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PHYS245
Electricity and Electronics Syllabus - Classes - Assignments - Laboratory - Resources Class 36: Answers to Assignment |
Chapter 8
Exercise 36:
Hint: Zi = R1 - jXc and Zf = R2.
Ans:
a) The gain for this high-pass filter is - jwR2C / (1 + jwR1C).
b) The gain at high frequencies is - R2/R1; 13.2 dB (inverting) for these components.
c) For 0.1 mF, the cutoff frequency is 884 Hz. For 1.0 mF, 88 Hz.
Exercise 41:
Hint: Zf = R2 / (1 + jwR2C).
Ans:
a) The gain for this low-pass filter is - (R2/R1) / (1 + jwR2C).
b) The gain at low frequencies is - R2/R1; 30 dB (inverting) for these components. The cutoff frequency is 4980 Hz.
Exercise 42:
Hint: Zi = R1- jwR1C1 and Zf = R2 / (1 + jwR2C2).
Ans:
The gain for this band-pass filter is - jwR2C1 / [(1 + jwR2C2)(1 + jwR1C1)].
For R1=R2=R and C1=C2=C this reduces to - jwRC / (1 + jwRC)2.
The magnitude of the gain is wRC / [1 + (wRC)2 ].
The magnitude of the gain has a maximum value of 1/2 when wRC=1. (Use a bit of differential calculus!) The resulting frequency is 160 Hz.
The cutoff frequencies are not clearly demarked as 'cutoffs' but we can consider the -3 dB frequencies; that is, the frequencies for which the gain is 1/[2 sqrt(2)]. Solving the resulting quadratic equation leads to a lower -3 dB frequency of 66 Hz and an upper -3 dB frequency of 384 Hz.
The quality factor of a bandpass filter is given as the ratio of the center frequency to the bandwidth (that is, the difference of the high and low -3 dB frequencies above). The Q of this circuit is thus 1/2.