Check if the system is linear
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4
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The system:
$$ T(x[n]) = ax[n] + bx[n-3] $$
For me it seems that the system is linear:
$$
begin{align}
T(alpha_1x_1[n]+alpha_2x_2[n]) & = a(alpha_1x_1[n]+alpha_2x_2[n]) + b(alpha_1x_1[n-3]+alpha_2x_2[n-3]) \
& = alpha_1(ax_1[n] + bx_1[n-3]) + alpha_2(ax_2[n] + bx_2[n-3]) \
& = alpha_1T(x_1[n])+alpha_1T(x_2[n])
end{align}$$
Thus it's linear, however in the presentation I got it says it's not linear. (without reasoning) Where I'm making the mistake (or maybe there is a mistake in the presentation)
linear-systems system-identification
add a comment |
up vote
4
down vote
favorite
The system:
$$ T(x[n]) = ax[n] + bx[n-3] $$
For me it seems that the system is linear:
$$
begin{align}
T(alpha_1x_1[n]+alpha_2x_2[n]) & = a(alpha_1x_1[n]+alpha_2x_2[n]) + b(alpha_1x_1[n-3]+alpha_2x_2[n-3]) \
& = alpha_1(ax_1[n] + bx_1[n-3]) + alpha_2(ax_2[n] + bx_2[n-3]) \
& = alpha_1T(x_1[n])+alpha_1T(x_2[n])
end{align}$$
Thus it's linear, however in the presentation I got it says it's not linear. (without reasoning) Where I'm making the mistake (or maybe there is a mistake in the presentation)
linear-systems system-identification
Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31
1
It could be hard as I don't have the access to the presentations. But it was in no context, just list of a few systems with the properties like linearity, time in-variance, causality, memorylessness, stability.
– sswwqqaa
Nov 28 at 19:35
Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The system:
$$ T(x[n]) = ax[n] + bx[n-3] $$
For me it seems that the system is linear:
$$
begin{align}
T(alpha_1x_1[n]+alpha_2x_2[n]) & = a(alpha_1x_1[n]+alpha_2x_2[n]) + b(alpha_1x_1[n-3]+alpha_2x_2[n-3]) \
& = alpha_1(ax_1[n] + bx_1[n-3]) + alpha_2(ax_2[n] + bx_2[n-3]) \
& = alpha_1T(x_1[n])+alpha_1T(x_2[n])
end{align}$$
Thus it's linear, however in the presentation I got it says it's not linear. (without reasoning) Where I'm making the mistake (or maybe there is a mistake in the presentation)
linear-systems system-identification
The system:
$$ T(x[n]) = ax[n] + bx[n-3] $$
For me it seems that the system is linear:
$$
begin{align}
T(alpha_1x_1[n]+alpha_2x_2[n]) & = a(alpha_1x_1[n]+alpha_2x_2[n]) + b(alpha_1x_1[n-3]+alpha_2x_2[n-3]) \
& = alpha_1(ax_1[n] + bx_1[n-3]) + alpha_2(ax_2[n] + bx_2[n-3]) \
& = alpha_1T(x_1[n])+alpha_1T(x_2[n])
end{align}$$
Thus it's linear, however in the presentation I got it says it's not linear. (without reasoning) Where I'm making the mistake (or maybe there is a mistake in the presentation)
linear-systems system-identification
linear-systems system-identification
edited Nov 28 at 16:52
Peter K.♦
16.7k83058
16.7k83058
asked Nov 28 at 16:04
sswwqqaa
1305
1305
Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31
1
It could be hard as I don't have the access to the presentations. But it was in no context, just list of a few systems with the properties like linearity, time in-variance, causality, memorylessness, stability.
– sswwqqaa
Nov 28 at 19:35
Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53
add a comment |
Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31
1
It could be hard as I don't have the access to the presentations. But it was in no context, just list of a few systems with the properties like linearity, time in-variance, causality, memorylessness, stability.
– sswwqqaa
Nov 28 at 19:35
Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53
Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31
Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31
1
1
It could be hard as I don't have the access to the presentations. But it was in no context, just list of a few systems with the properties like linearity, time in-variance, causality, memorylessness, stability.
– sswwqqaa
Nov 28 at 19:35
It could be hard as I don't have the access to the presentations. But it was in no context, just list of a few systems with the properties like linearity, time in-variance, causality, memorylessness, stability.
– sswwqqaa
Nov 28 at 19:35
Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53
Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53
add a comment |
4 Answers
4
active
oldest
votes
up vote
5
down vote
accepted
I believe there's either a mistake in the presentation or the presentation is using a different definition of linear.
For example, the system is linear in $x$ from a system perspective, but it's affine in $x[n]$ (and, therefore not linear) because of the $bx[n-3]$ offset.
On this site, we tend to go with the system definition rather than split hairs about linearity versus affine-ness.
4
The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context.
– Matt L.
Nov 28 at 17:26
1
Indeed, I have never heard about "being affine" in the ordinal variable.
– Laurent Duval
Nov 28 at 18:29
@MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it.
– Peter K.♦
Nov 28 at 18:43
I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
– Laurent Duval
Nov 28 at 20:17
1
@LaurentDuval 😅😂🤣😜
– Peter K.♦
Nov 28 at 23:43
add a comment |
up vote
4
down vote
[Note: it may happen that a teacher makes a oral mistake, that puzzles the audience. So here is an alternative explanation on this system being non-something]
This system is, as far as Peter K., Matt L. and I know, nicely linear. You already did the computations. With a little more work, among classical properties, it is also time-invariant, causal, stable.
The only basic property it does not possess is "to be memoryless", unless one of the constant $a$ or $b$ (including both) is zero (see for instance an extended conception of a memoryless system from What is a memory less system?).
2
It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter.
– Matt L.
Nov 28 at 20:42
1
hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-)
– Fat32
Nov 29 at 9:12
1
Go ahead and I'd suggest you to add the non-additive magma notion
– Laurent Duval
Nov 29 at 9:31
add a comment |
up vote
2
down vote
A system $mathcal{T}$ is linear if its response to a weighted sum of two signals equals the weighted sum of its individual responses to those two input signals:
$$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$
with arbitrary constants $alpha$ and $beta$, and arbitrary input sequences $x_1[n]$ and $x_2[n]$. A system $mathcal{T}$ satisfying $(1)$ is completely characterized by its impulse response $h[n]$, and its input-output relation can be formulated as a convolution of the input sequence with its impulse response:
$$mathcal{T}big{xbig}=sum_{k=-infty}^{infty}h[k]x[n-k]tag{2}$$
It is easily shown that the given system
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]tag{3}$$
satisfies $(1)$, and that it is characterized by the impulse response
$$h[n]=adelta[n]+bdelta[n-3]tag{4}$$
Clearly, its input-output relation $(3)$ can be written as a convolution sum $(2)$. Equivalently, the $mathcal{Z}$-transform of its response is given by the multiplication of the $mathcal{Z}$-transform of its input sequence and its transfer function $H(z)=mathcal{Z}{h[n]}$:
$$Y(z)=X(z)H(z)tag{5}$$
By contrast, an affine system does not satisfy $(1)$, and it cannot be characterized by an impulse response or, equivalently, by a transfer function. The given linear system $(3)$ could be made affine by adding a constant $c$ ($cneq 0$) to its output:
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$
This input-output relation cannot be formulated in terms of a convolution $(2)$, and it can easily be checked that the system $(6)$ doesn't satisfy $(1)$. Such a system is not linear, since part of the output (the constant $c$) does not depend on the input signal $x[n]$.
There is no reasonable definition of linearity according to which the given system $(3)$ could be classified as being non-linear.
add a comment |
up vote
-1
down vote
This system is, as far as Peter K., Laurent, and I know, nicely linear. It is also time-invariant, causal, and stable. Indeed it's LTI with impulse response $h[n] = a delta[n] + b delta[n-3]$.
The only basic property it does not possess is "to be memoryless", unless $a$ or $b$ is zero (see for instance an extended version from What is a memory less system?).
Maybe you need another folk copying your answer to get the first vote
– Laurent Duval
Nov 29 at 19:37
@LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-))
– Fat32
Nov 29 at 22:02
@LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
– Fat32
Dec 1 at 0:19
Sometimes, less is more. Lo siento
– Laurent Duval
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
I believe there's either a mistake in the presentation or the presentation is using a different definition of linear.
For example, the system is linear in $x$ from a system perspective, but it's affine in $x[n]$ (and, therefore not linear) because of the $bx[n-3]$ offset.
On this site, we tend to go with the system definition rather than split hairs about linearity versus affine-ness.
4
The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context.
– Matt L.
Nov 28 at 17:26
1
Indeed, I have never heard about "being affine" in the ordinal variable.
– Laurent Duval
Nov 28 at 18:29
@MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it.
– Peter K.♦
Nov 28 at 18:43
I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
– Laurent Duval
Nov 28 at 20:17
1
@LaurentDuval 😅😂🤣😜
– Peter K.♦
Nov 28 at 23:43
add a comment |
up vote
5
down vote
accepted
I believe there's either a mistake in the presentation or the presentation is using a different definition of linear.
For example, the system is linear in $x$ from a system perspective, but it's affine in $x[n]$ (and, therefore not linear) because of the $bx[n-3]$ offset.
On this site, we tend to go with the system definition rather than split hairs about linearity versus affine-ness.
4
The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context.
– Matt L.
Nov 28 at 17:26
1
Indeed, I have never heard about "being affine" in the ordinal variable.
– Laurent Duval
Nov 28 at 18:29
@MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it.
– Peter K.♦
Nov 28 at 18:43
I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
– Laurent Duval
Nov 28 at 20:17
1
@LaurentDuval 😅😂🤣😜
– Peter K.♦
Nov 28 at 23:43
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
I believe there's either a mistake in the presentation or the presentation is using a different definition of linear.
For example, the system is linear in $x$ from a system perspective, but it's affine in $x[n]$ (and, therefore not linear) because of the $bx[n-3]$ offset.
On this site, we tend to go with the system definition rather than split hairs about linearity versus affine-ness.
I believe there's either a mistake in the presentation or the presentation is using a different definition of linear.
For example, the system is linear in $x$ from a system perspective, but it's affine in $x[n]$ (and, therefore not linear) because of the $bx[n-3]$ offset.
On this site, we tend to go with the system definition rather than split hairs about linearity versus affine-ness.
answered Nov 28 at 16:55
Peter K.♦
16.7k83058
16.7k83058
4
The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context.
– Matt L.
Nov 28 at 17:26
1
Indeed, I have never heard about "being affine" in the ordinal variable.
– Laurent Duval
Nov 28 at 18:29
@MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it.
– Peter K.♦
Nov 28 at 18:43
I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
– Laurent Duval
Nov 28 at 20:17
1
@LaurentDuval 😅😂🤣😜
– Peter K.♦
Nov 28 at 23:43
add a comment |
4
The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context.
– Matt L.
Nov 28 at 17:26
1
Indeed, I have never heard about "being affine" in the ordinal variable.
– Laurent Duval
Nov 28 at 18:29
@MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it.
– Peter K.♦
Nov 28 at 18:43
I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
– Laurent Duval
Nov 28 at 20:17
1
@LaurentDuval 😅😂🤣😜
– Peter K.♦
Nov 28 at 23:43
4
4
The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context.
– Matt L.
Nov 28 at 17:26
The system is just a two-tap FIR filter. So I wouldn't say it's affine. It would be affine (if one was willing to make that distinction) if it were something like $y[n]=ax[n]+b$, i.e., just adding a constant (like a non-zero initial condition). That would make it non-linear in a signals&systems context.
– Matt L.
Nov 28 at 17:26
1
1
Indeed, I have never heard about "being affine" in the ordinal variable.
– Laurent Duval
Nov 28 at 18:29
Indeed, I have never heard about "being affine" in the ordinal variable.
– Laurent Duval
Nov 28 at 18:29
@MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it.
– Peter K.♦
Nov 28 at 18:43
@MattL. Understood! I've just seen some mathematicians get their knickers in a twist over systems like in the original post. Without knowing the source of the claim that it's not linear, it's hard to rebut it.
– Peter K.♦
Nov 28 at 18:43
I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
– Laurent Duval
Nov 28 at 20:17
I can't resist pedantry here :) maybe $a$ and $b$ are elements of a non-additive magma?
– Laurent Duval
Nov 28 at 20:17
1
1
@LaurentDuval 😅😂🤣😜
– Peter K.♦
Nov 28 at 23:43
@LaurentDuval 😅😂🤣😜
– Peter K.♦
Nov 28 at 23:43
add a comment |
up vote
4
down vote
[Note: it may happen that a teacher makes a oral mistake, that puzzles the audience. So here is an alternative explanation on this system being non-something]
This system is, as far as Peter K., Matt L. and I know, nicely linear. You already did the computations. With a little more work, among classical properties, it is also time-invariant, causal, stable.
The only basic property it does not possess is "to be memoryless", unless one of the constant $a$ or $b$ (including both) is zero (see for instance an extended conception of a memoryless system from What is a memory less system?).
2
It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter.
– Matt L.
Nov 28 at 20:42
1
hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-)
– Fat32
Nov 29 at 9:12
1
Go ahead and I'd suggest you to add the non-additive magma notion
– Laurent Duval
Nov 29 at 9:31
add a comment |
up vote
4
down vote
[Note: it may happen that a teacher makes a oral mistake, that puzzles the audience. So here is an alternative explanation on this system being non-something]
This system is, as far as Peter K., Matt L. and I know, nicely linear. You already did the computations. With a little more work, among classical properties, it is also time-invariant, causal, stable.
The only basic property it does not possess is "to be memoryless", unless one of the constant $a$ or $b$ (including both) is zero (see for instance an extended conception of a memoryless system from What is a memory less system?).
2
It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter.
– Matt L.
Nov 28 at 20:42
1
hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-)
– Fat32
Nov 29 at 9:12
1
Go ahead and I'd suggest you to add the non-additive magma notion
– Laurent Duval
Nov 29 at 9:31
add a comment |
up vote
4
down vote
up vote
4
down vote
[Note: it may happen that a teacher makes a oral mistake, that puzzles the audience. So here is an alternative explanation on this system being non-something]
This system is, as far as Peter K., Matt L. and I know, nicely linear. You already did the computations. With a little more work, among classical properties, it is also time-invariant, causal, stable.
The only basic property it does not possess is "to be memoryless", unless one of the constant $a$ or $b$ (including both) is zero (see for instance an extended conception of a memoryless system from What is a memory less system?).
[Note: it may happen that a teacher makes a oral mistake, that puzzles the audience. So here is an alternative explanation on this system being non-something]
This system is, as far as Peter K., Matt L. and I know, nicely linear. You already did the computations. With a little more work, among classical properties, it is also time-invariant, causal, stable.
The only basic property it does not possess is "to be memoryless", unless one of the constant $a$ or $b$ (including both) is zero (see for instance an extended conception of a memoryless system from What is a memory less system?).
edited 2 days ago
answered Nov 28 at 19:44
Laurent Duval
16k32058
16k32058
2
It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter.
– Matt L.
Nov 28 at 20:42
1
hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-)
– Fat32
Nov 29 at 9:12
1
Go ahead and I'd suggest you to add the non-additive magma notion
– Laurent Duval
Nov 29 at 9:31
add a comment |
2
It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter.
– Matt L.
Nov 28 at 20:42
1
hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-)
– Fat32
Nov 29 at 9:12
1
Go ahead and I'd suggest you to add the non-additive magma notion
– Laurent Duval
Nov 29 at 9:31
2
2
It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter.
– Matt L.
Nov 28 at 20:42
It's indeed linear and time-invariant, not affine. It would only be affine if in its output there were a constant term that is independent of the input signal, which is not the case here. It's just a plain FIR filter.
– Matt L.
Nov 28 at 20:42
1
1
hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-)
– Fat32
Nov 29 at 9:12
hmm let me put a (n almost) verbatim copy of your answer and wait to see if I get any upvotes :-)) and see if it is only you ;-)
– Fat32
Nov 29 at 9:12
1
1
Go ahead and I'd suggest you to add the non-additive magma notion
– Laurent Duval
Nov 29 at 9:31
Go ahead and I'd suggest you to add the non-additive magma notion
– Laurent Duval
Nov 29 at 9:31
add a comment |
up vote
2
down vote
A system $mathcal{T}$ is linear if its response to a weighted sum of two signals equals the weighted sum of its individual responses to those two input signals:
$$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$
with arbitrary constants $alpha$ and $beta$, and arbitrary input sequences $x_1[n]$ and $x_2[n]$. A system $mathcal{T}$ satisfying $(1)$ is completely characterized by its impulse response $h[n]$, and its input-output relation can be formulated as a convolution of the input sequence with its impulse response:
$$mathcal{T}big{xbig}=sum_{k=-infty}^{infty}h[k]x[n-k]tag{2}$$
It is easily shown that the given system
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]tag{3}$$
satisfies $(1)$, and that it is characterized by the impulse response
$$h[n]=adelta[n]+bdelta[n-3]tag{4}$$
Clearly, its input-output relation $(3)$ can be written as a convolution sum $(2)$. Equivalently, the $mathcal{Z}$-transform of its response is given by the multiplication of the $mathcal{Z}$-transform of its input sequence and its transfer function $H(z)=mathcal{Z}{h[n]}$:
$$Y(z)=X(z)H(z)tag{5}$$
By contrast, an affine system does not satisfy $(1)$, and it cannot be characterized by an impulse response or, equivalently, by a transfer function. The given linear system $(3)$ could be made affine by adding a constant $c$ ($cneq 0$) to its output:
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$
This input-output relation cannot be formulated in terms of a convolution $(2)$, and it can easily be checked that the system $(6)$ doesn't satisfy $(1)$. Such a system is not linear, since part of the output (the constant $c$) does not depend on the input signal $x[n]$.
There is no reasonable definition of linearity according to which the given system $(3)$ could be classified as being non-linear.
add a comment |
up vote
2
down vote
A system $mathcal{T}$ is linear if its response to a weighted sum of two signals equals the weighted sum of its individual responses to those two input signals:
$$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$
with arbitrary constants $alpha$ and $beta$, and arbitrary input sequences $x_1[n]$ and $x_2[n]$. A system $mathcal{T}$ satisfying $(1)$ is completely characterized by its impulse response $h[n]$, and its input-output relation can be formulated as a convolution of the input sequence with its impulse response:
$$mathcal{T}big{xbig}=sum_{k=-infty}^{infty}h[k]x[n-k]tag{2}$$
It is easily shown that the given system
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]tag{3}$$
satisfies $(1)$, and that it is characterized by the impulse response
$$h[n]=adelta[n]+bdelta[n-3]tag{4}$$
Clearly, its input-output relation $(3)$ can be written as a convolution sum $(2)$. Equivalently, the $mathcal{Z}$-transform of its response is given by the multiplication of the $mathcal{Z}$-transform of its input sequence and its transfer function $H(z)=mathcal{Z}{h[n]}$:
$$Y(z)=X(z)H(z)tag{5}$$
By contrast, an affine system does not satisfy $(1)$, and it cannot be characterized by an impulse response or, equivalently, by a transfer function. The given linear system $(3)$ could be made affine by adding a constant $c$ ($cneq 0$) to its output:
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$
This input-output relation cannot be formulated in terms of a convolution $(2)$, and it can easily be checked that the system $(6)$ doesn't satisfy $(1)$. Such a system is not linear, since part of the output (the constant $c$) does not depend on the input signal $x[n]$.
There is no reasonable definition of linearity according to which the given system $(3)$ could be classified as being non-linear.
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A system $mathcal{T}$ is linear if its response to a weighted sum of two signals equals the weighted sum of its individual responses to those two input signals:
$$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$
with arbitrary constants $alpha$ and $beta$, and arbitrary input sequences $x_1[n]$ and $x_2[n]$. A system $mathcal{T}$ satisfying $(1)$ is completely characterized by its impulse response $h[n]$, and its input-output relation can be formulated as a convolution of the input sequence with its impulse response:
$$mathcal{T}big{xbig}=sum_{k=-infty}^{infty}h[k]x[n-k]tag{2}$$
It is easily shown that the given system
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]tag{3}$$
satisfies $(1)$, and that it is characterized by the impulse response
$$h[n]=adelta[n]+bdelta[n-3]tag{4}$$
Clearly, its input-output relation $(3)$ can be written as a convolution sum $(2)$. Equivalently, the $mathcal{Z}$-transform of its response is given by the multiplication of the $mathcal{Z}$-transform of its input sequence and its transfer function $H(z)=mathcal{Z}{h[n]}$:
$$Y(z)=X(z)H(z)tag{5}$$
By contrast, an affine system does not satisfy $(1)$, and it cannot be characterized by an impulse response or, equivalently, by a transfer function. The given linear system $(3)$ could be made affine by adding a constant $c$ ($cneq 0$) to its output:
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$
This input-output relation cannot be formulated in terms of a convolution $(2)$, and it can easily be checked that the system $(6)$ doesn't satisfy $(1)$. Such a system is not linear, since part of the output (the constant $c$) does not depend on the input signal $x[n]$.
There is no reasonable definition of linearity according to which the given system $(3)$ could be classified as being non-linear.
A system $mathcal{T}$ is linear if its response to a weighted sum of two signals equals the weighted sum of its individual responses to those two input signals:
$$mathcal{T}big{alpha x_1+beta x_2big}=alphamathcal{T}big{x_1big}+betamathcal{T}big{x_2big}tag{1}$$
with arbitrary constants $alpha$ and $beta$, and arbitrary input sequences $x_1[n]$ and $x_2[n]$. A system $mathcal{T}$ satisfying $(1)$ is completely characterized by its impulse response $h[n]$, and its input-output relation can be formulated as a convolution of the input sequence with its impulse response:
$$mathcal{T}big{xbig}=sum_{k=-infty}^{infty}h[k]x[n-k]tag{2}$$
It is easily shown that the given system
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]tag{3}$$
satisfies $(1)$, and that it is characterized by the impulse response
$$h[n]=adelta[n]+bdelta[n-3]tag{4}$$
Clearly, its input-output relation $(3)$ can be written as a convolution sum $(2)$. Equivalently, the $mathcal{Z}$-transform of its response is given by the multiplication of the $mathcal{Z}$-transform of its input sequence and its transfer function $H(z)=mathcal{Z}{h[n]}$:
$$Y(z)=X(z)H(z)tag{5}$$
By contrast, an affine system does not satisfy $(1)$, and it cannot be characterized by an impulse response or, equivalently, by a transfer function. The given linear system $(3)$ could be made affine by adding a constant $c$ ($cneq 0$) to its output:
$$mathcal{T}big{xbig}=y[n]=ax[n]+bx[n-3]+ctag{6}$$
This input-output relation cannot be formulated in terms of a convolution $(2)$, and it can easily be checked that the system $(6)$ doesn't satisfy $(1)$. Such a system is not linear, since part of the output (the constant $c$) does not depend on the input signal $x[n]$.
There is no reasonable definition of linearity according to which the given system $(3)$ could be classified as being non-linear.
answered Nov 28 at 21:32
Matt L.
48.1k13683
48.1k13683
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This system is, as far as Peter K., Laurent, and I know, nicely linear. It is also time-invariant, causal, and stable. Indeed it's LTI with impulse response $h[n] = a delta[n] + b delta[n-3]$.
The only basic property it does not possess is "to be memoryless", unless $a$ or $b$ is zero (see for instance an extended version from What is a memory less system?).
Maybe you need another folk copying your answer to get the first vote
– Laurent Duval
Nov 29 at 19:37
@LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-))
– Fat32
Nov 29 at 22:02
@LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
– Fat32
Dec 1 at 0:19
Sometimes, less is more. Lo siento
– Laurent Duval
2 days ago
add a comment |
up vote
-1
down vote
This system is, as far as Peter K., Laurent, and I know, nicely linear. It is also time-invariant, causal, and stable. Indeed it's LTI with impulse response $h[n] = a delta[n] + b delta[n-3]$.
The only basic property it does not possess is "to be memoryless", unless $a$ or $b$ is zero (see for instance an extended version from What is a memory less system?).
Maybe you need another folk copying your answer to get the first vote
– Laurent Duval
Nov 29 at 19:37
@LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-))
– Fat32
Nov 29 at 22:02
@LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
– Fat32
Dec 1 at 0:19
Sometimes, less is more. Lo siento
– Laurent Duval
2 days ago
add a comment |
up vote
-1
down vote
up vote
-1
down vote
This system is, as far as Peter K., Laurent, and I know, nicely linear. It is also time-invariant, causal, and stable. Indeed it's LTI with impulse response $h[n] = a delta[n] + b delta[n-3]$.
The only basic property it does not possess is "to be memoryless", unless $a$ or $b$ is zero (see for instance an extended version from What is a memory less system?).
This system is, as far as Peter K., Laurent, and I know, nicely linear. It is also time-invariant, causal, and stable. Indeed it's LTI with impulse response $h[n] = a delta[n] + b delta[n-3]$.
The only basic property it does not possess is "to be memoryless", unless $a$ or $b$ is zero (see for instance an extended version from What is a memory less system?).
answered Nov 29 at 9:09
Fat32
13.9k31128
13.9k31128
Maybe you need another folk copying your answer to get the first vote
– Laurent Duval
Nov 29 at 19:37
@LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-))
– Fat32
Nov 29 at 22:02
@LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
– Fat32
Dec 1 at 0:19
Sometimes, less is more. Lo siento
– Laurent Duval
2 days ago
add a comment |
Maybe you need another folk copying your answer to get the first vote
– Laurent Duval
Nov 29 at 19:37
@LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-))
– Fat32
Nov 29 at 22:02
@LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
– Fat32
Dec 1 at 0:19
Sometimes, less is more. Lo siento
– Laurent Duval
2 days ago
Maybe you need another folk copying your answer to get the first vote
– Laurent Duval
Nov 29 at 19:37
Maybe you need another folk copying your answer to get the first vote
– Laurent Duval
Nov 29 at 19:37
@LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-))
– Fat32
Nov 29 at 22:02
@LaurentDuval 12 hours after the post and zero upvotes ! this's a disaster ! I should have been quick to copy (before you) PeterK instead ;-))
– Fat32
Nov 29 at 22:02
@LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
– Fat32
Dec 1 at 0:19
@LaurentDuval you see my copy is downvoted, whereas your copy is upvoted ;-)) this is plain injustice !
– Fat32
Dec 1 at 0:19
Sometimes, less is more. Lo siento
– Laurent Duval
2 days ago
Sometimes, less is more. Lo siento
– Laurent Duval
2 days ago
add a comment |
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Could you please share a document saying it is not linear? Maybe from the context we could get it
– Laurent Duval
Nov 28 at 18:31
1
It could be hard as I don't have the access to the presentations. But it was in no context, just list of a few systems with the properties like linearity, time in-variance, causality, memorylessness, stability.
– sswwqqaa
Nov 28 at 19:35
Maybe you could talk to the source to double check there is no misunderstanding
– Laurent Duval
Nov 28 at 20:53