help with trigonometric derivatives of 2 sine waves.

Thread Starter

sian_willcocks

Joined Mar 8, 2016
10
Hi

I am currently stuck on a particular question in maths that has completely thrown me.

I have been given:
Va = 2sin (314.2t)
Vb = 2sin (314.2t-70°)

The task is to determine a single wave from a combination (additive) of 2 waves.

I have completed the graph and written down the formula for the resulting wave;
Vc = 3.26sin (314.2t+36°)
This may not be correct but it must be in the form Vsin (wt-phi)

I am then asked to use trigonometry to derive an expression for the resultant wave form in the same form?

This has completely wiped me, the first thing that comes to mind is sina+b=sinacosb+cosasinb but have no idea if I am thinking along the right lines or what to do with it if I am!!

Please help!!

Sian.
 

WBahn

Joined Mar 31, 2012
25,918
Hi

I am currently stuck on a particular question in maths that has completely thrown me.

I have been given:
Va = 2sin (314.2t)
Vb = 2sin (314.2t-70°)

The task is to determine a single wave from a combination (additive) of 2 waves.

I have completed the graph and written down the formula for the resulting wave;
Vc = 3.26sin (314.2t+36°)
This may not be correct but it must be in the form Vsin (wt-phi)

I am then asked to use trigonometry to derive an expression for the resultant wave form in the same form?

This has completely wiped me, the first thing that comes to mind is sina+b=sinacosb+cosasinb but have no idea if I am thinking along the right lines or what to do with it if I am!!

Please help!!

Sian.
Your graphical answer is close, but you've made one big mistake.

Consider the value at t=0. Will Va be positive, negative, or zero? What about Vb? Based on that, what will Vc be? Now ask whether your expression for Vc will yield the same result.
 

RBR1317

Joined Nov 13, 2010
550
I have completed the graph...
Not sure what that graph may be, but if it is a phasor diagram then it should look something like the attached image. A phasor diagram will also make clear why your answer is wrong. When carefully drawn, the result can be read directly from the phasor diagram with slide rule accuracy. Not quite as good as a $9 scientific calculator (my Casio fx-260solar), but I got through four years of engineering school with just an inherited Keuffel & Esser.
 

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Thread Starter

sian_willcocks

Joined Mar 8, 2016
10
Your graphical answer is close, but you've made one big mistake.

Consider the value at t=0. Will Va be positive, negative, or zero? What about Vb? Based on that, what will Vc be? Now ask whether your expression for Vc will yield the same result.
My expression works when I tabulate it so I can't see where I have gone wrong?
 

WBahn

Joined Mar 31, 2012
25,918
My expression works when I tabulate it so I can't see where I have gone wrong?
That just tells me that you didn't check it the way I recommended.

At t=0, Va has an argument of 0° meaning that it will be zero.
At t=0, Vb has an argument of -70° meaning that it will be negative.

Therefore Vc = Va + Vb will be negative.

But at t=0, YOUR Vc will have an argument of +36° meaning that it will be positive.

Still don't believe me? Simply run the numbers!

At t=0, Va = 2*sin(0) = 0.0
At t=0, Vb = 2*sin(-70°) = -1.879

So at t=0, Vc = -1.879

But at t=0, YOUR Vc = 3.26 * sin(36°) = 1.916
 

Thread Starter

sian_willcocks

Joined Mar 8, 2016
10
That just tells me that you didn't check it the way I recommended.

At t=0, Va has an argument of 0° meaning that it will be zero.
At t=0, Vb has an argument of -70° meaning that it will be negative.

Therefore Vc = Va + Vb will be negative.

But at t=0, YOUR Vc will have an argument of +36° meaning that it will be positive.

Still don't believe me? Simply run the numbers!

At t=0, Va = 2*sin(0) = 0.0
At t=0, Vb = 2*sin(-70°) = -1.879

So at t=0, Vc = -1.879

But at t=0, YOUR Vc = 3.26 * sin(36°) = 1.916
When calculating Vb I get a positive answer of 1.87 as subtracting the minus 70 makes it a positive?

2 minuses make a positive right?
 

Thread Starter

sian_willcocks

Joined Mar 8, 2016
10
Where do you see two minuses?

What is (0 - 70°) ?

You are either adding -70 to 0, or you are subtracting 70 from 0.
I was given the equation 2sin(wt-phi) to work with though on the sheet I was given Vb is written as 2sin (314.2t-70°).

As I have been given phi as -70° I used 2sin (314.2t- -70°)
 

WBahn

Joined Mar 31, 2012
25,918
I was given the equation 2sin(wt-phi) to work with though on the sheet I was given Vb is written as 2sin (314.2t-70°).

As I have been given phi as -70° I used 2sin (314.2t- -70°)
How were you given that phi was -70°?

If you were specifically given Vb = 2sin(314.2t - 70°), on what basis can you change that?

If Vb = 2sin(wt - phi) and if Vb = 2sin(314.2t - 70°), then phi = +70°.

If you were separately given that, for Vb, phi = -70° then you have contradictory information and you need to confirm what is what with your instructor.
 

Thread Starter

sian_willcocks

Joined Mar 8, 2016
10
I am going to contact my tutor as I am now confused as to whether I have used an incorrect figure. I'm still looking for help on the second part! I though I had this part complete
 

WBahn

Joined Mar 31, 2012
25,918
I am going to contact my tutor as I am now confused as to whether I have used an incorrect figure. I'm still looking for help on the second part! I though I had this part complete
Follow the recommendation made by Alec_t in Post #3. The answer drops out of that in just a couple of lines of simple algebra.
 

WBahn

Joined Mar 31, 2012
25,918
I will give that a go. I've not used it before or even seen it before so this could be interesting!
A good rule is to not use identities that you haven't at least been through the proof of at least once. It's not the end of the world if that isn't the case, but you should want to take whatever opportunities you can to improve your math literacy.

Are you familiar with the more basic/common trig identities

\(
\sin\(A \, \pm \, B\) \; = \; \sin \( A \) \cos \( B \) \, \pm \, \cos \( A \) \sin \( B \)
\.
\cos\(A \, \pm \, B\) \; = \; \cos \( A \) \cos \( B \) \, \mp \, \sin \( A \) \sin \( B \)
\)

If so, then use them to start with the right hand side and see if you can reduce it to the left hand side.
 

MrAl

Joined Jun 17, 2014
7,668
Hi,

Has anyone mentioned this yet? If not here is a challenge...

Start with the general form for two sine (or cosine) waves with phase shifts pha and phb:
A*sin(x+pha)+B*sin(x+phb)

Expand that using known forms, then collect terms. With a little imagination we end up with two rotating vectors 90 degrees out of phase, which have properties where we can easily find both the amplitude and phase shift of any two added waves where they both have their individual phase shifts pha and phb as above.
It's also almost the same for the two added cosines form.
Very interesting to try if you havent already :)
 

RBR1317

Joined Nov 13, 2010
550
I've attached my graph and what I have so far.
I tried to recreate your graph, but got this instead. Since the function definition uses a combination of radian and degree measure, the graph uses a time scale such that 360 time units equals 2π radians. Also note that a negative phase shift moves the waveform to the right because it takes more time to overcome the negative phase shift.
 

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Thread Starter

sian_willcocks

Joined Mar 8, 2016
10
I have had confirmation from my tutor to use it as it's shown on the sheet, not the minus 70 he gave me so I will be starting again and creating a new graph and I shall go from there.

I shall be back if I am still completely lost!
 
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