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TEA [102]
3 years ago
10

9.19 Generate Bode magnitude and phase plots (straight-line approximations) for the following voltage transfer functions. (a) H(

ω) = 4×104(60+j6ω) (4+j2ω)(100+j2ω)(400+j4ω) (b) H(ω) = (1+j0.2ω)2(100+j2ω)2 (jω)3(500+jω) (c) H(ω) = 8×10−2(10+j10ω) jω(16−ω2 +j4ω) (d) H(ω) = 4×104ω2(100−ω2 +j50ω) (5+j5ω)(200+j2ω)3

Engineering
1 answer:
Norma-Jean [14]3 years ago
8 0

Answer:

attached below

Explanation:

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Two well-known NP-complete problems are 3-SAT and TSP, the traveling salesman problem. The 2-SAT problem is a SAT variant in whi
Hitman42 [59]

3-SAT ≤p TSP

If P ¹ NP, then no NP-complete problem can be solved in polynomial time.

both the statements are true.

<u>Explanation:</u>

  • 3-SAT ≤p TSP due to any  complete problem of NP to other problem by exits of reductions.
  • If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
  • If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
7 0
3 years ago
A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft
tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ;  f_{Lw}=6.6 mi/h

5' on right   ;    f_{Lc} = 0.4 mi/hr

3' on left   ;      no adjustment

3 lanes in each direction    f n = 3 mi/h

v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}

f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}

f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}

          = 0.877

v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}

       = 1,555 veh/hr/lane

FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}

      = (55 + 5) - 6.6 - 0.4 -3 -0

      = 50 mi/h

D = \dfrac{V_P}{s}

D = \dfrac{1555}{55} =28.27

level of service is D using speed flow curves and LOS for basic free moving of vehicle

5 0
4 years ago
In the engineering design and prototyping process, what is the advantage of drawings and symbols over written descriptions?
MrMuchimi

The advantages that can be associated to

drawings and symbols over written descriptions in engineering design and prototyping process are;

Communicate design ideas as well as technical information to engineers.

Symbols and drawings can be universal which means it is easy to interpret any where by professionals.

  • An engineering drawing serves as complex dimensional object and symbol use by engineer to communicate.

  • Drawings and symbols makes it easier to communicate design ideas and technical information to engineers and and how the process will go.

Therefore, drawings and symbols is universal to all engineer unlike written one.

Learn more at:

brainly.com/question/20925313?referrer=searchResults

4 0
3 years ago
Which band has an average of $3.58 per hour of parking?
Minchanka [31]
C it would be c because that has more and the others have less
6 0
2 years ago
Find: factor of safety (n)for point A and B by using both MSS and DE (you can neglect shear stress due to shear force and also n
gladu [14]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Explanation:

Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Attached below is the detailed solution

8 0
3 years ago
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