Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
250 Ω w i250 + V₁(t) i 500 500 Ω www x₁ = V₂ (+) and x₂ = 12 (4) V₁ (+) - 1250 (+) · 25012 - Vc (+) =0 (+) sin (2000+) a) KVL: 1,50 (+) = KCL: V₁(+)-vc(+) 250♫2 isto (t) + i₂ (+) + C.dv₂ (+) C.dv (+) = 0 1µF 0.5 H& → 1500 (+) = VC (+) and c = 1 µF = 1×10-6F 50052 X, <+> dx₁(+) H C == ₁₁ (+) = 1500 (+) = ↓ (iN(+) - X₁ (+)) 5000 The source voltage is v₁(t) = sin(2000t). We wish to obtain the current through the 250 resistor (1250) and the 500 2 resistor (1500). There is no initial energy stored in the circuit and the source turns on at time t = 0. (a) Find the state equation for this circuit. (b) Find the output equation. (c) Using your state and output equations, use MATLAB to plot the resistor currents for 0≤t ≤ 4 ms. Use a sufficient number of samples so that the plot appears smooth. = = 1 1x10-bp (Xx(t) - X₁ (+) 5000 (1x10 bp) (X₂ (+)- X₁ (+)) (-2000)x, (+) + (106) x₂ (+) - X₂ (+): dx₂ (t) = L· div(t) dt = = = Vect) -isto (+). 500 V₁ (+) - X₁ (+) - 500 X₂ (+) 0.5 H Yo.5 (V. (+) == x₁ (+)-(500)xx (+)) 2 (sin (2000) - X₁(+) -500x₂ (+)) 2 sin (2000t) -2x, (+) -(1000)x, (+)) A 54 Matux: x=Ax+Bu B dx, (+) -2000 106 X₁(+) 0 = dr (f) Sin (2000) -2 1800 Xx(+) b) Output Vastables: Y₁ = 1250 (+) - Y₁ = 1500 (+) = V₁ (+) - √c (+) 2502 Sin (2000) - x, (+) VC(+) 500 = X₁ (+) 50022 output Mataxi 3 ८ + D 4(+) = (x+Du (५.) 250 0 X₁ (+) = Sin(2000) Y₂(+) 0 500 Xx (+) 0 750-0

250 Ω w i250 + V₁(t) i 500 500 Ω www x₁ = V₂ (+) and x₂ = 12 (4) V₁ (+) - 1250 (+) · 25012 - Vc (+) =0 (+) sin (2000+) a) KVL: 1,50 (+) = KCL: V₁(+)-vc(+) 250♫2 isto (t) + i₂ (+) + C.dv₂ (+) C.dv (+) = 0 1µF 0.5 H& → 1500 (+) = VC (+) and c = 1 µF = 1×10-6F 50052 X, <+> dx₁(+) H C == ₁₁ (+) = 1500 (+) = ↓ (iN(+) - X₁ (+)) 5000 The source voltage is v₁(t) = sin(2000t). We wish to obtain the current through the 250 resistor (1250) and the 500 2 resistor (1500). There is no initial energy stored in the circuit and the source turns on at time t = 0. (a) Find the state equation for this circuit. (b) Find the output equation. (c) Using your state and output equations, use MATLAB to plot the resistor currents for 0≤t ≤ 4 ms. Use a sufficient number of samples so that the plot appears smooth. = = 1 1x10-bp (Xx(t) - X₁ (+) 5000 (1x10 bp) (X₂ (+)- X₁ (+)) (-2000)x, (+) + (106) x₂ (+) - X₂ (+): dx₂ (t) = L· div(t) dt = = = Vect) -isto (+). 500 V₁ (+) - X₁ (+) - 500 X₂ (+) 0.5 H Yo.5 (V. (+) == x₁ (+)-(500)xx (+)) 2 (sin (2000) - X₁(+) -500x₂ (+)) 2 sin (2000t) -2x, (+) -(1000)x, (+)) A 54 Matux: x=Ax+Bu B dx, (+) -2000 106 X₁(+) 0 = dr (f) Sin (2000) -2 1800 Xx(+) b) Output Vastables: Y₁ = 1250 (+) - Y₁ = 1500 (+) = V₁ (+) - √c (+) 2502 Sin (2000) - x, (+) VC(+) 500 = X₁ (+) 50022 output Mataxi 3 ८ + D 4(+) = (x+Du (५.) 250 0 X₁ (+) = Sin(2000) Y₂(+) 0 500 Xx (+) 0 750-0

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...
See similar textbooks
icon
Related questions
Question
100%

I worked on this problem for a long time but apparently the analysis was not right.  Can you show what the correct way is to solve?  (Only Part A and B, not C.)

250 Ω w i250 + V₁(t) i 500 500 Ω www x₁ = V₂ (+) and x₂ = 12 (4) V₁ (+) - 1250 (+) · 25012 - Vc (+) =0 (+) sin (2000+) a) KVL: 1,50 (+) = KCL: V₁(+)-vc(+) 250♫2 isto (t) + i₂ (+) + C.dv₂ (+) C.dv (+) = 0 1µF 0.5 H& → 1500 (+) = VC (+) and c = 1 µF = 1×10-6F 50052 X, <+> dx₁(+) H C == ₁₁ (+) = 1500 (+) = ↓ (iN(+) - X₁ (+)) 5000 The source voltage is v₁(t) = sin(2000t). We wish to obtain the current through the 250 resistor (1250) and the 500 2 resistor (1500). There is no initial energy stored in the circuit and the source turns on at time t = 0. (a) Find the state equation for this circuit. (b) Find the output equation. (c) Using your state and output equations, use MATLAB to plot the resistor currents for 0≤t ≤ 4 ms. Use a sufficient number of samples so that the plot appears smooth. = = 1 1x10-bp (Xx(t) - X₁ (+) 5000 (1x10 bp) (X₂ (+)- X₁ (+)) (-2000)x, (+) + (106) x₂ (+) - X₂ (+): dx₂ (t) = L· div(t) dt = = = Vect) -isto (+). 500 V₁ (+) - X₁ (+) - 500 X₂ (+) 0.5 H Yo.5 (V. (+) == x₁ (+)-(500)xx (+)) 2 (sin (2000) - X₁(+) -500x₂ (+)) 2 sin (2000t) -2x, (+) -(1000)x, (+)) A 54 Matux: x=Ax+Bu B dx, (+) -2000 106 X₁(+) 0 = dr (f) Sin (2000) -2 1800 Xx(+) b) Output Vastables: Y₁ = 1250 (+) - Y₁ = 1500 (+) = V₁ (+) - √c (+) 2502 Sin (2000) - x, (+) VC(+) 500 = X₁ (+) 50022 output Mataxi 3 ८ + D 4(+) = (x+Du (५.) 250 0 X₁ (+) = Sin(2000) Y₂(+) 0 500 Xx (+) 0 750-0
Transcribed Image Text:250 Ω w i250 + V₁(t) i 500 500 Ω www x₁ = V₂ (+) and x₂ = 12 (4) V₁ (+) - 1250 (+) · 25012 - Vc (+) =0 (+) sin (2000+) a) KVL: 1,50 (+) = KCL: V₁(+)-vc(+) 250♫2 isto (t) + i₂ (+) + C.dv₂ (+) C.dv (+) = 0 1µF 0.5 H& → 1500 (+) = VC (+) and c = 1 µF = 1×10-6F 50052 X, <+> dx₁(+) H C == ₁₁ (+) = 1500 (+) = ↓ (iN(+) - X₁ (+)) 5000 The source voltage is v₁(t) = sin(2000t). We wish to obtain the current through the 250 resistor (1250) and the 500 2 resistor (1500). There is no initial energy stored in the circuit and the source turns on at time t = 0. (a) Find the state equation for this circuit. (b) Find the output equation. (c) Using your state and output equations, use MATLAB to plot the resistor currents for 0≤t ≤ 4 ms. Use a sufficient number of samples so that the plot appears smooth. = = 1 1x10-bp (Xx(t) - X₁ (+) 5000 (1x10 bp) (X₂ (+)- X₁ (+)) (-2000)x, (+) + (106) x₂ (+) - X₂ (+): dx₂ (t) = L· div(t) dt = = = Vect) -isto (+). 500 V₁ (+) - X₁ (+) - 500 X₂ (+) 0.5 H Yo.5 (V. (+) == x₁ (+)-(500)xx (+)) 2 (sin (2000) - X₁(+) -500x₂ (+)) 2 sin (2000t) -2x, (+) -(1000)x, (+)) A 54 Matux: x=Ax+Bu B dx, (+) -2000 106 X₁(+) 0 = dr (f) Sin (2000) -2 1800 Xx(+) b) Output Vastables: Y₁ = 1250 (+) - Y₁ = 1500 (+) = V₁ (+) - √c (+) 2502 Sin (2000) - x, (+) VC(+) 500 = X₁ (+) 50022 output Mataxi 3 ८ + D 4(+) = (x+Du (५.) 250 0 X₁ (+) = Sin(2000) Y₂(+) 0 500 Xx (+) 0 750-0
Expert Solution
steps

Step by step

Solved in 2 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS