NAME ______________ Solutions ___________________
ES 442 Homework #4 Solutions
(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?
v (t )  cos  2 fC t     m(t )  cos  2 fC t   cos  2 fC t   
 m(t )  21 cos  2 fC t  2 fC t     21 cos  2  2 fC t    
m(t )
cos(  )  cos  2  2 fC t    
2 
The filter removes the cos(2  2 fC t   ) term, therefore

=
m(t )
cos(  )
2
(b) Next, using the result from part (a) above, what is the output of the filter when  is
equal to /2 radians?


m(t )
m(t )
cos    
cos   0
2
2
2
(c) How much phase shift  can be tolerated for a decrease no greater than 10% of the
magnitude at the output of the filter?
For a 10% decrease in response we require that
which corresponds to  = 25.8 degrees.
Homework 4
cos     0.9,
1
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)?
ANSWER: A periodic square wave can be expressed as a Fourier series which consists
of sinusoidal components which has a fundamental frequency sine wave (of frequency
f0) with the same period as the square wave. It also has odd-order harmonic sinusoids
containing 3 f0, 5 f0 , and so on. These harmonic frequencies can be filtered out leaving
the fundamental sinusoidal wave. The mixer operates using the fundamental sinusoidal
of the square wave.
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 fC t 
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.
Given  AM (t )  A 1  km(t )  cos C t  , therefore squaring this gives
A2
2
1  km(t )  1  cos(2  C t )

2
The LPF removes the cos(2  C t ) term, leaving the term
 2 AM (t )  A2 1  km(t )  cos2 C t  
 2out (t ) 
A2
2
1  km(t ) .

2
 2out (t ) 
Taking the square root of  2 AM (t ) gives
A2
A
A
2
1  km(t ) 

km(t )

2
2
2
Homework 4
2
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?
First, we begin with the signal passing through the upper AM modulator,
U (t )  Am(t )  cos C t 
And in the lower AM modulator, we have
L (t )   Am(t )  cos Ct 
At the summing node we have add u(t) and L(t) with the appropriate sign on the
adder at the output. Therefore,
AM (t )  Am(t )  cos C t     Am(t )  cos C t    2Am(t )  cos C t 
Based upon discussion in class it may not be clear that they add because of the minus
sign on the signal on the lower AM modulator. In that case they cancel and the result is
zero. Either answer is acceptable for this problem!
It would change the magnitude of the output by a small amount because they add
linearly.
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
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.
ANSWER: We first solve the problem using analytical expressions for the signals. The
signal at the input of the image rejection receiver is
(t )  ARF cos RF t   Aim cos imt 
Note that
RF  LO  LO  IM ,
thus,
RF  LO  IM
Each mixer forms the sum and difference frequencies with the LO frequency and RF
frequency (or image frequency). Starting with the upper mixer with a sin( LOt) LO drive
to the mixer, we have at point A as labeled on the diagram,
ARF
A
sin  LO  RF  t   im sin  LO  im  t 
2
2
A
A
 RF sin  LO  RF  t   im sin  LO  im  t  and
2
2
A
A
B (t )  RF cos  LO  RF  t   im cos  LO  im  t 
2
2
A
A
 RF cos  LO  RF  t   im cos  LO  im  t 
2
2
The low-pass filters remove the "sum of frequencies" terms
leaving us with
A
A
 A (t )  RF sin  LO  RF  t   im sin  LO  im  t  and
2
2
A
A
B (t )  RF cos  LO  RF  t   im cos  LO  im  t 
2
2
We can now write for  A (t )
 A (t ) 
ARF
A
sin  RF  LO  t   im sin  LO  im  t 
2
2
Homework 4
 A (t )  
5
Next we pass A(t) through the 90° phase shift (Hilbert transform) such where a positive
frequency is multiplied by –j and a negative frequency is multiplied by +j.
ARF
A
sin  RF  LO  t   im sin  LO  im  t 
2
2
In passing  A (t ) through the 90 degree phase shift gives at "C"
 A (t )  
ARF
A
cos  RF  LO  t   im cos  LO  im  t 
2
2
The key is that the signal components at "B" and "C" have the
same polarity, but the image components at "B" and "C" have
opposite polarity. This occurs because the 90 degree phase
shift distinguishes between LO  RF < 0 and LO  IM  0.
C (t ) 
This can also be analyzed graphically. It involves convolving the input spectrum with those of
sin(LOt) and cos(LOt) and then low pass filtering the results. This gives us the spectra at
points “A” and “B”, namely XA() and XB(), respectively. In XA(), j denotes
Homework 4
multiplication of the corresponding part of the spectrum by the same number. The 90°
phase shift operation multiplies the negative-frequency part of XA() by +j and the
positive-frequency part by –j, thereby producing the XC() spectrum. The sum of XC()
and XB() is therefore free from the image because of the cancellation. The figure
below is from Behzad Razavi, RF Microelectronics, Prentice Hall PTR, pages 140 and
141.
6
7
Homework 4