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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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Page 1: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

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)

Page 2: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.1)

Page 3: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.2)

Page 4: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.3)

Page 5: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

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)

Page 6: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

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):

= 2·f / 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

Page 7: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.9)

Page 8: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.10)

Page 9: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.11)

Page 10: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.12)

Page 11: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.13)

Page 12: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.14)

Page 13: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.15)

Page 14: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.16)

Page 15: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.17)

Page 16: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

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

.

Page 17: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.19)

Page 18: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.20)

Page 19: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.21)

Q 13.22)

Page 20: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.23)

Page 21: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

Q 13.24)

Page 22: Analog Integrated ircuit esign - University of Colorado ...tkalkur/ECE4340-5340/solutionsch13.pdfTony Chan Carusone Solutions contributed by: Tony Chan Carusone Yunzhi (Rocky) Dong

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