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Image attenuation in superheterodyne receivers is an important aspect of the receivers' overall performance. One of several approaches to image attenuation is to equip the receiver with two mixers driven by the local oscillator in quadrature. Each mixer's output signal is then separately filtered and processed by an image rejector. The rejector shifts the phase of the two filter outputs in such a manner that the target signals from the two filters add up, while the image signals cancel each other. The depth of image attenuation depends strongly on gain match and phase accuracy of the two mixers, filters and of the rejector.

Replacing the two separate post-mixer filters by a single polyphase filter¹ has several advantages, such as significant attenuation of the image versus the target signal even before their processing in the image rejector, less sensitivity to component mismatch than when using two separate filters, and symmetrical bandpass response at very low filter quality factor (Q). Those benefits were discussed in a previous RF Design paper².

The filter discussed here is shown in Figure 1, similar to the first figure in the June 2001 RF Design article. It is important to note that all voltage and current symbols throughout this paper represent magnitudes. The individual phase relations are respected in the vector diagrams.

The polyphase filter consists of two matched damped integrators, an inverting amplifier and of two matched crosscoupling elements. In the June 2001 RF Design article, it was assumed that the integrators' operational amplifiers have very large open-loop gain at the filter's operating frequency and that, therefore, the filter's input voltages are negligible fractions of the filter's output voltages. The subject of the previous article was a lucid explanation of the filter's operation and, mainly, a quantitative analysis of undesired image leak-through caused by any of the filter's passive component pair's mismatch.

Contrary to the previous article, we will assume in this paper that each component with suffix 1 matches perfectly the same component with suffix 2. However, contrary to the previous article, we assume here that the two operational amplifiers have limited, but matched open-loop gain in the filter's operating frequency band.

Because of the operational amplifiers' limited open-loop gain, non-zero voltages exist at the filter's current driven inputs. These error voltages are represented in Figure 1 by voltage sources v_i1 and v_i2 preceding the inputs of ideal operational amplifiers A₁ and A₂. The filter is usually built in differential topology (not shown in Figure 1 for clarity), thus, the inverting amplifier A₃ can be realized by simply crossing two leads. Therefore, inverting amplifier A₃ is considered here to be ideal.

As depicted in Figure 1, the filter input is fed by quadrature currents i_i1 and i_i2 (supplied by two mixers driven by a local oscillator in quadrature, not shown). The target signal is separated from the image signal by the unique phase difference between the two currents. If the mixers are arranged to deliver a target signal with the phase of i_i1 leading the phase of i_i2 and the image with i_i1 lagging i_i2, then the circuit in Figure 1 will act as a bandpass filter for the target and as an attenuator for the image.

As derived in the June 2001 RF Design article, the filter is in resonance when capacitor currents i_C are equal to crosscoupling currents i_R and input currents i_i are equal to feedback currents i_f (see Figure 1). Then no input current is being wasted and the filter's response is at its peak. From that the filter's resonance frequency is:

f_o = 1/(2πRC)

and its output voltage at resonance is:

v_o = i_iR_f

The filter's -3 dB bandwidth is:

f_b = 1/(πR_fC)

and, thus, the filter's quality factor is determined by:

Q = R_f/(2R)

If the operational amplifiers have insufficient open-loop gain in the filter's operating frequency band, input voltages v_i become a significant fraction of output voltages v_o.

As follows from Figure 1, the voltage v_C across the capacitors C, voltage v_R and v_f across resistors R and R_f are each a vector sum of output voltage v_o and input voltage v_i. The non-zero input voltages will cause error currents Δi_C = v_i/X_C, Δi_R = v_i/R, and Δi_f = v_i/R_f.

These error currents act in addition with the filter's input currents i_i and, thus, impact significantly the filter's resonance frequency, and also its transimpedance v_o/i_i.

When the filter is part of a complex system on a chip, the open-loop gain-bandwidth product of the opamps is limited by the chip's allowed power dissipation. Therefore, at frequencies around 10 MHz or more, it is difficult to build integrated opamps with high enough open-loop gain to keep the filter's input voltages negligibly small. But the open-loop gain of an integrated operational amplifier varies with process and temperature, and, thus, the filter's frequency response will also be process and temperature dependent. It is the purpose of the following sections to quantify this effect and to offer means of its minimization.

To illustrate the filter's deficiencies caused by the opamps' limited open-loop gain, the response of a sample filter is plotted in the following figures. To emphasize the limited gain's influence, the plots were generated with a low open-loop gain of A = 33.33, i.e. v_i/v_o = 0.03. The plot's y-axis is magnitude in Volts, not the usual dB, to make the comparison between the plots and the calculated expressions easier.

The parameters of the sample filter are (see Figure 1):

C = 1.59 pF

R = 10kΩ

R_f = 200kΩ

i_i = 5 µA.

With ideal operation amplifiers (v_i = 0) these would result in:

f_o = 1/(2πRC) = 10 MHz (resonance frequency)

f_b = 1/(πR_fC) = 1 MHz (-3 dB bandwidth)

Q = f_o/f_b = 10 (quality factor)

v_o = i_iR_f = 1 V (output voltage at f_o)

Since R_f/R = 2Q, the current if in feedback resistors R_f is always 2Q-times lower than the current i_R in resistors R or the current i_C in capacitors C. Therefore, the impact of non-zero input voltages v_i on the integrators' feedback branch, i.e. currents Δi_f, are relatively small compared to Δi_C and Δi_R and will be ignored in the following discussion.

The phase of an opamp's input voltage v_i relative to its output voltage v_o is a function of the opamp's complex gain. I will illustrate four corner cases, all with the filter in resonance: two when v_i and v_o are phase-parallel (0° or 180°), the other two, when they are in quadrature (± 90°). Because the polyphase filter is a linear circuit, one can then find the filter's response to an arbitrary case by superposition of properly sized corner cases.

The vector diagram with input voltages v_i and output voltages v_o of same phase is shown in Figure 2.

The case of v_i and v_o being in antiphase is easily derived from Figure 2 and is, therefore, not illustrated in a separate diagram. However, the calculations and the plots include that case, as well. The vector diagrams in this paper indicate only the individual phasor's angle, their length is not to scale.

The filter's resonance frequency with phase-parallel v_i and v_o is denoted as f_op.

As follows from Figure 1, and as shown in Figure 2, the voltage v_C across capacitors C is equal to the difference between output voltage v_o and the respective input voltage v_i. In effect:

v_C = v_o - v_i when v_o and v_i are of same phase (as in Figure 2), and

v_C = v_o + v_i when v_o and v_i are in antiphase (not illustrated).

When the filter is in resonance, the capacitor current is:

i_C = 2πCf_op(v_o ± v_i)

As follows from Figure 1 and as shown on the top half of Figure 2, the voltage across resistor R₂ (R₁) is the vector sum of output voltage -v_o1 (v_o2) and the opposite input voltage v_i2 (v_i1). The resulting resistor current can be decomposed into two components: the main current i_R = v_o/R, which is of same phase as the respective output voltage, and error current Δi_R = v_i/R, which is of same phase as the respective input voltage v_i.

The filter is in resonance when capacitor currents i_C are equal to main currents i_R. This yields:

2πCf_op(v_o ± v_i) = v_o/R

With:

1/(2πRC) = f_o (f_o being the resonance frequency with v_i = 0):

f_op/f_o = v_o/(v_o - v_i) when v_i and v_o are of same phase, and

f_op/f_o = v_o/(v_o + v_i) when v_i and v_o are in antiphase.

Using the opamps' open-loop gain A = v_o/v_i:

f_op/f_o = 1/(1 - 1/A) when v_i and v_o are of same phase, and

f_op/f_o = 1/(1 + 1/A) when v_i and v_o are in antiphase.

When the filter is in resonance, input currents i_i are of same phase as output voltages v_o (compare in Figure 2). The error currents Δi_R in resistors R are phase-parallel with v_i, thus with v_o, and therefore, are phase-parallel with ii. Error currents Δi_R combine with input currents i_i to create feedback current if. This is indicated by the arrows Δi_R1 and Δi_R2 in Figure 1 (disregard arrows Δi_C).

The output voltage with phase-parallel v_i and v_o is denoted as v_op:

v_op = i_fR_f = (i_i + Δi_R)R_f when v_o and v_i are of same phase, and:

v_op = i_fR_f = (i_i - Δi_R)R_f when v_o and v_i are in antiphase.

Using:

Δi_R = v_i/R and i_iR_f = v_o (v_o being the output voltage with v_i = 0) we get:

v_op = v_o ± v_iR_f/R.

With R_f/R = 2Q and v_i = v_op/A we get:

v_op = v_o ± 2Qv_op/A, and:

v_op/v_o = 1/(1 - 2Q/A) when v_o and v_i are of same phase, and:

v_op/v_o = 1/(1 + 2Q/A) when they are in antiphase.

Note that with v_p and v_i of same phase, the sample filter becomes unstable if loop gain drops to A = 20.

The vector diagram with input voltages v_i lagging output voltages v_o by 90° is shown in Figure 3.

The case of v_i leading v_o is easily derived from Figure 3 and is, therefore, not illustrated in a separate diagram. However, the calculations and the plots include that case, as well.

The filter's resonance frequency with v_i and v_o in quadrature is denoted as f_oq.

As follows from Figure 1 and as shown in Figure 3, the voltage v_R across resistors R is the difference between output voltage v_o and the respective input voltage v_i:

v_R = v_o - v_i when v_i lags v_o (as in Figure 3);

v_R = v_o + v_i when v_i leads v_o (not illustrated).

The resistor current is:

i_R = (v_o ± v_i)/R.

As follows from Figure 1, and as shown on the top half of Figure 3, the voltage across capacitors C₁ (C₂) is the vector sum of output voltage v_o1 (v_o2) and the respective input voltage v_i1 (v_i2). The resulting capacitor current can be decomposed into two components: the main current i_C = 2πf_oqCv_o, which is leading the respective output voltage vo, and error current Δi_C = 2πf_oqCv_i, which is leading the respective input voltage v_i.

The filter is in resonance when resistor currents i_R are equal to main currents i_C.

Thus, at resonance:

2πCf_oqv_o = (v_o ± v_i)/R.

With 1/(2πRC) = f_o (f_o being the resonance frequency with v_i = 0):

f_oq/f_o = (v_o - v_i)/v_o when v_i lags v_o, and

f_oq/f_o = (v_o + v_i)/v_o when v_i leads v_o.

Using the opamps' open-loop gain A = v_o/v_i:

f_oq/f_o = 1 - 1/A when v_i lags v_o (as in Figure 3), and

f_oq/f_o = 1 + 1/A when v_i leads v_o.

Again, as shown in Figure 3, input currents i_i are of same phase as output voltages v_o (compare the two halves of Figure 3). Input voltages v_i generate error currents Δi_C, each leading its respective driving voltage v_i by 90°. Error currents Δi_C are supplied by feedback resistor R_f, are phase-parallel with v_o, thus with i_i and, therefore, combine with i_i. This is indicated by the arrows Δi_C1 and Δi_C2 in Figure 1 (disregard arrows Δi_R).

The output voltage with v_i and v_o in quadrature is denoted as v_oq:

v_oq = (i_i + Δi_C)R_f when v_i lags v_o (as in Figure 3);

v_oq = (i_i - Δi_C)R_f when v_i leads v_o.

Using:

i_iR_f = v_o and Δi_C = 2πf_oqCv_i we get:

v_oq = v_o ± 2πf_oqR_fCv_i.

With πR_fC = 1/f_b:

v_oq = v_o ± 2v_if_oq/f_b.

Using the above derived expression for f_oq:

f_oq/f_b = (f_o/f_b)(1 ± 1/A) = Q(1 ± 1/A)

With v_i = v_oq/A we get:

v_oq = v_o ± 2Qv_oq(1 ± 1/A)/A.

From that:

v_oq/v_o = 1/[1 - 2Q(1 - 1/A)/A] when v_i lags v_o (as in Figure 3);

v_oq/v_o = 1/[1 + 2Q(1 + 1/A)/A] when v_i leads v_o.

Note that when v_i lags v_o, the sample filter becomes unstable if loop gain drops to A = 19.

Table 1 summarizes the above results.

The above results applied for the sample filter with A = 33.33 are listed in Table 2 and plotted in Figure 4 and Figure 5.

The differences between the plotted frequency responses are large. They represent corner cases for the open-loop gain of A = 33.33. Although process and temperature induced changes in a monolithic polyphase filter will never drive a filter from one corner case to another, the maximum filter response variations tolerable in a radio receiver are also very much smaller than the changes shown in the plots. Consequently, it is worthwhile to look for steps to stabilize the filter response by other means than by a power hungry, brute force increase of opamp open-loop gain.

When observing Figure 2 and Figure 3, one can see that input voltages v_i affect the resonance frequency and resonance amplitude of the filter by creating error currents Δi_R (of same phase as v_i) and error currents Δi_C (leading v_i by 90°). When v_i and v_o are phase-parallel (see Figure 2), error currents Δi_R cause a change in resonance amplitude and error currents Δi_C cause a change in resonance frequency.

When v_i and v_o are in quadrature (see Figure 3), error currents Δi_R cause a change in resonance frequency and error currents Δi_C cause a change in resonance amplitude. The effect of these error currents can be eliminated by generating their replicas Δi_Rx and Δi_Cx and feeding them into the filter's inputs³.

The replica currents will be:

Δi_Cx = 2πf_oCv_i, and

Δi_Rx = v_i/R, or to eliminate also the influence of v_i on the feedback branch:

Δi_Rx = v_i(1/R + 1/R_f).

Replica currents Δi_Rx can be generated by transconductors T_Rx driven by input voltage v_i directly and currents Δi_Cx by transconductors T_Cx driven by input voltage v_i via a 90° phase shifter. To produce proper values of currents Δi_x, transconductors T_x must have a transconductance of G_x = 1/R = 2πf_oC.

As it is well known, but not illustrated in this paper, currents i_R can also be made independent of v_i by replacing resistors R in Figure 1 by transconductors T_R, each controlled by its respective output voltage v_o. By making the transconductors' output conductance G_o = di_R/dv_i much smaller than their transconductance G_R = di_R/dv_o = 1/R, their output current i_R will be negligibly affected by input voltages v_i. Similarly, feedback resistors R_f can also be replaced by transconductors T_f.

The transconductance G_x required to generate replica currents Δi_x is the same as transconductance G_R needed for replacing resistors R, i.e. G_x = G_R. However, the output current capability of transconductors T_x can be A-times lower than that of T_R. Also, the matching requirement between T_x1 and T_x2 for the same degree of image suppression² is A-times lower than the matching requirement between T_R1 and T_R2. Therefore, transconductors T_x can easily be implemented by a simple long-tail differential transistor pair. The decision whether to eliminate the influence of input voltages v_i on currents i_R by using transconductors T_R or by using a combination of resistors R and transconductors T_Rx depends on circumstances not discussed here.

To implement the two 90° phase shifters required to generate Δi_Cx that would have a small phase error over the filter's passband is not easy. However, when observing Figure 2 and Figure 3, one can notice that the phase of Δi_C2 is the same as the phase of v_i1 and that the phase of Δi_C1 is 180° away from the phase of v_i2. Therefore, as shown in Figure 6, the 90° phase shifters can be eliminated by driving each transconductor T_Cx by the opposite opamp's input voltage v_i. The 180° phase difference required for driving T_Cx1 can be implemented again by crossing the leads in a differential circuit. Therefore, amplifier A₄ is considered to be ideal. Figure 6 depicts the filter also with transconductors T_Rx driven by and feeding the same filter input.

The stabilizing effect of replica currents Δi_x is illustrated in Figure 7 and is being discussed in a later paragraph.

In all the above discussions, the filter was driven by only one pair of input currents, such as i_i1 and i_i2, representing a target signal. In that case i_i1 leads i_i2 by 90° and their magnitudes are the same.

However, in an actual application of the filter, the filter will also be driven simultaneously by an uncorrelated image signal of the same frequency. The image signal will be represented by a second pair of input currents, such as i_m1 and i_m2, with i_m2 leading i_m1 by 90°. The filter input currents will be now the sum, such as i_s1 = (i_i1 + i_m1) and i_s2 = (i_i2 + i_m2) Because the magnitude and phase of image currents i_m are independent of target currents ii, sum currents i_s1 and i_s2 will be neither of equal magnitude, nor be in quadrature.

To make it worse, the source of an undesired image signal can be much closer to the receiver than the transmitter of the target signal. Therefore, receiver specifications usually require that the target signal be reliably readable in presence of a 60 dB stronger image signal. Thus, the ratio i_m/i_i could be as large as 1,000.

To allow verifying the proper function of the replica currents Δi_x in presence of an image signal, an ideal image rejector was added in Figure 6. The rejector consists of a phase shifter advancing the phase of v_o2 by 90° and a summing circuit. The latter generates v_r = (v_o1+ v_o2)/2 to maintain the resonance amplitude as v_r = i_iR_f in spite of each of its two inputs being v_o = i_fR_f. If v_o1 and v_o2 are balanced in magnitude and are in quadrature, target signals add in the summing circuit, while image signals cancel.

The unequal magnitude and non-quadrature phase relation between i_s1 and i_s2 raise a justifiable concern whether, in combination with low opamp gain, currents i_s will disturb the magnitude balance and quadrature phase of input voltages v_i1 and v_i2. If so, incorrect replica currents Δi_Cx would be generated in Figure 6. The resonance frequency and amplitude would change with v_i. A second valid concern could be raised whether the crosscoupled transconductors T_Cx with the non-zero input voltages v_i create some leakage between the two halves of the filter, which would deteriorate the image suppression in the rejector.

First, the image rejector needs to be calibrated using a rule derived in this authors' article, “Using Polyphase Filters as Image Attenuators,” RF Design, June 2001, pages 26 to 34². The rule says: if the resistor pair R₁, R₂ or the capacitor pair C₁, C₂ are mismatched by p percent, the image rejection ratio is (p/4) percent.

Thus, when applying image currents i_m to the mismatched filter with i_i = 0 and v_i = 0, the response at the output of the image rejector should be v_r = (p/400)i_mR_f. The blue curve in Figure 7 was generated by the sample filter mismatched by p = 0.5 percent, driven by i_m = 5 mA. The curve peaks at the correct value of v_r = 1.25 V.

Next, to test the above concerns, after removing the mismatch, the sample filter is driven with a target signal of i_i = 5 uA and simultaneously with the 60 dB stronger image signal of i_m = 5 mA. The input voltages v_i and the phase difference between i_i and i_m are then exercised through all four corner cases with all plots superimposed in Figure 7. There is no spread in resonance frequency and no deviation from the correct resonance amplitude of i_iR_f = 1 V. This proves first that the currents Δi_x generated by transconductors T_x in Figure 6 eliminate the errors caused by low opamp gain even in the presence of a 60 dB stronger image signal. Figure 7 further proofs that the crosscoupling via transconductors TC_x does not interfere with image rejection.

Finally, a caveat: the above tests were generated by a circuit simulator in which all elements were perfectly linear. The purpose of the tests was to prove the postulated principles. The degree to which the demonstrated results can be achieved in a practical circuit, depends on the linearity of its actual elements. The mixer/vector filter combination described in the article “An image-rejecting mixer and vector filter with 55 dB image rejection over process, temperature, and transistor mismatch,” in the IEEE Journal of Solid State Circuits January 2001 issue, pages 23 through 33⁴ included some of the above described measures and achieved an image rejection of -55 dB with operational amplifiers of open-loop gain of A = 30.

1. Jan Crols, Michiel Steyaert, “An Analog Integrated Polyphase Filter for a High Performance Low-IF Receiver,” Proceedings of the VLSI Circuit Symposium, Kyoto, Japan, June 1995, pp. 87-88.

2. Tom Hornak, “Using Polyphase Filters as Image Attenuators”, RF Design, June 2001, pp.26-34.

3. Tom Hornak, “Error Reduction in Quadrature Polyphase Filters with Low Open-Loop Gain Operational Amplifiers”, U.S. Patent No. 6,198,345.

4. Tom Hornak, Knud Knudsen, Andrew Grzegorek, Ken Nishimura, and William McFarland, “An Image-Rejecting Mixer and Vector Filter with 55 dB Image Rejection over Process, Temperature, and Transistor Mismatch,” IEEE Journal of Solid State Circuits, January 2001, pp. 23-33.

Tom Hornak received his M.S.E.E. from the Bratislava Slovak Technical University, and his Ph.D. in electrical engineering from the Czech Technical University in Prague, both in former Czechoslovakia.

He joined Hewlett-Packard Co.'s (www.hp.com) corporate research laboratories (HP Labs) in Palo Alto, California, in 1968. In his years in HP Labs his research covered high-speed analog-digital converters, high-speed fiber optic data communication systems, electronic instruments, and ICs for wireless communication.

He was a lecturer at the Czech Technical University in Prague, has published more than 50 papers, and holds more than 60 U.S. and foreign patents. After serving the last few years as one of HP Lab's principal scientists, he retired 1999. He was elected as an I.E.E.E. Fellow in 1985. Hornak can be reached by e-mail at thornak68@aol.com.