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Analog Integrated Circuit Design 2 nd Edition Chapter 13 Solutions Compiled by: Tony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong Ali Sheikholeslami & Khoman Phang (1 st ed. Solutions)
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Solutions Ch 13

Nov 17, 2015

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  • Analog Integrated Circuit Design 2nd Edition

    Chapter 13 Solutions

    Compiled by:

    Tony Chan Carusone

    Solutions contributed by:

    Tony Chan Carusone

    Yunzhi (Rocky) Dong

    Ali Sheikholeslami & Khoman Phang (1st ed. Solutions)

  • Q 13.1)

  • Q 13.2)

  • Q 13.3)

  • Q 13.4)

    The signal xc(t) has frequency content at 50kHz and 150kHz. Hence, it is bandlimited to fc =

    150kHz and must be sampled at least at its Nyquist rate: fs 2fc = 300 kHz

    Q 13.5)

  • Q 13.6)

    The continuous-time square pulse is given by

    ( ) ( ) ( 1ms)px t t t

    where is the unit step function defined in equation (13.2).

    The sampled signal is 10 consecutive samples with all other samples equalling zero. Hence,

    the spectrum is simply the summation of the corresponding 10 terms in equation (13.15):

    9

    0

    ( ) k

    k

    X z z

    Q 13.7)

    Equation (13.17):

    = 2f / fs f = fs / 2

    In this case: = /10 and fs = 10 MHz. Hence,

    f = (/10)(10 MHz) / 2 = 0.5 MHz

    Q 13.8)

    Recall that if x(n) X(z), then x(n-k) X(z)z-k. Hence,

    y(n) = x(n) + 2x(n-3)

    Y(z) = X(z) + 2X(z)z-3

  • Q 13.9)

  • Q 13.10)

  • Q 13.11)

  • Q 13.12)

  • Q 13.13)

  • Q 13.14)

  • Q 13.15)

  • Q 13.16)

  • Q 13.17)

  • Q 13.18)

    The transfer function of the block diagram in Fig. 13.12 is given by equation (13.29):

    From equation (13.26), the 3-dB frequency of H(z) is:

    In this case, = 2/50 and we must solve for a:

    cos(2/50) = 2 a/2 1/2a 0.5a2 1.008a + 0.5 = 0

    This has two solutions: a = 1.134 and 0.882

    Only the later results in a stable transfer function. Hence, a = 0.882.

    To ensure a dc gain of 2, we require (1) 2 2(1 ) 2(1 0.882) 0.2361

    bH b a

    a

    .

  • Q 13.19)

  • Q 13.20)

  • Q 13.21)

    Q 13.22)

  • Q 13.23)

  • Q 13.24)

  • Q 13.25)

    The spectrum of the D/A output is that of the sampled sinusoid, having images at

    100MHz 5MHzskf f k for all integers k , passed through the S/H sinc response.

    sin( / 2)

    ( )/ 2

    shH j

    The first image of the sinusoid will have the largest amplitude and will occur at 95sf f MHz.

    Hence, it is required that

    ( 2 ( )) 0.25 ( 2 )sh s shH j f f H j f

    sin(2 ( ) / 2) sin(2 / 2)

    0.25 0.252 ( ) / 2 2 / 2

    s

    s

    f f f

    f f f

    (assuming 1/ f )

    sin(2 ( ) / 2)

    0.252 ( ) / 2

    s

    s

    f f

    f f

    Solve numerically for:

    2 ( ) 2.48

    2.48 8.3ns2 ( )

    s

    s

    f f

    f f