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Blizzard [7]
2 years ago
11

Decide whether or not the Final Value Theorem is applicable to the following functions. If not, indicate why you cannot apply it

. If so, apply the Final Value Theorem. Note: you do not need to perform the inverse Laplace transform.a. X(s)= s/3s+7b. X(s)= 10/3s^2+7s+4
Engineering
1 answer:
LUCKY_DIMON [66]2 years ago
3 0

Answer:

a) Yes applicable, x(∞) = 0

b) Yes applicable, x(∞) = 0

Explanation:

Final Value Theroem indicates that final value of x(t) can be found by Laplace Transform X(s) as follows:

x(\infty)= \lim\limits_{s \to0} sX(s)

a) X(s)= \frac{s}{3s+7}

Since the pole is at left side, FVT is applicable

x(\infty)= \lim\limits_{s \to0} s\frac{s}{3s+7}=0

b) X(s)= \frac{10}{3s^2+7s+4}

\frac{10}{3s^2+7s+4}= \frac{10}{(3s+4 )(s+1)}

Since the poles are at left side, FVT is applicable

x(\infty)= \lim\limits_{s \to0} s \frac{10}{(3s+4 )(s+1)}=0

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(D)<br> 13. Describe the differences between an impact socket and a conventional socket.
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Answer:

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Explanation:

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2 years ago
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
Pavlova-9 [17]

Answer:

Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

Explanation:

FP Instructions=50*106=5300

INT  Instructions=110*106=11660

L/S  Instructions=80*106=8480

Branch  Instructions=16*106=1696

Calculating Execution Time:

Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4

New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

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3 years ago
Suppose the following two events occur at the same time: the Chicago Cubs win the World Series, and the workers who make Cubs me
Annette [7]

Answer:

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3 years ago
An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Giv
Anika [276]

The question is incomplete. The complete question is --

An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 1.01 cm and the young's modulus of aluminum is 69 GPa, determine the following :

1. the interference value, i

2. the radial pressure at the interference of the collar and the rod

3. the maximum effective stress in the collar

4. if the yield strength of aluminium is 200 MPa and assume a safety factor of 1.5, will the aluminium collar break

Solution:

Given :

Inner radius of the collar = 1 cm

So, inner diameter, $d_1$ = 2 cm

Outer radius of the collar = 12 cm

So, outer diameter, $d_2$ = 4 cm

The aluminium rod diameter, d = 1.01 cm

Now, from the figure, we can see that there will be no interference and so the rod will easily insert inside the collar.

1. So, the interference , i =0

2. The radial pressure is also 0.

3. There will be no stress developed. So the maximum effective stress is 0

4. The collar will not break

8 0
3 years ago
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