Page 446 - Analog and Digital Filter Design

Answers 44

40
7.3 R1=R4= - 270.902 kQ
=
2njOk470 lo-''
210.9k
R2 = R3 = ___ - 6.712 kR
-
40
2.5 loi3
R5 = = 6.313 kQ
40(2.5109 -2.4011O9)
2.5 io9 io4
R6 = ~ = 10.412 kQ
2.401 10'

Chapter 8

8.1 Delta, 6dB.

8.2 3 dB, because half the power goes into each load (-3 dB = half power).
The splitter circuit absorbs no power when the impedance of each load
is equal.
8.3 Because the passband of one filter section coincides with the stopband
of the other. In the passband, a filter presents the source with its load
impedance. The other filter section is connected in parallel with this
and must therefore present high impedance to avoid impedance
mismatch of the source.
8.4 At the -3dB point only half of the available power from the source
enters the filter. When both lowpass and highpass filters are
connected in parallel. half the power enters the lowpass filter and
half enters the highpass filter. All the power is thus absorbed and no
reflections occur-the source is matched to the load. At frequencies
below the -3 dB point, the lowpass filter absorbs a greater proportion
of power and the highpass filter absorbs less. Similarly. at frequencies
above the -3 dB point, the highpass filter absorbs more power and the
lowpass filter absorbs less. Thus the source power is absorbed at all
frequencies.

Chapter 9

9. I It is the unequal delay of some frequencies relative to others.
9.2 Square wave signals contain a fundamental frequency and all its odd
harmonics. Group delay causes some harmonic signals to be delayed
relative to the fundamental, so the waveform is distorted. The rise time
of the wave is slowed and the peak level contains ripple.