Description of the High Pass Filter Transfer Function …

This article continues our discussion of domain transfer functions and their role in the design and analysis of analog filters.

If you have read the previous articles in this series (on low pass transfer functions and [[poles and zeros]]), you are already familiar with several important concepts related to domain analysis s and analog filter theory. Let’s briefly review:

• We can generate an expression for the input-output behavior of a low-pass filter by analyzing the circuit in the s domain.
• The circuit’s V. OUT / VEN expression is the filter transfer function, and if we compare this expression with the standardized form, we can quickly determine two critical parameters, namely the cutoff frequency and the maximum gain.
• A transfer function can be written as a numerator polynomial divided by a denominator polynomial. The roots of the numerator polynomial are the zeros of the transfer function, and the roots of the denominator polynomial are the poles of the transfer function. Another way of saying this is that the zeros of the transfer function result in T (s) = 0 and the poles of the transfer function result in T (s) â†’ âˆž.
• The soundings cause the slope of the system’s Bode plot magnitude response to decrease by 20 dB / decade; zeros cause the slope to increase by 20 dB / decade.
• Surveys contribute â€“90 Â° of phase change, and zeros contribute + 90 Â° of phase change.

The high-pass transfer function

A first order RC high pass circuit is implemented as follows:

\$\$ frac {V_ {OUT}} {V_ {IN}} = frac {R} {frac {1} {sC} + R} = frac {sRC} {1 + sRC} = frac {s} {s + frac {1} {RC}} \$\$

The input-output behavior of a first-order high-pass filter can be described by the following standardized transfer function:

\$\$ T (s) HP = frac {a 1 s} {s + omega O} \$\$

Let’s compare this with the corresponding low pass expression:

\$\$ T (s) LP = frac {aO} {s + omegaO} \$\$

As you can see, the denominators are the same. In both cases, we have a pole at s = â€“Ï‰O, which means that both the low-pass filter and the high-pass filter will have the following characteristics:

• The magnitude response in Ï‰O will be 3 dB below the maximum magnitude response; with a passive filter, the maximum magnitude response is unity, in which case the enO value is â€“3 dB.
• The absolute value of the phase shift of the circuit enO will be 45 Â°.

Therefore, the response in O in these two circuits is very similar. However, the response at frequencies above and below 0 is influenced by the numerator of T (s), and the difference between the two numerators is what makes a low-pass filter very different from a high-pass filter. .

The effect of the numerator

The numerator of T (s) HP tells us two things: the initial slope of the magnitude response will be +20 dB / decade, and the maximum magnitude will be one1. Let’s take a closer look at these two features.

Initial slope

Since we now have the variable s in the numerator, we will have a transfer function zero at any value of s that makes the numerator equal to zero. In the case of a first-order high-pass filter, the entire numerator is multiplied by s, so zero is s = 0.

How does a zero at s = 0 affect the magnitude and phase response of a real circuit? First, let’s consider the magnitude. We know that a zero will cause the slope of the Bode plot curve to increase by 20 dB / decade. However, this rise occurs at Ï‰ = 0 rad / s (or Æ’ = 0 Hz), and here’s the catch: the horizontal axis of the Bode plot never reaches 0 Hz. It is a logarithmic axis, which means that the frequency decreases from 10 Hz to 1 Hz, to 0.1 Hz, to 0.01 Hz, and so on. It never reaches 0 Hz. Consequently, we never see the corner frequency of zero at Ï‰ = 0 rad / s.

Instead, the magnitude curve simply begins with a slope of +20 dB / decade. The magnitude continues to increase until the frequency of the pole; the pole reduces the slope by 20 dB / decade, resulting in a response that becomes flat (that is, slope = 0 dB / decade) and remains flat as Ï‰ increases towards infinity.

Maximum gain

All we need is a bit of mathematical manipulation to see that the maximum gain of a high pass filter is equal to 1. From the general form of the magnitude response of the high pass filter, we know that the gain cannot decrease as Ï‰ increases towards infinity. Therefore, we can find the maximum gain by evaluating T (s) for s â†’ âˆž. In the denominator, we have s + Ï‰O. Something that is added to infinity is infinity, so in this case, we can simplify T (s) as follows:

\$\$ T (s â†’ âˆž) = frac {a_ {1} s} {s} \$\$

The s in the numerator and the s in the denominator cancel out, so

High pass filter phase response

As mentioned earlier, a zero contributes + 90 Â° of phase shift to a system’s phase response, with + 45 Â° of phase shift at zero frequency. The phase shift reaches + 90 Â° at a frequency that is one decade above the zero frequency, but a high pass filter has a zero at Ï‰ = 0 rad / s, and you cannot specify a frequency that is one decade per above 0 rad / s: Again, we are dealing with a logarithmic scale, which means that the horizontal axis will never reach 0 rad / s, nor will it ever reach a frequency that is a decade above 0 rad / s (a frequency of this type) doesn’t really exist: 0 rad / s Ã— 10 = 0 rad / s).