Negative and Complex Frequencies?

Hi, I have been working with active filters for the past couple weeks now and have come across something that has stumped me.

For example, we have been dealing with second-order low pass filters and I just have a couple questions regarding this.

When asked to find the -3dB frequency I took the magnitude of the transfer function and set it equal to the low pass gain (A) divided by √2. When squaring all of this, expanding, and simplifying, this brings a quartic function that has to be solved with four roots. When I solve for these roots, I receive two complex conjugate frequencies, one positive frequency, and one negative frequency (equal in magnitude to the positive frequency).

1. What do these complex and negative frequencies mean? Looking at the bode plot the rolloff is -40 dB/decade but how does this work? Shouldn't there be a double positive root (pole) frequency?

2. Also, at what value of Q does a function start peaking at the oscillation frequency?

Thanks for the help,
JP

A negative frequency simply means that the frequency has been shifted from a reference point. In the usual case, the reference point is 0 Hz.

Here is a simple example. Suppose you have an RF carrier at 100kHz being modulated by a 5kHz signal. You will end up with a signal at 95 kHz and one at 105kHz. When viewed with respect to the reference at 100kHz, the 95 kHz is a signal at -5kHz. Now, you can demodulate such a signal by mixing with a 100kHz signal. Again, you will end up with a -5kHz signal and a 5kHz signal. Thus negative frequencies do exist, as strange as it may seem.

A complex frequency is a way of saying that the signal is phase shifted from zero phase. If your signal amplitude is x + iy, then the phase angle is the angle whose tangent is y/x, that is, it is the inverse tangent of y/x. For example, if x is equal to y, the phase angle is 45 degrees.

OK let's take the case of a simple unity DC gain, under-damped second order low pass filter with a transfer function of the form

\(G(s)=\frac{9}{s^2+3s+9}\)

to find the real valued -3dB frequency one would solve

\(|{\frac{9}{(j\omega)^2 +3j\omega+9}}|=\frac{1}{\sqr 2}\)

Which would eventually lead to a quartic equation in ω with roots at

3.8160589
- 3.8160589
2.3584541i
- 2.3584541i

Click to expand...

We know the solution can only have one real positive value that can satisfy the physical situation. So we discard all but the positive real root which implies the -3dB point occurs when

ω=3.8160589 rad/ sec

While the other roots are mathematically correct, they have no physical meaning in this situation.

t_n_k said:

OK let's take the case of a simple unity DC gain, under-damped second order low pass filter with a transfer function of the form

\(G(s)=\frac{9}{s^2+3s+9}\)

to find the real valued -3dB frequency one would solve

\(|{\frac{9}{(j\omega)^2 +3j\omega+9}}|=\frac{1}{\sqr 2}\)

Which would eventually lead to a quartic equation in ω with roots at

We know the solution can only have one real positive value that can satisfy the physical situation. So we discard all but the positive real root which implies the -3dB point occurs when

ω=3.8160589 rad/ sec

While the other roots are mathematically correct, they have no physical meaning in this situation.

Click to expand...

Cool. Thanks! I am still a little confused though if you actually interpret the bode plot.

Say we have the function:

\(G(s)=\frac{4}{s^2+4s+4} = \frac{4}{(s+2)(s+2)}\)

In this case, Q = 0.5 which means that this is a critically damped system and will not experience peaking, thus, no oscillation. There are two real identical roots which each contribute to the bode plot by bringing the rolloff to -40 dB/decade.

In the case where there is peaking, like the example that you provided, what is the other pole that is providing the means of making this -40 dB/decade. If there is only one positive pole that we found to be +3.18 rad/s, why isn't the rolloff -20 dB/decade?

Sorry for the continual questions. I appreciate the help as my TAs and prof have trouble explaining this stuff...
JP

The complex poles for the LPF function I gave are different to the -3dB solution

roots(s^2+3*s+9)
ans =

- 1.5 + 2.5980762i
- 1.5 - 2.5980762i

Click to expand...

They comprise a complex conjugate pair in the left hand complex plane. With respect to drawing the Bode plot one can't treat either of the conjugate pair in isolation. One must recognize that this is an under-damped case and that the simple break point Bode plot drawing 'rules' that one applies to poles lying on the negative real axis of the complex plane, can't be applied to complex poles. The single break point for the 2nd order under-damped case can be 'extracted' from the second order quadratic relationship. The damping factor (or Q) has a significant bearing on the "peakiness" of the response. The peak response doesn't occur at a fixed frequency but varies according to the damping.The break point is therefore better 'defined' as the frequency at which the phase shift crosses -90°.

If one re-casts the quadratic 's' domain function to resemble the form

\(T^2s^2+2\zeta Ts+1\)

Where \(\zeta\) is the damping factor, then the frequency break-point is at ω=1/T [radians/sec]. The attenuation above the break-point is asymptotic with a -40dB per decade slope.