Fundamentals of Microelectronics

(b)

(a)

Figure 14.49 Locations of poles for (a) second-order, and (b) n-th order Butterworth filter.

Example 14.25

Using a Sallen and Key topology as the core, design a Butterworth filter for the response derived in Example 14.24.

Solution

With n = 3 and !0 = 2 (1:45 MHz), the poles appear as shown in Fig. 14.50(a). The complex conjugate poles p1 and p3 can be created by a second-order SK filter, and the real pole p2 by a simple RC section. Since

For example, R3 = 1 k and C3 = 109:8 pF. Figure 14.50(b) shows the resulting design.

Exercise

If the 228-pF capacitor incurs an error of 10%, determine the error in the value of f1.

The Butterworth response is employed only in rare cases where no ripple in the passband can be tolerated. We typically allow a small ripple (e.g., 0.5 dB) so as to exploit responses that provide a sharper transition slope and hence a greater stopband attenuation. The Chebyshev response is one such example.

14.5.2 Chebyshev Response

The Chebyshev response provides an “equiripple” passband behavior, i.e., with equal local maxima (and equal local minima). This type of response specifies the magnitude of the transfer function as:

der. We consider !0 as the “bandwidth” of the filter. These polynomials are expressed recursively as

As illustrated in Fig. 14.51(a), higher-order polynomials experience a greater number of fluctuations between 0 and 1 in the range of 0 !=!0 1, and monotonically rise thereafter. Scaled by 2, these fluctuations lead to n ripples in the passband of jHj [Fig. 14.51(b)].

Example 14.26

Suppose the filter required in Example 14.24 is realized with a third-order Chebyshev response. Determine the attenuation at 2 MHz.

Solution

For a flatness (ripple) of 0.45 dB in the passband:

and hence

(b)

Figure 14.51 (a) Behavior of Chebyshev polynomials, (b) secondand third-order Chebyshev responses.

Also, !0 = 2 (1 MHz) because the response departs from unity by 0.45 dB at this frequency. It follows that

Remarkably, the stopband attenuation improves by 9.7 dB if a Chebyshev response is employed.

Exercise

How much attenuation can be obtained if the order if raised to four?

Let us summarize our understanding of the Chebyshev response. As depicted in Fig. 14.52, the magnitude of the transfer function in the passband is given by

Sec. 14.5 Approximation of Filter Response 765

revealing the attenuation at frequencies greater than !0. In practice, we must determine n so as to obtain required values of !0, ripple, and stopband attenuation.

Example 14.27

A Chebyshev filter must provide a passband ripple of 1 dB across a bandwidth of 5 MHz and an attenuation of 30 dB at 10 MHz. Determine the order of the filter.

Solution

We set !0 to 2 (5 MHz) and write

Exercise

If the order is limited to three, how much attenuation can be obtained at 10 MHz.

With the order of the response determined, the next step in the design is to obtain the poles and hence the transfer function. It can be shown [1] that the poles are given by

Example 14.28

Using two SK stages, design a filter that satisfies the requirements in Example 14.28.

Solution

With = 0:509 and n = 4, we have

Figure 14.53(a) plots the pole locations. We note that p1 and p4 fall close to the imaginary axis and thus exhibit a relatively high Q. The SK stage for p1 and p4 is characterized by the following transfer function:

j ω

p 1

p 2

228 pF

94.3 pF

p 4

Figure 14.53

= 0:986!2 ; (14.187)

0

s2 + 0:28!0s + 0:986!02

indicating that

Equation (14.61) suggests that such a high Q can be obtained only if C1=C2 is large. For example, with R1 = R2, we require

If R1 = R2 = 1 k , then (14.61) and (14.62) translate to

Figure 14.53(b) shows the overall design. The reader is encouraged to compute the sensitivities.

Exercise

Repeat the above example if capacitor values must exceed 50 pF.

14.6 Chapter Summary

The frequency response of filters has dependencies on various component values and, therefore, suffers from “sensitivity” to component variations. A well-designed filter ensures a small sensitivity with respect to each component.

First-order passive or active filters can readily provide a low-pass or high-pass response, but their transition band is quite wide and stopband attenuation only moderate.

Second-order filters have a greater stopband attenuation and are widely used. For a well- pbehaved frequency and time response, the Q of these filters is typically maintained below

22.

Continuous-time passive second-order filters employ RLC sections, but they become impractical at very low frequencies (because of large physical size of inductors and capacitors).

Active filters employ op amps, resistors, and capacitors to create the desired frequency response. The Sallen and Key topology is an example.

Second-order active (biquad) sections can be based on integrators. Examples include the KHN biquad and the Tow-Thomas biquad.

Biquads can also incorporate simulated inductors, which are derived from the “general impedance converter” (GIC). The GIC can yield large inductor or capacitor values through the use of two op amps.

The desired filter response must in practice be approximated by a realizable transfer function. Possible transfer functions include Butterworth and Chebyshev responses.

The Butterworth response contains n complex poles on a circle and exhibits a maximally-flat behavior. It is suited to applications that are intolerant of any ripple in the passband.

The Chebyshev response provides a sharper transition than Butterworth at the cost of some ripple in the passband and stopbands. It contains n complex poles on an ellipse.

Problems

1.Determine the type of response (low-pass, high-pass, or band-pass) provided by each network depicted in Fig. 14.54.

Figure 14.54

2.Derive the transfer function of each network shown in Fig. 14.54 and determine the poles and zeros.

3.We wish to realize a transfer function of the form

where a and b are real and positive. Which one of the networks illustrated in Fig. 14.54 can satisfy this transfer function?

4.In some applications, the input to a filter may be provided in the form of a current. Compute the transfer function, Vout=Iin, of each of the circuits depicted in Fig. 14.55 and determine the poles and zeros.

Figure 14.55

5.For the high-pass filter depicted in Fig. 14.56, determine the sensitivity of the pole and zero frequencies with respect to R1 and C1.

6.Consider the filter shown in Fig. 14.57. Compute the sensitivity of the pole and zero frequencies with respect to C1, C2, and R1.

C1

VinVout

R1

Figure 14.56

C2

VinVout

R1

C1

Figure 14.57

7.We wish to achieve a pole sensitivity of 5% in the circuit illustrated in Fig. 14.58. If R1 exhibits a variability of 3%, what is the maximum tolerance of L1?

L1

VinVout

R1

Figure 14.58

8. The low-pass filter of Fig. 14.59 is designed to contain two real poles.

L1

VinVout

C1 R1

Figure 14.59

(a)Derive the transfer function.

(b)Compute the poles and the condition that guarantees they are real.

(c)Calculate the pole sensitivities to R1, C1, and L1.

9.Explain what happens to the transfer functions of the circuits in Figs. 14.17(a) and 14.18(a) if the pole and zero coincide.

10.For what value of Q do the poles of the biquadratic transfer function, (14.23), coincide? What is the resulting pole frequencies?

11.Provepthat the response expressed by Eq. (14.25) exhibits no peaking (no local maximum) if

Q 2=2.

12.Prove that the response expressed by Eq. (14.25) reaches a normalized peak of Q=p1 , (4Q2),1 if Q > p2=2. Sketch the response for Q = 2, 4, and 8.

13.Prove that the response expressed by Eq. (14.28) reaches a normalized peak of Q=!n at ! = !n. Sketch the response for Q = 2, 4, and 8.

14.Consider the parallel RLC tank depicted in Fig. 14.26. Plot the location of the poles of the circuit in the complex plane as R1 goes from very small values to very large values while L1 and C1 remain constant.

15.Repeat Problem 14 if R1 and L1 remain constant and C1 varies from very small values to very large values.