# Solucion alexander ch 14

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Problem 14.1 12/10/1999
[pic]
[pic][pic], where [pic]

[pic] [pic]

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that [pic]. Thus, the sketches of H and ( are shown below.

Problem 14.2
[pic][pic], where [pic]

[pic] [pic]

The frequency response is identical to theresponse in Example 14.1 except that [pic]. Hence the response is shown below.

Problem 14.3
a) The Thevenin impedance across the second capacitor where [pic] is taken is
[pic]
[pic]

[pic]

[pic]
[pic][pic]

b) [pic]

There are no zeros and the poles are at
[pic][pic]
[pic][pic]Problem 14.4
a) [pic]

[pic]

[pic][pic]

b) [pic]

[pic][pic]

Problem 14.5
a) [pic]
[pic][pic]

b) [pic]

[pic]
[pic][pic]

Problem 14.6
a) Using current division,
[pic]
[pic]
[pic][pic]

b) We apply nodal analysis to thecircuit below.

[pic]

But [pic]

[pic]
[pic]
[pic]

[pic]
[pic]
[pic][pic]

Problem 14.7
a) [pic]
[pic]
[pic][pic]

b) [pic]
[pic]
[pic][pic]

c) [pic]
[pic]
[pic][pic]

Problem 14.8
a)[pic]
[pic][pic], [pic][pic]

b) [pic]
[pic][pic], [pic][pic]

c) [pic]
[pic][pic], [pic][pic]

d) [pic]
[pic][pic], [pic][pic]

Problem 14.9
[pic]

[pic]
[pic]

The magnitude and phase plots are shown below.

Problem 14.10
[pic]

[pic][pic]

The magnitude and phase plots are shown below.

Problem 14.11
[pic]

[pic]
[pic]

The magnitude and phase plots are shown below.

Problem 14.12
[pic]

[pic]
[pic]
[pic]

The magnitude and phase plots are shown below.

Problem 14.13
[pic]

[pic][pic]

The magnitudeand phase plots are shown below.

Problem 14.14
[pic]

[pic]
[pic]

The magnitude and phase plots are shown below.

Problem 14.15
[pic]

[pic]
[pic]
where [pic]
[pic]

The magnitude and phase plots are shown below.

Problem 14.16
[pic]

[pic]
[pic]The magnitude and phase plots are shown below.

Problem 14.17
[pic]

[pic]
[pic]

The magnitude and phase plots are shown below.

Problem 14.18
[pic]

[pic]
[pic]
[pic]

The magnitude and phase plots are shown below.

Problem 14.19 [pic]

A zero of slope [pic] at [pic]
A pole of slope[pic] at [pic]
A pole of slope [pic] at [pic]

Hence,
[pic]

[pic][pic]

Problem 14.20

A zero of slope [pic] at the origin[pic]
A pole of slope [pic] at [pic]
A pole of slope [pic] at [pic]

Hence,
[pic]

[pic][pic]

Problem 14.21 The phase plot is decomposed as shown below. 12/10/1999

[pic]
where[pic] is a constant since [pic].

Hence, [pic][pic], where [pic]

Problem 14.22 [pic]

[pic][pic]

[pic]
[pic]
[pic]
[pic][pic]

[pic]
[pic]
[pic]
[pic][pic]

[pic]
[pic]
[pic][pic]

[pic]
[pic]
[pic][pic]...