# Math Question!

Discussion in 'Homework Help' started by Nishki, Apr 26, 2009.

1. ### Nishki Thread Starter Member

May 30, 2008
11
0
Hello,
I was wondering if anyone might be able to help me. I am stuck on a maths question that involves the addition of two sinusoidal waveforms. I have two maths exercise books but I can't really understand the examples in these books! I have already determined the resultant waveform using excel, but I am not sure how to represent this as an equation. Any help would be greatly appreciated!
Thanks

The two equations are -
V1 = 2.75sin((t/5)+0.15)
V2 = 1.55sin((t/5)+0.35)

2. ### PRS Well-Known Member

Aug 24, 2008
989
36
Are the phase angles given in radians?

3. ### Ratch New Member

Mar 20, 2007
1,068
4
Nishki,

sin(A+B) = sin(A)cos(B)+cos(A)sin(B) where t/5 and 0.35 are A and B respectively

V1 = 2.719120464*sin((1/5)*t)+.4109548644*cos((1/5)*t)

V2 = 1.456027705*sin((1/5)*t)+.5314916016*cos((1/5)*t)

V1 + V2 = 4.175148169*sin((1/5)*t)+.9424464660*cos((1/5)*t)

sqrt(4.175148169^2+.9424464660^2) = 4.280194805

V1+V2 = 4.280194805(4.175148169*sin((1/5)*t)+.9424464660*cos((1/5)*t)))/4.280194805

= 4.280194805(.9754575105*sin((1/5)*t)+.2201877505*cos((1/5)*t)))
= 4.280194805(cos(Arccos((.9754575105))*sin((1/5)*t)+sin(Arcsin(.2201877505))*cos((1/5)*t)))
= 4.280194805(sin((1/5)*t)*cos(0.2220069400)+cos((1/5)*t)*sin(0.2220069400))
=4.280194805*sin((1/5)*t+0.2220069400)

Ratch

4. ### Nishki Thread Starter Member

May 30, 2008
11
0
Sorry, should have made that clear! Yes they are.

5. ### Nishki Thread Starter Member

May 30, 2008
11
0
Okay, thank you very much Ratch. I have just checked the equation against my graph and you are spot on so thank you again!

6. ### PRS Well-Known Member

Aug 24, 2008
989
36
Ratch is right. I just checked it with the method of phasors. This is where you convert to complex numbers, add them, then turn the complex numbers back into sines. Using < for angle....

V1 = .41-j2.72
V2 = .527-j1.46
V1+V2 = .937-j4.18 = 4.28<-1.35 = 4.28cos(t/5 -1.35) = 4.28sin(t/5 +.22)

As you can see phasors make it simple.

7. ### Ratch New Member

Mar 20, 2007
1,068
4
PRS,

Agreed, but I am puzzled by some of your calculations, even though you got the right answer. Since both V1 and V2 are in the first quadrant, both the x and y components should be positive. Some of your components are negative and it appears that you have the x and y components swapped. Here are my calculations.

V1 = 2.75sin((t/5)+0.15) ==> 2.72 + j0.41

V2 = 1.55sin((t/5)+0.35) ==> 1.46 + j0.53

V1+V2 = 4.18 + j0.94 ==> 4.28/_0.22 ==> 4.28sin(t/5 + 0.22)

Ratch

8. ### PRS Well-Known Member

Aug 24, 2008
989
36
Funny you should ask. I was taught to use cosines when using this method. After I posted I thought to myself: Why not just keep it in sines? I intended to recalculate this with sines but you did it for me. Thanks. I just learned something.