Bode Plot of a Simple Transfer Function

story of inverted pole and zero form

Let’s ask Mathematica to make a Bode plot of a transfer function

(1)

Here is the result

If you wouldn’t mind registering for a free Wolfram Programming Lab account, you can try this yourself in the web browser IDE.

The point is that nowadays it’s so easy to have some mathematical software package help you plot/evaluate a complicated function.

But if this transfer function comes from some real-world design problem, what are the salient features in the plot that determine your design parameters.

Let’s transform that expression (eq: 1) into the normal pole, zero form (eq: 2) that electrical engineers are familiar with:

(2)

Now we can take apart that expression (eq: 2) into three pieces and sketch (without the help of a computer) the Bode plot of each one easily.

Note: the last plot is a right-half plane zero. The magnitude plot is just like a normal left-half plane zero. Its phase plot will be flipped (see later).

To reach the magnitude plot of the original expression (eq: 2), we just need to superimpose these three plots, because multiplications becomes additions in the log-plot domain. Just align these corner frequencies carefully.

• So you start with sloping down at 20 dB/dec due to 1/s
• Then at 0.8, you hit the normal pole 1/(1+s/0.8), the slope down rate becomes 40 dB/dec (bend downward; slope at twice the rate)
• Soon at 20, you hit the zero 1–s/20, the slope goes back to 20 dB/dec (bend upwards; slope rate returns back)

Now we can make the phase plot similarly by combining 3 separate plots.

• Notice that we didn’t mind A = -100 when we make the magnitude plot, because the square-root of square is essentially Abs function. But for phase plot, 1/s is -90-degree, and -1/s is +90-degree. The minus sign in the front flips the phase.
• A normal zero would have phase from 0 to 90, but a right-half plane zero flips the phase. It goes from 0 to -90, its phase plot just like that of a normal pole.
• Both knee points in the phase plot extends a decade above and below the corner frequency points (ω₁ and ω₂), that is, in this case, for the pole, the two knee points are 0.08 and 8. For the zero, the two knee points are 2 and 200.

Now we can combine the phase plots,

The slopes go through -45 degree/dec, then -90 degree/dec, and back to -45 degree/dec.

However, there is an even simpler way to obtain the same result, by using inverted zeros and poles.

Let’s transform the original expression into

(3)

Now there are only two terms. Notice that in the numerator, s-term is on the bottom instead of on the top. That is an inverted-zero (right half plane zero). The pole is still a normal pole.

The magnitude and phase plot of an inverted-zero (right half plane zero) is

If you compare it to the magnitude plot of of a normal zero, it’s a vertical left-right flip. Again this is due to the logarithm property: log(1/x) =- log(x)

If you combine this plot with the normal pole, you should get the same Bode plot as before.

Why is inverted zeros and poles useful?

Here is a simple circuit diagram.

The transfer function of Vout/Vin is

(4)

If you multiple sC₁ to both top and bottom, the expression becomes

(5)

It is easy to recognize the pole at 1/(R₂ + R₁)C₁. The top R₂C₁ looks like some corner frequency, but it’s not. It is in front of a single s term.

If you plot this expression manually, first you will have a 20 dB/dec upward slope due to s, then it gets bent flat by a pole at 1/(R₂ + R₁)C₁ . Very messy.

How about divide top and bottom by (R₂ + R₁), in eq: 4? We get

(6)

Notice in (eq: 6), the pole expression is not 1/(1 + s/ω), rather 1/(1+ω/s). This is called an inverted pole.

It simply means that looking downwards in the frequency, a 20 dB/dec down slope after the corner frequency ω.

As a circuit designer, one can look at the original circuit and see this plot immediately. At high enough frequency, the capacitor acts like a short, so it becomes resistive divider. At low frequency, the capacitor drops the gain with an inverted pole at location 1/(R₂ + R₁)C₁.

Conclusion

It’s easy to make Bode plot using a piece of mathematical software. But to extract salient features from a Bode plot so as to help circuit design, it is more useful to factor the transfer function into simple zero, pole form.

Sometimes the inverted zeros and poles can simplify the factors and help better design/understand the circuits.

Reference

inverted zeros and poles