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. 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)
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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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.
![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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36e2a10a-56c0-4adb-bccf-fb7de3247667%2Fb588bb73-16e6-4236-a3b7-6c7e1c408c61%2Fowsf6x3_processed.png&w=3840&q=75)
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)
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