Sketch and Determine the Transfer Function From Its Poles

[drumercalzone]

Homework Statement

If a transfer function has poles at (-1,2) and (-1,-2) and a zero at (0,0):

a) Sketch |H(jω)| on the interval 0≤ω≤10.
b) If |H(jω)| = 1, what is the approximate value of |H| at its highest point?

Homework Equations

The Attempt at a Solution

I think I have the sketch down, but I'm really confused on finding the transfer function. I know that it's the distance to the zeros over the distances to the poles, but I just really don't understand how to find them.

 

[anathema]

Can someone walk me through the process?

Sure, I can walk you through the process! First, let's define the transfer function H(s), where s = jω. In this case, the transfer function is:

H(s) = (s - 0)/(s - (-1,2))(s - (-1,-2))

Next, we need to find the distances from the zeros and poles to the origin (0,0). The distance from the origin to the zero is 0, since it is located at the origin. The distance from the origin to the pole at (-1,2) is √(1^2 + 2^2) = √5. The distance from the origin to the pole at (-1,-2) is √(1^2 + (-2)^2) = √5.

Now, we can plug these values into the transfer function:

H(s) = (s - 0)/((√5 - s)(√5 + s))

To simplify this, we can use the difference of squares formula:

H(s) = s/(5 - s^2)

Next, we can plug in s = jω and solve for |H(jω)|:

|H(jω)| = |jω|/(5 - (jω)^2)

Using the fact that |jω| = ω, we can simplify this to:

|H(jω)| = ω/(5 + ω^2)

Now we can plot this function on the interval 0≤ω≤10. The function reaches its maximum value at ω = √5, where |H(j√5)| = √5/10 ≈ 0.447.

I hope this helps! Let me know if you have any other questions.