# Homework Help: Bode Plot question

1. Apr 10, 2010

### Melawrghk

1. The problem statement, all variables and given/known data
I'm doing a question that requires a Bode plot (magnitude and phase).

3. The attempt at a solution
I got a transfer function of
$$H(\omega)=\frac{1+\frac{j*\omega}{3}}{1-0.1667(\omega)^{2}+0.4167j*\omega}$$

So there is a simple zero at 3 rad/s; complex pole at 2.45 rad/s; and no DC gain.
I am able to plot the magnitude plot correctly, however my phase isn't exactly right.

For phase I have essentially two contributions from the zero and the pole:

So combining the two, I get (w==omega, i'm too lazy to write it out):
0<w<0.245: 0 degrees,
w>30: slope of 0

However the solution my prof provided says:
0<w<0.245: 0 degrees,
w>24.5: slope of 0, flattens out at -90 degrees.

Am I wrong? If so, why?

2. Apr 11, 2010

### Staff: Mentor

The zero is at -3 rad/s. Does that make a difference, or was that just a simple typo?

3. Apr 11, 2010

### Melawrghk

Uh, neither. We're only dealing with positive frequencies (actually, absolute values, I suppose). We're calling them poles and zeros, but the values we're interested in is just the corner frequencies.

4. Apr 11, 2010

### Staff: Mentor

Sorry, then why do you say you have a zero at 3 rad/s?

5. Apr 11, 2010

### Melawrghk

It's just the terminology used. Corner frequency is like absolute value of zero/pole.