Electronic – Unable to verify Superposition Principle via Matlab

Superposition of dependent sources isn't prohibited: Superposition of Dependent Sources is Valid in Circuit Analysis.

The author has investigated the presentation of superposition in circuits texts by surveying twenty introductory books on circuit analysis. Fourteen explicitly state that if a dependent source is present, it is never deactivated and must remain active (unaltered) during the superposition process. The remaining six specifically refer to the sources as being independent in stating the principle of superposition. Three of these present an example circuit containing a dependent source which is never deactivated. The other three do not present an example in which dependent sources are present. From this limited survey, it is clear that circuits texts either state or imply that superposition of dependent sources is not allowed. The author contends that this is a misconception.

As a simple example using superposition of a dependent source consider the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

By superposition, we can write the equation for \$V_x\$ by inspection:

$$V_x = V_s\frac{R_2}{R_1 + R_2} + 5i_x R_1|| R_2 $$

We also have, by inspection

$$i_x = \frac{V_s - V_x}{R_1} $$

Thus

$$V_x = V_s\frac{R_2}{R_1 + R_2} + 5 \frac{V_s - V_x}{R_1}R_1|| R_2$$

It's just algebra from here. No need for node equations or mesh equations.

The key to successfully using superposition with dependent sources is the following:

Do not attempt to solve for a numeric answer until the superposition sum has been written.

not know what to do with the capacitor and the inductor in parallel

Either write the differential equations with two storage elements, or solve it as a phasor circuit with two impedances in parallel and convert back to time domain.

how does differing angular frequencies change the summation of the two sources?

It doesn't. A + B is still A + B. Since \$\sin(\omega_1 t)\$ and \$\sin(\omega_2 t)\$ are orthogonal functions when \$\omega_1\ne\omega_2\$, there's no way to simplify \$\sin(\omega_1 t) + \sin(\omega_2 t)\$; it's already the simplest form you'll be able to write.

due to the current division theorem, the voltage across the capacitor is also the same voltage across the inductor?

It has nothing to do with the current division theorem. It's simply the consequence of being connected in parallel and Kirchoff's voltage law. No matter what path you take from node X to node Y, you must get the same potential difference between X and Y.

Do the same for the current source and add the two?

For a basic rundown of how to solve circuits by superposition, see here.