Page 442 - Analog and Digital Filter Design

ANSWERS

Chapter 1

1.1 The ratio of output power to input power is 0.316.0 = 0.05. The “gain” is
1 O.log(O.05) = -13 dB, therefore the attenuation, or signal loss, is +i 3 dB.
Relative to the volt, an input voltage of 2 V = 20.1og(2) = +6 dBV. With
the attenuation being 13dB, the output voltage (in dBs) will be 6 - 13 =
-7dBV. The actual voltage is lo”(-7/20) = lo(-0.35) = 0.4467V.
1.2 24dB. The filter gives a 12dB per octave attenuation rate; 2MHz 1s an
octave above 1 MHz and 4 MHz is an octave above 2 MHz. Two
octaves x 12 dB = 24 dB.
1.3 A lOmW input signal has a level (in dBs) of 10dBm. At 2MHz the
attenuation is 12dB, so the output level is -2dBm. This is 10’(-2/10) =
0.63mW. At 4MHz the attenuation is 24dB, so the output level is -14
dBm. This is lo“(-14/10) = 0.04mW.
1.4 At the -3dB point the voltage across the output will be 0.7071 (liroot
of 2) times V,”. Therefore the voltage across the capacitor will be
7,071 V. The current through the capacitor also flows through the
resistor, and since they have equal impedance at the -3dB point, the
voltage across the resistor is also 7.071 V. The peak in resistor voltage
is 90” ahead of the peak in capacitor voltage. (a) 7.071 V, (b) 7.071 V.

Chapter 2

2.1 Lowpass and bandstop.

2.2 The passband describes a range of frequencies that allow signals to
pass with little or no attenuation. The stopband describes a range of
frequencies that attenuate signals by at least the design specification
limit. The skirt is the range of frequencies between the passband and
the stopband, where attenuation will be more than 3dB but less than
the stopband level.