Give me an A please!!!¡!!!!¡

Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and

lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1] x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is $\frac{1}{x[n] . x[n-1] x[n+1]}$

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

for x[n] = 2 $\pi$ , y[n] = cos(2 $\pi$ ) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

3 0

2 years ago

In the planning process of the product development life cycle what is it important to inventory

Verizon [17]

Your Answer would be A I believe.

6 0

2 years ago

Free brainlist because im new and i just want to but you have t friend me first

Amiraneli [1.4K]

Okay sure.

I’ll 1)chords
2)pulse
3)aerophone
4) the answer is C
5)rhythm

Pretty sure those are the answers

4 0

2 years ago

An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava

s344n2d4d5 [400]

Answer:

Among the different types of excavation protection system, as a way of preventing accidents against cave-ins, the sloping involves cutting back the trench wall at an angle inclined away from the excavation. Shoring requires installing aluminum hydraulic or other types of supports to prevent soil movement and cave-ins. Shielding protects workers by using trench boxes or other types of supports to prevent soil cave-ins (OSHA). In addition, the regulations do not allow employees to work on excavations where there is an accumulation of water. If this occurs, water on the site must be constantly removed by suitable equipment preventing water from accumulating. The entry of surface water into the excavations must also be prevented by means of diversion ditches, dam, or other suitable means.

Explanation:

3 0

3 years ago

A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni

arlik [135]

Answer:

The population size would be $p' = 5000$

The yield would be $MaxYield = 2082 \ fishes \ per \ year$

Explanation:

So in this problem we are going to be examining the application of a population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield, so basically we would be applying an engineering solution to fishing

So the current yield which is mathematically represented as

$\frac{dN}{dt} = \frac{2000}{1 \ year }$

Where dN is the change in the number of fish

and dt is the change in time

So in order to obtain the solution we need to obtain the rate of growth

For this we would be making use of the growth rate equation which is

$r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}$

Where N is the population of the fish which is given as 4,000 fishes

and K is the carrying capacity which is given as 10,000 fishes

r is the growth rate

Substituting these values into the equation

$r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]} =0.833$

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

$Max Yield = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year$

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by $p'$ would be

$p' = \frac{K}{2} = \frac{10,000}{2} = 5000\ fishes$

7 0

3 years ago

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