... BOOK 5
THE WORDS 7
WORD ANALYSIS 10 3
IDIOM AND USAGE 11 7
14 Vocabulary 4000
annul cancel
annular ring-shaped
anodyne pain soothing
anoint consecrate, apply ointment
anomalous abnormal
anonymity ... formed by
rearranging the letters of another word
analgesic pain-soother
The Words 11
Quiz 1 (Matching)
Match each word in the first column with its definition in the se...
... the
array). This becomes a second number inside the brackets, separated by
a comma.
Rapid Mobile
Enterprise
Development
for Symbian OS
An Introduction to OPL Application Design and
Programming
Ewan Spence
With
Phil ... Jipping
0470 844302 418pp 2002 Paperback
ã Programming for the Series 60 Platform and Symbian OS
Digia
0470 849487 550pp 2002 Paperback
ã Symbian OS C++ for...
... into the fiber link to boost the optical signal.
For example, let’s assume an 8,000-km long fiber link connecting two continents.
To
compensate the fiber loss of 0 .25 dB/km, we insert an optical ...
-vs(t>
,
which is (4p /2) 2 for a DC-balanced
ideal NRZ
signal. Thenoisepoweriscalculated as
vi(t),
whichcan be written
1 /2. (~~,~ +21 ,2, ~),
given equal probabili...
...
6.28
.50
Q
*
0 .3
pF
=
30 .6GHz
fA
2
(5 .33 )
and
A
.
fA
=
1.44 .30 .6 GHz
=
44
GHz. Fine, you can design this amplifier. What’s
the TIA bandwidth now?
(5 .34 )
Hurray, you’ve ... second- and third-order intermodulation
distortions
for
the two-tone case are [82]
IMD2
X
la21
.
X,
IMD3
x
31 4.
la31
.
X2,
(4.72)
(4. 73)
where
X
is the amplitude of one...
...
144
TRANSIMPEDANCE AMPLIFIERS
Fig.
5.26
Receivers for analog signals: (a) low-impedance front-end and
(b)
front-end with
matching transformer.
matching transformer [ 14] . The transformer ... value
10 mV
0.01
16
-
14. 07 .3
kQ
.380nA
=
34. 6dB.
A?
And for a typical lO-Gb/s system
(RT
=
1
kC2,
iimhA
=
1 ,40 0nA) we get
lOmV
0.0116. 14. 07. 1 kQ.l ,40 0nA
=...
... on
a
0. 05- dB
power penalty
(PP
=
1.0116). For a 2 .5- Gb/s SONET system (r
=
72), we obtain
the numerical value
2 .5 Gb/s
fLF
5
0.01
16.
___
-
-
64
kHz.
6.28 .72
And for a 10-Gb/s ... than the numbers de-
rived above (e.g.,
2 .5
kHz for 2 .5 Gb/s and 25 kHz for
I0
Gb/s). The cutoff frequency
usually is set by an external capacitor, for example, a co...
...
(b)
Fig.
6. 53
Circuits for Problems
6. 7
and
6. 8:
(a)
stage with series feedback and
(b)
regu-
lated cascode.
6. 5
6. 6
6. 7
6. 8
(GBWs).
How large is the ratio of the GBW for the total ... the NRZ signal to an
RZ
signal in the
optical domain.
233
2 56
OPTICAL TRANSMITTERS
For example, given a fiber attenuation of 0.25 dB/km, typical for an SMF operate...
... transmitter
and is identical to that derived for MAS in Section
6.2.6:
2nfLFr
PP=l+-
B'
(8. 17)
DRIVER
ClRCUlT
CONCEPTS
277
Fig.
8. 17
A
simple predriver with pulse-width control. ... nonlinear leakage
currents),
or
to permit higher optical output power levels,
or
both.
Optical Feedforward Linearization.
Optical feedforward linearization capitalizes
on the...
... Thus, the differential input voltage that
switches 0 .99 1~ to one output and
0.01Z~
to the other is VBE(O .99 fM)
-
VBE(O.011M)
=
VT(In 0 .99
-
In
0.01)
=
4.6V~
M
115
mV.
Therateofcurrentchangeisdi/dt ...
2.l(a)
f
=
c/A
=
( 299 .8Mm/s)/(1.55pm)
=
193 .4THz.
2.Nb)
A
f
=
c/h2
.
Ah
=
( 299 .8 R4m/s)/(1.55
pm)2
.
0.1 nm
=
12.48
GHz.
2.2
The linear expres...