EE140 Introduction to Communication Systems
Homework 5
Instructor: Professor Lixiang Lian
Due on 21 Oct 23:59 UTC+8
Notes:
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Please provide enough calculation process to get full marks.
Please submit your homework to Gradescope with entry code: 2K2265.
Please check carefully whether the question number on the gradescope corresponds to each
question.
1. Definition of FM and PM (15pts)
t
An FM modulator has output xc (t ) = 2cos(2 f c t + 2 f d m( )d ) with f d = 10Hz / V and input
0
3t , 0 t 1
m(t ) = 3, 1 t 2
0, otherwise
(1) (5pts) Determine the phase deviation in radians.
(2) (5pts) Determine the frequency deviation in hertz.
Given a PM modulator has output xc (t ) = 2cos(2 f c t + k p m(t )) with k p = 10rad / V and input as
before.
(3) (5pts) Determine the peak phase deviation in radians and the peak frequency deviation in hertz.
2. Difference between FM and PM (25pts)
Given an angle modulated signal with the expression:
Sm (t ) = 6cos(2*106 t + 40sin 500 t + 20sin1000 t + 10sin 2000 t )
(1) (10pts) Compute the power P, peak frequency deviation f in Hz, and the bandwidth B.
(2) (5pts) Is it possible to distinguish between FM and PM signals using this expression? What’s
the relationship between FM and PM?
(3) (5pts) If this is a phase modulated signal with a phase deviation constant k p = 5rad / V ,
compute the message signal m(t).
(4) (5pts) If this is a frequency modulated signal with a frequency deviation constant
f d = 10000Hz / V , compute the message signal m(t).
3. Narrowband FM (35pts)
An FM signal is given by s(t ) = Ac cos[c t + sin(mt )] .
(1) (5pts) Given that this is an NBFM signal (this condition holds for the following questions),
prove that it can be approximated as s(t ) s (t ) = Ac cos(c t ) − Ac sin(c t )sin(mt ) .
(2) (10pts) Compute the spectrum S ( ) and plot it.
(3) (10pts) Determine the envelope of the NBFM signal. Compute the ratio of the maximum to the
minimum value of this envelope.
(Hint: use approximation
1+ x
1
1 + x when x 1 .)
2
(4) (10pts) Determine the average power of the NBFM signal, and express it as a percentage of the
average power of the carrier wave.
4. NBFM to WBFM (25pts)
The following block diagram shows a WBFM generator from NBFM signal to transmit information
signal x(t ) = sin(2 f mt ) where f m = 1 kHz . The carrier frequency of the NBFM signal is given to
be f1 = 100 kHz at point 1 in the block diagram and k f = 20 rad / V . The required WBFM signal
is given to have carrier frequency of 140MHz and FM bandwidth BW 3 smaller than 60 kHz at point
3.
1) (3pts) Determine the frequency deviation f1 and FM bandwidth BW1 of the NBFM signal at
point 1.
2) (7pts) Determine the value of the integer multiplier M if frequencies of the mixers are given to
be f x = 5MHz and f y = M f x . (Be careful about the limitation on the WBFM bandwidth
given in the text.)
3) (5pts) Determine the center frequency of the BPF between points 1 and 2. (In other words,
determine the carrier frequency at node 2.)
4) (5pts) Find FM bandwidth BW2 at point 2.
5) (5pts) Determine the frequency deviation f3 and WBFM bandwidth BW3 at point 3.
Hint: 1) The value of the positive integer M directly gives the result. Do not use approximate noninteger values of M. 2) Use Carson’s rule for calculating bandwidth of both NBFM and WBFM.
Point 2
Point 1
x(t)
NDFM
Point 3
Multiplier
BPF
Multiplier
BPF
WBFM
M
? MHz
100
140 MHz
Signal
cos(2 f x t )
cos(2 f y t )
0
