hi all, 1. For a system, if the input function is u1(t), the output is y1(t) and if the input u2(t) the output y2(t). They ask if this system is linear or not. I simply look for the transfer function: H1(s)=Y1(s)/U1(s) and H2(s)=Y2(s)/U2(s). If H1(s)=H2(s) the system is linear, if not it nonlinear. But I am confused. What if the system is linear but not time invariant. Does the proof above still work? 2. For my specific case: u1(t)=tstep(t), y2(t)=(t+1)^3step(t+1). u2(t)=3(t-1)step(t-1), y2(t)=9step(t). Step(t) is the step function/ Is there any another simpler way to verify the linearity? 3. In control books, they often use Laplace transform to find the transfer function. Why don't they use Fourier Transform? For some problems, I feel FOurier Transform is more convenient because we can compute the transform in both sides (negative and positive). Thanks
linear/non-linear system
Started by ●August 26, 2006
Reply by ●August 26, 20062006-08-26
VijaKhara wrote:> hi all, > > 1. For a system, if the input function is u1(t), the output is y1(t) > and if the input u2(t) the output y2(t). They ask if this system is > linear or not. I simply look for the transfer function: > H1(s)=Y1(s)/U1(s) and H2(s)=Y2(s)/U2(s). If H1(s)=H2(s) the system is > linear, if not it nonlinear. But I am confused. What if the system is > linear but not time invariant. Does the proof above still work? > > 2. For my specific case: > u1(t)=tstep(t), y2(t)=(t+1)^3step(t+1). > u2(t)=3(t-1)step(t-1), y2(t)=9step(t). > > Step(t) is the step function/ > > Is there any another simpler way to verify the linearity? > > > 3. In control books, they often use Laplace transform to find the > transfer function. Why don't they use Fourier Transform? For some > problems, I feel FOurier Transform is more convenient because we can > compute the transform in both sides (negative and positive).Positive and negative time are useful for many studies, but not applicable to causal systems. The Laplace transform embraces causality. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 26, 20062006-08-26
For linearity try to follow the basic definition of it. T{.} is a linear system if y = T{x} follows 1. Ay = T{Ax}, scaling by a scalar A scales the output 2. y = T{x1 + x2} = y1 + y2, where y1 = T{x1} and y2 = T{x2} Regaring Laplace, I am not sure but this is my guess: Laplace transform is a more general form of fourier transform. In control systems we are inerested in detemining the stability of the system via looking at the poles of the system. Determining the stability using fourier is not possible. VijaKhara wrote:> hi all, > > 1. For a system, if the input function is u1(t), the output is y1(t) > and if the input u2(t) the output y2(t). They ask if this system is > linear or not. I simply look for the transfer function: > H1(s)=Y1(s)/U1(s) and H2(s)=Y2(s)/U2(s). If H1(s)=H2(s) the system is > linear, if not it nonlinear. But I am confused. What if the system is > linear but not time invariant. Does the proof above still work? > > 2. For my specific case: > u1(t)=tstep(t), y2(t)=(t+1)^3step(t+1). > u2(t)=3(t-1)step(t-1), y2(t)=9step(t). > > Step(t) is the step function/ > > Is there any another simpler way to verify the linearity? > > > 3. In control books, they often use Laplace transform to find the > transfer function. Why don't they use Fourier Transform? For some > problems, I feel FOurier Transform is more convenient because we can > compute the transform in both sides (negative and positive). > > Thanks
Reply by ●August 26, 20062006-08-26
VijaKhara wrote:> hi all, > > 1. For a system, if the input function is u1(t), the output is y1(t) > and if the input u2(t) the output y2(t). They ask if this system is > linear or not.There is only one possible answer. Think why (hint: start with the definition of a linear system). Regards, Andor
Reply by ●August 26, 20062006-08-26
Andor wrote:> VijaKhara wrote: > >> hi all, >> >> 1. For a system, if the input function is u1(t), the output is y1(t) >> and if the input u2(t) the output y2(t). They ask if this system is >> linear or not. > > There is only one possible answer. Think why (hint: start with the > definition of a linear system).You guys here have taught me the peril of categorical statements. Suppose the system (as all real systems do) exhibits linearity only for signals smaller that some value? It seems to me that the only possible answer is "maybe". Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 26, 20062006-08-26
Jerry Avins wrote:> Andor wrote: > > VijaKhara wrote: > > > >> hi all, > >> > >> 1. For a system, if the input function is u1(t), the output is y1(t) > >> and if the input u2(t) the output y2(t). They ask if this system is > >> linear or not. > > > > There is only one possible answer. Think why (hint: start with the > > definition of a linear system). > > You guys here have taught me the peril of categorical statements. > Suppose the system (as all real systems do) exhibits linearity only for > signals smaller that some value? It seems to me that the only possible > answer is "maybe".That's exactly the issue. It is not possible to verify linearity from knowing the system's response on just two input signals (u1 and u2). Therefore, if the problem is well posed, the system has to be non-linear. Consider this system T (for any u1, u2, y1 and y2): y = T(u) = = y1, if u = u1, and = y2, if u = u2, and = 0 for any other u. Clearly, the system has y1 as response to u1 and y2 as response to u2. Is it linear? Regards, Andor
Reply by ●August 26, 20062006-08-26
Andor wrote:> ... It is not possible to verify linearity from > knowing the system's response on just two input signals (u1 and u2). > Therefore, if the problem is well posed, the system has to be > non-linear.I would say that it's not possible to know if it is linear. the Scots trial verdict of "not proven" isn't the same as "not guilty". Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 26, 20062006-08-26
Jerry Avins said the following on 26/08/2006 17:59:> Andor wrote: > > >> ... It is not possible to verify linearity from >> knowing the system's response on just two input signals (u1 and u2). >> Therefore, if the problem is well posed, the system has to be >> non-linear. > > I would say that it's not possible to know if it is linear. the Scots > trial verdict of "not proven" isn't the same as "not guilty". >I think Andor's point might have been that because the original question was from a textbook or a homework assignment, it has probably been designed to have a definitive answer. As it is impossible to definitively say that the system is linear, the only remaining possibility is that the equations in the problem demonstrate that the system must be non-linear. -- Oli
Reply by ●August 26, 20062006-08-26
Oli Filth wrote:> Jerry Avins said the following on 26/08/2006 17:59: >> Andor wrote: >> >> >>> ... It is not possible to verify linearity from >>> knowing the system's response on just two input signals (u1 and u2). >>> Therefore, if the problem is well posed, the system has to be >>> non-linear. >> >> I would say that it's not possible to know if it is linear. the Scots >> trial verdict of "not proven" isn't the same as "not guilty". >> > > I think Andor's point might have been that because the original question > was from a textbook or a homework assignment, it has probably been > designed to have a definitive answer. As it is impossible to > definitively say that the system is linear, the only remaining > possibility is that the equations in the problem demonstrate that the > system must be non-linear.So this is another example of "You had better guess what I mean because I'm incapable of expressing it clearly?" Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 26, 20062006-08-26
Jerry Avins wrote:> Oli Filth wrote: > > Jerry Avins said the following on 26/08/2006 17:59: > >> Andor wrote: > >> > >> > >>> ... It is not possible to verify linearity from > >>> knowing the system's response on just two input signals (u1 and u2). > >>> Therefore, if the problem is well posed, the system has to be > >>> non-linear. > >> > >> I would say that it's not possible to know if it is linear. the Scots > >> trial verdict of "not proven" isn't the same as "not guilty". > >> > > > > I think Andor's point might have been that because the original question > > was from a textbook or a homework assignment, it has probably been > > designed to have a definitive answer. As it is impossible to > > definitively say that the system is linear, the only remaining > > possibility is that the equations in the problem demonstrate that the > > system must be non-linear. > > So this is another example of "You had better guess what I mean because > I'm incapable of expressing it clearly?"Not quie sure. Oli interpreted what I said correctly. However, I can't think of how to show, just using the two input/output pairs (u1,y1) and (u2, y2) that the system is non-linear. One can show that it is non-LTI, because u1 and u2 are, up to scaling and delay, the same signals. But non-linearity?