how to find steady state response?

Hi
I want to know how the terms in red block are solved.
I know that e^jwt=cos(wt)+j*sin(wt)
how to solve j term in j*sin(wt)?

In order to get to the final answer, it is not necessary to use the identity e^jwt = cos(wt)+j*sin(wt). The easiest way to solve this is with a software package such as Matlab (or the free equivalent, GNU Octave). However, if you must do everything by hand, then try something like this:

The first boxed term is:

\[\left(e ^{j0.2} - 0.1\right)\left(e ^{j0.2} - 0.5\right) = e ^{j0.4} - 0.5e ^{j0.2} - 0.1e ^{j0.2} + 0.05 \\ =
e ^{j0.4} - 0.6e ^{j0.2} + 0.05\]

We know (or you can look it up / prove it) that for any complex number x, it is always true that |1/x| = 1/|x|. Then, one way of computing the magnitude of x, |x|, is to remember that x multiplied by its complex conjugate, x^*, equals |x|²:

\[xx^*=|x|^2\]

Therefore:

\[| e ^{j0.4} - 0.6e ^{j0.2} + 0.05|=\sqrt{\left( e ^{j0.4} - 0.6e ^{j0.2} + 0.05\right)\left( e ^{-j0.4} - 0.6e ^{-j0.2} + 0.05\right)}\]

By multiplying out the brackets, you can rearrange this into the form:

\[=\sqrt{1.3625 - 1.26\left(\frac{1}{2}(e ^{j0.2} + e ^{-j0.2})\right) + 0.1\left(\frac{1}{2}(e ^{j0.4} + e ^{-j0.4})\right)}\]

Now, if we remember the very well known formula for cosine, cos(x) = (e^jx + e^-jx)/2, our equation becomes:

\[=\sqrt{1.3625 - 1.26\cos(0.2) + 0.1\cos(0.4)}\]

Using a calculator, we find this equals 0.46875. Therefore, inverting and multiplying by 3 gives 6.4001 (or 6.4, as in your solution).

You can follow a similar series of steps to find out the angle term. However, I would never do this type of calculation by hand, as it seems like a bit of a waste of time to me...

OK, in general we have:

y=1/{[exp(j*a)+b]*[exp(j*a)+c]}=1/[exp(2*j*a)+(b+c)*exp(j*a)+b*c]

let's focusing on the denominator. Since exp(j*a)=cos(a)+j*sin(a):

denominator=cos(2*a)+j*sin(2*a)+(b+c)*cos(a)+(b+c)*j*sin(a)+b*c

rearranging:

denominator=[cos(2*a)+(b+c)*cos(a)+b*c]+j*[sin(2*a)+(b+c)*sin(a)]

so now we have our function in the form:

y=1/(m+j*n)=m/(m^2+n^2)-j*n/(m^2+n^2) that is:

Real=m/(m^2+n^2)
Imag=-n/(m^2+n^2)

then the modulus will be: |y|=1/sqrt(m^2+n*2) while the phase will be <y=atan(-n/m)

of course m=cos(2*a)+(b+c)*cos(a)+b*c and n=sin(2*a)+(b+c)*sin(a)

numerically:

m=0.383
n=0.270

thus

|y|=1/sqrt(0.383^2+0.270^2)=2.14
<y=atan(-0.270/0.383)=-0.614

remember now you have to multiply the modulus by 3, in order to get the final solution.