LTI

Let a0, a1, . . . , aM be any finite sequence of real numbers.

Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
discrete signal.

Show that L is a linear time invariant system and that it�s unit response
function is h where,
  
         |
   h(m)= |     am if 0 <= m <= M,
         |     0 otherwise.

for the question above i was wondering if i can say sum of ak.[x(n)-k] and
then from there prove that is it time invariant and also linear.

shoma wrote:
> Let a0, a1, . . . , aM be any finite sequence of real numbers.
>
> Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
> discrete signal.
>
> Show that L is a linear time invariant system and that it's unit response
> function is h where,
>
>          |
>    h(m)= |     am if 0 <= m <= M,
>          |     0 otherwise.
>
> for the question above i was wondering if i can say sum of ak.[x(n)-k] and
> then from there prove that is it time invariant and also linear.

My first thought is (though I may be mistaken here) I think that
summability, like you refer to in the bottom of your post is more often
an indicator of stability, rather than linearity and time invariance.

My second thought is that while it may be possible to approach the
problem in this fashion, that your safer bet would be to apply the
standard tests for linearity and time invarience and 'prove' these
properties of system that way.  The reason I suggest this is that the
teacher and especially any graders are going to expect that approach
and if you deviate from what they expect, your grade may suffer.

>Let a0, a1, . . . , aM be any finite sequence of real numbers.
>
>Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
>discrete signal.
>
>Show that L is a linear time invariant system and that it�s unit
response
>function is h where,
>  
>         |
>   h(m)= |     am if 0 <= m <= M,
>         |     0 otherwise.
>
>for the question above i was wondering if i can say sum of ak.[x(n)-k]
and
>then from there prove that is it time invariant and also linear.
>
>
>

Prove that the system is homogenous and additive. Also prove the
time-shifting property and then claim that it is linear time invarient.
Always keep it simple.

>>Let a0, a1, . . . , aM be any finite sequence of real numbers.
>>
>>Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
>>discrete signal.
>>
>>Show that L is a linear time invariant system and that it�s unit
>response
>>function is h where,
>>  
>>         |
>>   h(m)= |     am if 0 <= m <= M,
>>         |     0 otherwise.
>>
>>for the question above i was wondering if i can say sum of ak.[x(n)-k]
>and
>>then from there prove that is it time invariant and also linear.
>>
>>
>>
>
>Prove that the system is homogenous and additive. Also prove the
>time-shifting property and then claim that it is linear time invarient.
>Always keep it simple.
>

anyway u be able to show me an example?
thanx in advance

shoma skrev:
> Let a0, a1, . . . , aM be any finite sequence of real numbers.
>
> Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
> discrete signal.

What does "*" mean? Multiplication? If so, element-wise or scalar?
Or does it mean convolution?

There is a lot missing in the statement of the problem.

> for the question above i was wondering if i can say sum of ak.[x(n)-k] and
> then from there prove that is it time invariant and also linear.

For most systems, linearity and time invariance are given properties.
Linearity of an operator L(x) is defined as

 L(ax+by) = aL(x)+bL(y)

while time invariance is defined (in my awkward notation) as

L(x(t))(t) = L(x(t+u))(t+u).

So basically, you have to define your operator L first, and then test
it 
for LTI properties. You can't test for LTI of an arbitrary L. 

Rune

>
>shoma skrev:
>> Let a0, a1, . . . , aM be any finite sequence of real numbers.
>>
>> Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
>> discrete signal.
>
>What does "*" mean? Multiplication? If so, element-wise or scalar?
>Or does it mean convolution?
>
>There is a lot missing in the statement of the problem.
>
>> for the question above i was wondering if i can say sum of ak.[x(n)-k]
and
>> then from there prove that is it time invariant and also linear.
>
>For most systems, linearity and time invariance are given properties.
>Linearity of an operator L(x) is defined as
>
> L(ax+by) = aL(x)+bL(y)
>
>while time invariance is defined (in my awkward notation) as
>
>L(x(t))(t) = L(x(t+u))(t+u).
>
>So basically, you have to define your operator L first, and then test
>it 
>for LTI properties. You can't test for LTI of an arbitrary L. 
>
>Rune
>
>

* is convolution and that is what makes it tricky i belive
btw thanx for the answer. but can u show me a sample i can't understand
this at all. i know what LTI is and i looked at it in my book but this
proof is diff than the systems in the book where u can check your answer.

shoma wrote:
> Let a0, a1, . . . , aM be any finite sequence of real numbers.
>
> Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
> discrete signal.

does the "*" symbol mean (linear) convolution?


> Show that L is a linear time invariant system and that it's unit response
> function is h where,
>
>          |
>    h(m)= |     am if 0 <= m <= M,
>          |     0 otherwise.
>
> for the question above i was wondering if i can say sum of ak.[x(n)-k] and
> then from there prove that is it time invariant and also linear.

so, let me get this straight, you are given the fact that the output of
a discrete-time system is the convolution of the input against the
finite sequence {a0, a1, . . . , aM}, and from that, you want to prove
that it is LTI?  that's the easy direction.

what's a bit more of female canine is to start with the axioms of
linearity {additive superposition is sufficient, you can prove the
scaling property from that} and time-invariancy, and show that you get
the convolution summation (or, even more of a bitch, the convolution
integral for continuous-time systems).

i presume this is something like homework (which is why no one is
telling you verbatim what to write).  it's a homework problem i have
never heard before (since we were always proving it the other way).  i
guess after settling this, the nice tidbit to remember is that "LTI" is
equivalent to "convolution", in both directions.  i had never thought
of the pedagogical value in such a problem, but it does have some.

r b-j

robert bristow-johnson skrev:
> shoma wrote:
> > Let a0, a1, . . . , aM be any finite sequence of real numbers.
> >
> > Define a system L by L(X) := X * (a0, a1, . . . , aM), where X is any
> > discrete signal.
>
> does the "*" symbol mean (linear) convolution?
>
>
> > Show that L is a linear time invariant system and that it's unit response
> > function is h where,
> >
> >          |
> >    h(m)= |     am if 0 <= m <= M,
> >          |     0 otherwise.
> >
> > for the question above i was wondering if i can say sum of ak.[x(n)-k] and
> > then from there prove that is it time invariant and also linear.
>
> so, let me get this straight, you are given the fact that the output of
> a discrete-time system is the convolution of the input against the
> finite sequence {a0, a1, . . . , aM}, and from that, you want to prove
> that it is LTI?  that's the easy direction.

You need to fix the coefficients in time. If you allow the {am}
coefficients
to vary, you have problems.

> what's a bit more of female canine is to start with the axioms of
> linearity {additive superposition is sufficient, you can prove the
> scaling property from that} and time-invariancy, and show that you get
> the convolution summation

Actually, I did that not very long ago. It wasn't very difficult
(tedious, yes,
but not difficult), and the funny thing was that only after doing that
could
I tell myself I understood convolution. Some 15 years after taking the
first class in DSP.

That exercise is well worth the effort.

> (or, even more of a bitch, the convolution
> integral for continuous-time systems).

Agreed. Here we need to deal with all sorts of limit operations etc.

> i presume this is something like homework (which is why no one is
> telling you verbatim what to write).  it's a homework problem i have
> never heard before (since we were always proving it the other way).  i
> guess after settling this, the nice tidbit to remember is that "LTI" is
> equivalent to "convolution", in both directions.

Don't agree with you here. "LTI" is a set of properties that some
systems
might possess, that others do not. Of course we might start quarreling
whether the term "convolution" makes sense for non-linear systems,
but at least a linear system can (formally) be time-variant, with a
time-variant impulse response. It is straight-forward to derive the
convolution sum for LTV systems the same way one does it for
LTI systems, using only the superposition principle. For LTI systems,
time invariance is used to find the impulse responce for a delayed
impulse, d(n-m), whereas for LTV systems each impulse response
must be specified independently.

So the convolution operator is still valid for LTV systems, maybe even
NLTV systems, depending on how we agree to define "convolution".

> i had never thought
> of the pedagogical value in such a problem, but it does have some.

Rune

Rune Allnor skrev:
> robert bristow-johnson skrev:
> > what's a bit more of female canine is to start with the axioms of
> > linearity {additive superposition is sufficient, you can prove the
> > scaling property from that} and time-invariancy, and show that you get
> > the convolution summation
>
> Actually, I did that not very long ago. It wasn't very difficult
> (tedious, yes,
> but not difficult), and the funny thing was that only after doing that
> could
> I tell myself I understood convolution. Some 15 years after taking the
> first class in DSP.

One of the "groundbreaking" revelations to get through that
exercise was to set ut a slightly more elaborate notation than
what is usual.

Suppose we have the *sequence*

x[n],   n : -infinite ... +infinite                     [1]

Now, how do we denote the scalar value of the m't sample
in that sequence? The "obvious" choise is

x[m]                                                       [2]

for some fixed m. Then, the sequence can be expressed in terms
of scaled and delayed impulse functions D[n] as

x[n] = sum x[m]D[n-m]                           [3]

where the sum is taken ofer m from -infinite to +infinite.

What used to be confusing for me, was that [3] can be interpreted as
an element-wise multiplication between the *sequence* x[m] in [1]
and the delayed impulse D[n-m]. I never realy was able to wrap
my mind around that one, to get it to make sense.

So what I did, was to change the notation in [2] such that the
scalar value of the m'th sample in the sequence x[n] is denoted

x_m.                                                    [4]

The sequence x[n] is then defined as

x[n] ={x_n}, n = -infinite...+infinite           [5]

With this convention [3] takes the form

x[n] = sum x_mD[n-m]                           [6]

where it is obvious that this is not an element-wise multiplication
between two infinitely long sequences, but a scaled and delayed
impulse. With this notation in place, it is trivial to derive the
convolution sum formula using only the linear superposition
property.

So basically I wonder if the notational ambiguity in equation [3],
that is resolved in equation [6], might be a problem for more
people than me.

Just a thought.

Rune

>
>Rune Allnor skrev:
>> robert bristow-johnson skrev:
>> > what's a bit more of female canine is to start with the axioms of
>> > linearity {additive superposition is sufficient, you can prove the
>> > scaling property from that} and time-invariancy, and show that you
get
>> > the convolution summation
>>
>> Actually, I did that not very long ago. It wasn't very difficult
>> (tedious, yes,
>> but not difficult), and the funny thing was that only after doing that
>> could
>> I tell myself I understood convolution. Some 15 years after taking the
>> first class in DSP.
>
>One of the "groundbreaking" revelations to get through that
>exercise was to set ut a slightly more elaborate notation than
>what is usual.
>
>Suppose we have the *sequence*
>
>x[n],   n : -infinite ... +infinite                     [1]
>
>Now, how do we denote the scalar value of the m't sample
>in that sequence? The "obvious" choise is
>
>x[m]                                                       [2]
>
>for some fixed m. Then, the sequence can be expressed in terms
>of scaled and delayed impulse functions D[n] as
>
>x[n] = sum x[m]D[n-m]                           [3]
>
>where the sum is taken ofer m from -infinite to +infinite.
>
>What used to be confusing for me, was that [3] can be interpreted as
>an element-wise multiplication between the *sequence* x[m] in [1]
>and the delayed impulse D[n-m]. I never realy was able to wrap
>my mind around that one, to get it to make sense.
>
>So what I did, was to change the notation in [2] such that the
>scalar value of the m'th sample in the sequence x[n] is denoted
>
>x_m.                                                    [4]
>
>The sequence x[n] is then defined as
>
>x[n] ={x_n}, n = -infinite...+infinite           [5]
>
>With this convention [3] takes the form
>
>x[n] = sum x_mD[n-m]                           [6]
>
>where it is obvious that this is not an element-wise multiplication
>between two infinitely long sequences, but a scaled and delayed
>impulse. With this notation in place, it is trivial to derive the
>convolution sum formula using only the linear superposition
>property.
>
>So basically I wonder if the notational ambiguity in equation [3],
>that is resolved in equation [6], might be a problem for more
>people than me.
>
>Just a thought.
>
>Rune
>
>

thanx a bunch :)