Well the angle of the numerator is zero because it's just a real number.

And so I've just got the angle of the denominator so

since it's in the denominator I've got a minus sign here.

Minus arc tan of the imaginary over the real.

Imaginary part of the real is one-ninth, and that's the inverse tangent.

And that is equal to a degrees- 6.3 degrees.

So using these formulas up here with what

you just saw for the magnitude in angle,

then I have iout as my output is equal

to the amplitude of H times 10.

So it would be 0.99 cosine

same frequency 100 t and

then this angle, -6.3 degrees.

And that to shows how to use the transfer function to solve a particular problem

at a particular frequency.

If I wanted to go back and solve for a different input frequency,

different input amplitude or frequency, I don't have to resolve this whole problem.

I now have this relationship between the input and the output that I can use

very easily if I'm going to change the angles,

the angle or the amplitude, or even the frequency.