Javascript required
Skip to content Skip to sidebar Skip to footer
Home / Linear System Theory and Design Solution

Linear System Theory and Design Solution

  • Home
  • Engineering
  • Solution Manual for linear System Theory and Design by Chi-Tsong Chen 3rd edition

Solution Manual for linear System Theory and Design by Chi-Tsong Chen 3rd edition

  • View
    2.482

  • Download
    1.046

Embed Size (px)

Text of Solution Manual for linear System Theory and Design by Chi-Tsong Chen 3rd edition

  • Linear System Theory and Design SA01010048 LING QING

    1

    2.1 Consider the memoryless system with characteristics shown in Fig 2.19, in which u denotes the input and y the output. Which of them is a linear system? Is it possible to introduce a new output so that the system in Fig 2.19(b) is linear?

    Figure 2.19 Translation: 2.19 uy 2.19(b)

    Answer: The input-output relation in Fig 2.1(a) can be described as:

    uay *=

    Here a is a constant. It is a memoryless system. Easy to testify that it is a linear system. The input-output relation in Fig 2.1(b) can be described as:

    buay += *

    Here a and b are all constants. Testify whether it has the property of additivity. Let:

    buay += 11 *

    buay += 22 *

    then:

    buuayy *2)(*)( 2121 ++=+

    So it does not has the property of additivity, therefore, is not a linear system. But we can introduce a new output so that it is linear. Let:

    byz =

    uaz *= z is the new output introduced. Easy to testify that it is a linear system. The input-output relation in Fig 2.1(c) can be described as:

    uuay *)(=

    a(u) is a function of input u. Choose two different input, get the outputs:

    111 *uay =

  • Linear System Theory and Design SA01010048 LING QING

    2

    222 *uay =

    Assure:

    21 aa

    then:

    221121 **)( uauayy +=+

    So it does not has the property of additivity, therefore, is not a linear system. 2.2 The impulse response of an ideal lowpass filter is given by

    )(2

    )(2sin2)(

    0

    0

    tttt

    tg

    =

    for all t, where w and to are constants. Is the ideal lowpass filter causal? Is is possible to built the filter in the real world? Translation: tw to

    Answer: Consider two different time: ts and tr, ts < tr, the value of g(ts-tr) denotes the output at

    time ts, excited by the impulse input at time tr. It indicates that the system output at time ts is dependent on future input at time tr. In other words, the system is not causal. We know that all physical system should be causal, so it is impossible to built the filter in the real world.

    2.3 Consider a system whose input u and output y are related by

    >

    ==atfor

    atfortutuPty a 0

    )(:))(()(

    where a is a fixed constant. The system is called a truncation operator, which chops off the input after time a. Is the system linear? Is it time-invariant? Is it causal? Translation: a a Answer: Consider the input-output relation at any time t, ta:

    0=y

    Easy to testify that it is linear. So for any time, the system is linear. Consider whether it is time-invariable. Define the initial time of input to, system input is u(t), t>=to. Let to=to:

  • Linear System Theory and Design SA01010048 LING QING

    3

    =totherforattfortu

    ty0

    )()( 0

    Shift the initial time to to+T. Let to+T>a , then input is u(t-T), t>=to+T. System output:

    0)(' =ty

    Suppose that u(t) is not equal to 0, y(t) is not equal to y(t-T). According to the definition, this system is not time-invariant.

    For any time t, system output y(t) is decided by current input u(t) exclusively. So it is a causal system.

    2.4 The input and output of an initially relaxed system can be denoted by y=Hu, where H is some mathematical operator. Show that if the system is causal, then

    uHPPHuPyP aaaa ==

    where Pa is the truncation operator defined in Problem 2.3. Is it true PaHu=HPau? Translation: y=Hu H

    Pa 2.3PaHu=HPau

    Answer: Notice y=Hu, so:

    HuPyP aa =

    Define the initial time 0, since the system is causal, output y begins in time 0. If a0, we can divide u to 2 parts:

    =totherforatfortu

    tp0

    0)()(

    >

    =totherfor

    atfortutq

    0)(

    )(

    u(t)=p(t)+q(t). Pay attention that the system is casual, so the output excited by q(t) cant affect that of p(t). It is to say, system output from 0 to a is decided only by p(t). Since PaHu chops off Hu after time a, easy to conclude PaHu=PaHp(t). Notice that p(t)=Pau, also we have:

    uHPPHuP aaa =

    It means under any condition, the following equation is correct:

    uHPPHuPyP aaaa ==

    PaHu=HPau is false. Consider a delay operator H, Hu(t)=u(t-2), and a=1, u(t) is a step input begins at time 0, then PaHu covers from 1 to 2, but HPau covers from 1 to 3.

  • Linear System Theory and Design SA01010048 LING QING

    4

    2.5 Consider a system with input u and output y. Three experiments are performed on the system using the inputs u1(t), u2(t) and u3(t) for t>=0. In each case, the initial state x(0) at time t=0 is the same. The corresponding outputs are denoted by y1,y2 and y3. Which of the following statements are correct if x(0)0?

    1. If u3=u1+u2, then y3=y1+y2. 2. If u3=0.5(u1+u2), then y3=0.5(y1+y2). 3. If u3=u1-u2, then y3=y1-y2.

    Translation: u y u1(t), u2(t) u3(t)t>=0 x(0) y1,y2 y3 x(0)

    Answer: A linear system has the superposition property:

    0221102211

    022011 ),()(),()(

    )()(tttyty

    tttututxtx

    +

    ++

    In case 1:

    11 = 12 =

    )0()0(2)()( 022011 xxtxtx =+

    So y3y1+y2. In case 2:

    5.01 = 5.02 =

    )0()()( 022011 xtxtx =+

    So y3=0.5(y1+y2). In case 3:

    11 = 12 =

    )0(0)()( 022011 xtxtx =+

    So y3y1-y2. 2.6 Consider a system whose input and output are related by

    =

    =0)1(0

    0)1()1(/)()(

    2

    tuiftuiftutu

    ty

    for all t. Show that the system satisfies the homogeneity property but not the additivity property. Translation: ,,. Answer: Suppose the system is initially relaxed, system input:

    )()( tutp =

    a is any real constant. Then system output q(t):

  • Linear System Theory and Design SA01010048 LING QING

    5

    =

    =0)1(0

    0)1()1(/)()(

    2

    tpiftpiftptp

    tq

    =

    =0)1(0

    0)1()1(/)(2

    tuiftuiftutu

    So it satisfies the homogeneity property. If the system satisfies the additivity property, consider system input m(t) and n(t),

    m(0)=1, m(1)=2; n(0)=-1, n(1)=3. Then system outputs at time 1 are:

    4)0(/)1()1( 2 == mmr

    9)0(/)1()1( 2 == nns

    0)]0()0(/[)]1()1([)1( 2 =++= nmnmy

    1()1( sr +

    So the system does not satisfy the additivity property. 2.7 Show that if the additivity property holds, then the homogeneity property holds for all rational numbers a . Thus if a system has continuity property, then additivity implies homogeneity. Translation: a

    Answer: Any rational number a can be denoted by: nma /= Here m and n are both integer. Firstly, prove that if system input-output can be described

    as following: yx then: mymx Easy to conclude it from additivity. Secondly, prove that if a system input-output can be described as following: yx then:

    nynx //

    Suppose: unx / Using additivity:

    nuxnxn =)/(*

    So: nuy =

    nyu /=

  • Linear System Theory and Design SA01010048 LING QING

    6

    It is to say that:

    nynx //

    Then:

    nmynmx /*/*

    ayax It is the property of homogeneity. 2.8 Let g(t,T)=g(t+a,T+a) for all t,T and a. Show that g(t,T) depends only on t-T. Translation: t,T ag(t,T)=g(t+a,T+a) g(t,T) t-T Answer: Define:

    Ttx += Tty =

    So:

    2

    yxt += 2

    yxT =

    Then:

    )2

    ,2

    (),( yxyxgTtg +=

    )2

    ,2

    ( ayxayxg +++=

    )22

    ,22

    ( yxyxyxyxg +++++=

    )0,(yg=

    So:

    0)0,(),( =

    =

    xyg

    xTtg

    It proves that g(t,T) depends only

Linear System Theory and Design Solution

Source: https://dokumen.tips/engineering/solution-manual-for-linear-system-theory-and-design-by-chi-tsong-chen-3rd-edition.html