To determine Plot the maximum gain decibels versus α when ω = 0 for the circuit designed in Problem 15.61. Explanation Given data: The value of α varies from 0 to 1 in increment of 0.1 . From the Problem 15.61, The maximum gain value is 20 dB and the gain value is 17 dB . The value of the center frequency ( f o ) is 40 Hz . The value of the resistor ( R 1 ) is 11.1 k Ω . The value of the resistor ( R 2 ) is 100 k Ω . Formula used: Write the expression to calculate the angular center frequency. ω o = 2 π f o (1) Here, f o is the value of the center frequency in hertz. Write the expression to calculate the center frequency of the bass volume control circuit. ω o = 1 R 2 C 1 (2) Here, R 2 is the value of the resistor 2, and C 1 is the value of the capacitor 1. Calculation: The Bass volume control circuit is drawn as Figure 1. Write the transfer function equation of the bass volume control circuit. H ( j ω ) = − ( R 1 + α R 2 + j ω R 1 R 2 C 1 ) R 1 + ( 1 − α ) R 2 + j ω R 1 R 2 C 1 (3) The maximum gain is obtained when ω is equal to zero. That is, ω = 0 . Substitute 0 for ω in equation (3) to find H ( j 0 ) . H ( j 0 ) = − ( R 1 + α R 2 + j ( 0 ) R 1 R 2 C 1 ) R 1 + ( 1 − α ) R 2 + j ( 0 ) R 1 R 2 C 1 = − ( R 1 + α R 2 + 0 ) R 1 + ( 1 − α ) R 2 + 0 = − ( R 1 + α R 2 ) R 1 + ( 1 − α ) R 2 Take magnitude on both sides of the above equation to find | H ( j 0 ) | . | H ( j 0 ) | = | − ( R 1 + α R 2 ) R 1 + ( 1 − α ) R 2 | = | − ( R 1 + α R 2 ) | | ( R 1 + ( 1 − α ) R 2 ) | = ( − ( R 1 + α R 2 ) ) 2 ( R 1 + ( 1 − α ) R 2 ) 2 | H ( j 0 ) | = ( R 1 + α R 2 ) ( R 1 + ( 1 − α ) R 2 ) (4) The equation (4) is the magnitude of the transfer function of the circuit with a maximum gain that is at ω = 0 . Assume α = 1 and substitute in equation (4) to find | H ( j 0 ) | . | H ( j 0 ) | = ( R 1 + ( 1 ) R 2 ) ( R 1 + ( 1 − 1 ) R 2 ) = R 1 + R 2 R 1 + 0 | H ( j 0 ) | = R 1 + R 2 R 1 (5) Given that the maximum gain value is 20 dB , therefore the equation (5) becomes, 20 log ( R 1 + R 2 R 1 ) = 20 dB log ( R 1 + R 2 R 1 ) = 20 20 dB log ( R 1 + R 2 R 1 ) = 1 dB Simplify the above equation. R 1 + R 2 R 1 = 10 1 dB R 1 + R 2 R 1 = 10 R 1 + R 2 = 10 R 1 Rearrange the above equation to find R 2 . R 2 = 10 R 1 − R 1 = 9 R 1 Substitute equation (1) in equation (2). 2 π f o = 1 R 2 C 1 Rearrange the above equation to find C 1 . C 1 = 1 2 π f o R 2 Substitute 40 Hz for f o and 100 k Ω for R 2 in above equation to find C 1 . C 1 = 1 2 π ( 40 Hz ) ( 100 k Ω ) = 1 2 π ( 40 ) ( 100 × 10 3 ) Ω ⋅ Hz { ∵ 1 k = 10 3 } = 39.79 × 10 − 9 1 ( 1 F ) { ∵ 1 F = 1 1 Ω ⋅ 1 Hz } = 39.79 nF { ∵ 1 n = 10 − 9 } From the obtained capacitor value, the standard value is taken for the following calculation. C 1 = 39 nF Substitute 11.1 k Ω for R 1 and 100 k Ω for R 2 in equation (5) to find | H ( j 0 ) | . | H ( j 0 ) | = ( 11.1 k Ω ) + ( 100 k Ω ) 11.1 k Ω = 111.1 k Ω 11.1 k Ω = 10.009 Find the above value in decibel. | H ( j 0 ) | dB = 20 log ( 10.009 ) dB = 20.01 dB Calculate the center frequency of the circuit using the obtained standard values. Substitute 100 k Ω for R 2 and 39 nF for C 1 in equation (2) to find ω o . ω o = 1 ( 100 k Ω ) ( 39 nF ) = 1 ( 100 × 10 3 ) ( 39 × 10 − 9 ) Ω ⋅ F { ∵ 1 k = 10 3 , 1 n = 10 − 9 } = 1 ( 3.9 × 10 − 3 ) Ω ⋅ ( s Ω ) { ∵ 1 F = 1 s 1 Ω } = 256

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Chapter 15, Problem 63P

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Electric Circuits Plus Mastering Engineering with Pearson eText 2.0 - Access Card Package (11th Edition) (What's New in Engineering)

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