Explanation Given data: v s ( t ) = 10 e − 2 t u ( t ) (1) Formula used: Consider the general form of Fourier transform of f ( t ) is represented as F ( ω ) . F ( ω ) = ∫ − ∞ ∞ f ( t ) e − j ω t d t Calculation: Refer to Figure 18.45 in the text book for impedance circuit. Apply Fourier transform on both sides of equation (1) as follows. F [ v s ( t ) ] = F [ 10 e − 2 t u ( t ) ] V s ( ω ) = 10 2 + j ω By using source transformation technique, convert the voltage source to current source and modify the Figure 18.45 as shown in Figure 1. In Figure 1, as two 2 Ω resistors are connected in parallel, therefore the equivalent resistance of the parallel connected circuit is calculated as follows. R e q = ( 2 Ω ) ( 2 Ω ) 2 Ω + 2 Ω = 4 4 Ω = 1 Ω Modify Figure 1 as shown in Figure 2. Write the expression of current I , in Figure 2. I ( ω ) = ( V s ( ω ) 2 ) 1 + j ω = V s ( ω ) 2 ( 1 + j ω ) Substitute 10 2 + j ω for V s ( ω ) as follows. I ( ω ) = 10 2 ( 1 + j ω ) ( 2 + j ω ) = 5 ( 1 + j ω ) ( 2 + j ω ) Consider s = j ω to reduce complex algebra. Substitute s for j ω as follows. I ( s ) = 5 ( 1 + s ) ( 2 + s ) Take partial fraction for the equation. I ( s ) = A s + 1 + B s + 2 (2) Where A = ( s + 1 ) I ( s ) | s = − 1 Substitute 5 ( 1 + s ) ( 2 + s ) for I ( s ) as follows

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ISBN: 8220102801448

Author: Alexander

Publisher: YUZU

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Fourier Transform

In mathematics, the Fourier transformation is a mathematical transformation that rotates responsibilities by using region or time into tasks depending on the local or temporal frequency, such as the rendering of a musical track in step with existing volu…

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Chapter 18, Problem 45P

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Find the value of current i(t) in Figure 18.47.

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Chapter 18, Problem 44P

Chapter 18, Problem 46P

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4. (20 pts) You are given a channel with the following impulse response. Determine the set of equations that will be used to determine the coefficients of a Zero-Forcing Linear Equalizer. DO NOT SOLVE FOR THE COEFFICIENTS. Just show the set of equation that would be used to solved the coefficients. 0 m≤-31 -0.33 m = -2 .25 m = -1 h(mb) = 1 m = 0 -0.45 m = 1 0.5 m = 2 0 m≥3

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Chapter 18 Solutions

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS

Ch. 18.2 - Prob. 1PP Ch. 18.2 - Prob. 2PP Ch. 18.2 - Prob. 3PP Ch. 18.3 - Prob. 5PP Ch. 18.3 - Prob. 6PP Ch. 18.4 - Prob. 7PP Ch. 18.4 - Prob. 8PP Ch. 18.5 - (a) Calculate the total energy absorbed by a 1-...Ch. 18.5 - Prob. 10PP Ch. 18.8 - If a 2-MHz carrier is modulated by a 4-kHz...

Ch. 18.8 - Prob. 12PP Ch. 18 - Prob. 1RQ Ch. 18 - Prob. 2RQ Ch. 18 - The inverse Fourier transform of ej2+j is (a) e2t...Ch. 18 - Prob. 4RQ Ch. 18 - Prob. 5RQ Ch. 18 - Prob. 6RQ Ch. 18 - Prob. 7RQ Ch. 18 - Prob. 8RQ Ch. 18 - A unit step current is applied through a 1-H...Ch. 18 - Prob. 10RQ Ch. 18 - Prob. 1P Ch. 18 - Prob. 2P Ch. 18 - Prob. 3P Ch. 18 - Prob. 4P Ch. 18 - Prob. 5P Ch. 18 - Prob. 6P Ch. 18 - Prob. 7P Ch. 18 - Prob. 8P Ch. 18 - Prob. 9P Ch. 18 - Prob. 10P Ch. 18 - Prob. 11P Ch. 18 - Prob. 12P Ch. 18 - Prob. 14P Ch. 18 - Prob. 15P Ch. 18 - Prob. 16P Ch. 18 - Prob. 17P Ch. 18 - Given that F=Fft, prove the following results,...Ch. 18 - Prob. 19P Ch. 18 - Prob. 21P Ch. 18 - Prob. 22P Ch. 18 - Prob. 23P Ch. 18 - Prob. 24P Ch. 18 - Prob. 25P Ch. 18 - Prob. 26P Ch. 18 - Prob. 27P Ch. 18 - Prob. 28P Ch. 18 - Prob. 29P Ch. 18 - For a linear system with input x(t) and output...Ch. 18 - Prob. 31P Ch. 18 - Prob. 32P Ch. 18 - Prob. 33P Ch. 18 - Prob. 34P Ch. 18 - Prob. 35P Ch. 18 - Prob. 36P Ch. 18 - Prob. 37P Ch. 18 - Prob. 38P Ch. 18 - Prob. 39P Ch. 18 - Prob. 40P Ch. 18 - Prob. 41P Ch. 18 - Prob. 42P Ch. 18 - Prob. 43P Ch. 18 - Prob. 44P Ch. 18 - Prob. 45P Ch. 18 - Prob. 46P Ch. 18 - Prob. 47P Ch. 18 - Prob. 49P Ch. 18 - Prob. 51P Ch. 18 - Prob. 52P Ch. 18 - Prob. 53P Ch. 18 - Prob. 54P Ch. 18 - Prob. 55P Ch. 18 - Prob. 56P Ch. 18 - Prob. 57P Ch. 18 - Prob. 58P Ch. 18 - Prob. 59P Ch. 18 - Prob. 60P Ch. 18 - Prob. 61P Ch. 18 - Prob. 62P Ch. 18 - Prob. 63P Ch. 18 - Prob. 64P Ch. 18 - Prob. 65P Ch. 18 - Prob. 66P Ch. 18 - Given a signal g(t) = sinc(200 t), find the...Ch. 18 - Prob. 68CP Ch. 18 - Prob. 69CP

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