"Andor" <an2or@mailcircuit.com> wrote in message news:1121691416.593881.169750@f14g2000cwb.googlegroups.com...> Fred, > > you need to modify this definition of time invariant systems. Any > peridodic time-variant system with period n_0 also satisfies your > defintion of time-invariant. > > A nice example of a periodic time-variant system is a fixed-ratio > sample rate converter using a polyphase FIR with n_0 phases.Adding "for any value n_0" would seem to fix the loophole. That was probably implied by fred, but not explicitly stated.
Basic systems question: Is this system time-invariant?
Started by ●July 18, 2005
Reply by ●July 19, 20052005-07-19
Reply by ●July 19, 20052005-07-19
"Bernhard Holzmayer" <Holzmayer.Bernhard@deadspam.com> wrote in message news:dbg6ns$ng7$1@news01.versatel.de...> fred wrote: > >> Bernhard Holzmayer <Holzmayer.Bernhard@deadspam.com> wrote: >> >>> David Kirkland wrote: >>> >>> > kiki wrote: >>> >> y(t)=x(2*t - 2) - x(-3*t + 5 ) >>> >> >>> >> where x(t) is the input, y(t) is the output of the system. >>> >> >>> >> Is this system time-invariant? >>> >> >>> >> Thanks a lot! >>> >> >>> >> >>> > >>> > Nope >>> >>> Hi David, >>> >>> can you please explain this. >>> Since I thought that a system without storage elements (simply spoken) >>> is always time-invariant. >>> >>> In the above case, I'm not quite sure because of the signs. >>> However, just let's take x(t)=1 as a simple approach. >>> This results in: >>> y(t)=x(2*t - 2) - x(-3*t + 5 ) >>> y(t)=1-1 >>> which is certainly time-invariant. >>> >>> Am I wrong? >>> >>> Bernhard >> >> Yes. >> >> A definition of time invariance for discrete time systems is: >> >> A system is time invariant if we get the same output y[n] if we: >> 1) delay our input x[n] by n_0 and then put it through the system, and >> 2) we put x[n] through the system and then delay the result by n_0. >> >> So for a "system" that stretches the time step n by 2, ie y[n]=x[2n], >> the first processing path gives us y_1[n]=x[2(n-n_0)] and the second >> path gives us y[n]=x[2(n)-n_0], which is not equal to y_1[n]. So the >> system is not time invariant. >> >> fred >> > > I can agree to that theoretical information. > Thanks, Fred, for this. I guess I've learnt that long time ago... > > Then, my example with x(t)=1 would hold (==time-invariant). > Now, since nothing special is said about the source signal x(t), > what about the system, if it passes x(t)==1 as y(t)=0, which is certainly > a time-invariant result. > Does this give information about the system being time-variant or not, so > that we must say: it depends!? > Or does applying such a trivial signal just hide the system's properties? > (Like multiplying numbers with 0 before comparing just to find out that > all numbers are equal??) > No, I'm not joking. That's a real question, I'm just curious.What you say makes sense and the analogy with multiplying by zeros is a good one. If you constrain your problem enough, you can make it look time-invariant. Just like for a system that is linear for small inputs, but then non-linear for large inputs (like any real-world amplifier), if you only test it with small inputs, you would conclude that it is non-linear even though it strictly is not. But "it depends" is not the right answer. To be truly time-invariant means the system is time-invariant for _any_ input, not just a special one. You should instead say the system is time-variant but may be treated as time-invariant for certain special cases (e.g. for certain inputs or regions of operation). This approximation is sometimes useful in the real world.
Reply by ●July 19, 20052005-07-19
"Stan Pawlukiewicz" <spam@spam.mitre.org> wrote in message news:dbg9ee$l52$1@newslocal.mitre.org...> kiki wrote: >> y(t)=x(2*t - 2) - x(-3*t + 5 ) >> >> where x(t) is the input, y(t) is the output of the system. >> >> Is this system time-invariant? >> >> Thanks a lot! > Another way of looking at this is that the impulse response is > > delta(2*t-2) - delta(-3*t +5) > >What special about this impulse response make it non-time invariant? Thanks a lot!
Reply by ●July 19, 20052005-07-19
"kiki" <lunaliu3@yahoo.com> wrote in message news:dbf664$n0m$1@news.Stanford.EDU...> y(t)=x(2*t - 2) - x(-3*t + 5 ) > > where x(t) is the input, y(t) is the output of the system. > > Is this system time-invariant? > > Thanks a lot! >Can anybody say something about the x(-3*t+5) part? I understand x(a*t) make it non-time invariant for a not equal to 1. But I am not sure about the inverse time axis part... For example, Is this system time- invariant? y(t)=x(-t+5) How to use Fred's definition to show it? ps This is not a HW problem.
Reply by ●July 19, 20052005-07-19
kiki wrote: ...>>delta(2*t-2) - delta(-3*t +5) > > What special about this impulse response make it non-time invariant? > > Thanks a lot!I highlights in a simple way that delaying the input gives a result that is different from delaying the output. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●July 19, 20052005-07-19
Reply by ●July 19, 20052005-07-19
kiki wrote:> "Stan Pawlukiewicz" <spam@spam.mitre.org> wrote in message > news:dbg9ee$l52$1@newslocal.mitre.org... > >>kiki wrote: >> >>>y(t)=x(2*t - 2) - x(-3*t + 5 ) >>> >>>where x(t) is the input, y(t) is the output of the system. >>> >>>Is this system time-invariant? >>> >>>Thanks a lot! >> >>Another way of looking at this is that the impulse response is >> >>delta(2*t-2) - delta(-3*t +5) >> >> > > > What special about this impulse response make it non-time invariant? > > Thanks a lot! > >delay the input impulse by T seconds delta(2(t-T) -2) - delta(-3(t-T)+5) = delta(2t - 2 -2T) - delta(-3t +5 +3T) the output is not delayed by T seconds.
