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

[Stefff]

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

 

[haruspex]

You need to provide some attempt at a solution, or at the least an explanation of where you're stuck.

 

[Stefff]

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.

 

[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.

Then please post the working you have done.

 

[berkeman]

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.

 

[Stefff]

Then please post the working you have done.

x(t) = [1 + Æ©(from k= 1 to n) a_kq^-k -(1 + Æ©(k = 1 to m)b_kq^-k] * v(t)]/ 1 + [k=1]\sum[m] b_kq^-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(a_n-b_m)]\[1 + b_m*q^-m]

Thank you.

 

[haruspex]

x(t) = [1 + Æ©(from k= 1 to n) a_kq^-k -(1 + Æ©(k = 1 to m)b_kq^-k] * v(t)]/ 1 + [k=1]\sum[m] b_kq^-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(a_n-b_m)]\[1 + b_m*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 a_n and b_m? What happens when m and n are different?

 

[Stefff]

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 a_n and b_m? 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(a_n-b_m)]/[1 + b_m*q^-m]

i used the ellipsis because the sup and sub are equal i.e k, and tend to m and n.

 

[haruspex]

x(t)/v(t) =

[q^-1x(a-b)]/[1 + b*q^-1] + [q^-2x(a₂-b₂)]/[1 + b₂*q^-2] +...+ [q^-nx(a_n-b_m)]/[1 + b_m*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.

 

[bexter]

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 ?

 

[haruspex]

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.

 

[Stefff]

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.

 

[bexter]

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

 

[haruspex]

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:

Spoiler

e(t) and e(t-1) are independent

 

[bexter]

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.

 

[bexter]

I am not sure if I got that right or not. can you give me more hints if its not correct

 

[haruspex]

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?

 

[bexter]

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.

 

[haruspex]

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 σ_e².
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).

 

[bexter]

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

 

[bexter]

so I get for part c ---> answer is zero

 

[haruspex]

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-ω_iq^-1). We're told all the roots ω_i satisfy |ω_i| < 1. As a result, A^-1(q) = âˆ_iÆ©_n(-ω_iq^-1)ⁿ.

 

[bexter]

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.

 

[bexter]

(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

 

[haruspex]

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 q^k somewhere.