1. 51.
    0
    ibibik gibicem pilotmusun steward mı
    ···
  2. 52.
    0
    BRIEFING GUIDE FOR FLIGHT OPERATIONS
    INTRODUCTION
    Good morning / good afternoon / good day / good night to every one.
    First of all I would like to inform you that the communication between the flight crew and cabin crew will be in English.
    Our captains are...
    I would like to ( assign / appoint / station / distribute) the (numbering / numeration) of the crew according to our checklist duties;
    Mr/Mrs ... is the Cabin Chief 1
    ... cabin attendant No: 1
    Mr/Mrs ... is the Cabin Chief 2
    ... cabin attendant No:2
    We have ... F/C, ... B/C, ... C/C ,... Y/C passengers on board today.
    The details concerning our destination are stated / addressed on the line-based information documents.
    Please pay attention and be open to communication when / while interacting with the ground staff, crew members and passengers. Please use polite / appropriate professional and friendly language / sentences.
    The flight chefs duties are defined by the flight chef working procedure.
    We have an inexperienced cabin crew member so we will all help and guide him / her during the flight.
    We have a flight attendant who speaks ... language. He / she will translate all the announcements in the aircraft during the flight.
    All cabin attendants are responsible to provide / ensure flight safety and passenger / customer satisfaction.
    Cabin attendants are also responsible to maintain / execute / display and improve the quality and standarts of Turkish Airlines.
    Please make our passengers / customers feel valued / important / to us and well taken care of by showing good hospitality and polite manners.
    Be helpful and solve / approach the problems with a polite and friendly manner.
    ···
  3. 53.
    0
    COMPANY AND AVIATION RULES
    All Cabin crew are responsible of all general and additional provisions of Cabin Crew Manuel.
    All Cabin crew are responsible for updating and providing documents and their certificates.
    The commander has the authority to give the final decisions and each person on board the aircraft must obey all the orders and directives given by the commander.
    It should be considered that the Civil Aviation Authorities may carry out routine controls by coming to the airplane at any airport. Assist / accompany the staff at all times. (SAFA/SANA)

    EMERGENCY
    Emergency and technical checks / controls must be done / carried out according to the check list.
    Cabin crew must have full information of every emergency equipment according to the Emergency Equipment Plan (e.g. Med O2, Dr. Kit ... )
    The cleaning and catering personnel must be under surveillance at all times.
    Cabin crew awareness and being alert for unplanned, abnormal and unexpected emergency situations is very important.
    Cabin crew must check and match the ID cards of all staff entering the aircraft.
    Doctor and turbulence announcements must be made by any crew member within 20 seconds when necessary.
    Every cabin crew must join the preflight security search check and preflight must be done effectively. (ASSCL)
    Following the announcement cabin attendant No: 2 must perform the emergency exit briefing by using a safety card and report after completion.
    ···
  4. 54.
    0
    CATERING
    Cabin crew should coordinate with the catering staff and receive supplies in accordance with the catering manuel.
    Information concerning catering should be reported to the cabin chief.
    Speacial meals must be checked by referring the passenger list.
    Baby meals and child kits must be offered to related passengers.
    Meal and drink alternatives / choices must be presented / offered to the passengers.
    Cabin crew must serve the passenger with a polite behavior and a sincere smile.
    All food and drinks should be served with a napkin.
    Cabin crew must work safely, neatly and be silent in the galley areas.
    CLEANING
    Cleaning check must be done according to the checklist.
    Cabin crew should check that all standart cleaning material is delivered.
    All cleaning materials must be stored in their standart original places.
    Cabin crew are responsible for keeping the cabin, galley, crew rest and toilet areas clean during the entire flight. Check the cockpit rubbish periodically.
    Cleaning materials should be taken out for the cleaning staff for easy access.
    LAYOVER
    Cabin crew must obey the layover rules of our company.
    Commander and the cabin chief must at all times be informed if anyone leaves the hotel or if anyone gets ill / sick.
    Cabin crew behavior and manner of dressing should comply with our company rules.
    Cabin crew must be at the lobby area ten minutes before pick up time with an orderly fashion and a fresh look.
    Cabin crew should be in communication / contact with each other and the cockpit crew.
    ···
  5. 55.
    0
    EXTENTED RANGE
    Cabin crew must leave the cabin, galley, toilet areas and crew rest areas secure and clean. All material / stock should be placed according to their standart loading formation.
    During the in-between service, cold drinks will be served on a tray, and hot drinks will be served from the galleys, if asked for.
    Cabin crew shouldn’t eat their meals at the same time and never leave the cabin unattended.
    Cabin crew should inform / brief each other before going to the crew rest.
    The crew rest must be left neat and tidy after being used.
    Cabin crew should leave the crew rest looking fresh and ready.
    OTHER
    Cabin crew must at all times comply with the company uniform regulations and overall apperance.
    Customs and entry forms should be handed out to passengers before the service.
    During the flight, cabin crew should help passengers to fill out the forms if requested.
    Any special passenger information (e.g. UM, WCHR, Dr... ) should be shared by the cabin crew and special care and attention should be provided.
    Cabin crew must be sure that all infants are given loopbelts and collect them at the end of each flight.
    If there is an infant on board, baby bassinets should be placed and parents should be given instruction on safety issues. Remove the bassinet after the descending announcement.
    Cabin attendant No: 1 is responsible for filling out the delivery report form.
    Cabin crew must carry out their Cabin OK / Ready duties effectively and on time.
    Before descent cabin defects concerning maintenance and any other information concerning the whole flight should be reported to the cabin chief..
    Have a nice flight...
    ···
  6. 56.
    +1
    State-space systems
    Representations: From transfer function to state-space.
    Consider a linear, shift invariant, system:
    ! P(z) !
    y(k) u(k)
    We can express this as a transfer function,
    y(z) = P(z) u(z) =
    b(z)
    a(z)
    u(z)
    where a(z) and b(z) are polynomials, so,
    P(z) =
    b(z)
    a(z)
    =
    b0zm + b1zm−1 + · · · + bm
    zn + a1zn−1 + · · · + an
    .
    For causal systems the order of b(z) is less than or equal to the order of a(z). So m " n above.
    Assume for now that m < n,
    Roy Smith: ECE 147b 8: 2
    State-space systems
    Approach:
    • Represent the plant, P(z) (or P(s)) as an nth order differential equation.
    • Represent nth order differential equation as a 1st order matrix differential equation with
    dimension n.
    • Design methods now involve linear algebra.
    • Easy to handle large systems (with Matlab).
    • Easy to handle systems with multiple inputs and outputs.
    • Easy to simulate systems.
    x(k + 1) = Ax(k) + B u(k),
    y(k) = C x(k) +Du(k).
    A, B, C and D can be matrices. x(k) is a vector (state vector).
    Roy Smith: ECE 147b 8: 1
    State-space systems
    Chain of delays:
    Consider the first equation: !(z) =
    1
    a(z)
    u(z)
    We want to develop a chain of delay model to get !(k):
    First step: Write the expression for n delays:
    !(k) = z−1 !(k + 1)
    ...
    ...
    !(k + n − 1) = z−1!(k + n)
    In pictures ...
    " z−1 ! ! z−1 ! z−1 "
    !(k) !(k + 1) !(k + n − 2) !(k + n − 1) !(k + n)
    Second step: Express !(k + n) in terms of !(k), . . . , !(k + n − 1) and u(k).
    Roy Smith: ECE 147b 8: 4
    State-space systems
    Outline:
    1. Draw the system as an interconnected “chain of delays”,
    2. Relabel the signals in the system,
    3. Rewrite the input/output equations in terms of the new signals,
    4. Abbreviate the equations to a matrix form (state-space).
    Drawing a digital system block diagram in terms of delays is exactly the same as drawing a
    continuous system block diagram in terms of integrators.
    Split the system,
    b(z)
    1
    a(z)
    " ! "
    y(z) !(z) u(z)
    !(z) =
    1
    a(z)
    u(z) and y(z) = b(z) !(z).
    Roy Smith: ECE 147b 8: 3
    State-space systems
    Block diagram of: !(z) =
    1
    a(z)
    u(z)
    !(k + n) = u(k) − (a1!(k + n − 1) + · · · + an!(k)) .
    z−1 z−1
    a1
    an−1
    an
    ! ! ! ! ! #$
    %& ' +
    " "
    1. $
    %& +
    2. −
    '
    " "
    3. $
    %& +
    4.
    '
    "
    5.
    !(k) !(k + 1) !(k + n − 1) !(k + n) u(k)
    Roy Smith: ECE 147b 8: 6
    State-space systems
    Second step: Express !(k + n) in terms of !(k), . . . , !(k + n − 1) and u(k).
    To do this, write this as: !(z) a(z) = u(z),
    Expanding a(z) gives,
    !zn + a1zn−1 + · · · + an" !(z) = u(z)
    and expressing this in terms of the highest power of z gives:
    zn!(z) = u(z) − !a1zn−1 + · · · + an"!(z).
    Now write this is the time domain,
    !(k + n) = u(k) − (a1!(k + n − 1) + · · · + an!(k)) .
    This is the form we need to include in our block diagram.
    Roy Smith: ECE 147b 8: 5
    “Chain of delays” block diagram
    bm
    bm−1
    b0
    z−1 z−1
    a1
    an−1
    an
    ( ! ! ! ! ! ( )*
    +, + )*
    +, + -
    " "
    )*
    +, +
    6. −
    -
    " "
    )*
    +, +
    7.
    -
    "
    8.
    ! !
    )*
    +, +
    $
    -
    !
    $
    u(k)
    !(k + n − 1) !(k + n)
    !(k + 1)
    !(k + m)
    y(k) !(k)
    This could actually be constructed from shift registers, summers and multipliers.
    Roy Smith: ECE 147b 8: 8
    State-space systems
    Numerator term: y(z) = b(z) !(z)
    Expanding the b(z) polynomial gives,
    y(z) = b(z) !(z),
    = !b0zm + b1zm−1 + · · · + bm" !(z)
    and in the time domain this is,
    y(k) = b0!(k + m) + b1!(k + m − 1) + · · · + bm!(k).
    As m < n all of the signals !(k) to !(k + m) are available along to the top of the chain of
    delays.
    So y(k) is just a linear combination of signals from the previous block diagram.
    Roy Smith: ECE 147b 8: 7
    State-space representations
    Developing the matrix equations
    Relabel our intermediate variables,
    x1(k) = !(k + n − 1)
    x2(k) = !(k + n − 2)
    ...
    ...
    xn(k) = !(k)
    Now work out what happens to each of them at time k + 1:
    x1(k + 1) = !(k + n) =u(k) − (a1!(k + n − 1) + · · · + an!(k))
    = u(k) − (a1x1(n) + · · · + anxn(k))
    x2(k + 1) = !(k + n − 2 + 1) = x1(k)
    ... ...
    ...
    xn(k + 1) = !(k + 1) = xn−1(k)
    Now look at this as a matrix multiplication.
    Roy Smith: ECE 147b 8: 10
    State-space representations
    What if m = n ?
    Remove the highest order terms by polynomial division,
    b(z)
    a(z)
    = d +
    ˆb
    (z)
    a(z)
    .
    Now d is a constant and ˆb(z) has order m − 1 < n.
    We deal with
    ˆb
    (z)
    a(z)
    exactly as before and then add the constant, d, to the result.
    In pictures . . .
    ˆb
    (z)
    a(z)
    d
    ! ! ! ' !
    !
    9. )*
    +, +
    y(k) u(k)
    Roy Smith: ECE 147b 8: 9
    State-space representations
    Matrix equations
    x(k + 1) = Ax(k) + B u(k),
    where
    A =
    10. $$$%
    −a1 −a2 · · · −an−1 −an
    1 0 · · · 0 0
    ...
    . . . . . . ...
    0 · · · 1 0
    &(
    and B =
    11. $$$%
    1
    0...
    0
    &
    (
    This is the “state-update” equation (or sometimes just the “state”equation).
    What about the output y(k)?
    y(k) = b0!(k + m) +b1!(k + m − 1) · · · +bm!(k)
    = b0xn−m(k) +b1xn−m−1(k) · · · +bmxn(k)
    = C x(k)
    where C = )0 . . . 0 b0 . . . bm*. If m = n − 1 then there are no leading zeros in C.
    Roy Smith: ECE 147b 8: 12
    State-space representations
    Matrix equations
    x1(k + 1) = −a1 x1(k) −a2 x2(k) · · · −an−1 xn−1(k) −an xn(k) + u(k)
    x2(k + 1) = x1(k)
    x3(k + 1) = x2(k)
    ...
    xn(k + 1) = xn−1(k)
    or . . .
    12. $$$%
    x1(k + 1)
    x2(k + 1)
    ...
    xn(k + 1)
    &(
    =
    13. $$$%
    −a1 −a2 · · · −an−1 −an
    1 0 · · · 0 0
    ...
    . . . . . . ...
    0 · · · 1 0
    &
    (
    14. $$$%
    x1(k)
    x2(k)
    ...
    xn(k)
    &(
    +
    15. $$$%
    1
    0...
    0
    &
    (
    u(k)
    Define the state: x(k) =
    16. $$$%
    x1(k)
    x2(k)
    ...
    xn(k)
    &'''(
    to get the final equations.
    Roy Smith: ECE 147b 8: 11
    State-space systems
    Other domains:
    Discrete time,
    time invariant:
    x(k + 1) = Ax(k) + B u(k),
    y(k) = C x(k) + Du(k)
    Continuous time,
    time invariant:
    dx(t)
    dt
    = Ax(t) + B u(t),
    y(t) = C x(t) + Du(t)
    Nonlinear,
    time invariant:
    dx(t)
    dt
    = f(x(t), u(t)),
    y(t) = g(x(t), u(t))
    Nonlinear,
    time varying:
    dx(t)
    dt
    = f(t, x(t), u(t)),
    y(t) = g(t, x(t), u(t))
    Roy Smith: ECE 147b 8: 14
    State-space systems
    Equal numerator and denominator order case:
    ˆb
    (z)
    a(z)
    d
    ! ! ! ' !
    !
    17. )*
    +, +
    y(k) yˆ(k) u(k)
    We can calcuate the state-space representation for
    ˆb
    (z)
    a(z)
    :
    x(k + 1) = Ax(k) + B u(k),
    ˆy(k) = C x(k),
    and as, y(k) = ˆy(k) + du(k), we have,
    x(k + 1) = Ax(k) + B u(k),
    y(k) = C x(k) + du(k).
    Roy Smith: ECE 147b 8: 13
    State-space examples:
    Thermal system:
    Suppose that T1(t) is the output of interest and e(t) and T0(t) are both inputs.
    Now select state variables which will allow us to put this into the generic state-space matrix
    equation form.
    Try x(t) = +T1(t)
    T2(t),.
    Then, rearranging gives,
    dT1(t)
    dt
    =
    −k12
    m1c1
    T1(t) +
    k12
    m1c1
    T2(t)
    dT2(t)
    dt
    =
    k12
    m2c2
    T1(t) +
    −k12 − k20
    m2c2
    T2(t) +
    k20
    m2c2
    T0(t) +
    1
    m2c2
    e(t)
    So,
    A =#
    $$%
    dT1(t)
    dt
    dT2(t)
    dt
    &(
    = A +T1(t)
    T2(t), + B +T0(t)
    e(t) ,, where A =
    18. $$$%
    −k12
    m1c1
    k12
    m1c1
    k12
    m2c2
    −k12 − k20
    m2c2
    &
    '(
    Roy Smith: ECE 147b 8: 16
    State-space systems
    Examples: A thermal control system:
    Heater
    terminals
    e(t)
    T (t)
    T (t)
    T (t)
    2
    0
    1
    Ti(t) temperature of volume i
    e(t) energy into the volume 1
    mi mass of volume i
    ci specific heat for volume i
    kij thermal conductance for i, j interface
    We can derive the state-space representation directly from the thermal energy equations.
    For volume 1: m1c1
    dT1(t)
    dt
    = k12(T2(t) − T1(t))
    For volume 2: m2c2
    dT2(t)
    dt
    = −k12(T2(t) − T1(t)) − k20(T2(t) − T0(t)) + e(t)
    Roy Smith: ECE 147b 8: 15
    State-space examples:
    DC Motor connected to a rotational load
    L R
    J
    V Vemf in
    +
    +


    "
    i
    19.
    20. f
    The “back emf” is proportional to motor speed: Vemf = K
    d"
    dt
    .
    The motor has a series resistance, R, and inductance, L, so Vin = Vemf + L
    d i
    dt
    + Ri.
    The motor torque is proportional to the motor current: # = K" i.
    There is a friction torque, #f opposing the motor and proportional to its speed: #f = b
    d"
    dt
    .
    The rotational load has inertia J so the torque balance equation is: J
    d2 "
    dt2 + #f = #.
    Roy Smith: ECE 147b 8: 18
    State-space examples
    Exercises:
    1. What is B?
    2. What is C?
    3. What is D?
    4. What would C and D be if we had both T1(t) and T2(t) as outputs?
    5. Calculate the transfer functions from e(t) and T0(t) to T1(t).
    Roy Smith: ECE 147b 8: 17
    State-space examples:
    DC Motor connected to a rotational load
    We are interested in the model from the input voltage, Vin, to the rotor angle, ".
    Exercise:
    1. How many states are required for this model?
    2. Derive a state-space representation.
    3. What are the poles of the system?
    Roy Smith: ECE 147b 8: 19
    Tümünü Göster
    ···
  7. 57.
    0
    TÜM KABiN EKiPLERiNiN DiKKATiNE

    22. DÖNEM TOPLU iŞ SÖZLEŞMESi MADDE:99 3. GÖREV SINIRLANDIRMALARI D FIKRASI iLE iLGiLi AÇIKLAMA
    “UÇUCU EKiPLER ANA ÜSTE GÜN TANIMI SÜRESi iÇiNDE SADECE TEK BiR GÖREV SÜRESi VEYA DiĞER GÖREVLERLE GÖREVLENDiRiLEBiLiR. UÇUCU EKiPLERiN YEDEK EKiP REZERV (RM) GÖREVLERi YEREL SAATLE EN GEÇ SAAT 18.00’E KADAR YAYINLANMAK ZORUNDADIR. BU SAATTEN SONRA YAYINLANAN REZERV GÖREVLERi UÇUCU EKiPLERi BAĞLAMAZ.” ŞEKLiNDEDiR.
    BURADA AMAÇLANAN, GÜN iÇERiSiNDE ÇEŞiTLi NEDENLERLE OLUŞAN SEFER DEĞiŞiKLiKLERiNDEN EKiPLERiMiZiN ZAMANINDA BiLGi SAHiBi OLMASIDIR. BUNU SAĞLAMAK ÜZERE iŞLETME KOORDiNASYON BAŞKANLIĞI PLANLAMAYI TAMAMLAYARAK EN GEÇ SAAT 18:00’DE IVR SiSTEMiNDE YAYINLAMAKTADIR .
    BU NEDENLE; SiZLERiN SAAT 18.00’DEN ÖNCE, RZV GÖREVi iLE iLGiLi OLARAK IVR SiSTEMiNi ARAMANIZ DURUMUNDA; PLANLANMIŞ AMA HENÜZ IVR SiSTEMiNDE YAYINLANMAMIŞ GÖREVLERiNiZ iLE iLGiLi BiLGi ALAMAYACAĞINIZ BiLiNMELiDiR. RZV’DEN YAYINLANMIŞ HERTÜRLÜ GÖREViN ÖĞRENiLMESi, SAAT 18.00’DEN SONRA YAPILACAK ARAMALAR iLE MÜMKÜN OLUP, SiSTEM TARAFINDAN KAYITLARI TUTULMAKTADIR
    ···
  8. 58.
    -1
    nabıyonuz a.q ya,gidin başka yerde oynayın şimdi gelip kesicem topunuzu ha
    ···
  9. 59.
    0
    Surface Preparation and Volume Meshing

    * New "Repair Feature" mode in the Surface Repair Tool that makes it possible to fix problems in feature curves. This includes automatic detection of errors, a browser to move quickly between them, and automatic repair.
    * Multi-Region Imprint mode for automatic interfacing
    o Automatic detection of candidate pairs
    o Option to browse and imprint in sequence, or to imprint all
    o Imprint operation creates interfaces between regions
    * New parameter, "layer reduction percentage", which enables automatic reduction of layers to improves mesh quality in tight cavities.
    * Generalized Cylinder Mesher (optional with Polyhedral Mesher). This is a new volume mesher that will create extruded meshes in regions that can be characterized as generalized cylinders. Automatic detection of such regions is included.
    * JT Open import format available on Windows
    * The Select Objects dialog for selecting parts for regions etc has been improved to allow more advanced selection of objects, including better search capabilities
    * New Outward Subsurface option in extruder mesher which allows an alternative method of generating an extruded volume mesh
    * New interactive boundary selection method for trimmer wake refinement boundaries and wrapper contact prevention boundaries
    * The thin mesher now allows a relative or absolute thickness to be specified and is now defaulted to polygonal prisms and not triangular ones
    * A new cell set "invert" option has been added for volume meshes. The generation of a cell set also creates a field function for the set which can be useful when splitting a model etc.
    * It is now possible to color face normals in the surface repair offset tool

    Solvers

    * DFBI (6-DOF) solver can now superimpose an additional rigid body motion on the 6-DOF motion. This enables modeling of propellers
    * Significant improvement in parallel viewfactor calculations
    * General performance increase on Windows platform

    Physics

    * Multiple Passive Scalars
    * Equation of state models
    o Real gas models: extended to include IAPWS-IF97 for gaseous steam
    o IAPWS-IF97 added for modeling liquid water
    o User defined density: extends range of variable density models; compressible liquids can be modeled with this feature
    * Anisotropic porous energy: tensor profile input now possible for thermal conductivity and specific heat. Important for some heat exchanger applications
    * Heat exchanger enthalpy source: can now be activated at a point specified by the user
    * Large Eddy Simulation: Bounded Central Differencing solver provided for additional accuracy and stability
    * Aeroacoustics: several broadband noise source models have been implemented to estimate the local acoustic power generated by a turbulent flow field. The models are:
    o Curle model
    o Goldstein axisymmetric model
    o Linearized Euler Equation
    o Lilley model
    o Proudman model
    * Combustion modeling
    o New thermal NOx model that is not fuel dependent and does not require a pure fuel stream
    o Coal combustion modeling: implemented in the Lagrangian Multiphase framework; includes phase models for vaporization, devolatilization and char oxidation.
    * Lagrangian Multiphase
    o Two new injector types: cone and surface. Cone injector accepts inner and outer angle specification; surface injector accepts flux specification
    o General upgrade of injector values to accept field functions and tables
    * Eulerian Multiphase
    o Extended to turbulent flows with a multiphase form of the Standard K-Epsilon model. This includes high y+ wall functions
    o Additional phase interaction model for turbulent dispersion
    * Harmonic Balance method
    o Now has blade flutter capability: this displaces the mesh to account for blade deformation
    * Finite Volume Stress
    o Implicit coupling with small displacement finite volume stress
    o Now includes viscous shear traction load on solid from the fluid
    o Large displacement theory added that allows solid vertices to automatically move with the predicted solid displacement field. This enables prediction of the apparent stiffening of members in bending when parts are placed in tension, and buckling phenomena when parts are under compression
    * Convective velocity option for solid energy transport equation
    * VOF Waves now independent of DFBI (6-DOF) model
    * Cavitation model now coupled with standard saturation pressure model
    Tümünü Göster
    ···
  10. 60.
    0
    iNGiLiZCE BRiFiNGTE AŞAĞIDAKi ÖRNEK CÜMLELER KULLANILABiLiR
    Welcome on board. Our aircraft has ... over wing exits.
    Passengers occupying the exit row seats must put all their hand luggage and personal belongings in the over head bins.
    In case of an emergency, you will be responsible for observing the outside conditions, opening the exit and directing passengers to these exits.
    The safety cards in your seat pockets show how to open the exit in case of an emergency. Please study carefully your safety card.
    B737/400 -- Pull down the control handle. Remove the hatch from its frame and dispose it away from the exit
    B737/700, B737/800 -- Pull down the control handle and leave. The hatch rotates up and will be locked at the top of the exit.
    A319/320 -- Remove the handle cover.The handle light and the indicator illuminate. Pull down the control handle. Remove the hatch from its frame and dispose it away from the exit
    Can you take the responsibility? If not, I can change your seat.
    Thank you for your cooperation and I wish you a pleasant flight.
    ···
  11. 61.
    0
    olmamalıydı yani amk yeter ya
    ···
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    1 System Poles and Zeros
    The transfer function provides a basis for determining important system response characteristics
    without solving the complete differential equation. As defined, the transfer function is a rational
    function in the complex variable s = σ + jω, that is
    H(s) = bmsm + bm−1sm−1 + . . . + b1s + b0
    ansn + an−1sn−1 + . . . + a1s + a0
    (1)
    It is often convenient to factor the polynomials in the numerator and denominator, and to write
    the transfer function in terms of those factors:
    H(s) = N(s)
    D(s)
    = K
    (s − z1)(s − z2) . . . (s − zm−1)(s − zm)
    (s − p1)(s − p2) . . . (s − pn−1)(s − pn) , (2)
    where the numerator and denominator polynomials, N(s) and D(s), have real coefficients defined
    by the system’s differential equation and K = bm/an. As written in Eq. (2) the zi’s are the roots
    of the equation
    N(s) = 0, (3)
    and are defined to be the system zeros, and the pi’s are the roots of the equation
    D(s) = 0, (4)
    and are defined to be the system poles. In Eq. (2) the factors in the numerator and denominator
    are written so that when s = zi the numerator N(s) = 0 and the transfer function vanishes, that is
    lim
    s→zi
    H(s) = 0.
    and similarly when s = pi the denominator polynomial D(s) = 0 and the value of the transfer
    function becomes unbounded,
    lim
    s→pi
    H(s) = ∞.
    All of the coefficients of polynomials N(s) and D(s) are real, therefore the poles and zeros must
    be either purely real, or appear in complex conjugate pairs. In general for the poles, either pi = σi,
    or else pi, pi+1 = σi±jωi. The existence of a single complex pole without a corresponding conjugate
    pole would generate complex coefficients in the polynomial D(s). Similarly, the system zeros are
    either real or appear in complex conjugate pairs.
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