- #1

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a.cos(wt) + b.sin(wt) = M.cos(wt + ϕ)

Can you give me

Can you give me

**M**and**ϕ**in terms of**a**and**b**?- Thread starter hkBattousai
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- #1

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a.cos(wt) + b.sin(wt) = M.cos(wt + ϕ)

Can you give me**M** and **ϕ** in terms of **a** and **b**?

Can you give me

- #2

arildno

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cos(u+v)=cos(u)cos(v)-sin(u)sin(v)

That 's all you need.

That 's all you need.

- #3

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M = sqrt(a^2 + b^2)

and

ϕ = arctan(-b/a)

but I'm no sure.

Can anyone confirm it for me?

- #4

arildno

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No.

M = sqrt(a^2 + b^2)

and

ϕ = arctan(-b/a)

but I'm no sure.

Can anyone confirm it for me?

Try to do it for yourself, and we can correct whatever mistakes you make.

- #5

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Code:

```
M.cos(wt + ϕ) = a.cos(wt) + b.sin(wt)
cos(wt + ϕ) = (a/M).cos(wt) + (b/M).sin(wt)........(I)
cos(wt + ϕ) = cos(wt).cos(ϕ) - sin(wt).sin(ϕ)........(II)
From (I) and (II),
cos(ϕ) = (a/M)
sin(ϕ) = -(b/M)
cos^2(ϕ) + sin^2(ϕ) = (a^2 + b^2)/(M^2) = 1
We assume that M is always positive and we keep any negativity in the phase angle ϕ,
M = sqrt(a^2 + b^2)
sin(ϕ)/cos(ϕ) = tan(ϕ) = -(b/M)/(a/M) = -b/a
tan(ϕ) = -b/a ==> ϕ = arctan(-b/a)
```

- #6

arildno

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No.Is there anything wrong in my derivation?Code:`M.cos(wt + ϕ) = a.cos(wt) + b.sin(wt) cos(wt + ϕ) = (a/M).cos(wt) + (b/M).sin(wt)........(I) cos(wt + ϕ) = cos(wt).cos(ϕ) - sin(wt).sin(ϕ)........(II) From (I) and (II), cos(ϕ) = (a/M) sin(ϕ) = -(b/M) cos^2(ϕ) + sin^2(ϕ) = (a^2 + b^2)/(M^2) = 1 We assume that M is always positive and we keep any negativity in the phase angle ϕ, M = sqrt(a^2 + b^2) sin(ϕ)/cos(ϕ) = tan(ϕ) = -(b/M)/(a/M) = -b/a tan(ϕ) = -b/a ==> ϕ = arctan(-b/a)`

- #7

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I liked your way of "you must do it yourself if you want to success"... :)

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arildno

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It's true, isn't it?I liked your way of "you must do it yourself if you want to success"... :)

- #9

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In my question, this is a simple trigonometric identity, I expected a mathematician to write it for me since most of math guys have memorized and actively use these kind of identities.

- #10

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http://www.electro-tech-online.com/...transform-specific-frequency.html#post895294"

Final expression for v_{L}(t) greatly simplified by this trigonometric identity.

Final expression for v

Last edited by a moderator:

- #11

Landau

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t=0 gives a=M cos(ϕ),

t=pi/(2w) gives b=-M sin(ϕ).

Hence

a^2+b^2=M^2,

tan(ϕ)=-b/a.

- #12

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Wow, that's super simple, thanks!

t=0 gives a=M cos(ϕ),

t=pi/(2w) gives b=-M sin(ϕ).

Hence

a^2+b^2=M^2,

tan(ϕ)=-b/a.

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