A system is described by an equation that relates the output function y(t) and the input function x(t). Both x(t) and y(t) are functions of time. If the system is time invariant, it means that the mechanisms of the equation are not time dependent (the coefficients are constants). Ex: y(t)=3*x(t)+ [x(t)]^2 This means that y depends only "implicitly" on time, but not explicitly: y(x)= y(x(t)). y(t) can be written as a function of time only however: Ex: x(t)=2*t, then y(t)=3*(2t)+ [2t]^2 If the system is time variant, then the equation describing the relation between input and output has time t variable appearing as an explicit variable: Ex: y(t)=5*t+x(t) so y(t,x)=y(t, x(t)). y(t) can be written only as a function of time t too. x(t)=2t, then y(t)=5*t+2t Q: If I was ONLY given the function y(t) and was asked if it is the output of a time variant or invariant system, would I be able to tell? thanks fisico32
time variant or invariant output?
Started by ●October 17, 2009
Reply by ●October 17, 20092009-10-17
On Sat, 17 Oct 2009 12:30:43 -0500, fisico32 wrote:> A system is described by an equation that relates the output function > y(t) and the input function x(t). Both x(t) and y(t) are functions of > time. If the system is time invariant, it means that the mechanisms of > the equation are not time dependent (the coefficients are constants). > > Ex: y(t)=3*x(t)+ [x(t)]^2 > > This means that y depends only "implicitly" on time, but not explicitly: > y(x)= y(x(t)). > y(t) can be written as a function of time only however: Ex: x(t)=2*t, > then y(t)=3*(2t)+ [2t]^2 > > If the system is time variant, then the equation describing the relation > between input and output has time t variable appearing as an explicit > variable: > > Ex: y(t)=5*t+x(t) > > so y(t,x)=y(t, x(t)). > y(t) can be written only as a function of time t too. x(t)=2t, then > y(t)=5*t+2t > > Q: If I was ONLY given the function y(t) and was asked if it is the > output of a time variant or invariant system, would I be able to tell? > > thanks > fisico32Normal terminology in signals & systems is to call x(t) and y(t) _signals_. Yes, their values are functions of time, but you don't really care about the "functional" part nearly as much as you care about their behavior. Think of them as continuous vectors that "just are" more than as functions. A more general way to describe a time varying system is to define the system h as y(t) = h(x(t), t). In other words, h is some "thing" that acts on the input signal x(t) to generate the output signal y(t). The nice thing about the above definition is that you can immediately shift the input and output signals by some time t_s: y(t - t_s) =? h(x(t - t_s), t). If the above y and x _always_ match the non-shifted case for _all_ possible time shifts and _all_ possible input signals then the system is time invariant. Now, to answer your question: No. If you were given both the input and the output signals for all time, you could _sometimes_ determine that the system was either time varying or nonlinear. I don't think you could conclusively prove that the system was a linear time invariant system just from one sample x(t) and it's resulting sample y(t), however. -- www.wescottdesign.com
Reply by ●October 17, 20092009-10-17
Thank you, not sure why I got confused.... So, I guess the impulse response h , in the general case, could be written as h [delta(t-t0), t] If the system is time invariant, then h[ delta(t-t0) ] there is no time dependance t. So, for example, if the function h contains both t and t_0 (the location of the delta), in such a way that we ca always pair t and t_0 in a subtraction, then the system is time invariant: h(delta(t-t_0)) = 3+(t-t_0)^2. This system would be time invariant....... fisico32
Reply by ●October 17, 20092009-10-17
On Oct 17, 1:30=A0pm, "fisico32" <marcoscipio...@gmail.com> wrote:> A system is described by an equation that relates the output function y(t=)> and the input function x(t). Both x(t) and y(t) are functions of time. If > the system is time invariant, it means that the mechanisms of the equatio=n> are not time dependent (the coefficients are constants). > > Ex: y(t)=3D3*x(t)+ [x(t)]^2 > > This means that y depends only "implicitly" on time, but not explicitly: > y(x)=3D y(x(t)). > y(t) can be written as a function of time only however: > Ex: x(t)=3D2*t, then y(t)=3D3*(2t)+ [2t]^2 > > If the system is time variant, then the equation describing the relation > between input and output has time t variable appearing as an explicit > variable: > > Ex: y(t)=3D5*t+x(t) > > so y(t,x)=3Dy(t, x(t)). =A0 > y(t) can be written only as a function of time t too. > x(t)=3D2t, then y(t)=3D5*t+2t > > Q: If I was ONLY given the function y(t) and was asked if it is the outpu=t> of a time variant or invariant system, would I be able to tell? =A0A system is time invariant if and only if x(t) --> y(t) implies x(t-t0) --> y(t-t0) i.e, the only result of a time-shifted input is an output that is shifted by the same amount of time. Hope this helps. Greg
Reply by ●October 17, 20092009-10-17
On Oct 17, 4:00=A0pm, "fisico32" <marcoscipio...@gmail.com> wrote:> Thank you, not sure why I got confused.... > So, I guess the impulse response h , in the general case, could be writte=n> as > > h [delta(t-t0), t]> If the system is time invariant, then h[ delta(t-t0) ] there is no time =dependance t.> > So, for example, if the function h contains both t and t_0 (the location > of the delta), in such a way that we ca always pair t and t_0 in a > subtraction, then the system is time invariant: > > h(delta(t-t_0)) =3D =A03+(t-t_0)^2. This system would be time > invariant.......Linear systems are usually characterized by d(t-t0) --> h(t,t0) % d =3D Dirac delta, h =3D impulse response which reduces to d(t-t0) --> h(t-t0,0) when the linear sytem is time invariant. The response to an arbitrary input can then be written y(t) =3D INT(-inf,inf){ du h(t,u) x(u) } However, when the system in nonlinear, superposition does not apply and the concept of an impulse response is rather useless. Hope this helps. Greg
Reply by ●October 18, 20092009-10-18
On Oct 17, 8:05�pm, Tim Wescott <t...@seemywebsite.com> wrote:> On Sat, 17 Oct 2009 12:30:43 -0500, fisico32 wrote: > > A system is described by an equation that relates the output function > > y(t) and the input function x(t). Both x(t) and y(t) are functions of > > time. If the system is time invariant, it means that the mechanisms of > > the equation are not time dependent (the coefficients are constants). > > > Ex: y(t)=3*x(t)+ [x(t)]^2 > > > This means that y depends only "implicitly" on time, but not explicitly: > > y(x)= y(x(t)). > > y(t) can be written as a function of time only however: Ex: x(t)=2*t, > > then y(t)=3*(2t)+ [2t]^2 > > > If the system is time variant, then the equation describing the relation > > between input and output has time t variable appearing as an explicit > > variable: > > > Ex: y(t)=5*t+x(t) > > > so y(t,x)=y(t, x(t)). > > y(t) can be written only as a function of time t too. x(t)=2t, then > > y(t)=5*t+2t > > > Q: If I was ONLY given the function y(t) and was asked if it is the > > output of a time variant or invariant system, would I be able to tell? > > > thanks > > fisico32 > > Normal terminology in signals & systems is to call x(t) and y(t) > _signals_. �Yes, their values are functions of time, but you don't really > care about the "functional" part nearly as much as you care about their > behavior. �Think of them as continuous vectors that "just are" more than > as functions. > > A more general way to describe a time varying system is to define the > system h as > > y(t) = h(x(t), t). > > In other words, h is some "thing" that acts on the input signal x(t) to > generate the output signal y(t). �The nice thing about the above > definition is that you can immediately shift the input and output signals > by some time t_s: > > y(t - t_s) =? h(x(t - t_s), t). > > If the above y and x _always_ match the non-shifted case for _all_ > possible time shifts and _all_ possible input signals then the system is > time invariant. > > Now, to answer your question: > > No. �If you were given both the input and the output signals for all > time, you could _sometimes_ determine that the system was either time > varying or nonlinear. �I don't think you could conclusively prove that > the system was a linear time invariant system just from one sample x(t) > and it's resulting sample y(t), however. > > --www.wescottdesign.comIs it not true that if a system is nonlinear it's spectrum will have changed which is easily measurable as long as the change isn't too small to measure?
Reply by ●October 18, 20092009-10-18
On Sun, 18 Oct 2009 09:45:12 -0700, Fully Half Baked wrote:> On Oct 17, 8:05 pm, Tim Wescott <t...@seemywebsite.com> wrote: >> On Sat, 17 Oct 2009 12:30:43 -0500, fisico32 wrote: >> > A system is described by an equation that relates the output function >> > y(t) and the input function x(t). Both x(t) and y(t) are functions of >> > time. If the system is time invariant, it means that the mechanisms >> > of the equation are not time dependent (the coefficients are >> > constants). >> >> > Ex: y(t)=3*x(t)+ [x(t)]^2 >> >> > This means that y depends only "implicitly" on time, but not >> > explicitly: y(x)= y(x(t)). >> > y(t) can be written as a function of time only however: Ex: x(t)=2*t, >> > then y(t)=3*(2t)+ [2t]^2 >> >> > If the system is time variant, then the equation describing the >> > relation between input and output has time t variable appearing as an >> > explicit variable: >> >> > Ex: y(t)=5*t+x(t) >> >> > so y(t,x)=y(t, x(t)). >> > y(t) can be written only as a function of time t too. x(t)=2t, then >> > y(t)=5*t+2t >> >> > Q: If I was ONLY given the function y(t) and was asked if it is the >> > output of a time variant or invariant system, would I be able to >> > tell? >> >> > thanks >> > fisico32 >> >> Normal terminology in signals & systems is to call x(t) and y(t) >> _signals_. Yes, their values are functions of time, but you don't >> really care about the "functional" part nearly as much as you care >> about their behavior. Think of them as continuous vectors that "just >> are" more than as functions. >> >> A more general way to describe a time varying system is to define the >> system h as >> >> y(t) = h(x(t), t). >> >> In other words, h is some "thing" that acts on the input signal x(t) to >> generate the output signal y(t). The nice thing about the above >> definition is that you can immediately shift the input and output >> signals by some time t_s: >> >> y(t - t_s) =? h(x(t - t_s), t). >> >> If the above y and x _always_ match the non-shifted case for _all_ >> possible time shifts and _all_ possible input signals then the system >> is time invariant. >> >> Now, to answer your question: >> >> No. If you were given both the input and the output signals for all >> time, you could _sometimes_ determine that the system was either time >> varying or nonlinear. I don't think you could conclusively prove that >> the system was a linear time invariant system just from one sample x(t) >> and it's resulting sample y(t), however. >> >> --www.wescottdesign.com > > Is it not true that if a system is nonlinear it's spectrum will have > changed which is easily measurable as long as the change isn't too small > to measure?Any system can _change_ the spectrum of the input. A linear system can only change the amplitude of energy that was already in the input, a time- varying system can only convolve the input spectrum with it's own "time- varying-ness" spectrum and change the amplitude of energy that's already in the input, and will do so in a way that obeys superposition. A nonlinear system can do any damn thing it pleases with the spectrum of the input signal _and_ won't obey superposition. But the OP is asking if he can look at the output signal _only_. If he really means what he says, that he's just been handed a signal and told "here, this is the output of a system" then he has no clue about the linearity or time invariance of the system. If he's given the input _and_ the output, then he may be able to say "that system isn't LTI", but with just one input and one output signal I don't think he can say _for sure_ that the system is indeed LTI, or if not if it is nonlinear or time varying. -- www.wescottdesign.com
Reply by ●October 18, 20092009-10-18
Hello Tim, could you elaborate a bit more on your statements: "Any system can _change_ the spectrum of the input. A linear system can only change the amplitude of energy that was already in the input, a time- varying system can only convolve the input spectrum with it's own "time- varying-ness" spectrum and change the amplitude of energy that's already in the input, and will do so in a way that obeys superposition. A nonlinear system can do any damn thing it pleases with the spectrum of the input signal _and_ won't obey superposition." As you say, not only nonlinear systems generate new frequencies in the output spectrum. Also linear time-varying systems can do that. I know (even if don't completely own the concept yet) that the output spectrum Y(eta) is the convolution of the input spectrum X(omega) with a function that expresses the time variant property of the filter called the bifrequency function B(nu, omega), where the variable nu (also called Doppler shift)is measured in Hz and represents a measure of how rapidly the filter changes with time... What is happening to the input spectrum components? Are they amplitude and phase modulated in time or something like that? fisico32
Reply by ●October 18, 20092009-10-18
On Oct 18, 8:02�pm, Tim Wescott <t...@seemywebsite.com> wrote:> On Sun, 18 Oct 2009 09:45:12 -0700, Fully Half Baked wrote: > > On Oct 17, 8:05�pm, Tim Wescott <t...@seemywebsite.com> wrote: > >> On Sat, 17 Oct 2009 12:30:43 -0500, fisico32 wrote: > >> > A system is described by an equation that relates the output function > >> > y(t) and the input function x(t). Both x(t) and y(t) are functions of > >> > time. If the system is time invariant, it means that the mechanisms > >> > of the equation are not time dependent (the coefficients are > >> > constants). > > >> > Ex: y(t)=3*x(t)+ [x(t)]^2 > > >> > This means that y depends only "implicitly" on time, but not > >> > explicitly: y(x)= y(x(t)). > >> > y(t) can be written as a function of time only however: Ex: x(t)=2*t, > >> > then y(t)=3*(2t)+ [2t]^2 > > >> > If the system is time variant, then the equation describing the > >> > relation between input and output has time t variable appearing as an > >> > explicit variable: > > >> > Ex: y(t)=5*t+x(t) > > >> > so y(t,x)=y(t, x(t)). > >> > y(t) can be written only as a function of time t too. x(t)=2t, then > >> > y(t)=5*t+2t > > >> > Q: If I was ONLY given the function y(t) and was asked if it is the > >> > output of a time variant or invariant system, would I be able to > >> > tell? > > >> > thanks > >> > fisico32 > > >> Normal terminology in signals & systems is to call x(t) and y(t) > >> _signals_. �Yes, their values are functions of time, but you don't > >> really care about the "functional" part nearly as much as you care > >> about their behavior. �Think of them as continuous vectors that "just > >> are" more than as functions. > > >> A more general way to describe a time varying system is to define the > >> system h as > > >> y(t) = h(x(t), t). > > >> In other words, h is some "thing" that acts on the input signal x(t) to > >> generate the output signal y(t). �The nice thing about the above > >> definition is that you can immediately shift the input and output > >> signals by some time t_s: > > >> y(t - t_s) =? h(x(t - t_s), t). > > >> If the above y and x _always_ match the non-shifted case for _all_ > >> possible time shifts and _all_ possible input signals then the system > >> is time invariant. > > >> Now, to answer your question: > > >> No. �If you were given both the input and the output signals for all > >> time, you could _sometimes_ determine that the system was either time > >> varying or nonlinear. �I don't think you could conclusively prove that > >> the system was a linear time invariant system just from one sample x(t) > >> and it's resulting sample y(t), however. > > >> --www.wescottdesign.com > > > Is it not true that if a system is nonlinear it's spectrum will have > > changed which is easily measurable as long as the change isn't too small > > to measure? > > Any system can _change_ the spectrum of the input. �A linear system can > only change the amplitude of energy that was already in the input, a time- > varying system can only convolve the input spectrum with it's own "time- > varying-ness" spectrum and change the amplitude of energy that's already > in the input, and will do so in a way that obeys superposition. �A > nonlinear system can do any damn thing it pleases with the spectrum of > the input signal _and_ won't obey superposition. > > But the OP is asking if he can look at the output signal _only_. �If he > really means what he says, that he's just been handed a signal and told > "here, this is the output of a system" then he has no clue about the > linearity or time invariance of the system. �If he's given the input > _and_ the output, then he may be able to say "that system isn't LTI", but > with just one input and one output signal I don't think he can say _for > sure_ that the system is indeed LTI, or if not if it is nonlinear or time > varying. > > --www.wescottdesign.comOk maybe my idea of what is linear or not is a bit too simplistic. What I mean by change the spectrum is new frequencies are introduced not just changes in amplitude or shifts in time. Having said that fisico32 has said "not only nonlinear systems generate new frequencies in the output spectrum" but I don't know how that can happen if a system is said to be linear unless you deliberately generate new frequencies and add them to the output but then the overall effect is still nonlinear.
Reply by ●October 18, 20092009-10-18
Fully Half Baked wrote:> On Oct 18, 8:02 pm, Tim Wescott <t...@seemywebsite.com> wrote: >> On Sun, 18 Oct 2009 09:45:12 -0700, Fully Half Baked wrote: >>> On Oct 17, 8:05 pm, Tim Wescott <t...@seemywebsite.com> wrote: >>>> On Sat, 17 Oct 2009 12:30:43 -0500, fisico32 wrote: >>>>> A system is described by an equation that relates the output function >>>>> y(t) and the input function x(t). Both x(t) and y(t) are functions of >>>>> time. If the system is time invariant, it means that the mechanisms >>>>> of the equation are not time dependent (the coefficients are >>>>> constants). >>>>> Ex: y(t)=3*x(t)+ [x(t)]^2 >>>>> This means that y depends only "implicitly" on time, but not >>>>> explicitly: y(x)= y(x(t)). >>>>> y(t) can be written as a function of time only however: Ex: x(t)=2*t, >>>>> then y(t)=3*(2t)+ [2t]^2 >>>>> If the system is time variant, then the equation describing the >>>>> relation between input and output has time t variable appearing as an >>>>> explicit variable: >>>>> Ex: y(t)=5*t+x(t) >>>>> so y(t,x)=y(t, x(t)). >>>>> y(t) can be written only as a function of time t too. x(t)=2t, then >>>>> y(t)=5*t+2t >>>>> Q: If I was ONLY given the function y(t) and was asked if it is the >>>>> output of a time variant or invariant system, would I be able to >>>>> tell? >>>>> thanks >>>>> fisico32 >>>> Normal terminology in signals & systems is to call x(t) and y(t) >>>> _signals_. Yes, their values are functions of time, but you don't >>>> really care about the "functional" part nearly as much as you care >>>> about their behavior. Think of them as continuous vectors that "just >>>> are" more than as functions. >>>> A more general way to describe a time varying system is to define the >>>> system h as >>>> y(t) = h(x(t), t). >>>> In other words, h is some "thing" that acts on the input signal x(t) to >>>> generate the output signal y(t). The nice thing about the above >>>> definition is that you can immediately shift the input and output >>>> signals by some time t_s: >>>> y(t - t_s) =? h(x(t - t_s), t). >>>> If the above y and x _always_ match the non-shifted case for _all_ >>>> possible time shifts and _all_ possible input signals then the system >>>> is time invariant. >>>> Now, to answer your question: >>>> No. If you were given both the input and the output signals for all >>>> time, you could _sometimes_ determine that the system was either time >>>> varying or nonlinear. I don't think you could conclusively prove that >>>> the system was a linear time invariant system just from one sample x(t) >>>> and it's resulting sample y(t), however. >>>> --www.wescottdesign.com >>> Is it not true that if a system is nonlinear it's spectrum will have >>> changed which is easily measurable as long as the change isn't too small >>> to measure? >> Any system can _change_ the spectrum of the input. A linear system can >> only change the amplitude of energy that was already in the input, a time- >> varying system can only convolve the input spectrum with it's own "time- >> varying-ness" spectrum and change the amplitude of energy that's already >> in the input, and will do so in a way that obeys superposition. A >> nonlinear system can do any damn thing it pleases with the spectrum of >> the input signal _and_ won't obey superposition. >> >> But the OP is asking if he can look at the output signal _only_. If he >> really means what he says, that he's just been handed a signal and told >> "here, this is the output of a system" then he has no clue about the >> linearity or time invariance of the system. If he's given the input >> _and_ the output, then he may be able to say "that system isn't LTI", but >> with just one input and one output signal I don't think he can say _for >> sure_ that the system is indeed LTI, or if not if it is nonlinear or time >> varying. >> >> --www.wescottdesign.com > > Ok maybe my idea of what is linear or not is a bit too simplistic. > What I mean by change the spectrum is new frequencies are introduced > not just changes in amplitude or shifts in time. > Having said that fisico32 has said "not only nonlinear systems > generate new frequencies in the output spectrum" but I don't know how > that can happen if a system is said to be linear unless you > deliberately generate new frequencies and add them to the output but > then the overall effect is still nonlinear.To continue in a simplistic vein, imagine an amplifier with time-varying gain; the input is a single sinusoid of 2KHz, and the gain varies sinusoidally at a frequency of 50 Hz. What is the output spectrum? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
