Solving Question on Phasors: Transform into Exponential & Back to Sinusoidal

[Teczero]

Hi everyone!

I understand the general concept but I came across a function such as:

E(t)= 0.1 sin[10(pi)x]cos[6(pi)10^9t - Bz]

I'm supposed to combine them using phasors and I'm really confused how to

a) Transform it into the exponential function
b) Transform it back in the sinusoidal function

Thank you for the help =)

 

[Mr.Green]

First let us review how to covert a quantity written in phasor notation to time domain. As soon as you learn that it will be very simple to reverse the operation.

Assume that we have the phasor:

E(x, y, z) = A(x, y, z) exp^jф

Where:
-A is any real function of x, y, and z.
- ф is the phase reference.

The time domain of this phasor is e(x, y, z, t) and is given by:

e(x, y, z, t) = Re [ E(x, y, z) * exp^jωt] = Re [A(x, y, z) exp^j(ωt+ф)]
So,
e( x, y, z) = A(x, y, z) cos(ωt + ф)

Now to convert a time domain quantity to a phasor just remove the term [cos(ωt + ф)] and multiply A(x, y, z) by exp^jф

 

[Mene]

I would advise breaking down the problem into smaller steps to better understand and solve it. First, let's review the concept of phasors. A phasor is a complex number that represents the amplitude and phase of a sinusoidal function. It can be represented in the form Ae^(jθ), where A is the amplitude and θ is the phase angle.

To transform a sinusoidal function into its exponential form, we can use Euler's formula: e^(jθ) = cos(θ) + jsin(θ). This means that we can rewrite the function E(t) as:

E(t) = 0.1[e^(j10πx) - e^(-j10πx)][e^(j6π10^9t - Bz) + e^(-j6π10^9t + Bz)]

This is now in the form of A1e^(jθ1) + A2e^(jθ2), which can be represented as a phasor with two components: A1e^(jθ1) and A2e^(jθ2). To combine them, we can use the phasor addition formula: A1e^(jθ1) + A2e^(jθ2) = (A1cosθ1 + A2cosθ2) + j(A1sinθ1 + A2sinθ2).

To transform back into the sinusoidal function, we can use the inverse of Euler's formula: cos(θ) = (e^(jθ) + e^(-jθ))/2 and sin(θ) = (e^(jθ) - e^(-jθ))/2j. This means that we can rewrite the function as:

E(t) = 0.1[(e^(j10πx) + e^(-j10πx))/2][e^(j6π10^9t - Bz) + e^(-j6π10^9t + Bz))/2j]

Now we can simplify this expression to get the sinusoidal function again. I hope this helps and please let me know if you have any further questions.