5. Two Real Poles and a Zero

What do you get if you cascade the unit gain lowpass circuit of Section 1 and the unit gain highpass circuit of Section 2? What's left if you cut out the low frequencies and the high frequencies? Answer: you're left with the midrange, and the composite circuit is a bandpass filter. Here's the transfer function, with p₁ as the lowpass pole and p₂ as the highpass pole.

H(s) is not normalized to unit gain in this formulation; in fact, the gain varies with pole location up to a mazimum of unity in the midrange if the poles are far apart.

Here are a couple of circuits that are described by a transfer function of this form. Note that the poles of the left hand circuit are not -1/R₁C₁ and -1/R₂C₂, the poles of the two sections individually, because the second section loads the first (i.e., drains current from it), thereby shifting the pole location. As in the two real pole circuit of Section 4, you can use an op amp to make the transfer function equal to the product of the section transfer functions, if this is important to you.

The third example is a thermal system in which the output is the temperature T_c₁ in the outer chamber in response to changes in the ambient temperature T_a.

For the animation, we'll start with a large negative value of one of the poles and sweep it in to the origin, while the zero and the other pole remain fixed. The moving pole doesn't go into the right half plane, even though unstable systems still have step and impulse responses. Click here to see the animation and leave the animation window open as you read this section. Drag the slider back and forth with your mouse to get different parameter values and change all the graphs together.

Remember that time constants are negative reciprocals of the pole locations (e.g., t₁ = -1/p₁). You can see two different time constants when the poles are widely separated.

This is the simplest and cheapest form of bandpass filter. Clearly, it cuts out DC and very high frequencies, but the variation in H(w) means that only signals occupying a restricted range of frequencies will get through without appreciable distortion. The pole location must be chosen intelligently, since lowpassing a highpassed signal could result in a "no pass" filter - for example, when the lowpass pole p₁ is closer to the origin than the highpass pole p₂, as you saw at the end of the animation

The circuit is commonly used in demodulation of a bandpass signal: multiplication of the signal by a sine wave at the carrier frequency results in a component at or near 0 Hz and a component at twice the carrier frequency. Applying this circuit to the product eliminates the double frequency term and cuts out any DC bias.

In defining the impulse and step responses, we have to allow for the possibility that p₁=p₂, since a double pole produces a different functional form of the solution, one that sidesteps the division by zero that would result from use of the standard form. Both of these expressions were obtained by performing an inverse Laplace transform.