Please I on this problem, I can't figure out what to do

  • Thread starter Thread starter Stefff
  • Start date Start date
  • Tags Tags
    Figure
Click For Summary

Homework Help Overview

The discussion revolves around a problem involving filters and random variables in signal processing. The original poster presents a sequence of equations related to the output of a filter defined by polynomials in the forward shift operator. The problem includes multiple parts, such as finding relationships between different signals, determining coefficients, and computing expected values.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss substituting polynomials into equations to find relationships between signals. There are questions about the stability of the filter and the coefficients of the signals involved. Some participants express uncertainty about their approaches and seek clarification on specific steps.

Discussion Status

Several participants have shared their attempts at solving parts of the problem, particularly focusing on the relationship between x(t) and v(t) and the expected value computations. There is ongoing exploration of the implications of coefficients and the stability of the filter, with no explicit consensus reached yet.

Contextual Notes

Participants are working under the constraints of homework rules, which require them to show their attempts at solutions. There are also discussions about the implications of certain assumptions, such as the values of coefficients in the equations.

Stefff
Messages
5
Reaction score
0

Homework Statement



Let e(t), for t = 0,±1,±2, . . ., be a realization of an IID sequence of zero-mean random
variables with variance σe^2, and let v(t) be the output of the filter:
v(t) = (B(q)/A(q) )e(t)
where B(q) and A(q) are polynomials in the forward shift operator q (i.e. qv(t) = v(t + 1)),
given by:

B(q) = 1 +Ʃ(from k=1 to m) bk*q^-k ---- bk →b subscript k
A(q) = 1 +Ʃ(from k=1 to n ) ak*q^-k -----ak →a subscript k

and the solutions to A(z) = 0 and B(z) are all inside the unit circle.
(a) Let x(t) be the output of the following filter:
x(t) = ((A(q) − B(q))/B(q)) v(t)
(b) Write the relationship between x(t) and v(t) as a difference equation. What is the
coefficient of v(t) (i.e. at zero delay)? Is this filter stable?
(c) Compute the expected value E[x(t − k)e(t − k)], for k = 0, 1, 2, . . ..
(d) Let y(t) be defined as:
y(t) = (B(q)/A(q))x(t)
Can this signal be used as a “one-step†ahead prediction of v(t)? Justify your answer.
(e) Compute the expected value and autocorrelation of v(t) − y(t).

Homework Equations


3. The Attempt at a Solution [/b
 
Last edited:
Physics news on Phys.org
You need to provide some attempt at a solution, or at the least an explanation of where you're stuck.
 
Thank you Haruspex.
I did substituted the A(q) and B(q) into the x(t) equation in order to find the relationship between the x(t) and v(t) to get the coefficient of v(t). Although am not sure if that is right or not hence, i could not tell if the filter is stable or not. I could not compute the expected value either.

I am actually got stock in b.
 
Stefff said:
I did substituted the A(q) and B(q) into the x(t) equation in order to find the relationship between the x(t) and v(t) to get the coefficient of v(t). Although am not sure if that is right or not hence, i could not tell if the filter is stable or not. I could not compute the expected value either.

I am actually got stock in b.
Then please post the working you have done.
 
Stefff said:

Homework Statement



Let e(t), for t = 0,±1,±2, . . ., be a realization of an IID sequence of zero-mean random
variables with variance σe^2, and let v(t) be the output of the filter:
v(t) = (B(q)/A(q) )e(t)
where B(q) and A(q) are polynomials in the forward shift operator q (i.e. qv(t) = v(t + 1)),
given by:

B(q) = 1 +Ʃ(from k=1 to m) bk*q^-k ---- bk →b subscript k
A(q) = 1 +Ʃ(from k=1 to n ) ak*q^-k -----ak →a subscript k

and the solutions to A(z) = 0 and B(z) are all inside the unit circle.
(a) Let x(t) be the output of the following filter:
x(t) = ((A(q) − B(q))/B(q)) v(t)
(b) Write the relationship between x(t) and v(t) as a difference equation. What is the
coefficient of v(t) (i.e. at zero delay)? Is this filter stable?
(c) Compute the expected value E[x(t − k)e(t − k)], for k = 0, 1, 2, . . ..
(d) Let y(t) be defined as:
y(t) = (B(q)/A(q))x(t)
Can this signal be used as a “one-step†ahead prediction of v(t)? Justify your answer.
(e) Compute the expected value and autocorrelation of v(t) − y(t).


Homework Equations





The Attempt at a Solution


Check your PMs. You *must* show your efforts toward the solution when you post questions here.
 
haruspex said:
Then please post the working you have done.

x(t) = [1 + Æ©(from k= 1 to n) akq-k -(1 + Æ©(k = 1 to m)bkq-k] * v(t)]/ 1 + [k=1]\sum[m] bkq-k

the relationship between x(t) and v(t) is therefore

[q-1x(a-b)]\[1 + b*q-1] + [q-2x(a2-b2)]\[1 + b2*q-2] +...+ [q-nx(an-bm)]\[1 + bm*q-m]

Thank you.
 
Stefff said:
x(t) = [1 + Æ©(from k= 1 to n) akq-k -(1 + Æ©(k = 1 to m)bkq-k] * v(t)]/ 1 + [k=1]\sum[m] bkq-k

the relationship between x(t) and v(t) is therefore

[q-1x(a-b)]\[1 + b*q-1] + [q-2x(a2-b2)]\[1 + b2*q-2] +...+ [q-nx(an-bm)]\[1 + bm*q-m]

Thank you.
There are several things I don't understand in that last line.
First, there's no equals sign, so how does it express a relationship?
Second, there's no mention of v.
Third, what do the backslash symbols (\) represent?
Fourth, how can you use ellipsis (...) between a term that has a2 and b2 and a term that has an and bm? What happens when m and n are different?
 
haruspex said:
There are several things I don't understand in that last line.
First, there's no equals sign, so how does it express a relationship?
Second, there's no mention of v.
Third, what do the backslash symbols (\) represent?
Fourth, how can you use ellipsis (...) between a term that has a2 and b2 and a term that has an and bm? What happens when m and n are different?
x(t)/v(t) =

[q-1x(a-b)]/[1 + b*q-1] + [q-2x(a2-b2)]/[1 + b2*q-2] +...+ [q-nx(an-bm)]/[1 + bm*q-m]

i used the ellipsis because the sup and sub are equal i.e k, and tend to m and n.
 
Stefff said:
x(t)/v(t) =

[q-1x(a-b)]/[1 + b*q-1] + [q-2x(a2-b2)]/[1 + b2*q-2] +...+ [q-nx(an-bm)]/[1 + bm*q-m]

i used the ellipsis because the sup and sub are equal i.e k, and tend to m and n.
Yes, but if each goes up one at a time then they won't reach m and n together.
To clarify the difficulty, try writing that out in full with m = 2, n = 1.
 
  • #10
This is what i am doing for this question :

(b) Write the relationship between x(t) and v(t) as a difference equation. What is the
coefficient of v(t) (i.e. at zero delay)? Is this filter stable?

x(t) = ((A(q) − B(q))/B(q)) v(t)

x(t) * B(q) = (A(q) - B(q) ) * v(t)

bk *x(t-k) = ak*v(t-k) - bk*v(t-k)

for zero delay ----- k=zero.. substitute into the equation
b0 *x(t) = a0 * v(t) - b0 *v(t)
therefore the coef. of v(t) be ... ( a0 - b0 )

b0 * x(t) = v(t) (a0 - b0)

is that correct ?
 
  • #11
bexter said:
x(t) = ((A(q) − B(q))/B(q)) v(t)

x(t) * B(q) = (A(q) - B(q) ) * v(t)
I feel the LHS should be written B(q) * x(t). B(q) is an operator, so the order matters.
bk *x(t-k) = ak*v(t-k) - bk*v(t-k)

for zero delay ----- k=zero.. substitute into the equation
b0 *x(t) = a0 * v(t) - b0 *v(t)
therefore the coef. of v(t) be ... ( a0 - b0 )
But a0 = b0 = 1. Does that make things interesting?
And I would have thought the coefficient would be of the form x(t) = (coefficient)*v(t), so it would be a0/b0 - 1.
 
  • #12
Thank you for the suggestions, i believe the coefficient of v(t) = 0 if a0 = b0 =1. i.e at zero delay. Any suggestion on how to go about with the c? thank you.
 
  • #13
That sounds about right Steff. Now let's move to part c:

(c) Compute the expected value E[x(t − k)e(t − k)], for k = 0, 1, 2, . . ..

x(t) = ((A(q) − B(q))/B(q)) v(t)
where
v(t) = (B(q)/A(q) )e(t)
substituting v(t) into x(t) we get
x(t) = ((A(q) − B(q))/B(q))*(B(q)/A(q) )e(t) ---> x(t) = (((A(q) − B(q))/A(q))*e(t)
with a time shift to x(t) ---> x(t-k) the equation-> x(t-k) = (((A(q) − B(q))/A(q))*e(t-k)

Now for: e(t) with a time shift to e(t)---> e(t-k)

now we got both x(t-k) and e(t-k) ,,, multiply them together to get
x(t-k) = (((A(q) − B(q))/A(q))*e(t-k)
e(t-k)
we obtain:
x(t-k)e(t-k)= (((A(q) − B(q))/A(q))*e(t-k) *e(t-k) ----->(((A(q) − B(q))/A(q))*e^2(t-k)

take the expected value :
E[x(t-k)e(t-k)]=((A(q) − B(q))/A(q))*E[e^2(t-k)]
and E[e^2(t-k)] = σe^2 since the mean is zero ----- it is equal to σe^2

So the result is E[x(t-k)e(t-k)]=((A(q) − B(q))/A(q))* σe^2

check this, this is what I am doing for this question.
let me know if its correct
 
  • #14
bexter said:
now we got both x(t-k) and e(t-k) ,,, multiply them together to get
x(t-k) = (((A(q) − B(q))/A(q))*e(t-k)
e(t-k)
we obtain:
x(t-k)e(t-k)= (((A(q) − B(q))/A(q))*e(t-k) *e(t-k) ----->(((A(q) − B(q))/A(q))*e^2(t-k)
No, the A(q) etc. are operators. The '*' between (((A(q) − B(q))/A(q)) and e(t-k) does not represent ordinary multiplication, so you can't apply associativity like that.
Let's start with something a bit simpler: can you evaluate E[e(t) q-1 e(t)]?
Hint:
e(t) and e(t-1) are independent
 
  • #15
Okay, can you evaluate E[e(t) q-1 e(t)]?
with your hint : e(t) and e(t-1) are independent
then E[e(t)]E[e(t-1)] ---> and that is zero since the mean is zero.
 
  • #16
I am not sure if I got that right or not. can you give me more hints if its not correct
 
  • #17
bexter said:
Okay, can you evaluate E[e(t) q-1 e(t)]?
with your hint : e(t) and e(t-1) are independent
then E[e(t)]E[e(t-1)] ---> and that is zero since the mean is zero.
That's right.
When i first saw this question I was a bit non-plussed by the notion of dividing by an operator. So I did a bit of reading and it seems that x(t-k) = ((A(q) − B(q))/A(q))*e(t-k) means A(q)x(t-k) = (A(q) − B(q))*e(t-k), no more, no less. So I would favour working with that form.
So consider E[e(t-k) A(q) e(t-k)]. Can you evaluate that now?
 
  • #18
Same it would equal to zero
as
E[e(t-k) A(q) e(t-k)] = A(q)E[e(t-k)]E[(e(t-k)] ----> and E[e(t-k)] is zero.
 
  • #19
bexter said:
Same it would equal to zero
as
E[e(t-k) A(q) e(t-k)] = A(q)E[e(t-k)]E[(e(t-k)] ----> and E[e(t-k)] is zero.
No. E[e(t-k) e(t-k)] is not zero. As you posted earlier it would be σe2.
What in general will E[e(t-k)e(t-j)] evaluate to?
In E[e(t-k) A(q) e(t-k)], expand A(q).
 
  • #20
Perfect I see it now after expanding A(q) I would get σe2

What in general will E[e(t-k)e(t-j)] evaluate to?

zero for all j not equal to k

but if j=k then σe2
 
  • #21
so I get for part c ---> answer is zero
 
  • #22
bexter said:
so I get for part c ---> answer is zero
How do you get that? (I'm not saying it's wrong - I genuinely don't know.)

Btw, it occurred to me that another approach is to consider that polynomials can be factorised. E.g. A(q) could be written as âˆ(1-ωiq-1). We're told all the roots ωi satisfy |ωi| < 1. As a result, A-1(q) = âˆiÆ©n(-ωiq-1)n.
 
  • #23
this is the final equation I got for :
Need to compute E[x(t-k)e(t-k)]
x(t) = (SUM(ak*e(t-k)) - (SUM(bk*e(t-k) - (SUM(ak*x(t-k))

multiply x(t) by e(t):
x(t)e(t)= (SUM(ak*e(t-k)e(t)) - (SUM(bk*e(t-k)e(t)) - (SUM(ak*x(t-k)x(t))
then shift by j for both (x) and (e)
x(t-j)e(t-j)= (SUM(ak*e(t-k-j)e(t-j)) - (SUM(bk*e(t-k-j)e(t-j)) - (SUM(ak*x(t-k-j)x(t-j))
take the expected value:
E[x(t-j)e(t-j)]= E[(SUM(ak*e(t-k-j)e(t-j))] - E[(SUM(bk*e(t-k-j)e(t-j))] - E[(SUM(ak*x(t-k-j)e(t-j))]
lets look at each expectation by it self:
E[(SUM(ak*e(t-k-j)e(t-j))] = zero since k is never zero ...the shift difference between e(t-k-j)e(t-j)) is never the same
E[(SUM(bk*e(t-k-j)e(t-j))] similar to the above
E[(SUM(ak*e(t-k-j)e(t-j))] Since x(t) doesn't contain current values of e(t) it only contains past values . therefore for example E[e(t-1-j)e(t-j)] is zero.

so the result of the E[x(t-j)e(t-j)]= zero.
 
  • #24
(d) Let y(t) be defined as:
y(t) = (B(q)/A(q))x(t)
Can this signal be used as a ONE STEP ahead prediction of v(t)? Justify your answer

I got that

can you give me hints for

part c
 
  • #25
bexter said:
this is the final equation I got for :
Need to compute E[x(t-k)e(t-k)]
x(t) = (SUM(ak*e(t-k)) - (SUM(bk*e(t-k) - (SUM(ak*x(t-k))
I.e. x(t) = (A-B)e(t-k) + Ax(t-k)?
How do you arrive at that? I get:
Av(t) = Be(t)
Bx(t) = Av(t) - Bv(t) = Be(t) - (B/A)e(t)
To get a mix of t and t-k I would need to introduce a qk somewhere.
 

Similar threads

How should I find ## x_{2}(t) ## for this nonlinear integro-differential equation?
  • · Replies 6 ·
Replies
6
Views
3K
Signal Processing: Finding the auto-correlation
  • · Replies 2 ·
Replies
2
Views
2K
How to get general solution via Green's function?
  • · Replies 1 ·
Replies
1
Views
2K
Math modelling diffrential equation question?
  • · Replies 2 ·
Replies
2
Views
2K
How do we write a sinusoidal solution to a 2nd order DE as a sum of exponentials raised to complex roots?
  • · Replies 2 ·
Replies
2
Views
3K
How Do Vector Spaces of Linear Maps Differ from Standard Vector Spaces?
  • · Replies 9 ·
Replies
9
Views
2K
Applied directional derivative problem
  • · Replies 1 ·
Replies
1
Views
2K
One-Dimensional Wave Equation & Steady-State Temperature Distribution
  • · Replies 14 ·
Replies
14
Views
2K
[Linear Algebra] Show that H ∩ K is a subspace of V
  • · Replies 5 ·
Replies
5
Views
7K
Can anyone please verify/check if there's error in this word problem?
  • · Replies 4 ·
Replies
4
Views
2K