Active Filters

Filters A filter is a system that processes a signal in some desired fashion. A continuous-time signal or continuous signal of x(t) is a function of the continuous variable t. A continuous-time signal is often called an analog signal. A discrete-time signal or discrete signal x(kT) is defined only at discrete instances t=kT, where k is an integer and T is the uniform spacing or period between samples

Types of Filters There are two broad categories of filters: An analog filter processes continuous-time signals A digital filter processes discrete-time signals. The analog or digital filters can be subdivided into four categories: Lowpass Filters Highpass Filters Bandstop Filters Bandpass Filters

Analog Filter Responses H(f)H(f) f fcfc 0 H(f)H(f) f fcfc 0 Ideal “brick wall” filterPractical filter

Ideal Filters PassbandStopband Passband Stopband Lowpass FilterHighpass Filter Bandstop Filter PassbandStopband Bandpass Filter M(  )    c  c  c 1  c 1  c 2  c 2

There are a number of ways to build filters and of these passive and active filters are the most commonly used in voice and data communications.

Passive filters Passive filters use resistors, capacitors, and inductors (RLC networks). To minimize distortion in the filter characteristic, it is desirable to use inductors with high quality factors (remember the model of a practical inductor includes a series resistance), however these are difficult to implement at frequencies below 1 kHz. They are particularly non-ideal (lossy) They are bulky and expensive

Active filters overcome these drawbacks and are realized using resistors, capacitors, and active devices (usually op-amps) which can all be integrated: Active filters replace inductors using op-amp based equivalent circuits.

Op Amp Advantages Advantages of active RC filters include: reduced size and weight, and therefore parasitics increased reliability and improved performance simpler design than for passive filters and can realize a wider range of functions as well as providing voltage gain in large quantities, the cost of an IC is less than its passive counterpart

Op Amp Disadvantages Active RC filters also have some disadvantages: limited bandwidth of active devices limits the highest attainable pole frequency and therefore applications above 100 kHz (passive RLCfilters can be used up to 500 MHz) the achievable quality factor is also limited require power supplies (unlike passive filters) increased sensitivity to variations in circuit parameters caused by environmental changes compared to passive filters For many applications, particularly in voice and data communications, the economic and performance advantages of active RC filters far outweigh their disadvantages.

Bode Plots Bode plots are important when considering the frequency response characteristics of amplifiers. They plot the magnitude or phase of a transfer function in dB versus frequency.

The decibel (dB) Two levels of power can be compared using a unit of measure called the bel. The decibel is defined as: 1 bel = 10 decibels (dB)

A common dB term is the half power point which is the dB value when the P 2 is one- half P 1.

Logarithms A logarithm is a linear transformation used to simplify mathematical and graphical operations. A logarithm is a one-to-one correspondence.

Any number (N) can be represented as a base number (b) raised to a power (x). The value power (x) can be determined by taking the logarithm of the number (N) to base (b).

Although there is no limitation on the numerical value of the base, calculators are designed to handle either base 10 (the common logarithm) or base e (the natural logarithm). Any base can be found in terms of the common logarithm by:

Properties of Logarithms The common or natural logarithm of the number 1 is 0. The log of any number less than 1 is a negative number. The log of the product of two numbers is the sum of the logs of the numbers. The log of the quotient of two numbers is the log of the numerator minus the denominator. The log a number taken to a power is equal to the product of the power and the log of the number.

Poles & Zeros of the transfer function pole—value of s where the denominator goes to zero. zero—value of s where the numerator goes to zero.

Single-Pole Passive Filter First order low pass filter Cut-off frequency = 1/RC rad/s Problem : Any load (or source) impedance will change frequency response. v in v out C R

Single-Pole Active Filter Same frequency response as passive filter. Buffer amplifier does not load RC network. Output impedance is now zero. v in v out C R

Low-Pass and High-Pass Designs High Pass Low Pass

To understand Bode plots, you need to use Laplace transforms! The transfer function of the circuit is: R V in (s)

Break Frequencies Replace s with j  in the transfer function: where f c is called the break frequency, or corner frequency, and is given by:

Corner Frequency The significance of the break frequency is that it represents the frequency where A v (f) =  -45 . This is where the output of the transfer function has an amplitude 3-dB below the input amplitude, and the output phase is shifted by -45  relative to the input. Therefore, f c is also known as the 3-dB frequency or the corner frequency.

Bode plots use a logarithmic scale for frequency. where a decade is defined as a range of frequencies where the highest and lowest frequencies differ by a factor of One decade

Consider the magnitude of the transfer function: Expressed in dB, the expression is

Look how the previous expression changes with frequency: at low frequencies f<< f b, |Av| dB = 0 dB low frequency asymptote at high frequencies f>>f b, |Av(f)| dB = -20log f/ f b high frequency asymptote

Actual response curve High frequency asymptote  3 dB Low frequency asymptote Note that the two asymptotes intersect at f b where |Av(f b )| dB = -20log f/ f b

The technique for approximating a filter function based on Bode plots is useful for low order, simple filter designs More complex filter characteristics are more easily approximated by using some well-described rational functions, the roots of which have already been tabulated and are well-known.

Real Filters The approximations to the ideal filter are the: Butterworth filter Chebyshev filter Cauer (Elliptic) filter Bessel filter

Standard Transfer Functions Butterworth Flat Pass-band. 20n dB per decade roll-off. Chebyshev Pass-band ripple. Sharper cut-off than Butterworth. Elliptic Pass-band and stop-band ripple. Even sharper cut-off. Bessel Linear phase response – i.e. no signal distortion in pass-band.

Butterworth Filter The Butterworth filter magnitude is defined by: where n is the order of the filter.

From the previous slide: for all values of n For large  :

And implying the M(  ) falls off at 20n db/decade for large values of .

20 db/decade 40 db/decade 60 db/decade

To obtain the transfer function H(s) from the magnitude response, note that

Because s  j  for the frequency response, we have s 2  2. The poles of this function are given by the roots of

The 2n pole are: s k = e j[(2k-1)/2n]  n even, k = 1,2,...,2n e j(k/n)  n odd, k = 0,1,2,...,2n-1 Note that for any n, the poles of the normalized Butterworth filter lie on the unit circle in the s-plane. The left half-plane poles are identified with H(s). The poles associated with H(-s) are mirror images.

Recall from complex numbers that the rectangular form of a complex can be represented as: Recalling that the previous equation is a phasor, we can represent the previous equation in polar form: where

Definition: If z = x + jy, we define e z = e x+ jy to be the complex number Note: When z = 0 + jy, we have which we can represent by symbol:

The following equation is known as Euler’s law. Note that

This implies that This leads to two axioms: and

Observe that e j   represents a unit vector which makes an angle  with the positivie x axis.

Find the transfer function that corresponds to a third-order (n = 3) Butterworth filter. Solution: From the previous discussion: s k = e jk  /3, k=0,1,2,3,4,5

Therefore,

The roots are:

Using the left half-plane poles for H(s), we get which can be expanded to:

The factored form of the normalized Butterworth polynomials for various order n are tabulated in filter design tables.

n Denominator of H(s) for Butterworth Filter 1 s s s (s 2 + s + 1)(s + 1) 4 (s )(s s + 1) 5 (s + 1) (s s + 1)(s s + 1) 6 (s s + 1)(s s + 1 )(s s + 1) 7 (s + 1)(s s + 1)(s s + 1 )(s s + 1) 8(s s + 1)(s s + 1 )(s s + 1 )(s s + 1)

Frequency Transformations

So far we have looked at the Butterworth filter with a normalized cutoff frequency By means of a frequency transformation, we can obtain a lowpass, bandpass, bandstop, or highpass filter with specific cutoff frequencies.

Lowpass with Cutoff Frequency  u Transformation:

Highpass with Cutoff Frequency  l Transformation:

Bandpass with Cutoff Frequencies  l and  u Transformation: where

Bandstop with Cutoff Frequencies  l and  u Transformation:

Lect23 EEE Active Filters To achieve a gain > 1, we need to utilize an op-amp based circuit to form an active filter We can design all the common filter types using op-amp based active filters Recall for an ideal op-amp that V + = V – I + = I – = 0

Lect23 EEE Op Amp Model + – Inverting input Non-inverting input R in V+V+ V–V– + – A(V + –V – ) VoVo R out

Lect23 EEE Non-inverting Op-Amp Circuit + –+ – V in +–+– V out Z1Z1 Z2Z2

Lect23 EEE Inverting Op-Amp Circuit – + V in +–+– + – V out Z2Z2 Z1Z1

Lect23 EEE An Integrator – + V in + – V out R C +–+– Earlier in the semester, we saw this op-amp based integrator circuit. What type of filter does it create? Does this help you to understand time and frequency domain interrelations?

Lect23 EEE A Differentiator – + V in + – V out C R +–+– We also observed this op-amp based differentiator circuit. What type of filtering does it produce? How could we move the zero away from the origin?

Lect23 EEE Practical Example You are shopping for a stereo system, and the following specifications are quoted: Frequency range: –6 dB at 65 Hz and 22 kHz Frequency response: 75 Hz to 20 kHz ±3 dB What do these specs really mean? Note that the human hearing range is around 20 Hz to 20 kHz (audible frequencies) Could you draw a rough Bode (magnitude) plot for the stereo system?

Lect23 EEE MATLAB Filter Example How can we filter a real measurement signal? One method involves using a numerical algorithm called the Fast Fourier Transform (FFT) which converts a time domain signal into the frequency domain Start MATLAB, then download and run the ‘EEE202Filter.m’ file Can you observe the reduction in the high frequency components in both the time and frequency domain plots of the output signal?

Lect23 EEE MATLAB Filter Example (cont’d) Can we extend the knowledge acquired about transfer functions and filtering? How can we further reduce the high frequency component magnitudes? Show me. How can we remove more high frequencies from the output signal? Show me. Input Signal Fast Fourier Transform Freq. Domain Filtering Inverse FFT Output Signal