Find and express the results in polar form for the following:
z = 2(e^(-jω) - e^(j7ω)), where
ω = omega
j is the math equivalent of i = sqrt(-1)
With that pesky 7 in there, I'm at a loss for any way to combine the terms.
To find the result of z in polar form, we first need to simplify the expression and express it in the form a + bi.
Let's break down the given expression step by step:
z = 2(e^(-jω) - e^(j7ω))
Now, let's focus on the exponentials inside the parentheses.
Using Euler's formula, which states e^(jθ) = cos(θ) + j*sin(θ), we can rewrite the expression as:
z = 2(e^(-jω) - e^(j7ω))
z = 2[(cos(-ω) + j*sin(-ω)) - (cos(7ω) + j*sin(7ω))]
Next, let's expand and simplify the expression within the brackets:
z = 2[(cos(-ω) - cos(7ω)) + j*(sin(-ω) - sin(7ω))]
Now, let's simplify the trigonometric functions:
cos(-ω) = cos(ω) (cosine is an even function)
cos(7ω) = cos(-ω) = cos(ω) (cosine is a periodic function with period 2π)
sin(-ω) = -sin(ω) (sine is an odd function)
sin(7ω) = -sin(-ω) = -sin(ω) (sine is a periodic function with period 2π)
Thus, the expression becomes:
z = 2[(cos(ω) - cos(ω)) + j*(-sin(ω) - (-sin(ω)))]
z = 2[0 + j*0]
z = 0
So, the result of z is 0.
To express the result in polar form, we represent z as z = r * e^(jθ), where r is the magnitude of z and θ is the phase angle.
For z = 0, the magnitude (r) is 0, meaning that z lies at the origin, and the phase angle (θ) can be any angle since the magnitude is zero.
Therefore, the polar form of z = 0 is simply 0.