I am trying to understand what is meant by continuous phase FSK. Right now I am of the opinion that it means that a very quick change in frequency can take place as long as there is no discontinuity in the time waveform when the frequency change takes place. Is this a correct interpretation? Thanks, Brent
Question about Continuous Phase FSK
Started by ●September 3, 2010
Reply by ●September 3, 20102010-09-03
On Sep 3, 8:59�am, brent <buleg...@columbus.rr.com> wrote:> I am trying to understand what is meant by continuous phase FSK. > Right now I am of the opinion that it means that a very quick change > in frequency can take place as long as there is no discontinuity in > the time waveform when the frequency change takes place. �Is this a > correct interpretation? > > Thanks, > > BrentFrequency is the derivative of phase, so it can be discontinuous and you still get a continuous phase waveform. There are multiple ways to generate FSK, some which result in continuous phase and some that do not. What method is used depends on the system requirements and available resources. Jason
Reply by ●September 3, 20102010-09-03
On 09/03/2010 05:59 AM, brent wrote:> I am trying to understand what is meant by continuous phase FSK. > Right now I am of the opinion that it means that a very quick change > in frequency can take place as long as there is no discontinuity in > the time waveform when the frequency change takes place. Is this a > correct interpretation?There are two basic ways to implement FSK: Way one is to have two oscillators, differing by the desired frequency shift, and switch between them. Because they are free running, the phase will jump by an arbitrary amount when you switch between mark and space. The other way is to (effectively) have one oscillator, and to switch its frequency command*. Because it's one oscillator and it's perfect**, when you switch its frequency command the phase doesn't jump at all -- only the frequency changes. Given a continuous phase FSK signal you can make a coherent demodulator whose symbol length is greater than a bit length (e.g. MSK. See http://www.wescottdesign.com/articles/MSK/mskTop.html for more detail than you probably want). Because this phase information is lost with each bit transition with non-continuous FSK you can't do this. Further, the sharp phase discontinuities can significantly broaden the bandwidth of the signal. * There are other ways to achieve this effect, but let's stick with the conceptually simple one. ** I said _conceptually_ simple! -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●September 3, 20102010-09-03
brent <bulegoge@columbus.rr.com> writes:> I am trying to understand what is meant by continuous phase FSK. > Right now I am of the opinion that it means that a very quick change > in frequency can take place as long as there is no discontinuity in > the time waveform when the frequency change takes place. Is this a > correct interpretation? > > Thanks, > > BrentHi Brent, There are different ways of viewing continuous-phase FSK. To my way of thinking, the easiest is this: if we consider an FSK signal is in the form x(t) = A*cos(phi(t)), then a _continuous-phase FSK signal requires phi(t) to be continuous. Most of the time the term "phase" refers to the function theta in the argument: cos(w*t + theta). I seem to recall that in a more general (non-engineering) sense, the argument lambda to a trigonometric function cos(lambda) is also called the phase. In any case it is the argument that must be continuous, and it is fairly clear intuitively why this has advantages: a discontinuous argument would (generally) produce a discontinous x(t), and discontinuities require a very large (ideally, infinite) bandwidth. So continuous-phase FM requires less bandwidth for a similar noise (or error) performance. -- Randy Yates % "With time with what you've learned, Digital Signal Labs % they'll kiss the ground you walk mailto://yates@ieee.org % upon." http://www.digitalsignallabs.com % '21st Century Man', *Time*, ELO
Reply by ●September 3, 20102010-09-03
Randy Yates <yates@ieee.org> writes:> brent <bulegoge@columbus.rr.com> writes: > >> I am trying to understand what is meant by continuous phase FSK. >> Right now I am of the opinion that it means that a very quick change >> in frequency can take place as long as there is no discontinuity in >> the time waveform when the frequency change takes place. Is this a >> correct interpretation? >> >> Thanks, >> >> Brent > > Hi Brent, > > There are different ways of viewing continuous-phase FSK. To my way of > thinking, the easiest is this: if we consider an FSK signal is in the > form > > x(t) = A*cos(phi(t)), > > then a _continuous-phase FSK signal requires phi(t) to be continuous. > > Most of the time the term "phase" refers to the function theta > in the argument: > > cos(w*t + theta). > > I seem to recall that in a more general (non-engineering) sense, the > argument lambda to a trigonometric function cos(lambda) is also called > the phase. > > In any case it is the argument that must be continuous, and it is fairly > clear intuitively why this has advantages: a discontinuous argument > would (generally) produce a discontinous x(t), and discontinuities > require a very large (ideally, infinite) bandwidth. So continuous-phase > FM requires less bandwidth for a similar noise (or error) performance.Brent, this is wrong - please ignore this. Need more coffee. -- Randy Yates % "She has an IQ of 1001, she has a jumpsuit Digital Signal Labs % on, and she's also a telephone." mailto://yates@ieee.org % http://www.digitalsignallabs.com % 'Yours Truly, 2095', *Time*, ELO
Reply by ●September 3, 20102010-09-03
Randy Yates <yates@ieee.org> writes:> [...] > Brent, this is wrong - please ignore this. Need more coffee.OK, it is NOT wrong. Sorry! I got confused on the meaning of "continuous" for a few minutes. So what I originally wrote is correct. (Note that phi(t) may not be "differentiable" but it still can be "continuous" and thus a valid argument for a continuous-phase FSK signal.) -- Randy Yates % "My Shangri-la has gone away, fading like Digital Signal Labs % the Beatles on 'Hey Jude'" mailto://yates@ieee.org % http://www.digitalsignallabs.com % 'Shangri-La', *A New World Record*, ELO
Reply by ●September 3, 20102010-09-03
On Sep 3, 11:23�am, Randy Yates <ya...@ieee.org> wrote:> To my way of > thinking, the easiest is this: if we consider an FSK signal is in the > form > > � x(t) = A*cos(phi(t)), > > then a _continuous-phase FSK signal requires phi(t) to be continuous.More strongly, phi(t) should be a piecewise linear function as well as being continuous. That is, the graph of phi(t) should be a series of straight-line segments with the segments joining at their endpoints.> Most of the time the term "phase" refers to the function theta > in the argument: > > � cos(w*t + theta). >In such cases, theta is perhaps better referred to as the phase offset, but it is true that "phase" is very commonly used. In fact, for FSK perhaps it is best to write the signal as cos(w*t + phi(t)) where w is the nominal or average frequency (sum of Mark and Space frequencies divided by 2) and take phi(t) to be a piecewise linear continuous function where the line segments have positive or negative slopes (of equal magnitude). If phi(t) increases (or decreases) by pi/2 over a signaling interval, we have MSK; if the change is plus/minus pi, we have Sunde's FSK etc. Hope this helps --Dilip Sarwate
Reply by ●September 3, 20102010-09-03
On Sep 3, 12:16�pm, Tim Wescott <t...@seemywebsite.com> wrote:> On 09/03/2010 05:59 AM, brent wrote: > > > I am trying to understand what is meant by continuous phase FSK. > > Right now I am of the opinion that it means that a very quick change > > in frequency can take place as long as there is no discontinuity in > > the time waveform when the frequency change takes place. �Is this a > > correct interpretation? > > There are two basic ways to implement FSK: > > Way one is to have two oscillators, differing by the desired frequency > shift, and switch between them. �Because they are free running, the > phase will jump by an arbitrary amount when you switch between mark and > space. >Yes, this method would have discontinuities in the time waveform as I see it.> The other way is to (effectively) have one oscillator, and to switch its > frequency command*. �Because it's one oscillator and it's perfect**, > when you switch its frequency command the phase doesn't jump at all -- > only the frequency changes. >Yes, the frequency gets switched with the starting phase of F2 being at the same place (angle) as the ending phase of F1> Given a continuous phase FSK signal you can make a coherent demodulator > whose symbol length is greater than a bit length (e.g. MSK. �Seehttp://www.wescottdesign.com/articles/MSK/mskTop.htmlfor more detail > than you probably want). > > Because this phase information is lost with each bit transition with > non-continuous FSK you can't do this. �Further, the sharp phase > discontinuities can significantly broaden the bandwidth of the signal. >Thanks for the nice answer.> * There are other ways to achieve this effect, but let's stick with the > conceptually simple one. > > ** I said _conceptually_ simple! > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.html
Reply by ●September 3, 20102010-09-03
On Sep 3, 12:23�pm, Randy Yates <ya...@ieee.org> wrote:> brent <buleg...@columbus.rr.com> writes: > > I am trying to understand what is meant by continuous phase FSK. > > Right now I am of the opinion that it means that a very quick change > > in frequency can take place as long as there is no discontinuity in > > the time waveform when the frequency change takes place. �Is this a > > correct interpretation? > > > Thanks, > > > Brent > > Hi Brent, > > There are different ways of viewing continuous-phase FSK. To my way of > thinking, the easiest is this: if we consider an FSK signal is in the > form > > � x(t) = A*cos(phi(t)), > > then a _continuous-phase FSK signal requires phi(t) to be continuous. > > Most of the time the term "phase" refers to the function theta > in the argument: > > � cos(w*t + theta). > > I seem to recall that in a more general (non-engineering) sense, the > argument lambda to a trigonometric function cos(lambda) is also called > the phase. > > In any case it is the argument that must be continuous, and it is fairly > clear intuitively why this has advantages: a discontinuous argument > would (generally) produce a discontinous x(t), and discontinuities > require a very large (ideally, infinite) bandwidth. So continuous-phase > FM requires less bandwidth for a similar noise (or error) performance. > -- > Randy Yates � � � � � � � � � � �% "With time with what you've learned, > Digital Signal Labs � � � � � � �% �they'll kiss the ground you walk > mailto://ya...@ieee.org � � � � �% �upon."http://www.digitalsignallabs.com% '21st Century Man', *Time*, ELOThanks for the thoughtful input. Brent
Reply by ●September 3, 20102010-09-03
On Sep 3, 3:35�pm, dvsarwate <dvsarw...@gmail.com> wrote:> On Sep 3, 11:23�am, Randy Yates <ya...@ieee.org> wrote: > > > To my way of > > thinking, the easiest is this: if we consider an FSK signal is in the > > form > > > � x(t) = A*cos(phi(t)), > > > then a _continuous-phase FSK signal requires phi(t) to be continuous. > > More strongly, phi(t) should be a piecewise linear function as well > as being continuous. �That is, the graph of phi(t) should be a series > of straight-line segments with the segments joining at their > endpoints. > > > Most of the time the term "phase" refers to the function theta > > in the argument: > > > � cos(w*t + theta). > > In such cases, theta is perhaps better referred to as the phase > offset, but it is true that "phase" is very commonly used. �In > fact, for FSK perhaps it is best to write the signal as > > cos(w*t + phi(t)) > > where w is the nominal or average frequency (sum of Mark and > Space frequencies divided by 2) and take phi(t) to be a > piecewise linear continuous function where the line segments > have positive or negative slopes (of equal magnitude). �If > phi(t) increases (or decreases) by pi/2 over a signaling > interval, we have MSK; if the change is plus/minus pi, we > have Sunde's FSK etc. > > Hope this helps > > --Dilip SarwateThe modulation is for UAT (aviation system used to broadcast aircraft locations and weather info) system that I am looking into. It is at 978 MHz and it has a bit interval of .96 usec and the frequency shifts up or down by 312 KHz from 978 depending upon 1 or 0. As I think about this, the bit interval only has about 1/3 of a cycle in it, so while it is called FSK it seems like I need to think of it more in terms of phase shift than frequency shift.
