Expert: Paul Klarreich - 1/15/2006

Question
Hi Paul,

I am reviewing an assignment solution and am very puzzled how the solution goes from one term to the next as there is no detailed reduction. I can get the first term but the next is a mysterty how it is reduced. I was wondering if you could show me how they dervied the 2nd term. Here they are:
C = constant of intergration
k = phase lead
w = small omega

V1(t) = Ce^-((R/L)t) + V2Lw*((R*cos(wt)+Lw*sin(wt))/(R^2 + (Lw)^2))

next term:
V1(t) = Ce^-((R/L)t) + V2Lw*((sin(wt + k))/(sqrt(R^2 + (Lw)^2))),
where k = tan^-1(R/(Lw)) > 0

If you can explain and show this to me, I would greatly appreciate it. If the typed terms are too difficult to understand (because of all the parentheses) I am willing to forward a copy of the assignment solutions and indicate the question.

Thanks,
Brian Dickey

Answer
Hi, Brian,

You wrote:
------------------------
Hi Paul,

I am reviewing an assignment solution and am very puzzled how the solution goes from one term to the next as there is no detailed reduction. I can get the first term but the next is a mysterty how it is reduced. I was wondering if you could show me how they dervied the 2nd term. Here they are:
C = constant of intergration
k = phase lead
w = small omega

V1(t) = Ce^-((R/L)t) + V2Lw*((R*cos(wt)+Lw*sin(wt))/(R^2 + (Lw)^2))

next term:

V1(t) = Ce^-((R/L)t) + V2Lw*((sin(wt + k))/(sqrt(R^2 + (Lw)^2))),
where k = tan^-1(R/(Lw)) > 0

If you can explain and show this to me, I would greatly appreciate it. If the typed terms are too difficult to understand (because of all the parentheses) I am willing to forward a copy of the assignment solutions and indicate the question.

Thanks,
Brian Dickey
-------------------------------------
I don't usually do electrical engineering, but this one is cute enough.  Here is part of the expression (the only part that changes.)

STANDARD WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


R*cos(wt)+Lw*sin(wt)     
-------------------- =
  R^2 + (Lw)^2  

R*cos(wt)+Lw*sin(wt)          1
-------------------- ------------------
sqrt(R^2 + (Lw)^2)  sqrt(R^2 + (Lw)^2)

The first fraction in that product can be written as:

    R                  Lw
---------- cos(wt) + ---------- sin(wt)
sqrt(....)           sqrt(....)

Now make a right triangle like this: An acute angle is k, R is opposite k, and Lw is adjacent to k.  Then the hypotenuse is that square root.

   /|
  / |
 /  | R    <--- clumsy attempt at diagram
/k  |
+----+
 Lw

Then you can easily see that:
tan k = R/Lw and therefore  k = arctan(R/Lw)

sin k = R/sqrt(...), the first coefficient.
cos k = Lw/sqrt(...), the second coefficient.

So that expression becomes: (written again)

    R                  Lw
---------- cos(wt) + ---------- sin(wt)
sqrt(....)           sqrt(....)

sin k cos(wt) + cos k sin(wt)

and that matches one of those trigonometric identities:

sin(A + B) = cos A sin B + sin A cos B

making it sin(wt + k)

You can probably handle the rest.

Paul