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This part of the thesis, PAM4 signal filtering.

p - convolution noise

x - input signal (PAM4)

$$ E(p^4[n]x^2[n]) = $$

Where E - is Expected Value and p[n] = x[n]*e[n],

* - denotes convolution.

$$ =Ebigr[(sum_k^N x[n-k]e[k]sum_m^N x[n-m]e[m]sum_g^N x[n-g]e[g]sum_d^N x[n-d]e[d])x^2bigr] =$$

This expression has several option of solution, but for me important two of them:

1) k=m=g=d

2) k=m, g=d, k$ne$g

Tried to get solution to part - (2

$$ Ebigr[(sum_k^Nsum_m^N x[n-k]e[k] x[n-m]e[m]sum_g^Nsum_d^N x[n-g]e[g] x[n-d]e[d])x^2bigr] =$$

$$ frac1E(x^2[n])sum_k^Nsum_m^N Ebigr( x[n-k] x[n-m] e[k] e[m] x^2bigr)sum_g^Nsum_d^N Ebigr( x[n-g] x[n-d] e[d] e[g] x^2bigr) =$$

$$ frac1E(x^2[n])bigr(sum_k^N E(e^2[k]) E(x^2[n] x^2[n-k]) bigr)cdotbigr(sum_g^N E(e^2[g]) E(x^2[n] x^2[n-g]) bigr) =$$

$$ frac1E(x^2[n])bigr(E(x^4[n])E(e^2[0]) + sum_k,kne0^N sigma_x^4 cdot E(e^2[k]) bigr)cdotbigr(E(x^4[n])E(e^2[0]) + sum_g,gne0^N sigma_x^4 cdot E(e^2[g]) bigr) =$$

$$E(e^2[k])=E(e^2[g])=sigma_e^2$$

for every k and g

$$ frac1E(x^2[n])bigr(E(x^4[n])cdotsigma_e^2 + (N-1) cdot sigma_x^4 cdot sigma_e^2 bigr)cdotbigr(E(x^4[n])cdotsigma_e^2 + (N-1) cdot sigma_x^4 cdot sigma_e^2 bigr) =$$

$$ frac1E(x^2[n])cdot(sigma_e^2)^2cdotbigr(E(x^4[n]) + (N-1) cdot sigma_x^4 bigr)^2$$

My mentor says, that I have mistake in this solution. Maybe, someone can help?

convolution signal-processing

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edited Apr 1 at 20:04

Eric Wofsey

192k14218351

asked Apr 1 at 18:24

Nikolay KostishenNikolay Kostishen

11

New contributor

Nikolay Kostishen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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add a comment | 

0

$begingroup$

This part of the thesis, PAM4 signal filtering.

p - convolution noise

x - input signal (PAM4)

$$ E(p^4[n]x^2[n]) = $$

Where E - is Expected Value and p[n] = x[n]*e[n],

* - denotes convolution.

$$ =Ebigr[(sum_k^N x[n-k]e[k]sum_m^N x[n-m]e[m]sum_g^N x[n-g]e[g]sum_d^N x[n-d]e[d])x^2bigr] =$$

This expression has several option of solution, but for me important two of them:

1) k=m=g=d

2) k=m, g=d, k$ne$g

Tried to get solution to part - (2

$$ Ebigr[(sum_k^Nsum_m^N x[n-k]e[k] x[n-m]e[m]sum_g^Nsum_d^N x[n-g]e[g] x[n-d]e[d])x^2bigr] =$$

$$ frac1E(x^2[n])sum_k^Nsum_m^N Ebigr( x[n-k] x[n-m] e[k] e[m] x^2bigr)sum_g^Nsum_d^N Ebigr( x[n-g] x[n-d] e[d] e[g] x^2bigr) =$$

$$ frac1E(x^2[n])bigr(sum_k^N E(e^2[k]) E(x^2[n] x^2[n-k]) bigr)cdotbigr(sum_g^N E(e^2[g]) E(x^2[n] x^2[n-g]) bigr) =$$

$$ frac1E(x^2[n])bigr(E(x^4[n])E(e^2[0]) + sum_k,kne0^N sigma_x^4 cdot E(e^2[k]) bigr)cdotbigr(E(x^4[n])E(e^2[0]) + sum_g,gne0^N sigma_x^4 cdot E(e^2[g]) bigr) =$$

$$E(e^2[k])=E(e^2[g])=sigma_e^2$$

for every k and g

$$ frac1E(x^2[n])bigr(E(x^4[n])cdotsigma_e^2 + (N-1) cdot sigma_x^4 cdot sigma_e^2 bigr)cdotbigr(E(x^4[n])cdotsigma_e^2 + (N-1) cdot sigma_x^4 cdot sigma_e^2 bigr) =$$

$$ frac1E(x^2[n])cdot(sigma_e^2)^2cdotbigr(E(x^4[n]) + (N-1) cdot sigma_x^4 bigr)^2$$

My mentor says, that I have mistake in this solution. Maybe, someone can help?

convolution signal-processing

share|cite|improve this question

edited Apr 1 at 20:04

Eric Wofsey

192k14218351

asked Apr 1 at 18:24

Nikolay KostishenNikolay Kostishen

11

New contributor

Nikolay Kostishen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

This part of the thesis, PAM4 signal filtering.

p - convolution noise

x - input signal (PAM4)

$$ E(p^4[n]x^2[n]) = $$

Where E - is Expected Value and p[n] = x[n]*e[n],

* - denotes convolution.

$$ =Ebigr[(sum_k^N x[n-k]e[k]sum_m^N x[n-m]e[m]sum_g^N x[n-g]e[g]sum_d^N x[n-d]e[d])x^2bigr] =$$

This expression has several option of solution, but for me important two of them:

1) k=m=g=d

2) k=m, g=d, k$ne$g

Tried to get solution to part - (2

$$ Ebigr[(sum_k^Nsum_m^N x[n-k]e[k] x[n-m]e[m]sum_g^Nsum_d^N x[n-g]e[g] x[n-d]e[d])x^2bigr] =$$

$$ frac1E(x^2[n])sum_k^Nsum_m^N Ebigr( x[n-k] x[n-m] e[k] e[m] x^2bigr)sum_g^Nsum_d^N Ebigr( x[n-g] x[n-d] e[d] e[g] x^2bigr) =$$

$$ frac1E(x^2[n])bigr(sum_k^N E(e^2[k]) E(x^2[n] x^2[n-k]) bigr)cdotbigr(sum_g^N E(e^2[g]) E(x^2[n] x^2[n-g]) bigr) =$$

$$ frac1E(x^2[n])bigr(E(x^4[n])E(e^2[0]) + sum_k,kne0^N sigma_x^4 cdot E(e^2[k]) bigr)cdotbigr(E(x^4[n])E(e^2[0]) + sum_g,gne0^N sigma_x^4 cdot E(e^2[g]) bigr) =$$

$$E(e^2[k])=E(e^2[g])=sigma_e^2$$

for every k and g

$$ frac1E(x^2[n])bigr(E(x^4[n])cdotsigma_e^2 + (N-1) cdot sigma_x^4 cdot sigma_e^2 bigr)cdotbigr(E(x^4[n])cdotsigma_e^2 + (N-1) cdot sigma_x^4 cdot sigma_e^2 bigr) =$$

$$ frac1E(x^2[n])cdot(sigma_e^2)^2cdotbigr(E(x^4[n]) + (N-1) cdot sigma_x^4 bigr)^2$$

My mentor says, that I have mistake in this solution. Maybe, someone can help?

convolution signal-processing

share|cite|improve this question

edited Apr 1 at 20:04

Eric Wofsey

192k14218351

asked Apr 1 at 18:24

Nikolay KostishenNikolay Kostishen

11

New contributor

Nikolay Kostishen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 1 at 20:04

Eric Wofsey

192k14218351

asked Apr 1 at 18:24

Nikolay KostishenNikolay Kostishen

11

New contributor

Nikolay Kostishen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 1 at 20:04

Eric Wofsey

192k14218351

asked Apr 1 at 18:24

Nikolay KostishenNikolay Kostishen

11

New contributor

Nikolay Kostishen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 1 at 20:04

Eric Wofsey

192k14218351

edited Apr 1 at 20:04

Eric Wofsey

192k14218351

asked Apr 1 at 18:24

Nikolay KostishenNikolay Kostishen

11

New contributor

Nikolay Kostishen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 1 at 18:24

Nikolay KostishenNikolay Kostishen

11