# Understanding the First-Order High-Pass Filter Transfer Function

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

This article continues our discussion of s-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 various important concepts related to s-domain analysis and analog filter theory. Let’s briefly review:

- We can generate an expression for the input-to-output behavior of a low-pass filter by analyzing the circuit in the s-domain.
- The circuit’s V
_{OUT}/V_{IN}expression is the filter’s transfer function, and if we compare this expression to the standardized form, we can quickly determine two critical parameters, namely, cutoff frequency and 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 transfer-function zeros, and the roots of the denominator polynomial are the transfer-function poles. Another way of saying this is that transfer-function zeros result in T(s) = 0 and transfer-function poles result in T(s) → ∞.
- Polls 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.
- Polls contribute –90° of phase shift, and zeros contribute +90° of phase shift.

### 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-to-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 to the corresponding low-pass expression:

#### $$T(s)_{LP}=\frac{a_{O}}{s+\omega _{O}}$$

As you can see, the denominators are the same. In both cases, we have a pole at s = –ω_{O}, meaning that both the low-pass filter and the high-pass filter will have the following characteristics:

- The magnitude response at ω
_{O}will be 3 dB below the maximum magnitude response; with a passive filter, the maximum magnitude response is unity, in which case the value at ω_{O}is –3 dB. - The absolute value of the circuit’s phase shift at ω
_{O}will be 45°.

Thus, the response at ω_{O} in these two circuits is very similar. However, the response at frequencies above and below ω_{O} 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 a_{1}. Let’s take a closer look at these two characteristics.

#### Initial Slope

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

How does a zero at s = 0 affect the magnitude and phase response of an actual 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 increase occurs at ω = 0 rad/s (or ƒ = 0 Hz), and here’s the catch: the horizontal axis of the Bode plot never reaches 0 Hz. It’s 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 forth. It never gets to 0 Hz. Consequently, we never see the corner frequency of the zero at ω = 0 rad/s.

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

#### Maximum Gain

All we need is a bit of mathematical manipulation to see that the maximum gain of a high-pass filter will be equal to a_{1}. From the general shape of the high-pass filter magnitude response, we know that the gain cannot decrease as ω increases toward infinity. Thus, we can find the maximum gain by evaluating T(s) for s → ∞. In the denominator, we have s + ω_{O}. Something added to infinity is infinity, so in this case, we can simplify T(s) as follows:

#### $$T(s→∞)=\frac{a_{1}s}{s}$$

The s in the numerator and the s in the denominator cancel out, such that

#### $$T(s→∞)=a_{1}$$

### High-Pass Filter Phase Response

As mentioned above, a zero contributes +90° of phase shift to a system’s phase response, with +45° of phase shift at the zero frequency. The phase shift reaches +90° at a frequency that is one decade above the zero frequency, but a high-pass filter has a zero at ω = 0 rad/s, and you can’t specify a frequency that is one decade above 0 rad/s—again, we’re dealing with a logarithmic scale here, which means that the horizontal axis will never reach 0 rad/s, nor will it ever reach a frequency that is one decade above 0 rad/s (such a frequency doesn’t really exist: 0 rad/s × 10 = 0 rad/s).

The result of all this is that the high-pass filter phase response has an initial value of +90°. In other words, all low-frequency input signals will be shifted by +90°, and then the phase shift will begin to decrease as the input frequency approaches the pole frequency:

### Conclusion

We’ve examined the standard transfer function for a first-order high-pass filter, and we’ve seen how this transfer function leads to the characteristics of the high-pass magnitude and phase response.

In the next article, we’ll see that the low-pass transfer function and the high-pass transfer function can be combined into a general first-order transfer function, and we’ll also briefly consider the first-order all-pass filter.

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