From: kennish@kabuki.EECS.Berkeley.EDU (Ken A. Nishimura) Newsgroups: rec.radio.amateur.misc Subject: Ramblings about Intermod and the FT-530 (Warning: LONG) Date: 30 Mar 1994 01:00:03 GMT Greetings again. I've decided to put together a small Q/A blurb (now long) to help quell some common misconceptions about what intermod is, and what other things cause IM like behavior. Also, at the end are some random musings about the FT-530. I believe that all below is correct, if you wish to make a correction, or dispute facts, please E-mail me and I will consider. I will ignore flames. I hope to maybe add this (after editing / corrections if needed) to the FAQ. Q: What is distortion? A: Distortion is a process where where a non-linear signal path creates frequencies that were not there originally. If you have a single frequency and pass it through any linear network and then look at the output on a spectrum analyzer, you will see that frequency and only that frequency. If you pass that same signal through a non-linear system, you will note the original frequency plus harmonics at multiples of the original frequency. We call this harmonic distortion. The relative magnitudes of the harmonics are a function of the non-linearity, but in general, they tend to fall off as the order (frequency) of the harmonic rises. For those that are more mathematically inclined, if you plot the transfer function of the system and it is symmetric with respect to the y-axis, then you will get even order (2nd, 4th, 6th, etc.) harmonics. If it is odd symmetric, then you get odd (3rd, 5th, 7th, etc.) harmonics. If there is no symmetry, then you will get components at every multiple. Some of you may have imagined distortion coming from a circuit that altered the shape of a waveform. This is equivalent. A single frequency signal is a sine wave. By passing it through a non-linear network, one obtains something other than a pure sine wave. Now, if you take that waveform, and subtract from it the original sine wave, you get the difference or distortion products. If you now look at that, you will find (proof is Fourier analysis) that the distortion signal is a sum of sine waves at the harmonic frequencies that are scaled in proportion as dictated by the specific non-linearity. Those with a computer are encouraged to examine the relation between frequency and time domain analysis. Q: What is intermod? A: Intermodulation distortion is a specific case of distortion that results when a signal consisting of two or more distinct frequencies is passed through a non-linear network. Perhaps the easiest way to demonstrate this is through a bit of math. (Get your algebra and trig caps on...) Suppose we have a signal two signals, sin(w1t) and sin(w2t) which we shall denote A and B (ASCII sucks for equations) where w1 and w2 are two distinct frequencies. Let X be A+B. Note that addition is linear and the signal X, if examined on a spectrum analyzer, will still show only two frequencies, w1 and w2. Now, let us imagine we have a linear network whose transfer function is Y=K1*X. (This is a straight line...) Then the output Y is equal to K1*A + K1*B, or K1sin(w1t)+K1sin(w2t). Note that there are only two frequencies, w1 and w2. All is fine. Now substitute the linear network with say, a generalized third order polynomial, Y=K1*X+K2*X^2+K3*X^3. Now substitute X=A+B. Then expanding, we obtain: 2 2 3 2 2 3 Y=K1(A+B)+K2(A+2AB+B)+K3(A+3AB+3AB+B) Substituting the A=sin(w1t) and B=sin(w2t) and organizing by terms results in the following components: K2 (This is a DC term) + (K1+2.25K3)sin(w1t)+(K1+2.25K3)sin(w2t) (This is a linear term) + K2sin([w1+w2]t) (This is a sum of two frequencies term or 2nd order intermod) + K2sin([w1-w2]t) (This is a difference of two frequencies term/2nd IM) + 0.5K2sin(2w1t) (This is a 2nd harmonic term) + 0.5K2sin(2w2t) (This is also a 2nd harmonic term) + 0.25K3sin(3w1t) (This is a 3rd harmonic term) + 0.25K3sin(3w2t) (This is also a 3rd harmonic term) + 0.75K3sin([2w1+w2]t) (This is a 3rd order intermod term) + 0.75K3sin([2w1-w2]t) (This is also a 3rd order intermod term) + 0.75K3sin([2w2+w1]t) (This is yet another 3rd order intermod term) + 0.75K3sin([2w2-w1]t) (This is the last 3rd order intermod term) Note that apart from the second line, all the terms consist of frequencies that did not exist in the input signal. These are distortion products. Unlike the single frequency input case, there are terms which consist of sums and differences of the two frequencies, some of which consist of sums and differences of harmonics of the input frequencies. These are intermod products. For reasons that are too complex to go into here, most circuits exhibit 3rd order distortion less 2nd order distortion. Hence the test for 3rd order TTID (Twin Tone Intermod Distortion), where the term 2w1+/-w2 is set to be in the passband of interest. Example: w1=2pi*446MHz, w2=2pi*447MHz. The TTID product will appear at both 445 and 448 MHz. Intermod distortion is particularly troublesome since there are an almost infinite number of w1 and w2 combinations what will cause a tone to appear at the frequency of interest. Q: I'm getting spurious reception but I can't trace it to IM. What is it? A: If it is not harmonic distortion, it could be image response. Note that most all radio receivers use the hetrodyne method of detecting and processing the incoming RF. Briefly, to tune the radio, the incoming RF is mixed (multiplied) with a local oscillator (LO). If the LO frequency is at a frequency Fif, then a tone will appear at Frf +/- Flo. The system is designed so that the IF strip has a narrow bandpass frequency at Fif. Thus, by adjusting Flo using the synthesizer, signals at frequency Flo+/-Fif will be received by the system. Example: Fif=10.7MHz Flo=110 MHz. The system will receive RF signals at EITHER (110-10.7)=99.3 MHz OR (110+10.7)=120.7 MHz. Note that the system responds equally to two different frequencies, of which only one is desired. The false response the to non-desired signal is called image response. Note that this has nothing to do with harmonic or IM distortion, it's just the way mixers work. (There are image suppression mixers that use complex signals, but they are beyond the scope of this discussion, for now...) In order to reduce image response, the incoming RF must be filtered to remove signals at the image frequency before the mixer. This filter is commonly referred to as the image rejection filter. This makes life interesting for the RF designer.... Example: Let's talk about the two meter ham band, 144 to 148 MHz. Now, what are the choices for IF frequency? First, assume we have a fixed image rejection filter. Then, the filter must pass at least 144 to 148 MHz. This means that the IF frequency must be at least half the bandwidth of interest (144 to 148 MHz) or 2 MHz assuming a perfect image reject filter. In practice, as filters are quite non-selective, the IF is placed substantially higher than half the bandwidth of interest. The Yaesu FT-530 uses 15.25 MHz. This means that signals that are 30.5 MHz away from the desired signal could be imaged into the receiver. Note that the masses want and have gotten extended RX in HTs. The FT-530 covers 110 to 180 MHz, and with a fixed image reject filter would require a IF frequency of at least 35 MHz. (Low IF's are preferable as the bandpass filters for channel select are easier to get, and the circuits burn less power == long battery life). To use a 15.25 MHz IF, Yaesu uses a tunable image reject filter. The control voltage for the VCO within the VFO is used to vary the capacitance of a varactor (hyper-abrupt junction pn diode used for tuning). Thus, in theory, if the image reject filter is narrow, but tunable, things will be fine. In practice, a narrow tunable filter is hard, and expensive -- recall this is the primary reason superhet came around in the first place. So, in the FT-530, the image reject filter needs to be narrower than 30.5 Mhz and tune from 110 to 180 MHz. Not impossible, but not easy to do well. Similar examples could be made from the UHF side of the unit. Q: That sounds great, so what's the catch? A: Varactors are terribly non-linear. In many circuits that use varactors, the tuning voltage is in the tens of volts, so that the small RF signal will not disrupt the linearity of the diode. In an HT, the low battery voltage means that a low tuning voltage diode is used, which is much more non-linear. So, by having an unit with extended RX, you need a tunable image reject filter. This utilizes varactors, which are quite non-linear thus making the IM problem much worse. See? Q: Well, older HT's still suffer from IM... A: Sure, nothing is perfectly linear, and at some point, any RF circuit will IM distort. A useful figure of merit is the third order intercept point, or IM3. This is the input power level where the power of the fundamental or desired signal at the output is equal to the power of the intermod product. A by-product of the desire of the masses for infinite battery life is that the manufacturers are using lower and lower currents in the receive chain. This means that it is easier to overload the front end and drive it non-linear. So, independent of the extended RX, you can still intermod. Newer HT's with their lower current front ends will suffer more, just because they are more easily overloaded, not because of their extended RX, though that makes the problem a lot worse. I hope the above sheds some light and quells some of the myths of intermod, image response and harmonic distortion. I could go on forever (or at least for a very long time, but my boss would get mad....) However, I will include this little tidbit about the FT-530. Many people love their FT-530. I do, I think it's a great radio. But, this morning, I really looked at the schematic, and now I know why it behaves the way it does. Yes, it's really two radios in one. There is a VHF radio and a UHF radio. The VHF radio has a 15.25 MHz IF, a 455 KHz 2nd IF, and a FET mixer, and utilizes a varactor tuned image reject filter. The UHF radio is similar, except that its first IF is at 47.225 MHz and that it uses a BJT as a mixer. So far, OK. Now, one wonders how it does U/U and V/V receive. Well, there is a duplicate RF chain in each receiver, for the opposite "sex" -- so in the VHF receiver, there is a UHF RF amp. HOWEVER, the UHF RF amp in the VHF receiver (got it straight?) uses a FIXED image reject filter. This explains why it's response isn't as wide (receiving frequency range) as the true UHF receiver, and is relatively poor outside the ham bands. It also explains how come it's IM performance is BETTER than the true UHF side -- no varactors! The downside is that the LO for the UHF receive on the VHF side is set for the VHF IF, that is 15.25 MHz. Since the fixed image reject filter for this section is still quite wide (needs to pass the 420-450 range), the possibility for image response is quite high. (e.g. listen to 445 MHz on the VHF side, you really hear 475.5 MHz or 414.5 MHz, haven't figured out which -- still reading schizmos.) Similarly, receiving VHF on the UHF side is as interesting. Again, the image reject filter is fixed, but that isn't a big of a problem since the IF here is 47.225 MHz, which is greater than half the 110-180 MHz range. Again, no varactors, and yes, you get a FET mixer. The LO comes from the UHF PLL, but is divided down, so it may be a bit noisier. Because of the higher IF, VHF receive on the UHF side may be superior in both IM and image rejection than listening to VHF on the VHF side. Anyone try it? The rest of the radio seems OK, though it's obvious that the designers are looking at $$ when thinking. Well, I wrote this, so no measurements today, maybe tomorrow..... -Ken From: a-kevinp@microsoft.COM (Kevin Purcell, Rho) Newsgroups: rec.radio.amateur.misc Subject: Intermodulation Date: 19 Jul 93 20:38:12 GMT Intermodulation is when you get two signals modulating each other -- the problem comes when a multiple of one modulates a multiple of the other. Consider two frequencies, f(1) and f(2). If you want to make this a little more concrete say f(1) is 10.000Mhz and f(2) is 10.010Mhz. A specutrum analyser at the output of the ideal amplifier would look like this : | | | | | | | | | | | | | | | | | | | | ------------------------------------- f(1) f(2) frequency ---> For an amplifier with nonlinear behaviour harmonics of these two tones are generated and they are mixed by the nonlinearties in the amplifier. This produces spurious signals that are related to the input tones. These are the "intermodulation prodcuts". This is why a "two tone" test is used to measure the performance of amplifiers (and mixers) -- its is the simplest way to reproduce this behaviour. f(1) + f(2) we define not to be a problem as this is jsut linear mixer (or it may even be a desired mixer product!). Similarly with f(1) - f(2). Interesting things happend when you consider the higher intermod products. Such as the 3rd order products, like: 2f(1)-f(2) 2f(2)-f(1) 2f(1)+f(2) 2f(2)+f(1) These are third order because they involve the mixing of thre cooponents: f(2), f(2) (to give 2f(2)) and f(1). These are often the most important products. Fifth order products are the combinations of 3f(1) 3f(1)-2f(2) 3f(2)-2f(1) 3f(1)+2f(2) 3f(2)+2f(1) On a frequency analyser these look like: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | ------------------------------------- f(1) f(2) frequency ---> ^ ^ test tones ^ ^ third order ^ ^ fifth order ^ ^ seventh order The intermodulation products are spaced out by the same amount as the frequecny difference of the two test tones. Read the bit in ARRL handbook and any introductory comms books should give you a little more background. To calculate the amplitude of the IMD products you need to know the intercept point of the device (the input power at which the IMD signal is the same power output as the wanted signal), third order signals grow as the third power, fifth order as the fifth power (straight lines with slope 3 and 5 on a log power out vs log power in plot) and a little arithmetic. See my previous post (called TS-50). Kevin Purcell N7WIM / G8UDP a-kevinp@microsoft.com Sit simplex, stulte!