Linearization Problem: Solving for Vout in Original and Shifted Coordinates

[grothem]

Homework Statement

Linearize the equation

Vout = (10^v)*sin(x)

about x=0,0.1, and 1. Write the equation in both the original coordinates and the shifted (linearized) coordinates.

Homework Equations

y=f(a)+f'(a)(x-a)

The Attempt at a Solution

dVout/dx = (10^v)*cos(x)
evaluating that equation at x=0,0.1, and 1 gives me
10^v, 10^v*(.995), and 10^v*(.54) , respectively
therefore, Vout = 10^v*sin(0)+10^v*(x-0) for x=0, and similar for x=0.1, and 1. This doesn't seem right though, and I'm not sure how to write the equation in the shifted coordinates.

 

[mighty2000]

Hello,

Thank you for your post. To linearize the equation Vout = (10^v)*sin(x) about x=0, 0.1, and 1, we can use the Taylor series expansion. The general form of the Taylor series expansion for a function f(x) about a point a is given by:

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

Applying this to our equation Vout = (10^v)*sin(x), we get:

Vout = (10^v)*sin(0) + (10^v)*cos(0)(x-0) + (10^v)*(-sin(0))(x-0)^2/2! + (10^v)*(-cos(0))(x-0)^3/3! + ...

Simplifying this, we get:

Vout = 0 + (10^v)*(x) + 0 + 0 + ...

Therefore, the linearized equation about x=0 is given by:

Vout = (10^v)*x

Similarly, for x=0.1 and x=1, the linearized equations would be:

Vout = (10^v)*sin(0.1) + (10^v)*(cos(0.1))(x-0.1)

Vout = (10^v)*sin(1) + (10^v)*(cos(1))(x-1)

I hope this helps. Let me know if you have any further questions.