Explanation Given data: Refer to Figure 14.49 in the textbook. The value of the corner frequency before scaling ( ω ) is 1 rad s . The value of the cutoff frequency after scaling ( f ′ ) is 10 kHz . The value of the scaled capacitor ( C ′ ) is 15 nF . Formula used: Consider the equations used in magnitude and frequency scaling. Write the expression to calculate the scaled resistor 1. R 1 ′ = K m R 1 (1) Here, K m is the magnitude scaling factor, and R 1 is the value of the resistor 1. Write the expression to calculate the scaled resistor 2. R 2 ′ = K m R 2 (2) Here, R 2 is the value of the resistor 2. Write the expression to calculate the scaled inductor L ′ = K m K f L (3) Here, K f is the frequency scaling factor, and L is the value of the inductor. Write the expression to calculate the scaled capacitor 1. C ′ 1 = 1 K m K f C 1 (4) Here, C 1 is the value of the capacitor 1. Write the expression to calculate the scaled capacitor 2. C ′ 2 = 1 K m K f C 2 (5) Here, C 2 is the value of the capacitor 2. Write the general expression to calculate the scaled corner frequency. ω ′ = 2 π f ′ (6) Here, f ′ is the value of the scaled cutoff frequency. Write the expression to calculate the scaled corner frequency. ω ′ = K f ω (7) Here, ω is the value of the corner frequency. Rearrange the equation (7) to find K f . K f = ω ′ ω Substitute the equation (6) in above equation to find K f . K f = 2 π f ′ ω (8) Calculation: The given circuit is redrawn as Figure 1. Substitute 10 kHz for f ′ and 1 rad s for ω in equation (8) to find K f . K f = 2 π ( 10 kHz ) ( 1 rad s ) = 20 π × 10 3 Hz ( 1 1 s ) { ∵ 1 k = 10 3 } = 20 π × 10 3 ( 1 s ) ( 1 1 s ) { ∵ 1 Hz = 1 1 s } = 2 π × 10 4 Substitute 15 nF for C ′ , 2 π × 10 4 for K f and 1 F for C 1 in equation (4). ( 15 nF ) = 1 K m ( 2 π × 10 4 ) ( 1 F ) Rearrange the above equation to find K m

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

See similar textbooks

Question

Chapter 14.9, Problem 14PP

To determine

Scale the third-order Butterworth filter shown in Figure 14.49.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.8, Problem 13PP

Chapter 14.10, Problem 15PP

Students have asked these similar questions

1. For v(t)=2Σn-[8(t-n) + 28(t-n-0.4)], determine (a) (10%) a figure of v(t); (b) (5%) period To; (c) (10%) Fourier series in Form III; (d) (5%) Fourier transform; (e) (5%) total power.

5. In the figure, v(t) = m(t)ej2nfct where the message signal is m(t): = Acos (2πfmt) and the carrier signal is vc(t) = 2e−j(2nfct+0) where 0 is constant and 0 < fm

= cos (2π x 10t+ 0) where 0 is random with a probability density E [0, 2π), and f(0) = 0 otherwise. v,(t) passes through a linear filter below. 2. Consider a random process v(t) function f(0) = 1/(2) for vi(t)- H(f) vo(t) Determine (a) (5%) vo(t); (b) (10%) autocorrelation function of v(t); (c) (8%) power spectral density function of vo(t); (d) (7%) power of vo(t). 1 = H(f) 2πf2+1

Chapter 14 Solutions

Fundamentals of Electric Circuits

Ch. 14.2 - Obtain the transfer function VoVs of the RL...Ch. 14.2 - Prob. 2PP Ch. 14.4 - Draw the Bode plots for the transfer function...Ch. 14.4 - Sketch the Bode plots for H()=50j(j+4)(j+10)2 Ch. 14.4 - Construct the Bode plots for H(s)=10s(s2+80s+400)Ch. 14.4 - Obtain the transfer function H() corresponding to...Ch. 14.5 - A series-connected circuit has R = 4 and L = 25...Ch. 14.6 - A parallel resonant circuit has R = 100 k, L = 50...Ch. 14.6 - Calculate the resonant frequency of the circuit in...Ch. 14.7 - For the circuit in Fig. 14.40, obtain the transfer...

Knowledge Booster

Scale the third-order Butterworth filter shown in Figure 14.49.

Solution Summary: The author explains how to scale the third-order Butterworth filter shown in Figure 14.49.

Scale the third-order Butterworth filter shown in Figure 14.49.

Want to see the full answer?

Chapter 14 Solutions