This document provides the mathematical modeling and transfer function of an overhead crane. It defines the variables used such as position, cable length, and angle. It derives the kinetic and potential energies of the crane platform and pendulum and determines the Lagrangian. From this, it obtains the equations of motion for the crane position and cable angle. Small angle approximations yield a transfer function that relates the crane position to the applied force as a function of system parameters like masses and cable length.
Real Time Code Generation for Nonlinear Model Predictive ControlBehzad Samadi
This is a quick introduction to optimal control and nonlinear model predictive control. It also includes code generation for a NMPC controller. For a recorded webinar, follow this link: http://goo.gl/c5zFgN
Robust model predictive control for discrete-time fractional-order systemsPantelis Sopasakis
In this paper we propose a tube-based robust model predictive control scheme for fractional-order discrete-
time systems of the Grunwald-Letnikov type with state and input constraints. We first approximate the infinite-dimensional fractional-order system by a finite-dimensional linear system and we show that the actual dynamics can be approximated arbitrarily tight. We use the approximate dynamics to design a tube-based model predictive controller which endows to the controlled closed-loop system robust stability properties
This presentations includes the basic fundamentals of time series data forecasting. It starts with basic naive, regression models and then explains advanced ARIMA models.
Real Time Code Generation for Nonlinear Model Predictive ControlBehzad Samadi
This is a quick introduction to optimal control and nonlinear model predictive control. It also includes code generation for a NMPC controller. For a recorded webinar, follow this link: http://goo.gl/c5zFgN
Robust model predictive control for discrete-time fractional-order systemsPantelis Sopasakis
In this paper we propose a tube-based robust model predictive control scheme for fractional-order discrete-
time systems of the Grunwald-Letnikov type with state and input constraints. We first approximate the infinite-dimensional fractional-order system by a finite-dimensional linear system and we show that the actual dynamics can be approximated arbitrarily tight. We use the approximate dynamics to design a tube-based model predictive controller which endows to the controlled closed-loop system robust stability properties
This presentations includes the basic fundamentals of time series data forecasting. It starts with basic naive, regression models and then explains advanced ARIMA models.
Computation of electromagnetic_fields_scattered_from_dielectric_objects_of_un...Alexander Litvinenko
Tools for electromagnetic scattering from objects with uncertain shapes are needed in various applications.
We develop numerical methods for predicting radar and scattering cross sections (RCS and SCS) of complex targets.
To reduce cost of Monte Carlo (MC) we offer modified multilevel MC (CMLMC) method.
Robust Fuzzy Output Feedback Controller for Affine Nonlinear Systems via T–S ...Mostafa Shokrian Zeini
This presentation concerns the design of a robust H_∞ fuzzy output feedback controller for a class of affine nonlinear systems with disturbance via Takagi-Sugeno (T–S) fuzzy bilinear model. The parallel distributed compensation (PDC) technique is utilized to design a fuzzy controller. The stability conditions of the overall closed loop T-S fuzzy bilinear model are formulated in terms of Lyapunov function via linear matrix inequality (LMI). The control law is robustified by H_∞ sense to attenuate external disturbance. Moreover, the desired controller gains can be obtained by solving a set of LMI.
Energy-Based Control of Under-Actuated Mechanical Systems - Remotely Driven A...Mostafa Shokrian Zeini
This presentation concerns the energy-based swing-up control for a remotely driven acrobot (RDA) which is a 2-link planar robot with the first link being underactuated and the second link being remotely driven by an actuator mounted at a fixed base through a belt.
Computation of electromagnetic_fields_scattered_from_dielectric_objects_of_un...Alexander Litvinenko
Tools for electromagnetic scattering from objects with uncertain shapes are needed in various applications.
We develop numerical methods for predicting radar and scattering cross sections (RCS and SCS) of complex targets.
To reduce cost of Monte Carlo (MC) we offer modified multilevel MC (CMLMC) method.
Robust Fuzzy Output Feedback Controller for Affine Nonlinear Systems via T–S ...Mostafa Shokrian Zeini
This presentation concerns the design of a robust H_∞ fuzzy output feedback controller for a class of affine nonlinear systems with disturbance via Takagi-Sugeno (T–S) fuzzy bilinear model. The parallel distributed compensation (PDC) technique is utilized to design a fuzzy controller. The stability conditions of the overall closed loop T-S fuzzy bilinear model are formulated in terms of Lyapunov function via linear matrix inequality (LMI). The control law is robustified by H_∞ sense to attenuate external disturbance. Moreover, the desired controller gains can be obtained by solving a set of LMI.
Energy-Based Control of Under-Actuated Mechanical Systems - Remotely Driven A...Mostafa Shokrian Zeini
This presentation concerns the energy-based swing-up control for a remotely driven acrobot (RDA) which is a 2-link planar robot with the first link being underactuated and the second link being remotely driven by an actuator mounted at a fixed base through a belt.
Design of a Lift Mechanism for Disabled PeopleSamet Baykul
DATE: 2019.01
In this project, a lift mechanism for especially disabled people has been designed. It is known as home lift, platform lift, vertical lift or through floor lift. These products operate by moving up and down. The lift mechanism consists of powertrain, linkage system, and a raising platform.
- Design of a a shaft, connecting rods, pins and weldings
- Static force analysis
- Building shear and moment diagrams
- Calculation of mechanical design parameters
Feedback Control of Dynamic Systems 8th Edition Franklin Solutions ManualDakotaFredericks
Full download : https://alibabadownload.com/product/feedback-control-of-dynamic-systems-8th-edition-franklin-solutions-manual/ Feedback Control of Dynamic Systems 8th Edition Franklin Solutions Manual
El presente tiene como finalidad desarrollar los respectivos problemas aplicando el método de Bresse.
Para efectos de dichos cálculos se ha empleado hojas lectrónicas, Cada problema
constituye su respectivo análisis en lo que a su tipo se refiere, capturas de la hoja
electrónica empleada con su respectivo gráfico y finalmente la captura hecha del
software H-CANALES V3 que comprueba el correcto desarrollo del mismo.
Body travel performance improvement of space vehicle electromagnetic suspensi...Mustefa Jibril
Electromagnetic suspension system (EMS) is mostly used in the field of high-speed vehicle. In this paper, a space exploring vehicle quarter electromagnetic suspension system is modelled, designed and simulated using linear quadratic optimal control problem. Linear quadratic Gaussian and linear quadratic integral controllers are designed to improve the body travel of the vehicle using bump road profile. Comparison between the proposed controllers is done and a promising simulation result have been analyzed.
A short report that briefly illustrates the differential formulation of the Vaiana-Rosati Model of hysteresis. In such a report, you can also find the related matlab codes.
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
Instead of buying and hoping for the best, we can use data science to help us predict which products may be good fits for us. It includes various function programs to do the above mentioned tasks.
Data file handling has been effectively used in the program.
The automated cosmetic shop management system should deal with the automation of general workflow and administration process of the shop. The main processes of the system focus on customer's request where the system is able to search the most appropriate products and deliver it to the customers. It should help the employees to quickly identify the list of cosmetic product that have reached the minimum quantity and also keep a track of expired date for each cosmetic product. It should help the employees to find the rack number in which the product is placed.It is also Faster and more efficient way.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
Explore the innovative world of trenchless pipe repair with our comprehensive guide, "The Benefits and Techniques of Trenchless Pipe Repair." This document delves into the modern methods of repairing underground pipes without the need for extensive excavation, highlighting the numerous advantages and the latest techniques used in the industry.
Learn about the cost savings, reduced environmental impact, and minimal disruption associated with trenchless technology. Discover detailed explanations of popular techniques such as pipe bursting, cured-in-place pipe (CIPP) lining, and directional drilling. Understand how these methods can be applied to various types of infrastructure, from residential plumbing to large-scale municipal systems.
Ideal for homeowners, contractors, engineers, and anyone interested in modern plumbing solutions, this guide provides valuable insights into why trenchless pipe repair is becoming the preferred choice for pipe rehabilitation. Stay informed about the latest advancements and best practices in the field.
RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...
Lelt 240 semestre i-2021
1.
2. ii
ÍNDICE
PREVIOS.................................................................................................................................................................... 1
CAMPOS ELÉCTRICOS (I – 2021) ....................................................................................................................... 1
CAMPOS MAGNÉTICOS (I – 2021)...................................................................................................................... 2
MODELOS MATEMÁTICOS (I – 2021)................................................................................................................ 3
MODELOS MECÁNICOS (I – 2021) ..................................................................................................................... 7
MODELOS ELÉCTRICOS (I – 2021)................................................................................................................... 10
MEDIDA DE LA RESISTENCIA Y LA LEY DE OHM (I – 2021) ......................................................................... 12
EXAMEN DE TEORÍA Y SIMULACIÓN EN MULTISIM (I – 2021).................................................................... 15
RECUPERATORIOS................................................................................................................................................. 23
RECUPERATORIO (I-2021)............................................................................................................................... 23
3. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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PREVIOS
CAMPOS ELÉCTRICOS (I – 2021)
1. (100%) Considere 2 alambres semicirculares de radios 2R y R, ubicados en los cuadrantes primero -
segundo y tercero - cuarto respectivamente. Ambos tienen una densidad lineal de carga positiva lambda.
Determinar el vector campo eléctrico en el origen de coordenadas. Finalmente expresar las unidades del
campo eléctrico en función de las unidades fundamentales del Sistema Internacional.
Del gráfico (para una semicircunferencia de radio r cualquiera): ( ) ( )
sin ... 1
P
dE dE j
→ →
= −
Pero sabemos:
( )
2
;
d
dq dq
dE k s r ds rd
ds
r
= = = ⎯⎯⎯
→ = reemplazando en ( )
1 :
( ) ( ) ( ) ( )
2 2 2
sin sin sin sin
P
dq k ds k rd k d
dE k j j j j
r
r r r
→ → → → →
= − = − = − = −
( ) ( ) ( ) ( )
( )
0
0
sin cos cos cos 0
P
k k k
E j d j j
r r r
→ → → →
= − = − − = − − +
2
P
k
E j
r
→ →
= −
Para el semicírculo “2R”: 1
2
2
P
k k
E j j
R R
→ → →
= − = −
Para el semicírculo “R”: 2
2 2
P
k k
E j j
R R
→ → →
= − − =
1 2
P P P P
k
E E E E j
R
→ → → → →
= + → =
X
Y
X
Y
𝜃
𝜃
4. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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2 3
N V kg m N kg m
C m s A s C s A
= = → =
CAMPOS MAGNÉTICOS (I – 2021)
1. (40%) Deducir la ecuación que permite calcular la F.E.M. inducida de una maquina eléctrica. Usted debe
ver por conveniente que datos necesita.
( )
cos
B dA BA
= =
Ley de Faraday:
( )
( )
cos
d d
fem BA t
dt dt
= − = −
( )
( ) ( )
sin sin
fem BA t fem BA t
= − − → =
Para N espiras:
( )
sin
fem NBA t
=
2. (40%) Considere un circuito RL serie alimentado por una fuente de tensión E. Determinar la corriente
eléctrica como una función del tiempo. ¿Qué conclusión puede deducir de este resultado?
Por la Ley de Voltajes de Kirchhoff:
||
di
E Ri L L
dt
= +
( ) ( ) ( ) ( )
0
0
E R
RI s L sI s i LI s s
s L
= + − = +
( ) ( )
1
1 1 1 1 1 1
||
L L
L R R
I s I s L
R R
R
E s E R s R
s s
s s L L
L
−
= = − → = −
+ +
+
( ) ( )
( ) ( )
1 1 1
1
R R
t t
L L
E
i t e i t e
E R R R
− −
= − = −
N
B
S
Rotor
Estator
𝑅
𝑖ሺ𝑡ሻ
R1
1kΩ
L1
1mH
𝐿
𝐸
5. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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CONCLUSION: En C.C. una inductancia se comporta como un
cortocircuito.
3. (20%) Deducir la ecuación dimensional de la inductancia L.
De: 2
1
2
U LI
=
2
2 2
2
1 ;
m
J H A J N m N m kg H A
s
= = → = =
2 2
2 2 2
2 2 2 2
kg m M L
H L L M L T I
s A T I
− −
= → = =
MODELOS MATEMÁTICOS (I – 2021)
1. (100%) Determinar el modelo matemático y la función de transferencia de una grúa viajera.
Variables que se utilizan durante el desarrollo del modelo de la grúa viajera:
x : Posición del carro (grúa).
l : Longitud del cable (malacate).
: Ángulo del cable con respecto al eje vertical.
x
X
Y
x1
U1
m
CARRO
MALACATE
(POLIPASTO)
𝜃
6. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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1
U : Fuerza aplicada al carro.
Por otra parte, los parámetros del sistema son:
M : Masa de la plataforma (carro).
m : Masa de la carga.
Como el sistema tiene dos grados de libertad. Las coordenadas son x y .
Por energías:
La energía cinética de la plataforma (carro) es: 2 2
1
1 1
2 2
C
Ec Mv Mx
= =
A su vez para encontrar la energía cinética del péndulo, se hace uso de la longitud del cable l y el ángulo del
cable para encontrar la posición del péndulo, esto es:
sin ; cos
x y
p l p l
= =
Hallando las velocidades:
( ) ( )
sin cos ; cos sin
x y
d d
v l l v l l
dt dt
= = = = −
La velocidad total en el eje x, es: cos
xtotal
v x l
= +
Por lo tanto, el vector de velocidad del péndulo es:
cos
sin
p
x l
v
l
+
=
−
Desarrollando la anterior ecuación:
( ) ( )
2 2
2 2 2 2
cos sin 2 cos
p
v x l l x lx l
= + + − = + +
La energía cinética del péndulo es: ( )
2 2 2
1
2 cos
2
p
Ec m x lx l
= + +
La energía cinética del sistema está dada por:
( )
2 2 2 2
1 1
2 cos
2 2
C p
Ec Ec Ec Mx m x lx l
= + = + + +
La energía potencial del sistema es C p
Ep Ep Ep
= + , la energía potencial de la grúa es cero y la energía
potencial del péndulo es: ( )
cos
p
Ep mg l l
= −
Encontramos el lagrangiano: L Ec Ep
= −
7. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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( ) ( )
2 2 2 2
1 1
2 cos cos
2 2
L Ec Ep Mx m x lx l mg l l
= − = + + + − −
( ) 2 2 2
1 1
cos cos
2 2
L M m x ml lmx mgl mgl
= + + + + −
Aplicando las ecuaciones de LaGrange:
1 ; 0
d L L d L L
U
dt x x dt
− = − =
Hallando las derivadas: 0 ; sin sin
L L
mlx mgl
x
= = − −
( ) ( ) ( )
2
cos cos sin
L d L
M m x ml M m x ml
x dt x
= + + → = + + −
( )
2 2
cos cos sin
L d L
mlx ml ml x x ml
dt
= + → = − +
Por lo tanto, las ecuaciones de movimiento de la grúa viajera son:
( ) ( )
2
1 cos sin
U M m x ml
= + + −
( ) ( )
2
0 cos sin sin sin
ml x x ml mlx mgl
= − + − − −
Despejando la aceleración de la grúa y la aceleración del cable que sostiene a la carga:
( ) ( ) ( )
2
1 cos sin cos sin sin sin
;
U ml x x x g
x
M m l
− − − − + − −
= =
+
Modelo matemático:
( )
2
1 cos sin
cos sin
U ml
x
M m
x g
l
− −
=
+
− −
=
Asumiendo que las variaciones de y son muy pequeñas se obtiene que sin
= , cos 1
= y 2
0
=
en las anteriores ecuaciones:
( )
( )
1 1
0
;
U ml U ml x g
x x
M m M m l
− − − − −
= → = =
+ +
8. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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Reemplazando x en :
( )
( )
1
1
U ml
g
M m U ml g M m
l M m l
−
− −
+ − + − +
= =
+
Aplicando
L (LaPlace):
( ) ( ) ( ) ( ) ( ) ( )
2
1
0 0
Ml s s s U s g M m s
− − = − − +
Considerando valores iniciales nulos: ( ) ( )
0 0 0
= =
( ) ( ) ( ) ( )
( )
( ) ( )
2
1 2
1
1
s
Mls g M m s U s H s
U s Mls g M m
+ + = − → = = −
+ +
Función de transferencia:
( )
( )
2
1
lM
H s
g
s M m
lM
= −
+ +
Otra forma:
Sumatoria de fuerzas en ‘x’ y ‘y’ del carrito:
( )
1 sin 1
x x Cable
F Ma Mx U F Mx
= = → + =
( )
0 cos 0 2
y Cable
F N F Mg
= → − − =
Sumatoria de fuerzas en ‘x’ y ‘y’ de la carga:
( )
1 1 1
sin 3
x x Cable
F Ma Mx F Mx
= = → − =
( )
1 1 1
cos 4
y y Cable
F ma my F mg my
= = → − =
Y
X
N
Mg
U1
FCable
𝜃
D.C.L del carro:
𝜃
FCable
mg
Y
X
D.C.L de la grúa:
9. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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De la figura de la grúa viajera tenemos:
1 1
sin ; cos
x x l y l
= + =
En ( )
3 y ( )
4 :
( )
( ) ( ) ( )
2 2
sin sin cos sin cos 5
Cable
F m x l mx lm lm
− = + − + = − +
( )
( ) ( )
2
cos cos sin 6
Cable
F mg m
− = +
( )
5 en ( ) ( )
1 6
:
( ) ( ) ( )
2 2
1 1
sin cos sin cos
U mx lm lm Mx x M m U lm lm
− + − = → + = + −
( )
( )
( ) ( )
2
2
2 2
sin cos
cos cos sin sin cos sin cos
sin
mx lm lm
mg lm l g x
− +
− − = + → + = − −
Modelo matemático:
( )
2
1 cos sin
cos sin
U ml
x
M m
x g
l
− −
=
+
− −
=
MODELOS MECÁNICOS (I – 2021)
1. (100%) Considerando un péndulo invertido, determinar:
a. Su modelo matemático.
b. La función de transferencia.
c. De forma analítica demostrar que este sistema es inestable.
u
Y
𝜃
L
x
X
m
M
10. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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D.C.L de la barra:
Centro de gravedad del péndulo:
( ) ( )
sin 1 ; cos 2
cg cg
x x L y L
= + =
Aplicando la Segunda Ley de Newton al movimiento rotacional con referencial al
centro de gravedad del péndulo:
( )
sin cos 3
cg
NETO
T I I VL HL I
= = − =
Sumatoria de fuerzas respecto al eje “x” y “y”:
( ) ( ) ( ) ( )
2 2
2 2
4 ; 5
x y
x P cg y P cg
d d
F ma H m x F ma V mg m y
dt dt
= = = − =
Reemplazando ( )
1 en ( )
4 y ( )
2 en ( )
5 :
( ) ( ) ( ) ( )
2 2
2 2
sin 6 ; cos 7
d d
H m x L V mg m L
dt dt
= + − =
El objetivo es mantener el péndulo sobre la vertical; considerando a muy pequeño, es posible linealizar
las ecuaciones ( )
3 , ( )
6 y ( )
7 al sustituir sin
= y cos 1
= :
( ) ( ) ( ) ( )
8 ; 0 9 ; 10
VL HL I V mg V mg H m x L
− = − = = = +
Reemplazando ( )
9 y ( )
10 en ( )
8 :
( ) ( ) ( )
2
11
mgL mL x L I I mL mLx mgL
− + = + + =
D.C.L. del carrito:
Sumatoria de fuerzas respecto al eje “x”:
( )
12
x C
F Ma u H Mx
= − =
Reemplazando ( )
10 en ( )
12 :
( ) ( ) ( )
13
u m x L Mx M m x mL u
− + = + + =
a. Modelo matemático:
( )
( )
2
I mL mLx mgL
M m x mL u
+ + =
+ + =
m
L
mg
H
V
𝐿
cos
𝜃
𝐿 sin 𝜃
𝜃
u
M
H
11. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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De ( )
13 :
( ) ( )
14
u mL
M m x mL u x
M m
−
+ + = =
+
Reemplazando ( )
14 en ( )
11 y debido a que la masa se encuentra posicionado en el extremo superior del
péndulo, se puede suponer que el momento de inercia es igual a cero ( )
0
I = :
( )
0
2 2
||
u mL
I mL mLx mgL mL mL mgL M m
M m
−
+ + = + = +
+
( ) ( ) ( ) ( )
1
2
||
mL M m mL u mL mgL M m mL
−
+ + − = +
( ) ( ) ( )
||
L M m u mL g M m ML g M m u L
+ + − = + − + = −
( ) ( ) ( ) ( ) ( ) ( )
0
0
2
0 0
ML s s s g M m s U s
− − − + = −
( ) ( ) ( )
( )
( ) ( )
2
2
1
s
MLs g M m s U s
U s MLs g M m
− + = − = −
− +
b. Función de transferencia:
( )
2
1 1
H s
M m
ML
s g
ML
= −
+
−
De la función de transferencia
( )
2
1 1
H s
M m
ML
s g
ML
= −
+
−
igualamos el denominador a cero para obtener los polos:
( )
2
0 .
M m M m
ML s g s g k s k cte
ML ML
+ +
− = = = =
12. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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Graficando en el Plano-s:
Como existe un valor en la región inestable, se demuestra
matemáticamente que el péndulo invertido es un SISTEMA
INESTABLE.
Analíticamente podemos decir que el péndulo invertido es
un SISTEMA INESTABLE, ya que el péndulo puede caer en
cualquier momento a menos que se aplique una fuerza de
control ( )
u adecuada.
MODELOS ELÉCTRICOS (I – 2021)
De forma completa deducir la Función de Transferencia:
1.
( ) ( ) ( ) ( ) ( )
2 2
1
1
y t Ri t i t y t
R
= → =
L.V.K.:
Malla 1:
( ) ( ) ( )
1
1 2
1
2
di
x t L i i dt
dt C
= + −
Malla 2:
( ) ( ) ( )
1 2
1
0 3
y t i i dt
C
= + −
( ) ( )
2 3
+ :
( ) ( ) ( )
1
4
di
x t L y t
dt
= +
Derivando ( )
3 :
( ) ( ) ( ) ( )
2 1 1
1 1 1
0 ' '
y t i i y t y t i
C CR C
= + − = + −
( ) ( ) ( ) ( ) ( ) ( )
1
1
1 1
' '' ' 5
di
i t Cy t y t Cy t y t
R dt R
= + → = +
Región inestable
Región inestable
Región estable
Región estable
+k
-k
𝑗𝜔
𝜎
L2 1mH
C1 1µF
R2
1kΩ
V2
12V
𝐿
𝑅
𝐶 𝑦ሺ𝑡ሻ
𝑥ሺ𝑡ሻ
+
−
𝑖1ሺ𝑡ሻ 𝑖2ሺ𝑡ሻ
13. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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( )
5 en ( )
4 :
( ) ( ) ( ) ( )
'' '
L
x t LCy t y t y t
R
= + +
Modelo matemático:
( ) ( ) ( ) ( )
'' '
RLCy t Ly t Ry t Rx t
+ + =
Aplicando
L :
( ) ( ) ( ) ( ) ( ) ( ) ( )
2
0 ' 0 0
RLC s Y s sy y L sY s y RY s RX s
− − + − + =
Considerando valores iniciales nulos: ( ) ( )
0 ' 0 0
y y
= =
( ) ( ) ( )
2
RLCs Ls R Y s RX s
+ + =
( )
( )
( )
( )
2
2
1
1 1
Y s R LC
H s H s
X s RLCs Ls R
s s
RC LC
= = =
+ +
+ +
2.
( )
0
0 2 2 2
2
1
1
de
e i dt i C
C dt
= → =
L.V.K.:
Malla 2:
( )
1 2 2 1 2 2 0 2
C R C C
V V V V R i e
= + → = +
Malla 1:
( )
1 1 1
1 1 3
i R C i C
e V V e R i V
= + → = +
( )
1 ( )
2 en ( )
3 :
( )
0
1 1 2 2 0 1 1 2 2 0 4
i i
de
e R i R i e e R i R C e
dt
= + + → = + +
De la 2da malla:
( ) ( ) ( )
2 2 1
2
1 2 2 2 2 1 2 2 2
1 2 1 2
1 1 1 1
0 0 || ' 0
R C C
di
V V V i i dt R i i dt i i R i
C C C dt C
= + − → − + + = → − + + =
R3 1kΩ R4 1kΩ
C2 1µF C3 1µF
C4 1µF C5 1µF
𝑒𝑖ሺ𝑡ሻ 𝑒0ሺ𝑡ሻ
𝑅1 𝑅2
𝐶1 𝐶2
𝑖1ሺ𝑡ሻ 𝑖2ሺ𝑡ሻ
14. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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Reemplazando ( )
1 en la anterior ecuación:
( ) ( )
2
0 0 0 0 0
1 2 2 2 2 1 1 2 1 2 2 2
1 1 2
1 1 1
0 5
de de de de d e
d
i C R C C i C C C R C
C C dt dt dt C dt dt dt
− + + = → = + +
( )
5 en ( )
4 :
( )
2
0 0
1 2 1 2 1 1 2 2 2 0
2
o
i
d e de de
e C C R R R C C R C e
dt dt
dt
= + + + +
Modelo matemático:
( )
2
0
1 2 1 2 1 1 2 2 2 0
2
o
i
d e de
C C R R R C C R C e e
dt
dt
+ + + + =
Aplicando
L :
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
2
1 2 1 2 0 0 1 1 2 2 2 0 0 0
0 ' 0 0
o i
C C R R s E s se e R C C R C sE s e E s E s
− − + + + − + =
Considerando valores iniciales nulos: ( ) ( )
0 0
0 ' 0 0
e e
= =
( ) ( ) ( ) ( ) ( )
2
1 2 1 2 0 1 1 2 2 2 0 0 i
C C R R s E s R C C R C sE s E s E s
+ + + + =
( )
( )
( ) ( )
0
2
1 2 1 2 1 1 2 2 2
1
1
i
E s
H s
E s C C R R s R C C R C s
= =
+ + + +
( )
( )
1 2 1 2
1 1 2 2 2
2
1 2 1 2 1 2 1 2
1
1
C C R R
H s
R C C R C
s s
C C R R C C R R
=
+ +
+ +
MEDIDA DE LA RESISTENCIA Y LA LEY DE OHM (I – 2021)
1. (20%) De forma completa deducir la Ley de Ohm y la Ecuación Dimensional del Ohmio.
( )
1
J E
=
Pero sabemos que
1
;
i V
J E
A L
= = = en ( )
1 :
1
i V L
V i V Ri
A L A
= → = =
15. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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Ecuación Dimensional de Ohmio:
2
2
2 2 2 3 2
2
P W J Nm m kg
P i R R
A sA sA s A
i
= → = → = = = =
2 1 3 2
L M T I
− −
=
2. (50%) Un fluido de densidad fluye por un conjunto de dos tuberías horizontales de diámetros D y d
( )
D d
, las velocidades en ambas tuberías son constantes e iguales a 1
v y 2
v respectivamente.
Determinar:
Una ecuación que permita calcular la Resistencia Hidrodinámica del sistema de tuberías.
Utilizando como referencia su resultado anterior, determinar las unidades de la Resistencia Hidrodinámica
en función de las unidades fundamentales del Sistema Internacional.
a)
Resistencia hidrodinámica:
( )
1
H
P R Q
=
Ecuación de Bernoulli:
2 2
1 1 1 2 2 2
1 1
2 2
P gh v P gh v
+ + = + +
Tubería horizontal: 1 2
h h
=
( )
( ) ( )
2 2 2 1 1 1
2 2 2 2 1 1
1 2 2 1
2 1 2 1
2 2 2
v v A v v A v Q v Q
P P v v
A A A A
− = − = − = −
Caudal: 1 2
Q Q Q
= =
( )
2 1 2 1 2 1
1 2 2 2 2 2
2 2
2 2
2
2
4 4
v v v v v v
P P Q Q P Q
d D d D
d D
− = − = − → = −
Comparando ( )
1 y ( )
2 :
2 1
2 2
2
H
v v
R
d D
= −
𝐷 𝑑
𝑣1 𝑣2
16. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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b)
3
2 5 4
1
1
H H
Kg m
mKg Kg
m s
R R
m m s m s
= = =
3. De forma completa y sin utilizar resultados conocidos deducir las unidades de la Resistencia Térmica en
el Sistema Internacional. Luego realizar una analogía Eléctrica – Mecánica – Hidráulica.
T H x
H kA k
x A T
= → =
2 3
W m J N kg m
k
s m K s K
m K s K
= = = =
T
H x x
T T H R H
kA kA
= → = =
3
2
2
3
T T
x m s K
R R
kg m
k A kg m
m
s K
= = =
Analogía eléctrica – mecánica – hidráulica:
E
V
i
R
=
T
T
H
R
=
H
P
Q
R
=
R1
1kΩ
∆𝑉 𝑅𝐸 𝑖ሾ𝐴ሿ
R1
1kΩ
∆𝑇 𝑅𝑇 𝐻ሾ𝑊ሿ
R1
1kΩ
∆𝑃 𝑅𝐻 𝑄ሾ𝑚3
𝑠
Τ ሿ
17. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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EXAMEN DE TEORÍA Y SIMULACIÓN EN MULTISIM (I – 2021)
1. Aplicando el método matricial de las tensiones en los nodos, calcular la corriente que circula por la
resistencia de 3.
Teórica:
11 12 1 1
2 2 2 1 2 1
21 22 2 2
G G V i
G V I
G G V i
= → =
1
2
1 1 1 1 10 20
5 2 3 3 5 3
1 1 1 1 20
3 3 4 2 3
V
V
+ + − +
=
− + + −
1
2
860
7.107
121
3.967
480
121
V
V
V
V
V
V
=
−
−
1 2 20 3 1 2 3
0 20 3 0
V
V V V V V V i
− − − = → − − − =
1 2
3 3
860 480
20
20 360
121 121
2.975
3 3 121
V V
i i A A
− − −
− −
= = = − −
Experimental:
R1
5Ω
R2
3Ω
R3
2Ω
R4
4Ω
V1
10V
V2
20V
R5
2Ω
XMM1
XMM2
XMM3
R1
5Ω
R2
3Ω
R3
2Ω
R4
4Ω
V1
10V
V2
20V
R5
2Ω
𝑉1 𝑉2
𝑉3
𝑖3𝛺
18. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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2. En el circuito mostrado calcular la potencia máxima que el circuito activo puede suministrar a la resistencia
variable 4
R .
Teórica:
Hallando ab
R ( ab Th N
R R R
= = ):
R1
25Ω
V1
36V
R2
20Ω
R3
15Ω
V2
48V
R4
1kΩ
Key=A
50 %
R1
25Ω
V1
36V
R2
20Ω
R3
15Ω
V2
48V
𝑎
𝑏
𝑖
19. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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1
3
1 2
1 1
ab
R R
R R
−
= + +
1
1 1
15 26.111
25 20
ab ab
R R
−
= + + =
Hallando la corriente:
( )
1
1 2 1
1 2
36
0.8
25 20
V
R R i V i i A
R R
+ = − → = − = − =
+ +
Hallando ab
V :
1ra forma: ( )
20 36 48 20 0.8 36 48 28
ab V V ab
V V V V V V
= − + = − + =
2da forma: ( )
25 48 25 0.8 48 28
ab V ab
V V V V V
= − + = − + =
Para potencia máxima:
4 26.111
ab
R R
= =
( )
max max
4
28
7.506
4 4 26.111
ab
V
P P W
R
= = =
Experimental:
R1
25Ω
V1
36V
R2
20Ω
R4
26.111Ω
R3
15Ω
V2
48V
XWM1
V
I
25Ω
R2
20Ω
R3
15Ω
20. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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3. Las constantes R y L de una bobina se pueden obtener conectándola en serie con una resistencia
conocida
0 10
R = y midiendo la tensión en la bobina x
V , en la resistencia 0
V y la tensión total T
V . Si se
conoce que:
22.4
x
V V
= ,
0 20
V V
= ,
36
T
V V
= . Calcular las constantes R y L de la bobina.
Teórica:
Hallando la corriente del circuito:
0
0
20
2
10
V
i i A
R
= = → =
Hallando las impedancias:
22.4
11.2
2
x
x x x x
V
V Z i Z Z
i
= → = = → =
36
18
2
T
T T T x
V
V Z i Z Z
i
= → = = → =
Pero también sabemos que las impedancias en su forma vectorial son:
( )
0 0 0
; 0 ;
x T
Z R jX Z R j Z R R jX
= + = + = + +
Forma escalar:
( )
2 2 2 2
2 2 2 2 2 2 2
0 0 0 0
; ;
x x T T
Z Z R X Z Z R Z Z R R X
= = + = = = = + +
Restando 2 2
T x
Z Z
− :
( )
2
2 2 2 2 2 2 2 2 2 2
0 0 0
2
T x
Z Z R R X R X R RR R X R X
− = + + − − = + + + − −
R0
10Ω
L
32.014mH
V1
36Vrms
50Hz
0°
R
4.928Ω
𝑖
𝑉0
𝑉
𝑥
21. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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( )
2 2 2 2 2 2
2 2 2 0
0 0
0
18 11.2 10
2 4.928
2 2 10
T x
T x
Z Z R
Z Z RR R R R
R
− − − −
− = + → = = =
De 2 2 2
x
Z R X
= + :
2 2 2 2
11.2 4.928 10.058
x
X Z R X
= − = − → =
( )
10.058
2 32.014
2 2 50
X
X fL L L mF
f
= → = = =
Experimental:
R0
10Ω
L
32.01425571mH
V1
36Vrms
50Hz
0°
R
4.928Ω
XMM1
XMM2
22. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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4. Dos ramas de impedancias
15 y ( )
8 2
j
− se conectan en paralelo. Calcular la potencia activa que
consumen las dos resistencias, si la potencia activa total consumida en el circuito es
2000 W .
1 1 2 2
V i Z i Z
= =
( )
2
2
1 2
2 1
8 2 2 17
15 15
i Z
i Z
+ −
= = =
( )
2 2
1 1 1 1
15 1
P R i i
= =
( )
2 2
2 2 2 2
8 2
P R i i
= =
( )
( )
1
2
:
2
2
1 1
2 2
15 15 2 17 17
8 8 15 30
P i
P i
= = =
R1
15Ω
C1 1.592mF
V1
104.163Vrms
50Hz
0°
R2
8Ω
𝑖1
𝑖2
−𝑗2
23. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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( )
( )
1
2 1
1 2 2
34000
30 723.404
3 47
17
60000
2000 4 1276.596
47
P W W
P P
P P P W W
=
=
+ = =
Experimental:
Cálculos adicionales:
2 1
1 1 1 1
1
34000
17
47 20
15 141
P
P R i i A A
R
= → = = =
( )
1 1
17 17
20 15 300 104.1684397
141 141
V i Z V V
= = → = =
Z1
Z=A-jB
R1
15Ω
V1
104.1684397Vrms
50Hz
0°
XWM1
V
I
XWM2
V
I
XWM3
V
I
25. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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RECUPERATORIOS
RECUPERATORIO (I-2021)
Determinar la función de transferencia y el diagrama de bloques de:
1. (Amplificador Operacional)
0
1 2 3 1
1 3
; ;
i A A A
e e e e de
i i i C
R R dt
− −
= = =
0
4 5 2
2
;
A de
e
i i C
R dt
= = −
Nodo B:
( )
0 0
4 5 2 2 2
2
1
A
A
de de
e
i i C e R C
R dt dt
= → = − → = −
Nodo A:
( )
0 0
1 2 3 4 1 1
1 3 2 1 1 2 3 3
1 1 1
2
i A A i
A A A
A
e e e e e e
de e de
i i i i C e C
R R dt R R R R R R dt
− −
= + + → = + + → = + + − +
( )
1 en ( )
2 :
2
1 2 1 3 2 3 0 0
2 2 0 2 1 2 2
1 1 2 3 3
1
i
e R R R R R R de d e
R C e R C C
R R R R dt R dt
+ +
= − − −
Modelo matemático:
2
0 1 2 1 3 2 3 0
2 1 2 2 0
2
1 3 3 1
1 1
i
d e R R R R R R de
R C C C e e
R R dt R R
dt
+ +
+ + = −
U1
OPAMP_3T_VIRTUAL
C6 1µF
C7 1µF
R5
1kΩ
C8
1µF
R6
1kΩ
R7
1kΩ
C9
1µF
𝑅1 𝑅2
𝑅3
𝐶1
𝐶2
𝑒𝑖ሺ𝑡ሻ 𝑒0ሺ𝑡ሻ
Amplificador Operacional
R8 1kΩ C10 1µF C11 1µF
C12 1µF
𝑅
𝐶 𝑒0ሺ𝑡ሻ
𝑒𝑖ሺ𝑡ሻ
Circuito RC Serie
U1
OPAMP_3T_VIRTUAL
C6 1µF
C7 1µF
R5
1kΩ
C8
1µF
R6
1kΩ
R7
1kΩ
C9
1µF
𝑅1
𝑅2
𝑅3
𝐶1
𝐶2
𝑒𝑖ሺ𝑡ሻ 𝑒0ሺ𝑡ሻ
𝑖1ሺ𝑡ሻ
𝑖2ሺ𝑡ሻ
𝑖3ሺ𝑡ሻ
𝑖4ሺ𝑡ሻ
𝑖5ሺ𝑡ሻ
26. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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Aplicando
L (LaPlace):
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 1 2 1 3 2 3
2 1 2 0 0 0 2 0 0 0
1 3 3 1
1 1
0 ' 0 0 i
R R R R R R
R C C s E s se e C sE s e E s E s
R R R R
+ +
− − + − + = −
Considerando valores iniciales nulos: ( ) ( )
0 0
0 ' 0 0
e e
= =
( ) ( ) ( ) ( )
2 1 2 1 3 2 3
2 1 2 0 2 0 0
1 3 3 1
1 1
i
R R R R R R
R C C s E s C sE s E s E s
R R R R
+ +
+ + = −
( )
( )
( )
0 1
2 1 2 1 3 2 3
2 1 2 2
1 3 3
1
1
i
E s R
H s
E s R R R R R R
R C C s C s
R R R
−
= =
+ +
+ +
Función de transferencia:
( ) 1 2 1 2
2 1 2 1 3 2 3
1 2 3 1 2 3 1 2
1
1
R R C C
H s
R R R R R R
s s
R R R C R R C C
= −
+ +
+ +
De:
2
0 1 2 1 3 2 3 0
2 1 2 2 0
2
1 3 3 1
1 1
i
d e R R R R R R de
R C C C e e
R R dt R R
dt
+ +
+ + = −
2 2
0 1 2 1 3 2 3 0 0 0
0 0
2 2
1 2 3 1 2 3 1 2 1 2 1 2
1 1
i i
d e R R R R R R de d e de
e e A Be Ce
R R R C dt R R C C R R C C dt
dt dt
+ +
+ + = − → + + = −
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
1 0 1 0 2
2 0 2 0 0 0
'
' '' ' i
x t e t x t e t x t
x t e t x t e t Ae t Be t Ce t
= = =
= = = − − −
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( )
( )
( )
( )
1 1 2 1 1
2 0 0 2 2
0 0 1 0
' i
i
x t x t x t x t x t
e t
x t Ae t Be t Ce t x t x t
A B C
= +
= −
= − − − − −
Ec. de salida:
( )
( )
1
2
0 1
x t
y
x t
=
27. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
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Diagrama de Bloques:
Otra forma:
De:
( ) ( )
( )
0
2 2 2 2 0
0 1 2 1 3 2 3
1 1 1 1 1 1
1 2 3 3 2 3
||
1 1 1
||
A A
A
i A i
de
e R C L E s R C sE s
dt
e R R R R R R
de
e e R R R C L E s R C s
R R R R dt R R
= − → = −
+ +
= + + − + → = +
( ) ( )
1
0
3
A
R
E s E s
R
−
( ) ( )
0
2 2
1
A
E s E s
R C s
= −
( ) ( ) ( )
2 3 1 2
0
1 2 3 1 1 2 1 3 2 3 1 2 3 1 1 2 1 3 2 3
A i
R R R R
E s E s E s
R R R C s R R R R R R R R R C s R R R R R R
= +
+ + + + + +
𝑒𝑖 𝑒0
A
B
C
_
_
+
+
𝑥′2 𝑥2 𝑥1
𝑒𝑖 𝑒0
𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3
𝑅1𝑅2𝑅3𝐶1
1
𝑅1𝑅2𝐶1𝐶2
1
𝑅2𝑅3𝐶1𝐶2
−
−
+
+
𝑥′2 𝑥2 𝑥1
28. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
26
Diagrama de Bloques:
2. Circuito RC serie
( )
( )
( )
0
0
1
1
de t
e t idt i C
C dt
= → =
L.V.K.:
( ) ( )
1
2
i
e t Ri idt
C
= +
( )
1 en ( )
2 :
( )
( )
( )
0
0
i
de t
e t RC e t
dt
= +
Aplicando
L (LaPlace):
( ) ( ) ( ) ( ) ( )( ) ( )
' 0 1
i o i
E s RC sE s e E s E s RCs E s
= − + → + =
Considerando condiciones iniciales nulas: ( )
0 0
o
e =
( )( ) ( ) ( )
( )
( )
0 1
1
1
i
i
E s
E s RCs E s H s
E s RCs
+ = → = =
+
Función de Transferencia:
( )
1
1
RC
H s
s
RC
=
+
−
1
𝑅2𝐶2𝑠
𝑅2𝑅3
𝑅1𝑅2𝑅3𝐶1𝑠 + 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3
𝑅1𝑅2
𝑅1𝑅2𝑅3𝐶1𝑠 + 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3
𝐸𝑖ሺ𝑠ሻ 𝐸𝐴ሺ𝑠ሻ 𝐸0ሺ𝑠ሻ
+
+
R8 1kΩ C10 1µF C11 1µF
C12 1µF
𝑅
𝐶 𝑒0ሺ𝑡ሻ
𝑒𝑖ሺ𝑡ሻ 𝑖ሺ𝑡ሻ
29. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
27
De:
( )
( )
( )
( )
( ) ( )
0 0
0 0
1 1
i i
de t de t
e t RC e t e t e t
dt dt RC RC
= + → + =
( )
( )
( ) ( )
0
1 0 1 ; i
de t
x e t x u t e t
dt
= → = =
( )
1 1
1 1
x x u t
RC RC
= − +
Diagrama de Bloques:
Otra forma:
De:
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
0 0
0 0
1 1
||
||
1
||
R R
i R i R
e t i t dt L E s I s
C Cs
v t Ri t L V s RI s
e t V e t L E s V s E s R I s
Cs
= → =
= → =
= + → = + = +
Función de Transferencia:
( )
( )
( )
( )
( )
( )
0
1 1
1
1
i
I s
E s Cs RC
H s H s
E s
s
R I s
RC
Cs
= = =
+
+
De:
( ) ( ) ( ) ( ) ( ) ( ) ( )
0 0
1 1
; ;
R i R
V s E s E s I s V s E s I s
R Cs
= − = =
𝑒𝑖ሺ𝑡ሻ 𝑒0ሺ𝑡ሻ
𝑥′1 𝑥1
1
𝑅𝐶
1
𝑅𝐶
+
−
30. LABORATORIO DE CIRCUITOS ELÉCTRICOS [LELT – 240] | [MAY BER] 77712474
28
Diagrama de Bloques:
𝐸𝑖ሺ𝑠ሻ 𝐸0ሺ𝑠ሻ
𝑉𝑅ሺ𝑠ሻ 𝐼ሺ𝑠ሻ
1
𝑅
1
𝐶𝑠
+
−