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Semenov [28]
3 years ago
6

For each part of this problem, you are given the description of a continuous-time system. Determine if the given system is: (i)

Linear (iv) Causal (ii) Time-Invariant (v) Stable (iii) Memoryless (vi) Invertible
Mathematics
1 answer:
Mazyrski [523]3 years ago
7 0

Hello, you haven't provided any continuous-time system, therefore, I am going to explain to you how to solve this problem in general and you can apply this to your system or any continuous-time system.

Answer step-by-step explanation:

  • Linear: A system is linear if it obeys the principle of superposition Z{aX1(t)+bX2(t)}=aZ{X1(t)}+bZ{X2(t)}=aY1(t)+bY2(t), where Z is an operator which maps the input into output (Z is the system itself)
  • Causal: A system is causal if it doesn't depend on future inputs Y(t)=Z(x(t)) up to time t
  • Time-Invariant: A system is time-invariant if a time-shift on the input causes a equivalent shift on the output, if Y(t)=Z(x(t)) then Y(t-t0)=Z(x(t-t0))
  • Stable: A system is stable if every bounded input produces a bounded output, abs(Y(t))<B if abs(X(t))<A / abs(A) < inf and abs(B) < inf
  • Memoryless: A system is memoryless if its output at time t is dependent only on the input at that time, Y(t) at time t depends only on X(t) at time t
  • Invertible: A system is invertible if we can get back the input by passing the output through another system (No information loss)
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garik1379 [7]

Answer:

These problems are an example of equations with two unknowns. The way these equations are solved is that we write these equations one under the another.

If both equations have, such is the case here, same parts, we can simply cancel the same parts out and subtract the rest of equatuons. That way, we are left with only one unknown (the other one was eliminated), which makes it easy to solve.

After we have found the value of an unknown, we just plug it back into any of the starting equations and solve for the second unknown.

2. Adult ticket costs $12 and child ticket costs $14.

3. Adult ticket costs $10 and child ticket costs $5.

4. One daylily costs $9 and one bush of ornamental grass costs $2.

5. A van can carry 15 and a bus can carry 56 students.

Step-by-step explanation:

2. If we mark the price of one adult ticket with x and the price of one child ticket with y, we get that:

- first day: 7x + 12y = $252

- second day: 7x + 10y = $224

Now, we can make a system:

7x + 12y = 252

7x + 10y = 224

We can now subtract these two equations and 7x will cancel out, so we get:

12y - 10y = 252 - 224

2y = 28

y = 14

Now, we can plug the value of y into any of the two equations:

7x + 10y = 224

7x + 140 = 224

7x = 84

x = 12

3. Similarly, if we mark the price of one adult ticket with x and the price of one child tickey with y, we'll get a system:

x + 12y = 70

x + 9y = 55

Again, if we subtract these two, x will cancel out, so we have:

12y - 9y = 70 - 55

3y = 15

y = 5

Now, we plug the value of y into any of the two equations, and we get:

x + 9y = 55

x + 45 = 55

x = 10

4. Using the same principle, we can mark the price of one daylily with x and the price of one bunch of ornamental grass with y, we'll get a system:

12x + 11y = 130

12x + 12y = 132

Again, we subtract so that 12x cancel out and we get:

11y - 12y = 130 - 132

-y = -2

If we get minuses on both sides, we can simply multiply both sides with -1 and we get:

y = 2

Again, we plug y:

12x + 12y = 132

12x + 24 = 132

12x = 108

x = 9

5. If we mark number of students in a van with x and the number of students in a bus with y, we get a system:

2x + 12y = 702

2x + y = 86

As you've probably already noticed the pattern, we subtract equations and cancel 2x out to get:

12y - y = 702 - 86

11y = 616

y = 56

Once again, we plug the value of y into any equation:

2x + y = 86

2x + 56 = 86

2x = 30

x = 15

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galina1969 [7]
Cross multiply:

30n = 2400.  Then n = 80.
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