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!
Basic systems question: Is this system time-invariant?
Started by ●July 18, 2005
Reply by ●July 18, 20052005-07-18
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
Reply by ●July 18, 20052005-07-18
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! >> >> > > NopeHi 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
Reply by ●July 18, 20052005-07-18
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? > > BernhardYes. 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 --
Reply by ●July 18, 20052005-07-18
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! >> > > NopeI hope the instructor asks him why not, instead of just yes/no. One rarely helps anyone by doing their homework. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●July 18, 20052005-07-18
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. Bernhard
Reply by ●July 18, 20052005-07-18
Bernhard Holzmayer wrote: ...> Then, my example with x(t)=1 would hold (==time-invariant).... You weren't paying attention to Fred. Time invariance requires that all coefficients of t itself be 1 or 0. 2*t doesn't work, as Fred explained. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●July 18, 20052005-07-18
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. Regards, Andor
Reply by ●July 18, 20052005-07-18
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)
Reply by ●July 18, 20052005-07-18
Bernhard Holzmayer wrote:> 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. > > BernhardSometimes whether a system is time invariant (or linear, for that matter) depends on how you look at it. Generally if you say x(t) then you mean that x can vary with time, in which case the system is time-varying. You can choose the special case for x(t) = 1 for all time, but then the notation 'x(t)' is deceptive. -- ------------------------------------------- Tim Wescott Wescott Design Services http://www.wescottdesign.com
