On Mar 13, 11:02�am, Tim Wescott <t...@seemywebsite.com> wrote:
> On Fri, 13 Mar 2009 02:46:38 -0700, Maurizio wrote:
> > I'd like to ask some question about superhet receiver and related
> > problem of image frequency.
>
> > Suppose that receiver is composed by following schema:
>
> > [s(t)] ---> [Circuit RF] ---> [MIXER] ---> [IF]
>
> > s(t) is input signal eg. DSB-SC modulated, having bandwidth of BT and
> > carrier frequency fc.
>
> > Here I'd like to know why the RF circuit cannot filter the s(t) signal
> > using bandwidth of BT, instead of delegate this task to IF circuit.
>
> Radios that tune the RF stages then go straight to a detector are called
> "tuned RF". �They were very popular as AM receivers at the dawn of the
> broadcast era, and in fact if you do a web search on "TRF" plus "radio"
> you'll get some very nice pictures of brass-era radios with those pretty
> globular tubes.
>
> But there are a number of reasons why this doesn't work well in practice.
>
> First, building a good filter is hard enough; building a good narrow
> filter that tunes is practically black magic.
>
> Second, the bandwidth of a bandpass filter with respect to it's center
> frequency determines the quality factor (Q) and precision required of
> it's components. �Coils for bandpass filters generally have unloaded Q's
> of between 10 and 100; given that you generally want a filter with a
> loaded Q that's not more than about 1/3 of your unloaded Q, this leaves
> you with a bandwidth to carrier ratio of about 30:1 if you're using the
> World's Best Coil. �Add the requirement to have a multi-stage filter with
> all the stages tracking accurately, and you'll begin to understand why a
> superheterodyne radio, with it's one tunable element in the oscillator,
> is a Good Thing.
>
> Third, you can build a nice accurate, narrow-band filter using crystals
> or other piezoelectric resonators -- even the cheap ones have Q's in the
> tens of thousands, and it's easy to get them with precisions in the tens
> or 100's of ppms of center frequency. �But you can't tune a crystal
> filter, so if you want to use one in a tunable radio it has to go in the
> IF. �Ditto for other techniques like SAW filters or mechanical resonators.
>
> Fourth, if you were to build a 1-frequency radio you could, in theory,
> use a crystal filter in it's front end. �But crystal filters are not
> trivial to design, and they're big, and they're expensive. �So if you
> wanted to build a line of 1-frequency radios that could work on different
> channels for different customers you'd have this huge engineering and
> inventory expense.
>
> Fifth, if you put all of your gain at one frequency it gets very hard to
> have high gain without making the whole darn radio oscillate. �Using a
> superheterodyne topology allows you to apportion the gain between stages,
> which makes it easier to build a stable receiver. �If you're using a
> superhet anyway, why not use a fixed IF?
>
> There's more reasons, but I'm too lazy to put them down. �If you really
> want to know, get your hands on the ARRL Handbook (the RSGB Handbook is
> also good, I believe, and I'm sure that other countries have good
> equivalents in their local languages -- you may want to check to see what
> your country's amateur radio community has to offer). �Hayward's books on
> RF circuit design and receiver design are also informative.
>
> > Now I'd like to suppose that at receiver input there are two signal. For
> > simplify the analyze of the circuit I'd like to suppose that the carrier
> > frequency of the two signals are: fc and fc1 where fc1 = fLO + fIF.
>
> > fLO = Local Frequency
> > fIF = Intermediate Frequency
>
> > These two signals will be passed from RF to Mixer because I'm supposing
> > that BRF (the bandwidth of RF circuit) is large enough.
>
> > Mixer is operating with frequency fLO = fc + fIF
>
> > At output mixer I'm expecting four rect signals, having amplitude half
> > respect input signal and frequency translated.
>
> I assume you mean "signals with rectangular spectra" here.
>
> > For rect center at fc, I'm expecting:
>
> > fc + fLO = fc + fc + fIF = 2fc + fIF
> > fc - fLO = fc - fc - fIF = - fIF
>
> > For rect centered at fc1, I'm expecting:
>
> > fc1 + fLO = fLO + fIF + fLO = 2fLO + fIF = 2fc + 2fIF + fIF = 2fc + 3fIF
> > fc1 - fLO = fLO + fIF - fLO = fIF
>
> > Now if I've understand correctly the frequency image problem is present
> > because two rect centred at fIF e -fIF interfering each other.
>
> That is correct. �Once those two signal are added together, subsequent
> signal processing has no way of distinguishing them.
>
> > My problem is the following: In previous course the band pass filter was
> > symmetric so for instance if I've a band pass filter with bandwidth of
> > fBP center in fCBP frequency, in my amplitude/frequency diagram that
> > illustrate frequency response of that filter, I have 2 rects on centred
> > in fCBP and other centred in -fCBP. If I think to put white noise signal
> > B0/2 input of filter, in output I'll have:
>
> > B02/2 if �-fCBP - fBP < f < - fCBP �+ fBP
>
> > and
>
> > B02/2 if �fCBP - fBP < f < fCBP �+ fBP
>
> > Now the component of negative frequency is no longer present. And this
> > confuse me.
>
> Your statement confuses me, because the filter that you describe _would_
> pass a signal at -fCBP as readily as it passes one at +fCBP.
>
> > Furthermore, previous course, negative frequency was not interfering
> > with positive so in previous example of superhet receiver, the two rects
> > centred in fIF and -fIF cannot interfering each other.
>
> Either you misunderstood the material in the previous course, or they
> were simplifying things so you could wrap your brain around easily
> digestible chunks.
>
> There _are_ such things as image-reject mixers, that let you pass the
> upper or lower image unmolested, with the unwanted image much attenuated
> (30dB is easy, 40dB is practical, 60dB is the stuff of lab bench
> experiments that totally fall apart when the ambient temperature changes,
> etc.).
>
> I hope this helps. �If you really want to know, I'm going to reiterate my
> advise about the ARRL Handbook -- University classes teach you all sorts
> of wonderful theory, but they leave the problem of making it work to
> you. �The Handbooks are light on theory, but chock-full of practice. �
> With the Handbook plus some University courses you're well equipped to go
> out and design radios.
>
> --http://www.wescottdesign.com
Actually, there were superhet radios with TRF front ends (I had one in
the fifties). They didn't last long.
Also, you can have a tunning stage that has a constant Q. You have
the vary the inductor, not the capacitor. There was a military radio
with a tunned inductor front-end (Collins Radio I think) that had a
roller on the inside of an air-core inductor). That didn't last long
either.
The reason these came and went so quickly is because of all the things
Tim listed!!!
Maurice (no cracks about age) Givens
Reply by Tim Wescott●March 13, 20092009-03-13
On Fri, 13 Mar 2009 02:46:38 -0700, Maurizio wrote:
> I'd like to ask some question about superhet receiver and related
> problem of image frequency.
>
> Suppose that receiver is composed by following schema:
>
> [s(t)] ---> [Circuit RF] ---> [MIXER] ---> [IF]
>
> s(t) is input signal eg. DSB-SC modulated, having bandwidth of BT and
> carrier frequency fc.
>
> Here I'd like to know why the RF circuit cannot filter the s(t) signal
> using bandwidth of BT, instead of delegate this task to IF circuit.
Radios that tune the RF stages then go straight to a detector are called
"tuned RF". They were very popular as AM receivers at the dawn of the
broadcast era, and in fact if you do a web search on "TRF" plus "radio"
you'll get some very nice pictures of brass-era radios with those pretty
globular tubes.
But there are a number of reasons why this doesn't work well in practice.
First, building a good filter is hard enough; building a good narrow
filter that tunes is practically black magic.
Second, the bandwidth of a bandpass filter with respect to it's center
frequency determines the quality factor (Q) and precision required of
it's components. Coils for bandpass filters generally have unloaded Q's
of between 10 and 100; given that you generally want a filter with a
loaded Q that's not more than about 1/3 of your unloaded Q, this leaves
you with a bandwidth to carrier ratio of about 30:1 if you're using the
World's Best Coil. Add the requirement to have a multi-stage filter with
all the stages tracking accurately, and you'll begin to understand why a
superheterodyne radio, with it's one tunable element in the oscillator,
is a Good Thing.
Third, you can build a nice accurate, narrow-band filter using crystals
or other piezoelectric resonators -- even the cheap ones have Q's in the
tens of thousands, and it's easy to get them with precisions in the tens
or 100's of ppms of center frequency. But you can't tune a crystal
filter, so if you want to use one in a tunable radio it has to go in the
IF. Ditto for other techniques like SAW filters or mechanical resonators.
Fourth, if you were to build a 1-frequency radio you could, in theory,
use a crystal filter in it's front end. But crystal filters are not
trivial to design, and they're big, and they're expensive. So if you
wanted to build a line of 1-frequency radios that could work on different
channels for different customers you'd have this huge engineering and
inventory expense.
Fifth, if you put all of your gain at one frequency it gets very hard to
have high gain without making the whole darn radio oscillate. Using a
superheterodyne topology allows you to apportion the gain between stages,
which makes it easier to build a stable receiver. If you're using a
superhet anyway, why not use a fixed IF?
There's more reasons, but I'm too lazy to put them down. If you really
want to know, get your hands on the ARRL Handbook (the RSGB Handbook is
also good, I believe, and I'm sure that other countries have good
equivalents in their local languages -- you may want to check to see what
your country's amateur radio community has to offer). Hayward's books on
RF circuit design and receiver design are also informative.
> Now I'd like to suppose that at receiver input there are two signal. For
> simplify the analyze of the circuit I'd like to suppose that the carrier
> frequency of the two signals are: fc and fc1 where fc1 = fLO + fIF.
>
> fLO = Local Frequency
> fIF = Intermediate Frequency
>
> These two signals will be passed from RF to Mixer because I'm supposing
> that BRF (the bandwidth of RF circuit) is large enough.
>
> Mixer is operating with frequency fLO = fc + fIF
>
> At output mixer I'm expecting four rect signals, having amplitude half
> respect input signal and frequency translated.
I assume you mean "signals with rectangular spectra" here.
> For rect center at fc, I'm expecting:
>
> fc + fLO = fc + fc + fIF = 2fc + fIF
> fc - fLO = fc - fc - fIF = - fIF
>
> For rect centered at fc1, I'm expecting:
>
> fc1 + fLO = fLO + fIF + fLO = 2fLO + fIF = 2fc + 2fIF + fIF = 2fc + 3fIF
> fc1 - fLO = fLO + fIF - fLO = fIF
>
> Now if I've understand correctly the frequency image problem is present
> because two rect centred at fIF e -fIF interfering each other.
That is correct. Once those two signal are added together, subsequent
signal processing has no way of distinguishing them.
> My problem is the following: In previous course the band pass filter was
> symmetric so for instance if I've a band pass filter with bandwidth of
> fBP center in fCBP frequency, in my amplitude/frequency diagram that
> illustrate frequency response of that filter, I have 2 rects on centred
> in fCBP and other centred in -fCBP. If I think to put white noise signal
> B0/2 input of filter, in output I'll have:
>
> B02/2 if -fCBP - fBP < f < - fCBP + fBP
>
> and
>
> B02/2 if fCBP - fBP < f < fCBP + fBP
>
> Now the component of negative frequency is no longer present. And this
> confuse me.
Your statement confuses me, because the filter that you describe _would_
pass a signal at -fCBP as readily as it passes one at +fCBP.
> Furthermore, previous course, negative frequency was not interfering
> with positive so in previous example of superhet receiver, the two rects
> centred in fIF and -fIF cannot interfering each other.
Either you misunderstood the material in the previous course, or they
were simplifying things so you could wrap your brain around easily
digestible chunks.
There _are_ such things as image-reject mixers, that let you pass the
upper or lower image unmolested, with the unwanted image much attenuated
(30dB is easy, 40dB is practical, 60dB is the stuff of lab bench
experiments that totally fall apart when the ambient temperature changes,
etc.).
I hope this helps. If you really want to know, I'm going to reiterate my
advise about the ARRL Handbook -- University classes teach you all sorts
of wonderful theory, but they leave the problem of making it work to
you. The Handbooks are light on theory, but chock-full of practice.
With the Handbook plus some University courses you're well equipped to go
out and design radios.
--
http://www.wescottdesign.com
Reply by John●March 13, 20092009-03-13
On Mar 13, 5:46�am, Maurizio <galia...@yahoo.com> wrote:
> I'd like to ask some question about superhet receiver and related
> problem of image frequency.
>
> Suppose that receiver is composed by following schema:
>
> [s(t)] ---> [Circuit RF] ---> [MIXER] ---> [IF]
>
> s(t) is input signal eg. DSB-SC modulated, having bandwidth of BT and
> carrier frequency fc.
>
> Here I'd like to know why the RF circuit cannot filter the s(t) signal
> using bandwidth of BT, instead of delegate this task to IF circuit.
>
Tunable RF filters exist, but they are expensive, large, and don't
have a constant BW. You'd still need an IF filter.
John
Reply by Maurizio●March 13, 20092009-03-13
I'd like to ask some question about superhet receiver and related
problem of image frequency.
Suppose that receiver is composed by following schema:
[s(t)] ---> [Circuit RF] ---> [MIXER] ---> [IF]
s(t) is input signal eg. DSB-SC modulated, having bandwidth of BT and
carrier frequency fc.
Here I'd like to know why the RF circuit cannot filter the s(t) signal
using bandwidth of BT, instead of delegate this task to IF circuit.
Now I'd like to suppose that at receiver input there are two signal.
For simplify the analyze of the circuit I'd like to suppose that the
carrier frequency of the two signals are: fc and fc1 where fc1 = fLO +
fIF.
fLO = Local Frequency
fIF = Intermediate Frequency
These two signals will be passed from RF to Mixer because I'm
supposing that BRF (the bandwidth of RF circuit) is large enough.
Mixer is operating with frequency fLO = fc + fIF
At output mixer I'm expecting four rect signals, having amplitude half
respect input signal and frequency translated.
For rect center at fc, I'm expecting:
fc + fLO = fc + fc + fIF = 2fc + fIF
fc - fLO = fc - fc - fIF = - fIF
For rect centered at fc1, I'm expecting:
fc1 + fLO = fLO + fIF + fLO = 2fLO + fIF = 2fc + 2fIF + fIF = 2fc +
3fIF
fc1 - fLO = fLO + fIF - fLO = fIF
Now if I've understand correctly the frequency image problem is
present because two rect centred at fIF e -fIF interfering each other.
My problem is the following: In previous course the band pass filter
was symmetric so for instance if I've a band pass filter with
bandwidth of fBP center in fCBP frequency, in my amplitude/frequency
diagram that illustrate frequency response of that filter, I have 2
rects on centred in fCBP and other centred in -fCBP. If I think to put
white noise signal B0/2 input of filter, in output I'll have:
B02/2 if -fCBP - fBP < f < - fCBP + fBP
and
B02/2 if fCBP - fBP < f < fCBP + fBP
Now the component of negative frequency is no longer present. And this
confuse me.
Furthermore, previous course, negative frequency was not interfering
with positive so in previous example of superhet receiver, the two
rects centred in fIF and -fIF cannot interfering each other.
I'd like to thank you very mush for help!
Regards
Maurizio