark

hi, thanks for the reply. I cant work out how to post an imagine in my reply? hopefully the link above shows an image of my laptop screen!?

I'm unsure where and how to input the figures in to the graphing calculator to produce the curve?

thanks

Original post by richjpark

If you click on the "tv arrow" in the top right, you get an url which you can post.

But you need to plot the t-Vs sinusoidal curve and the Vs=5, neither of which youve done / or explained what youve tried to do.

You could start by plotting a simple sinusoidal curve

Vs = sin(t)

and take it from there.

(edited 1 month ago)

Original post by mqb2766

If you click on the "tv arrow" in the top right, you get an url which you can post.

But you need to plot the t-Vs sinusoidal curve and the Vs=5, neither of which youve done / or explained what youve tried to do.

You could start by plotting a simple sinusoidal curve

Vs = sin(t)

and take it from there.

But you need to plot the t-Vs sinusoidal curve and the Vs=5, neither of which youve done / or explained what youve tried to do.

You could start by plotting a simple sinusoidal curve

Vs = sin(t)

and take it from there.

I think I'm out of my depth here! I've never used a graphing calculator before so no idea what figures to put where and in what format. I've got a meeting arranged with a tutor in a few weeks so hopefully they can show me how to use the graphing calculator then.

thanks for your help

Original post by richjpark

I think I'm out of my depth here! I've never used a graphing calculator before so no idea what figures to put where and in what format. I've got a meeting arranged with a tutor in a few weeks so hopefully they can show me how to use the graphing calculator then.

thanks for your help

thanks for your help

To plot

v = sin(t)

youd just type

sin(x)

in the left hand plane. So

https://help.desmos.com/hc/en-us/articles/4407512915469-Getting-Started-Creating-Your-First-Graph

should be enough to do it. But if youre feeling that unsure, then discuss with your tutor.

(edited 1 month ago)

Original post by richjpark

I think I'm out of my depth here! I've never used a graphing calculator before so no idea what figures to put where and in what format. I've got a meeting arranged with a tutor in a few weeks so hopefully they can show me how to use the graphing calculator then.

thanks for your help

thanks for your help

Hello

I am stuck on the same part of this question. Did you manage to figure out how to draw the two cycles?

Original post by thomas0611

Hello

I am stuck on the same part of this question. Did you manage to figure out how to draw the two cycles?

I am stuck on the same part of this question. Did you manage to figure out how to draw the two cycles?

Have you tried using desmos and doing a basic sin graph as a start? You can post your attempt using the "Share Graph" button in the top right of desmos

Original post by mqb2766

Have you tried using desmos and doing a basic sin graph as a start? You can post your attempt using the "Share Graph" button in the top right of desmos

I've been using 'Graph' software but I have input the equation into desmos to share, as follows:https://www.desmos.com/calculator/ykizzynoj2

Thanks

Original post by thomas0611

Hello

I am stuck on the same part of this question. Did you manage to figure out how to draw the two cycles?

I am stuck on the same part of this question. Did you manage to figure out how to draw the two cycles?

Hi,

Think I cracked it. I put...

y = 8sin (2 π f x - π / 4)

In the the left hand panel. You have to replace f with the figure given for the frequency (make sure its in Hz not kHz).

This should produce a sinewave. I took 2 cycles to mean zoom to an extent on the graph that shows 2 full complete cycles of the sinewave.

Hope that helps!

(edited 2 weeks ago)

Original post by richjpark

Hi,

Think I cracked it. I put...

y = 8sin (2 π f x - ( π / 4)

In the the left hand panel. You have to replace f with the figure given for the frequency (make sure its in Hz not kHz).

This should produce a sinewave. I took 2 cycles to mean zoom to an extent on the graph that shows 2 full complete cycles of the sinewave.

Hope that helps!

Think I cracked it. I put...

y = 8sin (2 π f x - ( π / 4)

In the the left hand panel. You have to replace f with the figure given for the frequency (make sure its in Hz not kHz).

This should produce a sinewave. I took 2 cycles to mean zoom to an extent on the graph that shows 2 full complete cycles of the sinewave.

Hope that helps!

Hello

https://www.desmos.com/calculator/olh25pm4og

Does this look similar to your sine wave?

Thanks

Original post by thomas0611

No, sorry, my fault, I typed it wrong in my reply. It should be y = 8sin (2 π f x - π / 4).

You rightly substituted f for 500000 but add it directly in to the equation. So it should be y = 8sin (2 π 500000 x - π / 4).

On the settings tab change the x axis values to -1x10^-6 and 1x10^-6.

Send a link to what that produces

I think i have done it correctly now.

https://www.desmos.com/calculator/8fj1tbrhl1

Did you add notes explaining the sine wave?

Appreciate your help.

https://www.desmos.com/calculator/8fj1tbrhl1

Did you add notes explaining the sine wave?

Appreciate your help.

Original post by thomas0611

I think i have done it correctly now.

https://www.desmos.com/calculator/8fj1tbrhl1

Did you add notes explaining the sine wave?

Appreciate your help.

https://www.desmos.com/calculator/8fj1tbrhl1

Did you add notes explaining the sine wave?

Appreciate your help.

The - pi/4 needs to be inside the sin argument as its a horizontal shift, not a vertical one.

Original post by thomas0611

I think i have done it correctly now.

https://www.desmos.com/calculator/8fj1tbrhl1

Did you add notes explaining the sine wave?

Appreciate your help.

https://www.desmos.com/calculator/8fj1tbrhl1

Did you add notes explaining the sine wave?

Appreciate your help.

Try playing about with the y axis figures too. I tried -10 and 10 and it came out better. Yeah I added some annotations, I found this website useful...

https://www.electronics-tutorials.ws/accircuits/sinusoidal-waveform.html

No problem 👍

Original post by richjpark

Hi,

Think I cracked it. I put...

y = 8sin (2 π f x - π / 4)

In the the left hand panel. You have to replace f with the figure given for the frequency (make sure its in Hz not kHz).

This should produce a sinewave. I took 2 cycles to mean zoom to an extent on the graph that shows 2 full complete cycles of the sinewave.

Hope that helps!

Think I cracked it. I put...

y = 8sin (2 π f x - π / 4)

In the the left hand panel. You have to replace f with the figure given for the frequency (make sure its in Hz not kHz).

This should produce a sinewave. I took 2 cycles to mean zoom to an extent on the graph that shows 2 full complete cycles of the sinewave.

Hope that helps!

Yours has the same problem mentioned in #14

Original post by mqb2766

The - pi/4 needs to be inside the sin argument as its a horizontal shift, not a vertical one.

Would that mean adding more brackets like this y = 8sin ((2pi*500000 x - π / 4))

Original post by thomas0611

Would that mean adding more brackets like this y = 8sin ((2pi*500000 x - π / 4))

You dont need both sets, but you do need the sin(2pi...-pi/4)

Original post by mqb2766

You dont need both sets, but you do need the sin(2pi...-pi/4)

Do i need to keep the same equation but rearrange where the pi/4 goes?

y = 8sin (2pi*500000 x - π / 4)

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