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Gwar [14]
2 years ago
10

Disk storage channel. Binary data storage with a thin-film disk can be approximated by an input-dependent additive white Gaussia

n noise channel where the noise n has a variance dependent on the transmitted (stored) input. The noise has the following input dependent density: 2 2ơ 2To2 2ơ 2Ta2 0 and σ-31ơ1. The channel inputs are equally-likely. a. For either input, the output can take on any real value. On the same graph, plot the two possible output probability density functions (pdf's). i.e. Plot the output pdf for 0 and the output pdf for x -1. Indicate (qualitatively) the decision regions on your graph. 2 72 2πσ 2 2πσ 2 0

Mathematics
1 answer:
schepotkina [342]2 years ago
6 0

Answer:

Step-by-step explanation:

You might be interested in
What is the solution to this eqution? 3x +17 + 5x = 7x +10
ddd [48]

Answer:

-7

Step-by-step explanation:

3x +17 + 5x = 7x +10

3x +17 + 5x - 7x = +10-17

x=-7

4 0
2 years ago
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
2 years ago
A 700 g dry fruit pack costs ₹216 .It contains some almonds and the rest cashew kernel.If almonds cost ₹288 per kg and cashew ke
slamgirl [31]

Answer:

Almonds = 400grams

Cashew = 300 grams

Step-by-step explanation:

Let the grams of cashew be represented by = C

the grams of almonds be represented by = A

A 700 g dry fruit pack costs ₹216 .It contains some almonds and the rest cashew kernel

C + A = 700g ......Equation 1

C = 700 - A

If almonds cost ₹288 per kg and cashew kernel cost ₹336 per kg

Let's convert them to grams

1 kg = 1000grams

For almonds

= 288/1000 = 0.288g

For cashew

= 336/1000 = 0.336g

Hence,

0.288 × A + 0.336g × C = ₹216

0.288A + 0.336C = 216

From Equation 1, substitute 700 - A for C in Equation 2

0.288A + 0.336(700 - A) = 216

0.288A + 235.2 - 0.336A = 216

Collect like terms

0.288A - 0.336A = 216 - 235.2

- 0.048A = -19.2

A = -19.2/-0.048

A = 400 grams

Substitute 400g for a in Equation 1

C + A = 700g ......Equation 1

= C + 400g = 700g

C = 700g - 400g

C = 300g

Therefore, the number of grams of Almonds = 400grams

Cashew = 300 grams

6 0
2 years ago
Please help!! I need to get this done! Im begging you, brainliest and points!
Viefleur [7K]

Answer:

<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>

Step-by-step explanation:

See procedure below;

First let us make these two functions ( y = -x + 1 , y = x^2 + x - 2 ) equivalent;

- x + 1  = x^2 + x - 2,

-x - x + 1 + 2 - x^2 = 0,

- 2x + 3 - x^2 = 0,

x^2 + 2x - 3 = 0,

Now factor the simplified equation;

x^2 + 2x - 3 = 0,

( x^2 - x ) + ( 3x - 3 ) = 0,

x ( x - 1 ) + 3 ( x - 1 ) = 0,

( x - 1 )( x + 3 ) = 0,

And solve for x;

x - 1 = 0, and x + 3 = 0,

x = 1, and x = - 3

Now substitute this value of x into the two functions as to receive the y  values for each x - value;

y = - ( 1 ) + 1, <em>y = 0 for x = 1</em>,

y = ( - 3 )^2 - 3 - 2, y = 9 - 3 - 2, <em>y = 4 for x = - 3</em>,

<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>

8 0
1 year ago
On Low Budget Airlines, the maximum weight of the luggage a passenger can bring without charge is 50 pounds. Mary Ellen has deci
densk [106]
It's actually legit 50 pounds because you sound up if it's 5 or more but if it's less it stays the same
7 0
2 years ago
Read 2 more answers
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