In working with complex exponentials we need some simple rules and short cuts for doing the algebra.
Remember that a complex number z is a vector in the 2-dimensional plane. It has a length (also called "magnitude") |z| and an angle ∠(z) measured counterclockwise from the real axis. You need to know the following facts:
Re(a+jb)=a; Im(a+jb)=b
|a+jb|=a2+b2 ; ∠(a+jb)=atan2(b,a)
(atan2(y,x) is the same operation as arctan(yx), except our solution is from −π to π to retain phase information. For more information see here).
|z1z2|=|z1||z2| ; |z1z2|=|z1||z2|
∠(z1z2)=∠(z1)+∠(z2); ∠(z1z2)=∠(z1)−∠(z2)
|ejθ|=1; ∠(ejθ)=θ
Now let's practice on the problem before us. We have a complex particular solution:
vPS(t)=Vi11+jωRCejωt
Let's get its magnitude. The complex number before us is a product of factors, so its magnitude is the product of the magnitude of the factors. Vi is a real number, so its magnitude is its absolute value. The second factor is a quotient, so its magnitude is the quotient of the magnitudes of the numerator and the denominator. The third factor is a unit vector; so its magnitude is 1.
To find the magnitude of the complex particular solution vPS(t), we need to calculate the magnitude of each factor separately and then multiply them together.
Let's break down the given complex number vPS(t) = Vi(1+ jωRCejωt) and calculate the magnitude:
1. Magnitude of Vi:
The magnitude of a real number Vi is simply its absolute value. So, |Vi| = |Vi|.
2. Magnitude of (1+ jωRCejωt):
Here, we have a sum of two complex numbers: 1 and jωRCejωt. To find the magnitude, we need to find the absolute value of each of these complex numbers and then add them.
For the first complex number 1, its magnitude is |1| = 1.
For the second complex number jωRCejωt, we can use the properties of magnitudes mentioned earlier. The magnitude of a product is equal to the product of the magnitudes. In this case, we have a product of j, ωRC, and ejωt.
The magnitude of j is 1, the magnitude of ωRC is simply its absolute value |ωRC|, and the magnitude of ejωt is 1. So, the magnitude of jωRCejωt is |jωRCejωt| = |j| × |ωRC| × |ejωt| = |ωRC|.
Putting it all together, the magnitude of (1+ jωRCejωt) is |1+ jωRCejωt| = 1 + |ωRC|.
3. Magnitude of ejωt:
Here, we have a unit vector ejθ, where θ is the argument of the complex number. In this case, the argument is ωt. The magnitude of any unit vector is always 1. So, |ejωt| = 1.
Now, let's multiply the magnitudes of all the factors together to find the magnitude of vPS(t):
|vPS(t)| = |Vi(1+ jωRCejωt)|
= |Vi| × |1+ jωRCejωt| × |ejωt|
= |Vi| × (1 + |ωRC|) × 1
= |Vi|(1 + |ωRC|)
Therefore, the magnitude of the complex particular solution vPS(t) is |Vi|(1 + |ωRC|).