Electrical Engineering maths help with Desmos graphing calculator

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Original post by thomas0611
Do i need to keep the same equation but rearrange where the pi/4 goes?

y = 8sin (2pi*500000 x - π / 4)

The expression is correct, but your desmos wasnt as -pi/4 was not passed to sin(...) as it lay outside the brackets.
(edited 1 week ago)
Original post by mqb2766
The expression is correct, but your desmos wasnt as -pi/4 not passed to sin(...) as it lay outside the brackets.

Ok I understand you now. Is this how it should look?

https://www.desmos.com/calculator/fdo03jjhf1

Original post by thomas0611
Ok I understand you now. Is this how it should look?

https://www.desmos.com/calculator/fdo03jjhf1

Pretty much. You only need one set of brackets, but two sets has the same behaviour.

Tbh, plotting the signal should really get you to try and understand the signal/expression so what the 8 represents, the -pi/4, the frequency, so assuming you do, then you should be able to explain why the graph is correct.
Original post by mqb2766
Pretty much. You only need one set of brackets, but two sets has the same behaviour.

Tbh, plotting the signal should really get you to try and understand the signal/expression so what the 8 represents, the -pi/4, the frequency, so assuming you do, then you should be able to explain why the graph is correct.

Am I correct in saying:
8 represents the peak voltage
-pi/4 causes a phase shift resulting in the output signal to lag -45 degrees
frequency is cycling 500,000 times per second

Thanks
Original post by thomas0611
Am I correct in saying:
8 represents the peak voltage
-pi/4 causes a phase shift resulting in the output signal to lag -45 degrees
frequency is cycling 500,000 times per second

Thanks

Thats pretty much correct (lag is synomous with the negative sign), but it sounds a bit like textbook definitions rather than you understanding how it applies to the graph. The peak voltage and the phase shift were incorrect in the old graph, so just make sure you understand how they affect the signal. Just change the values a bit and see how the graph changes so if you did something like
https://www.desmos.com/calculator/dw9ifrb8yq
(if you add a variable in the graph formula, desmos automatically adds a slider) then you can see if you can predict/explain the effect of changing parameters.
(edited 1 week ago)
Original post by mqb2766
Thats pretty much correct (lag is synomous with the negative sign), but it sounds a bit like textbook definitions rather than you understanding how it applies to the graph. The peak voltage and the phase shift were incorrect in the old graph, so just make sure you understand how they affect the signal. Just change the values a bit and see how the graph changes so if you did something like
https://www.desmos.com/calculator/dw9ifrb8yq
(if you add a variable in the graph formula, desmos automatically adds a slider) then you can see if you can predict/explain the effect of changing parameters.

Thanks I’ll have a play around with the graph once I finish shift.

Did I not do the peak voltage and phase shift correct in my graph?
Original post by thomas0611
Thanks I’ll have a play around with the graph once I finish shift.

Did I not do the peak voltage and phase shift correct in my graph?

In the new one yes and if you understand what they mean, then you should see that.
In the old one no.