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Solution Manual For Communication Systems An Introduction to Signals and Noise in Electrical Communication 4th Edition By A. Bruce Carlson, Paul B. Crilly / Latest Version 2024 A+$12.99
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Solution Manual For Communication Systems An Introduction to Signals and Noise in Electrical Communication 4th Edition By A. Bruce Carlson, Paul B. Crilly / Latest Version 2024 A+
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Course
Communication Systems An Introduction to Signals
Institution
Communication Systems An Introduction To Signals
Solution Manual For Communication Systems An Introduction to Signals and Noise in Electrical Communication 4th Edition By A. Bruce Carlson, Paul B. Crilly / Latest Version 2024 A+
Chapter 2
2.1-1
0
0
0
/ 2 2 ( )
/ 2
0
sinc( )
0 otherwise
j j
T
j mnft j
n
T
Ae Ae n m
c e dt Ae m n
...
solution manual for communication systems an intro
an introduction to signals and noise in electrical
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Communication Systems An Introduction to Signals
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Solutions Manual
to accompany
Communication
Systems
An Introduction to Signals and Noise in
Electrical Communication
Fourth Edition
A. Bruce Carlson
Rensselaer Polytechnic Institute
Paul B. Crilly
University of Tennessee
Janet C. Rutledge
University of Maryland at Baltimore
,Solutions Manual to accompany
COMMUNICATION SYSTEMS: AN INTRODUCTION TO SIGNALS AND NOISE IN ELECTRICAL COMMUNICATION,
FOURTH EDITION
A. BRUCE CARLSON, PAUL B. CRILLY, AND JANET C. RUTLEDGE
The contents, or parts thereof, may be reproduced in print form solely for classroom use with COMMUNICATION SYSTEMS: AN
INTRODUCTION TO SIGNALS AND NOISE IN ELECTRICAL COMMUNICATION, provided such reproductions bear copyright
notice, but may not be reproduced in any other form or for any other purpose without the prior written consent of The McGraw-Hill
Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance
learning.
www.mhhe.com
,Chapter 2
2.1-1
Ae jφ Ae jφ n = m
∫− T
T
j 2 π ( m−n )f 0t jφ
cn = e dt = Ae sinc( m − n ) =
T0 0 otherwise
2.1-2
c0 v (t ) = 0
2 T 2π nt T 2π nt 2A πn
cn =
T0 ∫
0
A cos
T0
dt + ∫ ( − A)cos
T T0
dt =
πn
sin
2
n 0 1 2 3 4 5 6 7
cn 0 2A/π 0 2 A / 3π 0 2 A / 5π 0 2 A / 7π
arg cn 0 ±180° 0 ±180°
2.1-3
c0 = v (t ) = A / 2
2 T0 /2 2 At 2π nt A A
cn =
T0 ∫ 0
A−
T0
cos
T0
dt =
πn
sin π n −
(π n) 2
(cos π n − 1)
, 2 A sin (π − π n ) 2t / T0 sin (π + π n ) 2t / T0
T
2 T 2π t 2π nt
cn =
T0 ∫0
A cos
T0
cos
T0
dt =
T0 4(π − π n) / T0
+
4(π + π n ) / T0 0
A/2 n = ±1
[ sinc(1 − n) + sinc(1 + n )] =
A
=
2 0 otherwise
2.1-5
c0 = v (t ) = 0
2 T 2π nt A
cn = − j
T0 ∫0
A sin
T0
dt = − j
πn
(1 − cos π n )
n 1 2 3 4 5
cn 2A/π 0 2 A / 3π 2 A / 5π
arg cn −90° −90° −90°
2.1-6
c0 = v(t ) = 0
2 A sin (π − π n ) 2t / T0 sin ( π + π n ) 2t / T0
T
2 T 2π t 2π nt
cn = − j
T0 ∫0
A sin
T0
sin
T0
dt = − j
T0 4(π − π n ) / T0
−
4(π + π n )/ T0 0
m jA / 2 n = ±1
[sinc(1 − n ) − sinc(1 + n ) ] =
A
= −j
2 0 otherwise
2.1-7
1 T
v ( t) e− jnω0 t dt + ∫ v(t )e − jnω 0t dt ]
T0
cn = ∫
T0 0 T
T0 T
where ∫T
v(t )e − jnω0 t dt = ∫
0
v (λ + T0 /2) e− jnω 0λ e− jnω 0T d λ
T
= −e jnπ ∫ v (t )e − jnω0 t dt
0
since e jnπ = 1 for even n, cn = 0 for even n
2-2
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