Explanation Given data: The given function is, x ( t ) = cos t + ∫ 0 t e λ − t x ( λ ) d λ (1) Formula used: Write the general expression to find the convolution integral. y ( t ) = f 1 ( t ) ∗ f 2 ( t ) = ∫ 0 t f 1 ( λ ) f 2 ( t − λ ) d λ (2) Write the general expression to find the Laplace transform function. L [ cos ω t ] = s s 2 + ω 2 (3) L [ sin ω t ] = ω s 2 + ω 2 (4) L [ e − a t ] = 1 s + a (5) Here, s is the complex variable. Write the general expression for the Laplace transform. F ( s ) = L [ f ( t ) ] (6) Calculation: Assume the following to reduce the equation (1). y ( t ) = ∫ 0 t e λ − t x ( λ ) d λ y ( t ) = ∫ 0 t e − ( t − λ ) x ( λ ) d λ (7) Substitute the equation (7) in equation (1) to find x ( t ) . x ( t ) = cos t + y ( t ) (8) On comparing the equation (2) and equation (7). y ( t ) = x ( t ) ∗ e − t Substitute x ( t ) ∗ e − t for y ( t ) in equation (8) to find x ( t ) . x ( t ) = cos t + x ( t ) ∗ e − t (9) Apply the Laplace transform function of equation (3), (4), (5) (6) in equation (9). X ( s ) = s s 2 + 1 2 + X ( s ) ( 1 s + 1 ) X ( s ) − X ( s ) ( 1 s + 1 ) = s s 2 + 1 X ( s ) ( 1 − 1 s + 1 ) = s s 2 + 1 Simplify the equation as follows, X ( s ) ( s + 1 − 1 s + 1 ) = s s 2 + 1 X ( s ) ( s s + 1 ) = s s 2 + 1 X ( s ) = ( s s 2 + 1 ) ( s + 1 s ) X ( s ) = s + 1 s 2 + 1 (10) Case a: Reduce the equation as follows, X ( s ) = s s 2 + 1 + 1 s 2 + 1 Taking inverse Laplace transform for the above equation as follows. L − 1 [ X ( s ) ] = L − 1 [ s s 2 + 1 ] + L − 1 [ 1 s 2 + 1 ] x ( t ) = L − 1 [ s s 2 + 1 2 ] + L − 1 [ 1 s 2 + 1 2 ] x ( t ) = cos ( t ) + sin ( t ) Case b: Expand X ( s ) in equation (10) using partial fraction method. X ( s ) = s + 1 s 2 + 1 = s + 1 ( s + j ) ( s − j ) X ( s ) = A s + j + B s − j (11) Here, A and B are the constants. Find the constants by using residue method. Constant A : A = ( s + j ) X ( s ) | s + j = 0 (12) Substitute equation (10) in equation (12) to find the constant A . A = ( s + j ) × s + 1 s 2 + 1 | s + j = 0 = ( s + j ) × s + 1 ( s + j ) ( s − j ) | s = − j = s + 1 s − j | s = − j = − j + 1 − j − j Reduce the equation as follows, A = 1 − j − 2 j = 0

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Among all the electrical engineering students, this topic of convolution integral is very confusing. It is a mathematical operation of two functions f and g that produce another third type of function (f * g) , and this expresses how the shape of one is …

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Chapter 15, Problem 53P

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Find the value of x(t) using Laplace transform.

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Chapter 15, Problem 52P

Chapter 15, Problem 54P

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Find Vb and Va using Mesh analysis

1. The communication channel bandwidth is 25 MHz centered at 1GHz and has a noise power spectral density of 10^-9 W/Hz. The channel loss between the transmitter and receiver is 25dB. The application requires a bit rate of 200Mbps and BER of less than 10^-4. Excluding Mary FSK, Determine the minimum transmit power required.

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Chapter 15 Solutions

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS

Ch. 15.2 - Prob. 1PP Ch. 15.2 - Prob. 2PP Ch. 15.3 - Prob. 3PP Ch. 15.3 - Prob. 4PP Ch. 15.3 - Prob. 5PP Ch. 15.3 - Prob. 6PP Ch. 15.3 - Obtain the initial and the final values of...Ch. 15.4 - Prob. 8PP Ch. 15.4 - Find f(t) if F(s)=48(s+2)(s+1)(s+3)(s+4)Ch. 15.4 - Obtain g(t) if G(s)=s3+2s+6s(s+1)2(s+3)

Ch. 15.4 - Find g(t) given that G(s)=20(s+1)(s2+4s+13)Ch. 15.5 - Graphically convolve the two functions in Fig....Ch. 15.5 - Given g(t) and f(t) in Fig. 15.20, graphically...Ch. 15.5 - Use convolution to find vo(t) in the circuit of...Ch. 15.6 - Prob. 15PP Ch. 15.6 - Prob. 16PP Ch. 15 - Prob. 1RQ Ch. 15 - Prob. 2RQ Ch. 15 - Prob. 3RQ Ch. 15 - Prob. 4RQ Ch. 15 - Prob. 5RQ Ch. 15 - If F(s) = 1/(s + 2), then f(t) is (a) e2t u(t) (b)...Ch. 15 - Prob. 7RQ Ch. 15 - Prob. 8RQ Ch. 15 - Prob. 9RQ Ch. 15 - Prob. 10RQ Ch. 15 - Prob. 1P Ch. 15 - Prob. 2P Ch. 15 - Prob. 3P Ch. 15 - Prob. 4P Ch. 15 - Prob. 5P Ch. 15 - Prob. 6P Ch. 15 - Prob. 7P Ch. 15 - Prob. 8P Ch. 15 - Prob. 9P Ch. 15 - Prob. 10P Ch. 15 - Find F(s) if: (a) ft=6etcosh2t (b) ft=3te2tsinh4t...Ch. 15 - If g(t) = 4e 2t cos 4t, find G(s).Ch. 15 - Prob. 13P Ch. 15 - Prob. 14P Ch. 15 - Prob. 15P Ch. 15 - Prob. 16P Ch. 15 - Prob. 17P Ch. 15 - Prob. 18P Ch. 15 - Prob. 19P Ch. 15 - Prob. 20P Ch. 15 - Prob. 21P Ch. 15 - Prob. 22P Ch. 15 - Prob. 23P Ch. 15 - Design a problem to help other students better...Ch. 15 - Let F(s)=18(s+1)(s+2)(s+3) (a) Use the initial and...Ch. 15 - Determine the initial and final values of f(t), if...Ch. 15 - Prob. 27P Ch. 15 - Prob. 28P Ch. 15 - Prob. 29P Ch. 15 - Prob. 30P Ch. 15 - Find f(t) for each F(s): (a) 10ss+1s+2s+3 (b)...Ch. 15 - Prob. 32P Ch. 15 - Prob. 33P Ch. 15 - Prob. 34P Ch. 15 - Obtain f(t) for the following transforms: (a)...Ch. 15 - Prob. 36P Ch. 15 - Prob. 37P Ch. 15 - Prob. 38P Ch. 15 - Determine f(t) if: (a)...Ch. 15 - Show that...Ch. 15 - Prob. 41P Ch. 15 - Design a problem to help other students better...Ch. 15 - Prob. 43P Ch. 15 - Prob. 44P Ch. 15 - Given h(t) = 4e2tu(t) and x(t) = (t) 2e 2tu(t),...Ch. 15 - Given the following functions...Ch. 15 - A system has the transfer function...Ch. 15 - Find f(t) using convolution given that: (a)...Ch. 15 - Prob. 49P Ch. 15 - Prob. 50P Ch. 15 - Given that v(0) = 5 and dv(0)/dt = 10, solve...Ch. 15 - Prob. 52P Ch. 15 - Prob. 53P Ch. 15 - Design a problem to help other students better...Ch. 15 - Prob. 55P Ch. 15 - Solve for v(t) in the integrodifferential equation...Ch. 15 - Prob. 57P Ch. 15 - Given that dvdt+2v+50tv()d=4u(t) with v(0) = 1,...Ch. 15 - Solve the integrodifferential equation...Ch. 15 - Prob. 60P

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