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Chapter 10: Problem 4

The gain of a certain amplifier decreases by 6 dB when the frequency is reduced from I kHy to 10 Hz. The roll-off is (a) \(-3 \mathrm{~dB} / \mathrm{decode}\) (b) \(-6\) dB/decsie (c) \(-3\) dB/octave (d) \(-6\) dB/octave

Short Answer

Expert verified
The gain decreases by 6 dB when the frequency is reduced from 1000 Hz to 10 Hz. Calculating the number of decades change, we get Number of decades = log10(1000 Hz / 10 Hz) = log10(100) = 2. Then, the roll-off per decade = Gain change / Number of decades = -6 dB / 2 = -3 dB/decade. Thus, the correct answer is (a) \(-3 \mathrm{~dB} / \mathrm{decade}\).

Step by step solution

01

Find the number of octaves and decades change

Let's first find out how many octaves and decades the frequency has changed. For octaves, we can use the formula: Number of octaves = log2(Final frequency / Initial frequency) For decades, we can use the formula: Number of decades = log10(Final frequency / Initial frequency) Given: Initial frequency = 1 kHz = 1000 Hz Final frequency = 10 Hz Now, we can calculate the number of octaves and decades change.
02

Calculate the number of octaves and decades change

Number of octaves = log2(1000 Hz / 10 Hz) = log2(100) 鈮� 6.64 Number of decades = log10(1000 Hz / 10 Hz) = log10(100) = 2
03

Calculate the roll-off per octave and per decade

Given that the gain decreases by 6 dB. Now, we can calculate the roll-off per octave and per decade. Roll-off per octave = Gain change / Number of octaves = -6 dB / 6.64 鈮� -0.90 dB/octave Roll-off per decade = Gain change / Number of decades = -6 dB / 2 = -3 dB/decade
04

Compare calculated roll-off with given options

Now, we compare the calculated roll-off per octave and per decade with the given options: (a) \(-3 \mathrm{~dB} / \mathrm{decade}\) (b) \(-6\) dB/decade (c) \(-3\) dB/octave (d) \(-6\) dB/octave We can see that the calculated roll-off per decade (-3 dB/decade) matches option (a). Hence, the correct answer is (a) \(-3 \mathrm{~dB} / \mathrm{decade}\).

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