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Preliminary Laboratory (Prelab) Work (Week #1) 1. In the space provided below for prelab work, complete the following tasks for the circuit shown in figure 2. If you require more page area to complete your computations, feel free to add space as needed between the prelab work areas below. a) Find the inverse Laplace transform for equation 1 to generate an expression for the time domain voltage out(t) = xc(t) assuming nominal values of R₁ = 5.2[k], R2 = 47.0[k], C=0.01 [u], and a time-domain source voltage signal of Xs(t) u(t). (9 points) b) Sketch the response, Yout(t) = associated with the response. c) c(t) for t≥0. Include both ys(t) and yc(t) in your sketch, and indicate the time constant (3 points) Generate a transient response using the Pspice tool in the Cadence software package to verify the time domain plot completed in part 1b above. (3 points) Prelab Information 1. Laboratory Preliminary Discussion First-order Low-pass RC Filter Analysis The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques. vs(t) i₁(t) + R₁ ww V₁(t) 12(t) Lic(t) Vout(t) = V2(t) R₂ Vc(t) C Vc(t) VR2(t) = V2(t) + Vs(s) Figure 1: A first-order low-pass RC filter represented in the time domain. I₁(s) R1 W + V₁(s) V₂(s) 12(s) Ic(s) + Vout(S) == Vc(s) Vc(s) Zc(s) = = VR2(S) V2(s) Figure 2: A first-order low-pass RC filter represented in the s-domain.

Preliminary Laboratory (Prelab) Work (Week #1) 1. In the space provided below for prelab work, complete the following tasks for the circuit shown in figure 2. If you require more page area to complete your computations, feel free to add space as needed between the prelab work areas below. a) Find the inverse Laplace transform for equation 1 to generate an expression for the time domain voltage out(t) = xc(t) assuming nominal values of R₁ = 5.2[k], R2 = 47.0[k], C=0.01 [u], and a time-domain source voltage signal of Xs(t) u(t). (9 points) b) Sketch the response, Yout(t) = associated with the response. c) c(t) for t≥0. Include both ys(t) and yc(t) in your sketch, and indicate the time constant (3 points) Generate a transient response using the Pspice tool in the Cadence software package to verify the time domain plot completed in part 1b above. (3 points) Prelab Information 1. Laboratory Preliminary Discussion First-order Low-pass RC Filter Analysis The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques. vs(t) i₁(t) + R₁ ww V₁(t) 12(t) Lic(t) Vout(t) = V2(t) R₂ Vc(t) C Vc(t) VR2(t) = V2(t) + Vs(s) Figure 1: A first-order low-pass RC filter represented in the time domain. I₁(s) R1 W + V₁(s) V₂(s) 12(s) Ic(s) + Vout(S) == Vc(s) Vc(s) Zc(s) = = VR2(S) V2(s) Figure 2: A first-order low-pass RC filter represented in the s-domain.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Preliminary Laboratory (Prelab) Work (Week #1) 1. In the space provided below for prelab work, complete the following tasks for the circuit shown in figure 2. If you require more page area to complete your computations, feel free to add space as needed between the prelab work areas below. a) Find the inverse Laplace transform for equation 1 to generate an expression for the time domain voltage out(t) = xc(t) assuming nominal values of R₁ = 5.2[k], R2 = 47.0[k], C=0.01 [u], and a time-domain source voltage signal of Xs(t) u(t). (9 points) b) Sketch the response, Yout(t) = associated with the response. c) c(t) for t≥0. Include both ys(t) and yc(t) in your sketch, and indicate the time constant (3 points) Generate a transient response using the Pspice tool in the Cadence software package to verify the time domain plot completed in part 1b above. (3 points)
Transcribed Image Text:Preliminary Laboratory (Prelab) Work (Week #1) 1. In the space provided below for prelab work, complete the following tasks for the circuit shown in figure 2. If you require more page area to complete your computations, feel free to add space as needed between the prelab work areas below. a) Find the inverse Laplace transform for equation 1 to generate an expression for the time domain voltage out(t) = xc(t) assuming nominal values of R₁ = 5.2[k], R2 = 47.0[k], C=0.01 [u], and a time-domain source voltage signal of Xs(t) u(t). (9 points) b) Sketch the response, Yout(t) = associated with the response. c) c(t) for t≥0. Include both ys(t) and yc(t) in your sketch, and indicate the time constant (3 points) Generate a transient response using the Pspice tool in the Cadence software package to verify the time domain plot completed in part 1b above. (3 points)
Prelab Information 1. Laboratory Preliminary Discussion First-order Low-pass RC Filter Analysis The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques. vs(t) i₁(t) + R₁ ww V₁(t) 12(t) Lic(t) Vout(t) = V2(t) R₂ Vc(t) C Vc(t) VR2(t) = V2(t) + Vs(s) Figure 1: A first-order low-pass RC filter represented in the time domain. I₁(s) R1 W + V₁(s) V₂(s) 12(s) Ic(s) + Vout(S) == Vc(s) Vc(s) Zc(s) = = VR2(S) V2(s) Figure 2: A first-order low-pass RC filter represented in the s-domain.
Transcribed Image Text:Prelab Information 1. Laboratory Preliminary Discussion First-order Low-pass RC Filter Analysis The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques. vs(t) i₁(t) + R₁ ww V₁(t) 12(t) Lic(t) Vout(t) = V2(t) R₂ Vc(t) C Vc(t) VR2(t) = V2(t) + Vs(s) Figure 1: A first-order low-pass RC filter represented in the time domain. I₁(s) R1 W + V₁(s) V₂(s) 12(s) Ic(s) + Vout(S) == Vc(s) Vc(s) Zc(s) = = VR2(S) V2(s) Figure 2: A first-order low-pass RC filter represented in the s-domain.
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