NAME ____________________________________________
ES 442 Homework #4
(Spring 2016 – Due February 29, 2016 )
Print out homework and do work on the printed pages.
Textbook: B. P. Lathi & Zhi Ding, Modern Digital and Analog Communication
Systems, 4th edition, Oxford University Press, New york, 2009.
Problem 1 (Baseband Recovery) (20 points)
A frequency-translated baseband signal (frequency shifted by fc), denoted by m(t), is
v(t) = m(t)cos(2 fct). We can recover m(t) by multiplying v(t) by a local oscillator signal
given by cos(2 fct + ). You are asked to investigate the effect of the offset in phase
angle  .
(a) The modulation product of v(t) and cos(ct + ) is passed through a low-pass filter
rejecting the double-frequency term. After filtering, what is the signal output?
(b) Next, using the result from part (a) above, what is the output of the filter when  is
equal to /2 radians?
Homework 4
(c) How much phase shift  can be tolerated for a decrease no greater than 10% of the
magnitude at the output of the filter?
1
SCORE : ______ out of 100 points
Problem 2 Non-sinusoidal periodic waveform (15 points)
We know from Problem 1 that a baseband signal m(t) can be frequency translated
using a sinusoidal carrier signal. However, for this problem we only have available a
periodic square-wave signal p(t) of period 1/ fc.
Why can a square-wave signal p(t) still be used to successfully frequency translate
m(t)?
Problem 3 Baseband Recovery with Squarer/Square-Rooter (20
points)
A conventional AM signal of the form
AM (t )  A1 km(t )  cos  2 fCt 
Homework 4
is applied to the system shown below. Assuming |m(t)| < 1 for all time t, that m(t) is
band limited to within the range –B  f  B, and the carrier frequency is fc > 2B.
Prove that m(t) can be extracted from the system’s output.
2
Problem 4 Balanced Modulator (20 points)
The circuit shown below is known as a “balanced modulator.”
(a) The input to the upper AM modulator is baseband signal m(t) and the input to the
lower AM modulator is –m(t). Both AM modulators are assumed to be exactly identical
with respect to amplitude sensitivity. What is the output AM(t) of the balanced
modulator?
Homework 4
(b) Suppose the AM modulators are not identical – assume the upper AM modulator’s
amplitude output is 10% greater than that of the lower AM modulator. How would this
change the answer you came up with in part (a)?
3
Problem 5 Image Rejection Receiver (25 points)
First, we introduce the concept of an “image frequency” (IM). In the figure below we
see two signal spectra; one is the desired RF band signal (centered around RF) and an
unwanted or image RF signal (centered around IM). Both are centered about the local
oscillator (LO) frequency in the frequency domain. The distinction between the signal
and the images is possible because the two spectra lie on different sides of the LO
frequency. When RF signals are presented to a mixer or modulator with LO at
frequency LO, the IF (intermediate frequency) output translates both the RF and image
spectra to a frequency band centered at frequency IF. In the below figure note they
are positioned on top of each other around the IF frequency and this is the problem in
signal reception and detection.
Homework 4
It is possible to build an image rejection receiver mixer or demodulator to solve this
problem. The general principle behind an image rejection architectures is to process
the signal and the image differently which allows for cancellation of the image signal by
its negated replica. One possible image rejection receiver is shown below (this is
known as the Hartley receiver). Both the RF desired band and the RF image signal
band (i.e., spectra) enter the receiver at the RF input. For this problem, show that only
the RF desired band appears about IF at the output of the receiver. To show this
start with an input signal of in(t) = Acos(2 fRF t) + Bcos(2 fIM t) where the first
term is the desired RF signal and the second is the image signal.
4
Homework 4
Hint for this problem: The 90° phase shift is implemented by changing a cos( t) to a
sin(t) and a sin(t) to a - cos(t) – that is, in the time domain. That is equivalent to
taking spectral components in the frequency domain and multiplying by –j for positive 
components, and multiplying by +j for negative  components. This is presented in
most books as the result of the Hilbert transform. Remember that multiplying by plus or
minus j (square root of -1) is equivalent to a phase shift by plus or minus ninety
degrees.
5