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Problem with Solution - Analysis and Design of Engineering System | ES 205, Assignments of Civil Engineering

Material Type: Assignment; Class: Analysis & Design of Engr Syst; Subject: Engineering Science; University: Rose-Hulman Institute of Technology; Term: Spring 2005;

Typology: Assignments

Pre 2010

Uploaded on 08/19/2009

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Answer to HW Problem 21.2 ES205
For ω = 500 rad/s, )50.1500cos(66.0)( += tteo
For ω = 1000 rad/s, )1000cos(10)( tteo=
For ω = 1500 rad/s, )45.11500cos(2.1)( = tteo
Do you see why this circuit is called a “bandpass filter” ?
00.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-10
-5
0
5
10
Time (s)
Input and output (V)
Fig 23.2(a): w = 500 rad/s
Inp ut
Response
00.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-10
-5
0
5
10
Time (s)
Input and output (V)
Fig 23.2(a): w = 1000 rad/s
Inp ut
Response
00.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-10
-5
0
5
10
Time (s)
Input and output (V)
Fig 23.2(a): w = 1500 rad/s
Inp ut
Response
Note that input and output are identical.

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Answer to HW Problem 2 1 .2 ES

For ω = 500 rad/s, eo ( t )= 0. 66 cos( 500 t + 1. 50 )

For ω = 1000 rad/s, eo ( t )= 10 cos( 1000 t )

For ω = 1500 rad/s, e o ( t )= 1. 2 cos( 1500 t − 1. 45 )

Do you see why this circuit is called a “bandpass filter”?

Time (s)

Input and output (V)

Fig 23.2(a): w = 500 rad/s

Input Response

Time (s)

Input and output (V)

Fig 23.2(a): w = 1000 rad/s

Input Response

Time (s)

Input and output (V)

Fig 23.2(a): w = 1500 rad/s

Input Response

Note that input and output are identical.