B. The Super-Heterodyne Receiver

Transcription

1 10/9/2007 The Super Heterodyne Receiver notes 1/2 B. The Super-Heterodyne Receiver The super-het is by ar the most popular receiver architecture in use today. HO: The Super-Heterodyne Receiver Q: So how do we tune a super-het? To what requency should we set the local oscillator? A: HO: Super-Heterodyne Tuning Another vital element o a super-het receiver is the preselector ilter. HO: The Preselector Filter Q: So what should this preselector ilter be? How should we determine the required order o this ilter? A: HO: The Image and Third-Order Signal Rejection Q: I have heard o some receivers being described as upconversion receivers, what exactly are they? A: HO: Up-Conversion There are many variants o the basic super-het receiver that can improve receiver perormance. HO: Advanced Receiver Designs

2 10/9/2007 The Superhet Receiver 1/9 The Super-Heterodyne Receiver Note that the homodyne receiver would be an excellent design i we always wanted to receive a signal at one particular signal requency ( s, say): antenna narrow-band ampliier ( = ) G s 1 A Fixed-Frequency Homodyne Receiver narrow-band ilter T T ( s ) ( ) = 1 s 0 î ( t ) narrow-band detector/ demodulator No tuning is required! Moreover, we can optimize the ampliier, ilter, and detector perormance or one and only one signal requency (i.e., s ). Q: Couldn t we just build one o these ixed-requency homodyne receivers or each and every signal requency o interest?

3 10/9/2007 The Superhet Receiver 2/9 A: Absolutely! And we sometimes (but not oten) do. We call these receivers channelized receivers. ( = ) G s 1 1 T ( = 1 ) 1 î ( t ) 1 ( = ) G s 1 2 T ( = 2 ) 1 î ( t ) 2 ( = ) G s 1 3 T ( = 3 ) 1 î ( t ) 3 ( = ) G sn 1 T ( = N ) 1 în ( t ) A Channelized Receiver

4 10/9/2007 The Superhet Receiver 3/9 But, there are several important problems involving channelized receivers. They re big, power hungry, and expensive! For example, consider a design or a channelized FM radio. The FM band has a bandwidth o = 20 MHz, and a channel spacing o 200 khz. Thus we ind that the number o FM channels (i.e., the number o possible FM radio stations) is: 20 MHz = 100 channels!!! 200 khz Thus, a channelized FM radio would require 100 homodyne receivers! Q: Yikes! Aren t there any good receiver designs!?! A: Yes, there is a good receiver solution, one developed more than 80 years ago by Edwin Howard Armstrong! In act, is was such a good solution that it is still the predominant receiver architecture used today. Armstrong s approach was both simple and brilliant: Instead o changing (tuning) the receiver hardware to match the desired signal requency, we should change the signal requency to match the receiver hardware!

5 10/9/2007 The Superhet Receiver 4/9 Q: Change the signal requency? How can we possibly do that? A: We know how to do this! We mix the signal with a Local Oscillator! We call this design the Super-Heterodyne Receiver! A super-heterodyne receiver can be viewed as simply as a ixed requency homodyne receiver, proceeded by a requency translation (i.e., down-conversion) stage. Frequency Translation ( Stage) Fixed Heterodyne Rx ( Stage) acosω t s ( = ) G 1 T T ( ) ( ) = 1 0 î ( t ) tuning Acos ω t = s A Simple Super-Het Receiver Design

6 10/9/2007 The Superhet Receiver 5/9 The ixed homodyne receiver (the one that we match the signal requency to), is known as the stage. The ixedrequency that this homodyne receiver is designed (and optimized!) or is called the Intermediate Frequency (). Q: So what is the value o this Intermediate Frequency?? How does a receiver design engineer choose this value? A: Selecting the requency value is perhaps the most important choice that a super-het receiver designer will make. It has many important ramiications, both in terms o perormance and cost. * We will discuss most o these ramiications later, but right now let s simply point out that the should be selected such that the cost and perormance o the () ampliier, () ilter, and detector/demodulator is good. * Generally speaking, as we go lower in requency, the cost o components go down, and their perormance increases (these are both good things!). As a result, the requency is typically (but not always!) selected such that it is much less (e.g., an order o magnitude or more) than the signal requencies we are attempting to demodulate. * Thereore, we typically use the mixer/ to down-convert the signal requency rom its relatively high requency to a relatively low requency.

7 10/9/2007 The Superhet Receiver 6/9 We are thus generally interested in the second-order mixer term. As a result, we must tune the so that s = that is, i we wish to demodulated the signal at requency s! For example, say there exits radio signals (i.e., radio stations) at 95 MHz, 100 MHz, and 103 MHz. Likewise, say that the requency selected by the receiver design engineer is = 20 MHz. We can tune to the station at 95 MHz by setting the Local Oscillator to 95-20=75 MHz: W/Hz W/Hz T ( ) (MHz) 95 (MHz) G ( = 20) 1 T T ( ) ( ) = î ( t ) = 75MHz tuning W/Hz (MHz)

8 10/9/2007 The Superhet Receiver 7/9 Or, we could tune to the station at 103 MHz by tuning the Local Oscillator to =83 MHz: W/Hz W/Hz T ( ) (MHz) 95 (MHz) G ( = 20) 1 T T ( ) ( ) = î ( t ) = 83MHz W/Hz tuning (MHz) Q: Wait a second! You mean we need to tune an oscillator. How is that any better than having to tune an ampliier and/or ilter? A: Tuning the is much easier than tuning a band-pass ilter. For an oscillator, we just need to change a single value its carrier requency! This can typically be done by changing a single component value (e.g., a varactor diode)

9 10/9/2007 The Superhet Receiver 8/9 Contrast that to a ilter. We must somehow change its center requency, without altering its bandwidth, roll-o, or phase delay. Typically, this requires that every reactive element in the ilter be altered or changed as we modiy the center requency (remember all those control knobs!). Q: What about the ilter? I understand that its center requency is equal to the receiver Intermediate Frequency ( ), but what should its bandwidth be? A: Remember, we want only one signal (the desired signal we tuned to) to appear at the demodulator, so the ilter bandwidth should be just wide enough to allow or the desired signal bandwidth. I.E.,: s = s Q: What about the ilter roll-o? How much stop-band attenuation is required by the ilter? A: The most problematic signals or the ilter are the signals (e.g., radio stations) on either side o the desired signal requency s. These signals in adjacent channels are by deinition very close in requency to s, and thus are the most diicult to attenuate.

10 10/9/2007 The Superhet Receiver 9/9 W/Hz Receiver Selectivity T ( ) 1 channel spacing + 1 channel spacing The attenuation o adjacent channels by the ilter (in db) deines selectivity o the receiver. Typically this value is between 30 db and 60 db. A 1934 advertisement enticing men to enter the glamorous and lucrative ield o radio engineering.

11 10/9/2007 Superhet Tuning 1/8 Super-Het Tuning Say we wish to recover the inormation encoded on a radio signal operating at a requency that we shall call s. Recall that (typically) we must down-convert this signal to a lower requency (i.e., < s ), by tuning the requency to a requency such that this second-order equation: s = is satisied. Note or a given s and, there are two possible solutions or the value o requency : =± s = ± 0 0 = Thereore, the two down-conversion solutions are: = + OR = s s In other words, the requency should be set such that it is:

12 10/9/2007 Superhet Tuning 2/8 1) a value higher than the desired signal requency (i.e., = s + ). 2) or, a value lower than the desired signal requency (i.e., = s ). Note in the irst case, the requency will be higher than the signal requency ( > s ). We call this solution high-side tuning. W/Hz High-Side Tuning < < s 0 s For second case, the requency will be lower than the signal requency ( < s ). We call this solution low-side tuning. W/Hz Low-Side Tuning < < s 0 s

13 10/9/2007 Superhet Tuning 3/8 For example, consider again the FM band. Say a radio engineer is designing an FM radio, and has selected an requency o 30 MHz. Since the FM band extends rom 88 MHz to 108 MHz (i.e.,88 MHz 108 MHz ), the radio engineer has two choices or bandwidth. I she chooses high-side tuning, the bandwidth must be = 30MHz higher than the bandwidth, i.e.,: 88 MHz + < < 108 MHz MHz < < 138 MHz s Alternatively, she can choose low-side tuning, with an bandwidth o: 88 MHz < < 108 MHz 58 MHz < < 78 MHz Q: Which o these two solutions should she choose? A: It depends! Sometimes high-side tuning is better, other times low-side is the best choice. We shall see later that this choice aects spurious signal suppression. In addition, this choice aects the perormance o our Local Oscillator (). Let s look at the last consideration now. We ll be positive and look at the advantages o each solution:

14 10/9/2007 Superhet Tuning 4/8 Advantages o low-side tuning: For low-side tuning, the will operate at lower requencies, which generally results in: 1. Lower cost. 2. Slightly greater output power. 3. Lower phase-noise 4. Most importantly, lower requency generally means better requency accuracy. Advantages o high-side tuning: For high-side tuning, the will require a smaller percentage bandwidth, which generally results in: 1. Lower cost. 2. Lower phase-noise Q: Percentage bandwidth? Jut what does that mean? A: Percentage bandwidth is simply the bandwidth, normalized to its center (i.e., average) requency: % bw center requency

15 10/9/2007 Superhet Tuning 5/8 For our example, each local oscillator solution (low-side and high-side) has a bandwidth o =20 MHz (the same width as the FM band!). However, the center (average) requency o each solution is o course very dierent. For low-side tuning: = 68 MHz 2 And thus the percentage bandwidth is: For high-side tuning: 20 % bandwidth = = % = 128 MHz 2 And thus the percentage bandwidth is a ar smaller value o: 20 % bandwidth = = % 128 A really wide bandwidth is generally not speciied in terms o its % bandwidth, but instead in terms o the ratio o its highest and lowest requency. For our examples, either:

16 10/9/2007 Superhet Tuning 6/8 78 = 134. or = Again, a smaller value is generally better. I the bandwidth is exceptionally wide, this ratio can approach or exceed the value o 2.0. I the ratio is equal to 2.0, we say that the has an octave bandwidth ( do you see why?). Generally speaking, it is diicult to build a single oscillator with a octave or greater bandwidth. I our receiver design requires an octave or greater bandwidth, then the typically must be implemented using multiple oscillators, along with a microwave switch. For example, an oscillator with a bandwidth rom 2 to 6 GHz might be implemented as: 2 GHz to 3.5 GHz tuning Microwave Switch Acos ω t 3.5 GHz to 6.0 GHz

17 10/9/2007 Superhet Tuning 7/8 Q: You said that a lower requency would provide better accuracy. Why is that? I thought that long-term stability (in ppm) would be relatively constant with respect to requency. A: Expressed in parts-per-million (ppm), it is! But recall that ppm is essentially a percentage (i.e., geometric) error, whereas the importance value or receiver design is the absolute (ie., arithmetic) error in Hz! Again consider the example. Say each solution (high-side and low-side) has a stability o ± 1.0 ppm (i.e., 1 Hz MHz ). For the low-side solution, this means an absolute error ε o: ε ± 1 Hz = 68 MHz = ± 68 Hz MHz Whereas or high-side, the error is: ε ± 1Hz = 128 MHz = ± 128 Hz MHz The high-side solution has nearly twice as much error! Q: How much accuracy do we need?

18 10/9/2007 Superhet Tuning 8/8 A: Remember, the must convert the desired signal to precisely the receiver requency. I we are o a little then all or part o the desired signal might miss some o the narrowband ilter! W/Hz T ( ) ε A designer rule o thumb is that the absolute error must be less then 10% o the ilter bandwidth: ε < 01.

19 10/9/2007 The Preselector Filter 1/14 The Preselector Filter Say we wish to tune a super-het receiver to receive a radio station broadcasting at 100 MHz. I the receiver uses and requency o = 30 MHz, and uses high-side tuning, we must adjust the local oscillator to a requency o = 130 MHz. s = 100MHz G ( = 30) 1 T T ( ) ( ) = î ( t ) tuning = 130MHz Thus, the desired signal will be down-converted to the requency o 30 MHz. But beware, the desired radio station is not the only signal that will appear at the output o the mixer at 30 MHz!

20 10/9/2007 The Preselector Filter 2/14 Q: Oh yes, we remember. The mixer will create all sorts o nasty, non-ideal spurious signals at the mixer port. Among these are signals at requencies: 1 st order: = 100MHz, = 130MHz 2 nd order: 2 = 200MHz, 2 = 260MHz, + = 230MHz 2 = 70MHz, 3 rd order: Right? 2 = 160MHz, 3 = 300MHz, 3 = 390MHz, 2 + = 330MHz, + 2 = 360MHz A: Not exactly. Although it is true that all o these products will exist at the mixer port they will not pose any particular problem to us as radio engineers. The reason or this is that there is a narrow-band ilter between the mixer port and the demodulator! Look at the requencies o the spurious signals created. They are all quite a bit larger than the ilter center requency o 30MHz. All o the spurious signals are thus rejected by the ilter none (eectively) reach the detector/demodulator!

21 10/9/2007 The Preselector Filter 3/14 W/Hz W/Hz Spectrum, showing products appearing at the mixer port (MHz) s = 100MHz G ( = 30) 1 T T ( ) ( ) = î ( t ) tuning = 130MHz W/Hz T ( ) Spectrum ater ilter has attenuated spurious signals. (MHz) Now, look again at the statement I just made: But beware the desired radio station is not the only signal that will appear at the output o the mixer AT 30 MHz! In other words, there can be spurious signals that appear precisely at our requency o 30 MHz.

22 10/9/2007 The Preselector Filter 4/14 The ilter will not o course ilter these out (ater all they re at 30 MHz!), but instead let them pass through unimpeded to the demodulator. The result demodulated signal î ( t ) is an inaccurate, distorted mess! Q: I m just totally baled! Where do these unilterable signals come rom? How are they produced? A: The answer is a proound one an incredibly important act that every radio engineer worth his or her salt must keep in mind at all times: The electromagnetic spectrum is ull o radio signals. We must assume that the antenna delivers signals operating at any and all requencies! In other words, we are only interested in a signal at 100 MHz; but that does not mean that other signals don t exist. You must always consider this act!

23 10/9/2007 The Preselector Filter 5/14 Q: But I m still conused. How do all these signals cause multiple signals precisely at our requency? A: Remember, each o the signals will mix with the drive signal, and thus each signal will produce its very own set o mixer products (1 st order, 2 nd order, 3 rd order, etc.) Here s the problem some o these mixer products might lie at our requency o 30 MHz! To see which input signal requencies will cause this problem, we must reverse the process o determining our mixer output products. * Recall earlier we started with known values o desired signal requency (e.g., s =100 MHz) and tuning requency (e.g., =130 MHz), and then determined all o the spurious signal requencies created at the mixer port. * But now, we start with a known tuning requency (e.g., =130 MHz), and a known value o the receiver (e.g., =30 MHz), and then we try to determine the requency o the signal that would produce a spurious signal at precisely our receiver.

24 10/9/2007 The Preselector Filter 6/14 For example, let s start with the 3 rd order product 2. In order or this product to be equal to the receiver requency o 30 MHz, we ind that: = =± 30 2 = 130± ± 30 = 2 = 50, 80 Thus, when attempting to listen to a radio station at s =100 MHz by tuning the to =130 MHz we ind that radio stations at both 50 MHz and 80 MHz could create a 3 rd order product at 30 MHz precisely at our ilter center requency! But the bad news continues there are many other mixer products to consider: 2 2(130) = = ± 30 = = 290, 230

25 10/9/2007 The Preselector Filter 7/ (130) + = = 30 = = 230 Q: What?! A radio station operating at a negative requency o -230 MHz? Does this have any meaning? A: Not in any physical sense! We ignore any negative requency solutions they are not a concern to us = = = 2 = 50 Again, a negative solution that we can ignore. 3 3 = = 3 = 10

26 10/9/2007 The Preselector Filter 8/14 OK, that s all the 3 rd order products, now let s consider the second-order terms: 130 = = ± 30 = = 100, 160 * Note that this term is the term created by an ideal mixer. As a result, we ind that one o the signals that will create a mixer product at 30 MHz is = 100 MHz the requency o the desired radio station! * However, we ind that even this ideal mixer term causes problems, as there is a second solution. An signal at 160 MHz would likewise result in a mixer product at 30 MHz even in an ideal mixer! We will ind this second solution to this ideal mixer (i.e., down-conversion) term can be particularly problematic in receiver design. As such, this solution is given a speciic name the image requency. For this example, 160 MHz is the image requency when we tune to a station at 100 MHz.

27 10/9/2007 The Preselector Filter 9/ = = = = = 2 = 15 No problem here! Finally, we must consider one 1 st order term: = 30 In other words, an signal at 30 MHz can leak through the mixer (recall mixer isolation) and appear at the port ater that there s no stopping it until it reaches the demodulator! In summary, we have ound that that: 1. An signal (e.g., radio station) at 30 MHz can cause a 1 st -order product at our ilter requency o 30 MHz. 2. signals (e.g., radio stations) at either 15 MHz or 160 MHz can cause a 2 nd -order product at our ilter requency o 30 MHz.

28 10/9/2007 The Preselector Filter 10/14 3. signals (e.g., radio stations) at 10 MHz, 50MHz, 80 MHz, 230 MHz, or 290 MHz can cause a 3 rd -order product at our ilter requency o 30 MHz. W/Hz W/Hz Many other spurious signals at other req. are likewise created, but not shown! 30 30MHz G ( = 30) 1 T T W/Hz ( ) ( ) = î ( t ) tuning = 130MHz T ( ) ω 30MHz Q: Arrrg! I now see the problem! There is no way to separate the spurious signals at the requency o 30 MHz rom the desired station at 30 MHz. Clearly, your hero E.H. Armstrong was wrong about this Super-Heterodyne receiver design!

29 10/9/2007 The Preselector Filter 11/14 A: Armstrong wrong!?! NEVER!!! There is an additional element o Armstrong s super-het design that we have not yet discussed. The preselector ilter. The ONLY way to keep the mixer rom creating spurious signals at the receiver is to keep the signals that produce them rom the mixer port! O course, we must simultaneously let the desired station reach the mixer. Q: Hmmm A device that lets signals pass at some requencies, while rejecting signals at other requencies sounds like a microwave ilter! A: That s correct! By inserting a preselector ilter between the antenna and the mixer, we can reject the signals that create spurious signals at our center requency, while allowing the desired station to pass through to the mixer unimpeded.

30 10/9/2007 The Preselector Filter 12/14 W/Hz W/Hz T ( = ) = 130MHz W/Hz Preselector Filter tuning 30MHz Q: So how wide should we make the pass-band o the preselector ilter? A: The pass-band o the preselector ilter must be just wide enough to allow any and all potential desired signals to pass through.

31 10/9/2007 The Preselector Filter 13/14 * Consider our example o s = 100 MHz. This signal is smack-dab in the middle o the FM radio band, and so let s assume it is an FM radio station (i it were, it would actually be at requency or 99.9 MHz). * I we are interested in tuning to one FM station, we might be interested in tuning into any o the others, and thus the preselector ilter pass-band must extend rom 88 MHz to 108 MHz (i.e., the FM band). * Note we would not want to extend the pass-band o the preselector ilter any wider than the FM band, as we are (presumably) not interested in signals outside o this band, and those signals could potentially create spurious signals at our center requency! As a result, we ind that the preselector ilter eectively deines the bandwidth o a superheterodyne receiver. Q: OK, one last question. When calculating the products that could create a spurious signal at the center requency, you neglected the terms, 2 and 3. Are these terms not important?

32 10/9/2007 The Preselector Filter 14/14 A: They are actually very important! However, the value o is not an unknown to be solved or, but in act was (or our example) a ixed value o = 130MHz. Thus, 2 = 260MHz, and 3 = 390MHz none o these are anywhere near the center requency o 30 MHz, and so these products are easily rejected by the ilter. However, this need not always be true! * Consider, or example, the case were we again have designed a receiver with an center requency o 30 MHz. This time, however, we desire to tune to radio signal operating at 60 MHz. * Say we use low-side tuning in our design. In that case, the signal requency must be = = 30MHz. * Yikes! You must see the problem! The Local Oscillator requency is equal to our center requency ( = ). The signal will leak through mixer (recall mixer isolation) and into the, where it will pass unimpeded by the ilter to the demodulator (this is a very bad thing). Thus, when designing a receiver, it is unathomably important that the requency, along with any o its harmonics, lie nowhere near the center requency!

33 10/9/2007 Image and Third Order Product Rejection 1/12 Image and Third-Order Signal Rejection Recall in a previous handout the example where a receiver had an requency o = 30 MHz. We desired to demodulate a radio station operating at = 100 MHz, so we set the to a requency o = 130 MHz (i.e., high-side tuning). s We discovered that signals at many other requencies would likewise produce signals at precisely the receiver requency o 30 MHz a very serious problem that can only be solved by the addition o a preselector ilter. Recall that this preselector ilter must allow the desired signal (or band o signals) to pass through unattenuated, but likewise must suiciently reject (i.e., attenuate) all the signals that could create spurious signals at the requency. We ound or this example that these annoying signals reside at requencies: 10 MHz, 15 MHz, 30 MHz, 80 MHz, 160 MHz, 230 MHz, and 290 MHz Note that the most problematic o these signals are the two at 80 MHz and 160 MHz.

34 10/9/2007 Image and Third Order Product Rejection 2/12 Q: Why do these two signals pose the greatest problems? A: Because the requencies 80 MHz and 160 MHz are the closest to the desired signal requency o 100 MHz. Thus, they must be the closest to the pass-band o the preselector ilter, and so will be attenuated the least o all the signals in the list above. As a result, the 30 MHz mixer products produced by the signals at 80 MHz and 160 MHz will be likely be larger than those produced by the other problem requencies they are the ones most need to worry about! Let s look closer at each o these two signals. Image Frequency Rejection We determined in an earlier handout that the radio requency signal at 160 MHz was the image requency or this particular example. Recall the image requency is the other solution to the (ideal) second-order mixer term =! I.E.: = = ± = ±

35 10/9/2007 Image and Third Order Product Rejection 3/12 For low-side tuning, the desired signal is (by deinition) the solution that is greater than : = s s = + (low-side tuning) And thus or low-side tuning the image signal is the solution that is less than : image = Using the act that or low-side tuning = s, we can likewise express the image requency as: = image ( ) = s = 2 s W/Hz Low-Side Tuning < < < image s 0 image s

36 10/9/2007 Image and Third Order Product Rejection 4/12 And thus in summary: = image = 2 s (low-side tuning) Similarly, or high-side tuning, the desired signal is (by deinition) the solution that is less than : = s + s = (high-side tuning) And thus or high-side tuning the image signal is the solution that is greater than : image = + Using the act that or high-side tuning = s +, we can likewise express the image requency as: = + image ( ) = + + s = + 2 s

37 10/9/2007 Image and Third Order Product Rejection 5/12 W/Hz High-Side Tuning < < < s image 0 And thus in summary: s image = + image = + 2 s (high-side tuning) Note or both high-side and low-side tuning, the dierence between the desired signal and its image requency is 2 : = 2 image This is a very important result, as is says that we can increase the distance between a desired signal and its image requency by simply increasing the requency o our receiver design!

38 10/9/2007 Image and Third Order Product Rejection 6/12 For example, again consider the FM band (88 MHz to 108 MHz). Say we decide to design an FM radio with an o 2O MHz, using high-side tuning. Thus, the bandwidth must extend rom: 88 + < < < < < < 128 The image bandwidth is thereore: < < image < < image 128 < < 148 image Thus, the preselector ilter or this FM radio must have pass-band that extends rom 88 to 108 MHz, but must also suiciently attenuate the image signal band extending rom 128 to 148 MHz. Note that 128 MHz is very close to 108 MHz, so that attenuating the signal may be very diicult. Q: By how much do we need to attenuate these image signals?

39 10/9/2007 Image and Third Order Product Rejection 7/12 A: A very good question; one that leads to a very important point. Since the image requency creates the same secondorder product as the desired signal, the conversion loss associated with each signal is precisely the same (e.g. 6 db)! As a result, the signal created by image signals will typically be just as large as those created by the desired FM station. This means that we must greatly attenuate the image band, typically by 40 db or more! Q: Yikes! It sounds like we might require a ilter o very high order!?! A: That s certainly a possibility. However, we can always reduce this required preselector ilter order i we simply increase our design requency! To see how this works, consider what happens i we increase the receiver requency to = 40MHz. For this new, the bandwidth must increase to: 88 + < < < < < < 148 The new image bandwidth has thereore increased to:

40 10/9/2007 Image and Third Order Product Rejection 8/ < < image < < image 168 < < 188 image Note this image band is now much higher in requency than the FM band and thus much more easily iltered! =20 image rejection FM band Image band or =20 Image band or =40 =40 image rejection T ( ) (MHz) The amount by which the preselector attenuates the image signals is known as the image rejection o the receiver. For example, i the preselector ilter attenuates the image band by at least 50 db, we say that the receiver has 50 db o image rejection.

41 10/9/2007 Image and Third Order Product Rejection 9/12 So by increasing the receiver requency, we can either get greater image rejection rom the same preselector ilter order, or we can reduce the preselector ilter order while maintaining suicient image rejection. But be careul! Increasing the requency will also tend to increase cost and reduce detector perormance. 3 rd -Order Signal Rejection In addition to the image requency (the other solution to the second order term = ), the other radio signals that are particularly diicult to reject are the solutions to the 3 rd -order product terms 2 = and 2 =. There are our possible solutions (two or each term): + 1 = 3 = = 4 = 2 2 Each o these our solutions represents the requency o a radio signal that will create a 3 rd -order product precisely at the receiver requency, and thus all our must be adequately rejected by the preselector ilter!

42 10/9/2007 Image and Third Order Product Rejection 10/12 However, solutions 1 and 4 will typically* be the most problematic (i.e., closest to the desired requency band). For instance, in our original example, the problem signal at 80 MHz is the term 1 (i.e., 1 = 80 MHz ). For low-side tuning, where = s, these 3 rd -order solutions can likewise be expressed as: s 1 = 3 = 2s 2 s 2 2 = 4 = 2 s 3 2 (low-side tuning) And or high-side tuning, where = s +, these 3 rd -order solutions can likewise be expressed as: s = 3 = 2s s 2 = 4 = 2s + 2 (high-side tuning) * Note that typically does not mean always.

43 10/9/2007 Image and Third Order Product Rejection 11/12 O course, in a good receiver design the preselector ilter will attenuate these problematic signals beore they reach the mixer port. The amount by which the preselector attenuates these 3 rd -order solutions is known as the 3 rd -order signal rejection o the receiver. Q: By how much do we need to attenuate these signals? A: Since these signals produce 3 rd -order mixer products, the signal power produced is generally much less than that o the (2 nd order) image signal product. As a result, we can at times get by with as little as 20 db o 3 rd -order signal rejection but this depends on the mixer used. Q: Just 20 db o rejection? It sounds like achieving this will be a piece o cake at least compared with satisying the image rejection requirement! A: Not so ast! Oten we will ind that these 3 rd -order signals will be very close to the desired band. In act (i we re not careul when designing the receiver) these 3 rd -order signals can lie inside the desired band then they cannot be attenuated at all!

44 10/9/2007 Image and Third Order Product Rejection 12/12 Thus, rejecting these 3 rd order radio signals can be as diicult (or even more diicult) than rejecting the image signal. Q: We ound earlier that by increasing the requency, we could make the image rejection problem much easier. Is there a similar solution to improving 3 rd order signal rejection? A: Yes there is but you won t like this answer! Look at these solutions or the 3 rd -order mixer terms: and: 4 = 2 s 3 (low-side tuning) s = (high-side tuning) 2 From these solutions, it is evident that some 3 rd -order solutions can be moved away rom the desired band (thus making them easier to ilter) by decreasing the requency. This solution o course is exactly opposite o the method used to improve image rejection. Thus, there is a conlict between the two design goals. It is your job as a receiver designer to arrive at the best possible design compromise, providing both suicient image and 3 rd -order signal rejection. Radio Engineering is not easy!

45 10/9/2007 Up Conversion 1/7 Up-Conversion Typically, we down-convert a desired signal at requency s to a lower Intermediate Frequency (), such that: < s This down-conversion is a result o the ideal 2 nd -order mixer term: s = < s Recall, however, that there is a second ideal 2 nd -order mixer term: s + = > s Note that the resulting requency is greater than the original signal requency s. This term produces an upconversion s to a higher requency. The tuning solution or this up-conversion term is (given s and ): = s > Note that unlike its down-conversion counterpart, the upconversion term only has one solution!

46 10/9/2007 Up Conversion 2/7 Q: So, there is no such thing as high-side tuning or low-side tuning or up-conversion? A: Yes and no. There is only one tuning solution, so we do not choose whether to implement a high-side solution ( > s ) or a low-side solution ( < s ). Instead, the single solution = s will choose or us! This solution or will either be greater than s (i.e., highside), or less than s (i.e., low-side). Hopeully is apparent to you that a low-side solution will result i 2 < s <, whereas a high-side solution must occur i 0 < < 2. s Occasionally receivers are designed that indeed use upconversion instead o down conversion! Q: But wouldn t the requency or these receivers be very high?? A: That s correct! The o an up-conversion receiver might be in the range o 1-6 GHz or even higher! Q: But that would seemingly increase cost and reduce perormance. Why would a receiver designer do that? A: Let s examine the requencies ( ) o any signals that would create spurious responses precisely at an up-conversion

47 10/9/2007 Up Conversion 3/7 receiver, given this receiver ( ), and given that the is tuned to requency. They are: 1 st order = 2 nd order = ± = 2 3 nd order = = = = = = = 2 2 Since we know that > s and > we can conclude that the terms above which are most problematic (i.e., they might be close to desired signal requency s!) are:

48 10/9/2007 Up Conversion 4/7 = = = = = 2 = 2 2 Inserting the up-conversion tuning solution ( = s ) into these results, we can determine the problematic requencies in terms o requency and desired signal requency s : 2 + s = = = s 2 s = = 2s = 2 2 There are some important things to note about these requencies: 1) The only 2 nd -order term is = 2. In other words, there is no image requency! This o course is a result o the act that the up-conversion term + = has only one solution. s 2) These terms are much dierent than those deemed important or the down-conversion receiver.

49 10/9/2007 Up Conversion 5/7 3) As the value o the receiver requency becomes large (i.e., > 2 ) we ind that the requency s o all these problematic signals likewise become large becomes negative). (2 s As a result, these spurious-signal causing signals can be (i they exist) at much higher requencies than the desired signal s they can be easily iltered out by a preselector ilter, and thus the spurious signal (i.e., image and 3 rd -order) suppression can be very good or up-conversion receivers. For example, consider a desired signal bandwidth o: 05. GHz 06. GHz s Say we design an up-conversion receiver with an o 3.0 GHz. The problematic signals are thus: = 1GHz = 1 5. GHz s 325. GHz 33. GHz GHz 2 0. GHz s 2 2 s 27. GHz 275. GHz

50 10/9/2007 Up Conversion 6/7 Note that the requencies o all these potentially spurcreating signals are a signiicant distance rom the desired signal band o 05. GHz 06. GHz. s Thus, a receiver designer can easily attenuate these problematic signals with a preselector ilter whose passband extends rom 0.5 GHz to 0.6 GHz. Q: Wow! Why don t we just design receivers with very high requencies? A: Remember, increasing the receiver will in general increase cost and reduce perormance (this is bad!). In addition, note that increasing the center requency o the ilter ( ) while the bandwidth remains constant decreases the percentage bandwidth o this ilter. Recall there is a practical lower limit on the percentage bandwidth o a bandpass ilter, thus there is a practical upper limit on receiver requency, given an bandwidth! For example, consider a desired signal with a bandwidth o: = 10 MHz s s The receiver ilter, thereore would likewise require a bandwidth o = 10 MHz. I it is impractical or the

51 10/9/2007 Up Conversion 7/7 bandpass ilter to have a percentage bandwidth less than 0.2%, then: > And so: 10 MHz > = =5.0 GHz In other words, or this example, the receiver must be less than 5.0 GHz in order or the ilter to be practical.

52 10/9/2007 Advanced Receiver Designs 1/11 Advanced Receiver Designs So, we know that as our requency increases, the rejection o image and (some) other spurious signals will improve. But, as our requency decreases, the cost and perormance o our receiver and demodulator will improve. Q: Isn t there some way to have it both ways? Can t we have our cake and eat it too? A: Yes, there is (sort o)! To achieve exceptional image and 3 rd order product rejection, and enjoy the cost and perormance beneits o a low requency, receiver designers oten employ these two advanced receiver architectures.

53 10/9/2007 Advanced Receiver Designs 2/11 1. Selectable Preselection Instead o implementing a single preselector ilter, we can use a bank o selectable preselector ilters: S P 4 T S P 4 T ilter Preselector ilters Tuning and Control In other words, we use multiple preselector ilters to span the desired receiver bandwidth. This is particularly useul or wideband receiver design. Q: Why? How is this useul? What good is this design? A: Consider an example. Say we have been tasked to design a receiver with an bandwidth extending rom 8 GHz to 12 GHz. A standard receiver design might implement a single preselector ilter, extending rom 8 GHz to 12 GHz.

54 10/9/2007 Advanced Receiver Designs 3/11 Instead, we could implement a bank o preselector ilters that span the bandwidth. We could implement 2, 3, 4, or even more ilters to accomplish this. Let s say we use our ilters, each covering the bandwidths shown in the table below: Bandwidth Filter #1 Filter #2 Filter #3 Filter #4 8-9 GHz 9-10 GHz GHz GHz Say we wish to receive a signal at 10.3 GHz; we would tune the local oscillator to the proper requency, AND we must select ilter #3 in our ilter bank. Thus, all signals rom GHz would pass through to the port o the mixer a band that includes our desired signal at 10.3 GHz. However, signals rom 8-10 GHz and GHz will be attenuated by ilter #3 ideally, little signal energy rom these bands would reach the port o the mixer. I we wish

55 10/9/2007 Advanced Receiver Designs 4/11 to receive a signal in these bands, we must select a dierent ilter (as well as retune the requency). As a result, signals over just 1GHz o bandwidth reach the port o the mixer, as opposed to the single ilter design wherein a signal spectrum 4GHz wide reaches the mixer port! Q: Again I ask the question: How is this helpul? A: Let s say this receiver design likewise implements low-side tuning. I we wish the tune to a signal at 12 GHz (i.e., = 12 GHz ), we ind that the image requency lies at: s = 12GHz 2 image O course, we need the preselector ilter to reject this image requency. I we receiver design used just one preselector itler (rom 8 to 12 GHz), then the image signal requency must be much less than 8 GHz (i.e., well outside the image ilter passband). As a result, the receiver requency must be: 8GHz 12GHz 2 8GHz GHz 2 4GHz 2GHz

56 10/9/2007 Advanced Receiver Designs 5/11 In other words, the 4.0 GHz bandwidth results in a requirement that the receiver Intermediate Frequency () be much greater than 2.0 GHz. This is a pretty darn high! Instead, i we implement the bank o preselector ilters, we would select ilter #4, with a passband that extends rom 12 GHz down to 11 GHz. As a result, image rejection occurs i: 11GHz 12GHz 2 11GHz GHz 2 1GHz 05. GHz In other words, since the preselector ilter has a much narrower (i.e., 1GHz) bandwidth than beore (i.e., 4GHz), we can get adequate image rejection with a much lower requency (this is a good thing)! Moreover, this improvement in spurious signal rejection likewise applies to other order terms, including that annoying 3 rd -order term! Thus, implementing a bank o preselector ilters allows us to either:

57 10/9/2007 Advanced Receiver Designs 6/11 1. Provide better image and spurious signal rejection at a given requency. 2. Lower the requency necessary to provide a given level o image and spurious signal rejection. As we increase the number o preselector ilters, the image and spurious signal rejection will increase and/or the required requency will decrease. But beware! Adding ilters will increase the cost and size o your receiver! 2. Dual Conversion Receivers A dual conversion receiver is another great way o achieving exceptional image and spurious rejection, while maintaining the beneits o a low requency. In this architecture, instead o employing multiple preselector ilters, we employ multiple (i.e. two) ilters! As the name implies, a dual conversion receiver converts the signal requency twice. As a result, this receiver architecture implements two Local Oscillators and two mixers.

58 10/9/2007 Advanced Receiver Designs 7/11 Preselector Filter (wideband) 1 st Filter (relatively narrowband) 2 nd Filter (narrowband) Tuning T T ( 1 ) ( ) = T T ( 2 ) ( ) = = ± Q: Two requency conversions! Why would we want to do that? A: The irst mixer/local oscillator converts the signal to the irst requency 1. The value o this irst requency is selected to optimize the suppression o the image requency and all other signals that would produce spurious signals (e.g., 3 rd order products) at the irst. Optimizing spurious signal suppression generally results in an requency 1 that is very high much higher than a typical requency. Q: But won t a high requency result in reduced component and demodulator perormance, as well as higher cost? A: That s why we employ a second conversion!

59 10/9/2007 Advanced Receiver Designs 8/11 The second mixer/local oscillator simply down converts the signal to a lower ( 2 ) a requency where both component perormance and cost is good. Q: What about spurious signals produced by this second conversion; don t we need to worry about them? A: Nope! The irst conversion (i designed properly) has adequately suppressed them. The irst ilter (like all ilters) is relatively narrow band, thus allowing only the desired signal to reach the port o the second mixer. We then simply need to down-convert this one signal to a lower, more practical requency! Now, some very important points about the dual-conversion receiver. Point 1 The irst must be tunable just like a normal super-het local oscillator. However, the second has a ixed requency there is no need or it to be tunable! Q: Why is that? A: Think about it. The signal at the port o the second mixer must be precisely at requency 1 (it wouldn t have made it through the irst ilter otherwise!). We need to down-convert this

60 10/9/2007 Advanced Receiver Designs 9/11 signal to a second requency o 1, thus the second requency must be: or: ( ) 2 = high-side tuning ( ) 2 = 1 2 low-side tuning Either way, no tuning is required or the second! This o course means that we can use, or example, a crystal or dielectric resonator oscillator or this second. Point 2 Recall the criteria or selecting the irst is solely image and spurious signal suppression. Since the second conversion reduces the requency to a lower (i.e., lower cost and higher perormance) value, the irst requency 1 can be as high as practicalble. In act, the irst requency can actually be higher than the signal! In other words, the irst conversion can be an upconversion. For example, say our receiver has an bandwith that extends rom 900 MHz to 1300 MHz. We might choose a irst

61 10/9/2007 Advanced Receiver Designs 10/11 at 1 =2500 MHz, such that the irst mixer/ must perorm an up-conversion o as much as 1600 MHz. Q: Say again; why would this be a good idea? A: Remember, we ound that an extremely high irst will make the preselector s job relatively easy all signals that would produce spurious signals at the irst are well outside the preselector bandwidth, and thus are easily and/or greatly suppressed. But be careul! Remember, the signals that cause spurious signals when up-converting are not the usual suspects we ound when downconverting. You must careully determine all oending signals produced rom all mixer terms (1 st, 2 nd, and 3 rd order)! Point 3 The bandwidth o the irst ilter ( 1 ) should be narrow, but not as narrow as the bandwidth o the second ilter ( 2): > 1 2 Remember, the second ilter will have a much lower center requency, and so it will generally have a much larger percentage bandwidth, and typically better perormance (e.g.,

62 10/9/2007 Advanced Receiver Designs 11/11 insertion loss) than the higher requency bandpass ilter required or the irst. Thus, designers generally rely on the second to provide the requisite selectivity. In act, cascading two ilters with the same 3dB bandwidth is a bad idea, as the 3dB bandwidth eectively becomes a 3+3=6db bandwidth! A designer rule-o-thumb is to make the irst ilter bandwidth about 10 times that o the second: One last point. The astute receiver designer will oten ind that a combination o these two architectures (multiple preselection and dual conversion) will provide an elegent, eective, and cost eicient solution!