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KengaRu [80]
3 years ago
15

For the following three (3) questions, assume that the link layer uses the flag bytes with byte stuffing approach for framing, w

ith the following details:
The beginning of a data frame is indicated by the special flag bytes: DLESTX
The end of a data frame is indicated by the special flag bytes: DLEETX
The special escape bytes DLE are used for escaping accidental occurrences of either the flag bytes or the escape bytes within the data.
For simplicity, assume that no other header/trailer information is added to the data.
The following byte streams represent data that needs to be framed by the link layer on the sender’s side. Derive the resultant byte streams by adding necessary flag bytes and performing byte stuffing.
1. PZDLEAFRGSTXGYK
2. PARSDLEETXZKPUM
3. ASTXDLELEFTDAOY
Engineering
1 answer:
pav-90 [236]3 years ago
3 0

I know it's either 1 2 or 3..

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Please answer fast. With full step by step solution.​
lina2011 [118]

Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

or

<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

Recall that for |<em>z</em>| < 1, we have

\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n

Replace <em>z</em> with 1/<em>z</em> to get

\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}

so that by substitution, we can write

\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}

Now condense <em>f(z)</em> into one series:

\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}

\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}

So, the inverse <em>Z</em> transform of <em>f(z)</em> is \boxed{6+2^{2-n}}.

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