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

## Homework Statement

Let e(t), for t = 0,Â±1,Â±2, . . ., be a realization of an IID sequence of zero-mean random
variables with variance Ïƒe^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 +Æ©(from k=1 to m) bk*q^-k ---- bk â†’b subscript k
A(q) = 1 +Æ©(from k=1 to n ) ak*q^-k -----ak â†’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) âˆ’ 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 âˆ’ k)e(t âˆ’ k)], for k = 0, 1, 2, . . ..
(d) Let y(t) be defined as:
y(t) = (B(q)/A(q))x(t)
(e) Compute the expected value and autocorrelation of v(t) âˆ’ y(t).

## Homework Equations

3. The Attempt at a Solution [/b

Last edited:

haruspex
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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) in to 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
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I did substituted the A(q) and B(q) in to 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
Mentor

## Homework Statement

Let e(t), for t = 0,Â±1,Â±2, . . ., be a realization of an IID sequence of zero-mean random
variables with variance Ïƒe^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 +Æ©(from k=1 to m) bk*q^-k ---- bk â†’b subscript k
A(q) = 1 +Æ©(from k=1 to n ) ak*q^-k -----ak â†’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) âˆ’ 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 âˆ’ k)e(t âˆ’ k)], for k = 0, 1, 2, . . ..
(d) Let y(t) be defined as:
y(t) = (B(q)/A(q))x(t)
(e) Compute the expected value and autocorrelation of v(t) âˆ’ y(t).

## The Attempt at a Solution

Check your PMs. You *must* show your efforts toward the solution when you post questions here.

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.

haruspex
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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?

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.

haruspex
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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.

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) âˆ’ 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
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x(t) = ((A(q) âˆ’ 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.

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.

That sounds about right Steff. Now lets move to part c:

(c) Compute the expected value E[x(t âˆ’ k)e(t âˆ’ k)], for k = 0, 1, 2, . . ..

x(t) = ((A(q) âˆ’ 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) âˆ’ B(q))/B(q))*(B(q)/A(q) )e(t) ---> x(t) = (((A(q) âˆ’ B(q))/A(q))*e(t)
with a time shift to x(t) ---> x(t-k) the equation-> x(t-k) = (((A(q) âˆ’ 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) âˆ’ B(q))/A(q))*e(t-k)
e(t-k)
we obtain:
x(t-k)e(t-k)= (((A(q) âˆ’ B(q))/A(q))*e(t-k) *e(t-k) ----->(((A(q) âˆ’ B(q))/A(q))*e^2(t-k)

take the expected value :
E[x(t-k)e(t-k)]=((A(q) âˆ’ B(q))/A(q))*E[e^2(t-k)]
and E[e^2(t-k)] = Ïƒe^2 since the mean is zero ----- it is equal to Ïƒe^2

So the result is E[x(t-k)e(t-k)]=((A(q) âˆ’ B(q))/A(q))* Ïƒe^2

check this, this is what I am doing for this question.
let me know if its correct

haruspex
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now we got both x(t-k) and e(t-k) ,,, multiply them together to get
x(t-k) = (((A(q) âˆ’ B(q))/A(q))*e(t-k)
e(t-k)
we obtain:
x(t-k)e(t-k)= (((A(q) âˆ’ B(q))/A(q))*e(t-k) *e(t-k) ----->(((A(q) âˆ’ B(q))/A(q))*e^2(t-k)
No, the A(q) etc. are operators. The '*' between (((A(q) âˆ’ 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

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.

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

haruspex
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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) âˆ’ B(q))/A(q))*e(t-k) means A(q)x(t-k) = (A(q) âˆ’ 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?

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
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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 Ïƒe2.
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).

Perfect I see it now after expanding A(q) I would get Ïƒe2

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 Ïƒe2

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

haruspex
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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-Ï‰iq-1). We're told all the roots Ï‰i satisfy |Ï‰i| < 1. As a result, A-1(q) = âˆiÆ©n(-Ï‰iq-1)n.

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.

(d) Let y(t) be defined as:
y(t) = (B(q)/A(q))x(t)

I got that

can you give me hints for

part c

haruspex
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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.