MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 6 Solutions
Issued,October 21,2003 Due,October 29,2003
Exercise for home study,
O&W 6.49
Solution,
(a) To determine the time constants of the di?erential equation P6.49-1 in O&W,we
perform a Fourier transform of the equation,By linearity,the Fourier transform of the
equation is the Fourier transform of each of the individual equation terms,This gives
Y (j?)
X(j?)
=
9
(j?)
2
+ 11(j?) + 10
= H(j?),(1)
The time constants are the zeros of the denominator (c
1
=?1 and c
2
=?10),
(b) Equation 1 can be rewritten as
9
H(j?) =, (2)
(j? + 1)(j? + 10)
To make this a parallel interconnection of two?rst order systems,we do a partial
fraction expansion on Equation 2 to?nd
1
H(j?) = +
1
. (3)
j? + 1 j? + 10
By linearity,the inverse Fourier transform,h(t),is a parallel interconnection of two
rst order systems,h(t) is the sum of the inverse Fourier transform of each term in
Equation 3,Each term can be quickly determined using Table 4.2 of O&W,Thus,
u(t)? e
10t
h(t) = e
t
u(t). (4)
(c) The dominant time constant in a system with multiple decaying exponentials is the
time constant that takes the longest to die out,For this problem,the dominant time
constant is c
1
=?1,We can approximate this to Equation 1 as shown,
9 1
. (5)H(j?) =
(j?)
2
+ 11(j?) + 10
j? + 1
1
This approximation has a maximum di?erence of 10% at? = 0 and the di?erence
decreases as? increases,
(d) We want to approximate the faster component of Equation 4 as an impulse with a
height equal to its?nal value,We will then check to see how this approximation
a?ects the overall h(t) of the entire second order system,
The faster component,h
f
(t)=?e
10t
u(t),Let
h
f
(t) denote the approximation,For
h
f
(t) = s
f
(?)?(t),we need to determine s
f
(?),The step response is de?ned as
t
s(t) = h(t)dt,
Thus,
1
e
10t
dt =s
f
(?) = h
f
(t)dt =,?
10
0
h
f
(t) =?
1
(t) and the approximation to the overall system is
10
h
f
(t) + h
s
(t) =?h(t) =
1
(t) + e
t
u(t),(6)
10
Equation 6 enables us to determine the frequency response,We determine,
1
H(j?) =?0.1 + =
0.1j? + 0.9
,(7)
j? + 1 j? + 1
We then do algebraic manipulations and an inverse Fourier transform to Equation 7
to get the di?erential equation of the approximation,This gives,
dy dx
+ y(t) = 0.9x(t)? 0.1,
dt dt
The original frequency response,H(j?) and the approximate frequency response
H(j?)
are shown below,
2
Exercise,Problem 6.49? |H(j?)| Exercise,Problem 6.49? This is the approximate to |H(j?)|
1 1
0.9 0.9
0.8 0.8
0.7 0.7
0.6 0.6
0.4 0.4
0.3 0.3
0.2 0.2
0.1 0.1
0
0 20 40 60 80
frequency (?)
0
100 0 20 40 60 80 100
frequency (?)
(j
)|
|H
approx
|H(j
)|
0.5 0.5
The two are shown on the same plot below,The plot of H(j?) is solid and the plot of
H(j?) is dashed,
3
Exercise,Problem 6.49? Both |H(j?)| and |H
approx
(j?| Magnification of |H(j?)| and |H
approx
(j?| at low frequencies
1 1
|H(j?)|
|H
appr
(j?)|
ox
|H(
|H
a
j?)|
pprox
(j?)|
0.9
0.9
0.8
0.8
0.7
|H(j
)| and |H
approx
(j
)|
|H(j
)| and |H
approx
(j
)|
0.6
0.5
0.4
0.3
0.7
0.6
0.5
0.2
0.4
0.1
0
0 10 20 30 40 50
frequency (?)
0.3
0 0.5 1 1.5 2
frequency (?)
From the graphs above,it is visible that for? < 1,H(j?) approximates H(j?) well,
As? increases beyond 1,the two diverge from each other until roughly? = 20,This
is due to the fact that the approximation to the fast part of the signal reaches its?nal
value instantaneously,hence it must include all fast frequency components,
We can also compare the step response of the original signal,s(t) with the step response
of the approximate signal,?s(t),
4
t
s(t) = h(t)dt (8)
t t
= e
t
dt? e
10t
dt (9)
o 0
= (0.9? e
t
+ 0.1e
10t
)u(t),(10)
t
s?(t) = h(t)dt (11)
t t
= e
t
dt? (?0.1)?(t)dt (12)
o 0
= (0.9? e
t
)u(t),(13)
The two step responses are plotted together below,
5
Exercise,Problem 6.49? Both s(t) and s
approx
)t) Magnification of s(t) and s
approx
(t)
1
0.4
0.5
0.6
0.7
0.8
0.9
s(t) and s
approx
(t)
s(t)
s
approx
(t)
0.4
s(t) and s
approx
(t)
0.35
0.3
0.25
0.2
0.15
0.1
0.3 0.05
0.2 0
0.1?0.05
0
0 2 4 6 8
time (sec)
0.1
10 0 0.1 0.2 0.3 0.4 0.5
time (sec)
From the graphs above,it is apparent that for t >,5 seconds,the two step responses
are nearly equal,Furthermore,a general rule of thumb is that an exponential nearly
reaches its?nal value after 5 time constants,Thus,for e
10t
u(t),it should nearly reach
its?nal value by t = 0.5 seconds,
Problems to be turned in,
Problem 1 O&W 4.35
s(t)
s
a
(t)
pprox
Solution,
6
(a) We need to?nd the magnitude,the phase and the impulse response of the continuous-
time LTI system with frequency response
H(j?) =
a? j?
,
a + j?
Determining the magnitude,
a
2
+?
2
H(j?) = = 1.| |?
a
2
+?
2
Determining the phase,
e
j arctan
a
e =
e
j<H(j?)
= e
2j arctan
j arctan
a
,
a
Determining the impulse response,
+ F
1
j?
h(t) = F
1
{H(j?)} = F
1
a
= h
1
(t) + h
2
(t),
a + j? a + j?
From table 4.2,h
1
(t) = ae
at
u(t),To determine h
2
(t),we do the following,
{
j? 1 1 dh
1
(t)
h
2
(t) = F
1
(j?)H
1
(j?)} =,
a + j?
} = F
1
{?
a
a dt
Using the product rule,
dh
1
(t)
= ae
at
(t)? a
2
e
at
u(t),
dt
Thus,h
2
(t) =?e
at
(t) + ae
at
u(t),Combining h
1
(t) and h
2
(t) gives,
u(t)? e
at
(t).h(t) = 2ae
at
Alternatively we can do a partial fraction expansion of H(j?),
2a
H(j?) =
j? + a
1,
Now,h(t) can be found by looking up each term above in the tables to give the same
u(t)? e
at
(t).result h(t) = 2ae
at
(b) We need to determine y(t) when a = 1 and x(t) = cos(t/
3) + cos(t) + cos(
3t),We
can use the convolution property,Y (j?) = X(j?) × H(j?),We have H(j?),we need
to determine X(j?),
1 1
X(j?) =?[?( ) +?(? + ) +?( 1) +?(? + 1)?
3
3
+?(
3) +?(? +
3)] (14)
=?[X
a
(j?) + X
b
(j?) + X
c
(j?)],(15)
7
3
Because the system is LTI,we can write the output as the sum of three terms,
Y (j?) = Y
a
(j?) + Y
b
(j?) + Y
c
(j?),(16)
where the?rst term is de?ned as
Y
a
(j?) = H(j?)X
a
(j?) (17)
a? j?
=?X
a
(j?) (18)
a + j?
= e
2j tan
1
a
X
a
(j?) (19)
1
=?e
2j arctan
1
3
) +?e
2j arctan
a
(
a
(? + ),(20)?
3
We now substitue a = 1 into Y
a
(j?),Also,because X
a
(j?) is the sum of two delta
functions,we can substitute? =
1
3
into the arctangent term in the?rst? term of
Y
a
(j?) and? =
1
into the arctangent term for the second? term of Y
a
(j?),This
3
gives,
Y
a
(j?) =?e
j
3
3
+
1
.
3
1
+?e
j
3
This is now recognizable as a cosine function in the time domain with a phase change
of
,Hence,
1?
y
a
(t) = cos t?,?
3
3
The other terms,Y
b
(j?) and Y
c
(j?) can be solved similarly to give cosine functions
with various phase changes,We get,
j
Y
b
(j?) =?e
j
2
( 1) +?e
2
(? + 1),(21)
y
b
(t) = cos t? = sin(t),(22)
2
j
2?
3 3
Y
c
(j?) =?e
j
2?
(
3) +?e?(? +
3),(23)
2?
y
c
(t) = cos
3t?, (24)
3
Summing all the terms,
1? 2?
y(t) = cos t? + sin(t) + cos
3t?,?
3
3 3
These two functions,y(t) and x(t) are plotted together below,They are nonperiodic
functions,Why are they nonperiodic? Because the three periods that sum to make
8
up the function are 3 di?erent irrational numbers and hence we will?nd no rational
number,T,where each term will have an integer number of waveforms in that T,The
period of the?rst term is T
a
= 2?
3,,the period of the second term is T
b
= 2?,and
the period of the last term is T
c
= 2?
1
.
3
Problem 1? Both x(t) and y(t)
x(t) and y(t)
3
2
1
0
1
2
3
x(t)
y(t)
10?8?6?4?2 0 2 4 6 8 10
time (pi*seconds)
Problem 2 O&W 6.28 (a)-(iii) and (v)
Solution,
16
(a)-(iii) To sketch the frequency response of H(j?) =
(j?+2)
4
,we recognize that this is a cascade
of 4 identical?rst order systems,It’s magnitude,H(j?) can be written as,| |
1 1 1 1
|H(j?) = H
1
(j?)
4
= ( )( )( )( )| | | |
0.5j? + 1 0.5j? + 1 0.5j? + 1 0.5j? + 1
|
1 1 1 1
= ( )( )( )( )?
1 + 0.25?
2
1 + 0.25?
2
1 + 0.25?
2
1 + 0.25?
2
Because the system is LTI,the bode plot of the cascade of the 4 identical?rst order
systems is equal to the sum of the bode plots for each?rst order system,The?rst
order system,20 log H
1
(j?) =?10 log(1 + 0.25?
2
) has the plot,| |
0 for 2
(25)?10 log(1 + 0.25?
2
)?
20 log(0.5?) for? ≤ 2
9
Therefore the bode plot of the cascade of the 4?rst order systems has the plot,
H(j?)
0 for 2
(26)| |?
80 log(0.5?) for? ≤ 2
The following graph illustrates the plot,The two straight-line approximations intersect
at? = 2,The solid line is the approximation to H(j?) and the dashed line is the real | |
H(j?),| |
Problem 2? Bode plot of approximate and real |H(j?)|
|H(j
)| and |H
approx
(j
)|
(dB)
0
50
100
150
|H
approx
(j?)|
|H(j?)|
10
1
10
0
10
1
10
2
frequency (?)
We can see from the graph and by calculation that at? = 2,the actual magnitude is
H(j?) =?40 log(1 + 0.25 2
2
) =?40 log(2) =?12 dB,| ←
The phase is next determined and sketched,Again expressing the function as the
cascade of 4 identical?rst order functions we have,
j?H(j?)
e =
e
j tan
1
(0)
e
j tan
1
(0)
e
j tan
1
(0)
e
j tan
1
(0)
e
j tan
1?
2
e
j tan
1?
2
e
j tan
1?
2
e
j tan
1?
2
= e
4j tan
1?
2
,
To plot this we can plot each of the?rst order phase plots and then sum them to get
the?nal fourth order plot,For the?rst order phase plot and making straight line
approximations,we have
H(j?)?
0
4
(log(
2
) + 1)
2
for? √ 0.2
for 0.2 <? < 20
for? → 20
(27)
10
By summing this plot with itself four times,we get the overall phase plot characteristics,
0 for? √ 0.2
H(j?)(log(
2
) + 1) for 0.2 <? < 20 (28)
for? 20?2? →
The approximation is plotted along with the exact function in the graph below,The
exact function,< H(j?) is shown as a dashed line and the approximation is shown as
a solid line,
Problem 2? Bode plot of <H(j?)? approximation and exact function
<H(j
) and <H
approx
(j
) in units of radians/
0.5
0
0.5
1
1.5
2
2.5
<H
approx
(j?)
<H(j?)
10
1
10
0
10
1
10
2
frequency (?)
(a)-(v) To sketch the Bode plots,we again break the function down into its?rst order systems,
nd the bode plot of each?rst order system and then add the two plots together to
get the overall plot,We need to determine the magnitude and phase of the function,
j?
10
1
H(j?) =
1 + j?
The magnitude is found by dividing the magnitude of the numerator with the magni-
tude of the denominator,
0.01?
2
+ 1
H(j?) =| |?
2
+ 1
The logarithm of H(j?) is then,| |
20 log H(j?) = 10 log (0.01?
2
+ 1)? 10 log (?
2
+ 1) (29)| |
11
= 20 log H
1
(j?) + 20 log H
2
(j?)| | | |
To plot 20 log H(j?) we determine the bode plot of each term in Equation 29 and | |
then sum the two plots to get the bode plot for the magnitude of H(j?),| |
20 log H
1
(j?)
10 log 1 = 0 for 10
(30)| |?
20 log 20 log 10 for? ≤ 10
20 log H
2
(j?)
10 log 1 = 0 for 1
(31)| |?
20 log? for? ≤ 1
The bode plots for the two terms are shown below separately,Below the two separate
plots is a plot of the addition of the two plots to give the bode plot for the magnitude
of H(j?),
Prob 2(v)? Bode plot of approximate and exact |H
1
(j?)| Bode plot of approximate and exact |H
2
(j?)|
40
35
30
25
20
15
10
5
0
5
|H
2
(j
)| and |H
2approx
(j
)|
(dB)
5
0
5
10
15
20
25
|H
1
(j
)| and |H
1approx
(j
)|
(dB)
|H
1approx
(j?)|
|H
1
(j?)|
|H
2approx
(j?)|
|H
2
(j?)|
10
1
10
1
10
1
10
0
10
2
10
0
10
2
10
1
frequency (?) frequency (?)
12
Bode plot of approximate and exact |H(j?)|
|H
(
j
)| and |H
(j
)|
(dB)
approx
5
0
5
10
15
20
25
|H
approx
(j?)|
|H
(
j?)|
10
1
10
0
10
1
10
2
frequency (?)
To plot the phase of H(j?),we plot the approximate phase of the numerator of H(j?)
and we plot the approximate phase of the denominator of H(j?) and we add the two
plots together to get the phase plot of H(j?),For the numerator,we have
for? √ 1
H
num
(j?)?
4
(log(
10
)? 1) +
for 1 <? < 100 (32)
2
for? 100
2
→
The approximation is plotted along with the exact function in the graph below,
Problem 2? Bode plot of numerator of <H(j?) (exact and approximate)
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
<H
num
(j
) and <H
num approx
(j
) in units of radians/
<H
num approx
(j?)
<H
num
(j?)
10
1
10
0
10
1
10
2
10
3
frequency (?)
13
For the denominator,we have
0 for? √,1
H
den
(j?)?
4
(log(?) + 1) for,1 <? < 10 (33)
for? 10
2
→
The approximation is plotted along with the exact function in the graph below,
Problem 2? Bode plot of denominator of <H(j?) (exact and approximate)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.1
0.2
<H
den
(j
) and <H
den approx
(j
) in units of radians/
<H
den approx
(j?)
<H
den
(j?)
10
2
10
1
10
0
10
1
10
2
10
3
frequency (?)
The sum of the phases for the numerator and the denominator gives the phase of the
overall function H(j?) and is shown below,
14
Problem 2? Bode plot of <H(j?) (exact and approximate)
10
2
10
1
10
0
10
1
10
2
10
3
0
0.2
0.4
0.6
0.8
1
<H(j
) and <H
approx
(j
) in units of radians/
<H
approx
(j?)
<H(j?)
frequency (?)
Problem 3 O&W 6.32 (b)
Solution,
The system of Figure P6.32 consists of a cascade of a compensator with a a system with
1
the frequency response,H
2
(j?) =
j?+50
,We want the log magnitude of the cascade of the
compensator with H
2
(j?) to have the following specs,
(a) It should have a slope of +20 dB/decade for 0 <? < 10,
(b) It should be between +10 and +30 dB for 10 <? < 100,
(c) It should have a slope of -20 dB/decade for 100 <? < 1000,
(d) It should have a slope of -40 dB/decade for? > 1000,
To get those specs we need to sum the bode plot of H
2
(j?) with the bode plot of the
compensator,C(j?),
The bode plot of the magnitude of H
2
(j?) is as shown below,
20 log H
2
(j?)
20 log 50 for?√ 50
(34)| |?
20 log? for? 50→
The approximation is plotted along with the exact function in the graph below,
15
Bode plot of approximate and exact |H
2
(j?)| for Problem 3
30
35
|H
2
(j
)| and |H
2 approx
(j
)|
(dB)
40
45
50
55
60
65
|H
2approx
(j?)|
|H
2
(j?)|
10
0
10
1
10
2
10
3
frequency (?)
The above bode plot needs to be cascaded with some compensator function,C(j?) to
give us the specs listed above,We will determine C(j?) to be some cascaded combination
of a constant and one or more of the four frequency responses,H
a
(j?) = j?,H
b
(j?) =
j?
,
1
H
c
(j?) =
1+j
k
and H
d
(j?) = (1 + j?ω
k
) with appropriate ω
k
’s selected,
C
k
To meet the?rst and second spec,we need H
a
(j?) = j? and H
c
(j?) =
1+j
k
,By
selecting ω
k
= 0.1,there will be +20 dB/decade for? < 10 due to H
a
(j?) and there will be 0
dB/decade for? > 10 due to the cascade of H
a
(j?) and H
c
(j?),H
a
(j?) will meet the?rst
spec because it is a continuous slope of +20 dB/decade and H
c
(j?) will ‘turn o?’ the +20
dB/decade at an appropriate? and make the slope = 0 for frequencies greater than that?,
For the third spec,we don’t need to add anything since H
2
(j?) will give us -20 dB/decade
for? > 50,However we need to cascade another H
c
(j?) to get -40 dB/decade for? > 1000,
Choosing ω
k
= 0.001 for H
c
(j?) will cause the slope change from -20 dB/decade to -40
dB/decade to occur at? = 1000,
Thus far we have the following frequency response for the compensator,
C
k
j?
C(j?) =
(1 + j0.1?)(1 + j0.001?)
The last part is to determine the constant,C
k
of the compensator,Because 20 log H
2
(j?) =
20 log 50 =?34 dB for? < 50 and because the second spec requires the overall log
magnitude to be between +10 dB and +30 dB for 10 <? < 100 we select the log magnitude
of C(j?) at? = 10 to add to the -34 dB and give +29.9 dB,We choose to be close to the
16
1
upper bound of the spec (close to +30 dB as opposed to somewhere in the middle of +10
dB and +30 dB) because at? = 50,H
2
(j?) begins to decrease at a slope of -20 dB/decade,
Thus,we algebraically determine
34 + 20 log C
k
+ 20 log 10 = 29.9
yields C
k
= 156.3,The?nal compensator frequency response is
156.3j?
C(j?) =,
(1 + j0.1?)(1 + j0.001?)
The?nal frequency response for H(j?) is
156.3j?
H(j?) =,
(1 + j0.1?)(50 + j?)(1 + j0.001?)
The?nal frequency response for H(j?) is plotted as an approximation and with the exact
function in the graph below,
Bode plot of approximate and exact |H(j?)| for Problem 3
|H(j
)| and |H
approx
(j
)|
(dB)
30
20
10
0
10
20
30
40
|H
approx
(j?)|
|H(j?)|
10
1
10
0
10
1
10
2
10
3
10
4
frequency (?)
Problem 4 O&W 6.39 (j)
Solution,
We need to sketch the log magnitude and phase of the the function,
1
H(e
j?
) =,
(1? 0.25e
j?
)(1 + 0.75e
j?
)
17
1
We can write this as a cascade of two frequency responses,H
1
(e
j?
) =
(1?0.25e
j?
)
and
1
,ThenH
2
(e
j?
) =
(1+0.75e
j?
)
20 log H(e
j?
) = 20 log |H
1
(e
j?
) + 20 log H
2
(e
j?
)| | | | |
and
H(e
j?
) =?H
1
(e
j?
) +?H
2
(e
j?
)
We need to determine 20 log |H
1
(e
j?
) and 20 log H
2
(e
j?
), Using complex number manipu-| | |
lation,we get,
20 log H
1
(e
j?
) =?10 log ((1? 0.25 cos?)
2
+ 0.25
2
sin
2
)| |
and
20 log H
2
(e
j?
) =?10 log ((1 + 0.75 cos?)
2
+ 0.75
2
sin
2
)| |
The two bode plots were calculated using Matlab and are shown below separately,Note the
di?erent ordinate axis scales for each frequency response,The magnitude of H
1
(e
j?
) is much
less than the magnitude of H
2
(e
j?
),Note also that because the coe?cent in the denominator
is positive (+0.25) for H
1
(e
j?
) whereas it is negative (?0.75) for H
2
(e
j?
),their peak values
are shifted by? from each other,H
1
(e
j?
) has more low frequency components and H
2
(e
j?
)
has more high frequency components,
20log|H
1
(e
j?
)|? Prob,4 (O&W 6.39(j)) 20log|H
2
(e
j?
)|
3 14
12
2
10
8
1
20log|H
2
(e
j
)|
dB
20log|H
1
(e
j
)| dB
6
0 4
2
1
0
2
2
4
3?6
2?1 0 1 2?2?1 0 1 2
frequency normalized by? (?/?) frequency normalized by? (?/?)
The two functions are shown summed together to give the overall log magnitude of H(e
j?
),
18
20log|H(e
j?
)| for Problem 4 (O&W 6.39(j))
12
10
20log|H(e
j
)|
dB
8
6
4
2
0
2
4
2?1.5?1?0.5 0 0.5 1 1.5 2
frequency normalized by? (?/?)
To determine?H(e
j?
),we?nd and plot?H
1
(e
j?
) and?H
2
(e
j?
) and then sum the two
functions to get?H(e
j?
),
1 0.25 sin?
H
1
(e
j?
) =?
1? 0.25e
j?
=? tan
1
1? 0.25 cos?
and
1 0.75 sin?
H
2
(e
j?
) =? =? tan
1
,
1 + 0.75e
j?
1 + 0.75 cos?
Each of these phases are plot separately and summed together to give the overall phase of
H(e
j?
),
19
<H e
j?
<H e
j?
1 2
0.1 0.3
0.08
0.2
0.06
<H
1
(e
j
normalized by
(radians/
)
0.04
0.02
0
0.02
0.04
<H
2
(e
j
normalized by
(radians/
)
0.1
0
0.1
0.2
<H(e
j
normalized by
(radians/
)
0.06
0.3
0.08
0.1
2?1 0 1
frequency normalized by? (?/?)
0.4
2?2?1 0 1 2
frequency normalized by? (?/?)
<He
j?
0.25
0.2
0.15
0.1
0.05
0
0.05
0.1
0.15
0.2
0.25
2?1.5?1?0.5 0 0.5 1 1.5 2
frequency normalized by? (?/?)
Problem 5 O&W 6.43 (a)
Solution,
We need to transform the low pass?lter,H
lp
(e
j?
) shown in Figure P6.43 of O&W into a
high pass?lter,H
lp
(e
j?
),In this problem we accomplish this by modulating h
lp
[n],Speci?-
20
cally,h
hp
[n] = (? 1)
n
h
lp
[n],This is equivalent to shifting the frequency response by? since
F
from Table 5.1 of O&W we have e
j?
o
n
x[n]?≈ X(e
j(
o
)
) and we know that e
j?n
= (? 1)
n
,
Thus,
j()
H
hp
(e
j?
) = H
lp
e,
The frequency response,H
hp
(e
j?
) is shown below superimposed onto the original H
lp
(e
j?
):
H
j? j?
)
H
j?
)
H
j?
)
2?
lp
(e ),H
hp
(e
2?
lp
(e
hp
(e
Because a discrete signal is periodic with period = 2?,its low frequency components
occur periodically at 0,± 2?,± 4?,etc,Thus,its high frequency components occur halfway
between the low frequency components at ±?,± 3?,etc,The shifted frequency response
corresponds to a high pass?lter since the original signal had its peak magnitude centered
around 0,± 2?,etc,and the new signal is shifted by? from that original signal,
Problem 6 O&W 6.58 (a),(b)
Solution,
21
(a) 1,We need to?nd h
1
[n] as a function of h[n],To accomplish this we investigate
H
1
(e
j?
),From Figure P6.58 of O&W,we can write
j?
)
G(e
j?
)
H(e =
j?
)
,(35)
X(e
j?
)
R(e
j?
)
H(e = (36)
G(e
j?
)
R(e
j?
) = S(e
j?
),(37)
We can manipulate the equations to write,
R(e
j?
) = G(e
j?
)H(e
j?
) (38)
j?
)H(e
j?
)H(e
j?
).= X(e (39)
Thus,
H
1
(e
j?
) =
S(e
j?
)
(40)
j?
)X(e
j?
)H(e
j?
).= H(e (41)
From this function,we recognize by the convolution property,h
1
[n] = h[n] h[?n].←
We can show that the system has zero phase characteristic by noting that since
h[n] is real,H(e
j?
) = H
(e
j?
),Thus,
H
1
(e
j?
) = H(e
j?
)H(e
j?
) = H(e
j?
)H
(e
j?
) = |H(e
j?
)
2
.|
2,Because H
1
(e
j?
)| = |H(e
j?
)
2
,then the magnitude H
1
(e
j?
) = H(e
j?
)
2
and the | | | | |
phase?H
1
(e
j?
) = 0,
(b) 1,We look at the Fourier transforms to determine h
2
[n],Since y[n] = g[n] + r[?n],
Y (e
j?
) = G(e
j?
) + R(e
j?
),From Figure P6.58 of O&W we determine
j?
)X(e
j?
)Y (e
j?
) = G(e
j?
) + R(e
j?
) = H(e
j?
) + H(e
j?
)X(e
Thus,
H
2
(e
j?
) = H(e
j?
) + H(e
j?
)?≈ h
2
[n] = h[n] + h[?n],
To determine if |H
2
(e
j?
) has zero phase characteristic,we manipulate H
2
(e
j?
)| | |
to get it into polar form,We use the property in Table 5.1 that since h[n] is real,
then ≥e{H(e
j?
)} = ≥e{H(e
j?
)} and ∞m{H(e
j?
)} = {H(e
j?
)},Hence,?∞m
|H
2
(e
j?
) = {H(e
j?
)} + ∞m{H(e
j?
)} + ≥e{H(e
j?
)} + ∞m{H(e
j?
)}|(42)| |≥e
j?
)}= 2≥e{H(e (43)
= 2 H(e
j?
)| cos (< H(e
j?
)),(44)|
From this,we can see that |H
2
(e
j?
) has zero phase characteristic,|
2,Since we already put |H
2
(e
j?
) in polar form,we know that |
|H
2
(e
j?
) = 2 H(e
j?
) cos (< H(e
j?
)). (45)| | |
and we know that?H
2
(e
j?
) = 0,
22
Problem 7 O&W 7.22
Solution,
In this problem we need to?gure out a range of values for the sampling period,T,to recover
y(t) completely from y
p
(t),To do this we need to determine the bandwidth of the original
Y (j?) and use the sampling theorem,By the convolution property,Y (j?) = X
1
(j?)X
2
(j?),
The bandwidth of Y (j?) then will be the bandwidth of the smaller of the two bandwidths,
X
1
(j?) or X
2
(j?),Hence,Y (j?) = 0 for? > 1000?,Then,using the sampling theorem,| |
2?
s
= > 2?
m
= 2(1000?),
T
This gives the range of T as 0 < T < 0.001 seconds,
Problem 8 O&W 7.23
Solution,
(a) We need to sketch X
p
(j?) and Y (j?),In the frequency domain,X
p
(j?) =
1
X(j?) ←
2?
P(j?),We need to determine P(j?),Since P(j?) is periodic,we need to use the
periodic Fourier transform formula,That is
P(j?) = 2? a
k
( k?
o
),
k=
1
T
p(t)e
jk?
o
t
Here,?
o
=
2?
=
,We need to determine the a
k
’s using the formula a
k
=,
T T
a
A few are shown below:
1
2?
0
= (?(t)(t))dt = 0
2?
0
1
2?
1 1
t
dt = (1? 1 e
j?
) =
2?
·
j?
(?( )?(?))et t
a
1
=
2?
1
0
2?
1
t
dt = (1? 1 e
j2?
) = 0
2?
·
2j?
(?( )?(?))et t
a
2
=
2?
1
0
2?
1 1
t
dt = (1? 1 e
j3?
) =
2?
·
3j?
(?( )?(?))et t
a
3
=
2?
0
1
Thus,a
k
= 0 for k even and a
k
=
for k odd and
2
2?
P(j?) =?( k ) = (2k + 1)
kodd k=
From this Fourier transform for P(j?),we can sketch X
p
(j?) as copies of X(j?) scaled
1
by
and replicated at intervals of? = (2k + 1)
,for all k,This can be seen in the
gure below,
23
X
p
)
1/?
(j
4?/3?/2?///? 2?/? 3?/? 4?/?
H(j?) is a sum of two ideal unity gain bandpass?lters,Thus,Y (j?) is the part of
X
p
(j?) that is passed through H(j?),This is shown below,
Y?)
1/?
(j
4?/3?/2?///? 2?/? 3?/? 4?/?
(b) To recover
p
(t x
) we need to do two things,First,we need to multiply
with a cosine function,cos
x(t) from x
p
(t)
t,This will shift X
p
(j?) such that one of the copies of
X(j?) is centered around? = 0,Second,we send the shifted signal through a lowpass
lter,R(j?),to eliminate the extra copies of X(j?),To achieve this we have a?lter,
R(
2?
j?) with gain =?,bandwidth and centered around? = 0,This is shown below,
)
R(j
//? 2?/?
The overall system is shown below,
cos(?/?t)
x
p
)
x x(t) (t)
R(j
24
=
(c) To recover x(t) from y(t) we need to run Y (j?) through two parallel?lter systems.
The top parallel path will multiply ( ) by cos ty
t which will shift the demi-replicate of
X(j?) that is centered at? = over to? = 0,The shifted signal then passes through
the lowpass?lter,R(j?) described above in part (b) to eliminate the extra copies.
3?
The bottom parallel path will multiply ( ) by cos ty
t which will shift the demi-replicate
of X(j?) that is centered at? =
3?
over to? = 0,The shifted signal then passes
through the lowpass?lter,R(j?) described above in part (b) to eliminate the extra
copies,Thus,the two halves combine together to form a complete X(j?) and x(t) is
recovered,The overall system is shown below,
+
x
x
t/?)
)
)
x(t) y(t)
cos(
R(j
R(j
cos(3?t/?)
(d) To recover x(t) from x
p
(t) and y(t),X
p
(j?) cannot have any overlap in the copies of
X(j?),Because of this particular p(t),the copies of X(j?) are at? = (2k + 1)
for all
k,Thus,just looking at one interval to make sure the copies of X(j?) don’t overlap,
we have one copy of X(j?) centered at? =
and one copy of X(j?) centered at
3?
,(See Figure of X
p
(j?) above),For no overlap between these copies,
3?
+?
m
<
m
,
which gives
< or?
max
=,
m
m
25
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 6 Solutions
Issued,October 21,2003 Due,October 29,2003
Exercise for home study,
O&W 6.49
Solution,
(a) To determine the time constants of the di?erential equation P6.49-1 in O&W,we
perform a Fourier transform of the equation,By linearity,the Fourier transform of the
equation is the Fourier transform of each of the individual equation terms,This gives
Y (j?)
X(j?)
=
9
(j?)
2
+ 11(j?) + 10
= H(j?),(1)
The time constants are the zeros of the denominator (c
1
=?1 and c
2
=?10),
(b) Equation 1 can be rewritten as
9
H(j?) =, (2)
(j? + 1)(j? + 10)
To make this a parallel interconnection of two?rst order systems,we do a partial
fraction expansion on Equation 2 to?nd
1
H(j?) = +
1
. (3)
j? + 1 j? + 10
By linearity,the inverse Fourier transform,h(t),is a parallel interconnection of two
rst order systems,h(t) is the sum of the inverse Fourier transform of each term in
Equation 3,Each term can be quickly determined using Table 4.2 of O&W,Thus,
u(t)? e
10t
h(t) = e
t
u(t). (4)
(c) The dominant time constant in a system with multiple decaying exponentials is the
time constant that takes the longest to die out,For this problem,the dominant time
constant is c
1
=?1,We can approximate this to Equation 1 as shown,
9 1
. (5)H(j?) =
(j?)
2
+ 11(j?) + 10
j? + 1
1
This approximation has a maximum di?erence of 10% at? = 0 and the di?erence
decreases as? increases,
(d) We want to approximate the faster component of Equation 4 as an impulse with a
height equal to its?nal value,We will then check to see how this approximation
a?ects the overall h(t) of the entire second order system,
The faster component,h
f
(t)=?e
10t
u(t),Let
h
f
(t) denote the approximation,For
h
f
(t) = s
f
(?)?(t),we need to determine s
f
(?),The step response is de?ned as
t
s(t) = h(t)dt,
Thus,
1
e
10t
dt =s
f
(?) = h
f
(t)dt =,?
10
0
h
f
(t) =?
1
(t) and the approximation to the overall system is
10
h
f
(t) + h
s
(t) =?h(t) =
1
(t) + e
t
u(t),(6)
10
Equation 6 enables us to determine the frequency response,We determine,
1
H(j?) =?0.1 + =
0.1j? + 0.9
,(7)
j? + 1 j? + 1
We then do algebraic manipulations and an inverse Fourier transform to Equation 7
to get the di?erential equation of the approximation,This gives,
dy dx
+ y(t) = 0.9x(t)? 0.1,
dt dt
The original frequency response,H(j?) and the approximate frequency response
H(j?)
are shown below,
2
Exercise,Problem 6.49? |H(j?)| Exercise,Problem 6.49? This is the approximate to |H(j?)|
1 1
0.9 0.9
0.8 0.8
0.7 0.7
0.6 0.6
0.4 0.4
0.3 0.3
0.2 0.2
0.1 0.1
0
0 20 40 60 80
frequency (?)
0
100 0 20 40 60 80 100
frequency (?)
(j
)|
|H
approx
|H(j
)|
0.5 0.5
The two are shown on the same plot below,The plot of H(j?) is solid and the plot of
H(j?) is dashed,
3
Exercise,Problem 6.49? Both |H(j?)| and |H
approx
(j?| Magnification of |H(j?)| and |H
approx
(j?| at low frequencies
1 1
|H(j?)|
|H
appr
(j?)|
ox
|H(
|H
a
j?)|
pprox
(j?)|
0.9
0.9
0.8
0.8
0.7
|H(j
)| and |H
approx
(j
)|
|H(j
)| and |H
approx
(j
)|
0.6
0.5
0.4
0.3
0.7
0.6
0.5
0.2
0.4
0.1
0
0 10 20 30 40 50
frequency (?)
0.3
0 0.5 1 1.5 2
frequency (?)
From the graphs above,it is visible that for? < 1,H(j?) approximates H(j?) well,
As? increases beyond 1,the two diverge from each other until roughly? = 20,This
is due to the fact that the approximation to the fast part of the signal reaches its?nal
value instantaneously,hence it must include all fast frequency components,
We can also compare the step response of the original signal,s(t) with the step response
of the approximate signal,?s(t),
4
t
s(t) = h(t)dt (8)
t t
= e
t
dt? e
10t
dt (9)
o 0
= (0.9? e
t
+ 0.1e
10t
)u(t),(10)
t
s?(t) = h(t)dt (11)
t t
= e
t
dt? (?0.1)?(t)dt (12)
o 0
= (0.9? e
t
)u(t),(13)
The two step responses are plotted together below,
5
Exercise,Problem 6.49? Both s(t) and s
approx
)t) Magnification of s(t) and s
approx
(t)
1
0.4
0.5
0.6
0.7
0.8
0.9
s(t) and s
approx
(t)
s(t)
s
approx
(t)
0.4
s(t) and s
approx
(t)
0.35
0.3
0.25
0.2
0.15
0.1
0.3 0.05
0.2 0
0.1?0.05
0
0 2 4 6 8
time (sec)
0.1
10 0 0.1 0.2 0.3 0.4 0.5
time (sec)
From the graphs above,it is apparent that for t >,5 seconds,the two step responses
are nearly equal,Furthermore,a general rule of thumb is that an exponential nearly
reaches its?nal value after 5 time constants,Thus,for e
10t
u(t),it should nearly reach
its?nal value by t = 0.5 seconds,
Problems to be turned in,
Problem 1 O&W 4.35
s(t)
s
a
(t)
pprox
Solution,
6
(a) We need to?nd the magnitude,the phase and the impulse response of the continuous-
time LTI system with frequency response
H(j?) =
a? j?
,
a + j?
Determining the magnitude,
a
2
+?
2
H(j?) = = 1.| |?
a
2
+?
2
Determining the phase,
e
j arctan
a
e =
e
j<H(j?)
= e
2j arctan
j arctan
a
,
a
Determining the impulse response,
+ F
1
j?
h(t) = F
1
{H(j?)} = F
1
a
= h
1
(t) + h
2
(t),
a + j? a + j?
From table 4.2,h
1
(t) = ae
at
u(t),To determine h
2
(t),we do the following,
{
j? 1 1 dh
1
(t)
h
2
(t) = F
1
(j?)H
1
(j?)} =,
a + j?
} = F
1
{?
a
a dt
Using the product rule,
dh
1
(t)
= ae
at
(t)? a
2
e
at
u(t),
dt
Thus,h
2
(t) =?e
at
(t) + ae
at
u(t),Combining h
1
(t) and h
2
(t) gives,
u(t)? e
at
(t).h(t) = 2ae
at
Alternatively we can do a partial fraction expansion of H(j?),
2a
H(j?) =
j? + a
1,
Now,h(t) can be found by looking up each term above in the tables to give the same
u(t)? e
at
(t).result h(t) = 2ae
at
(b) We need to determine y(t) when a = 1 and x(t) = cos(t/
3) + cos(t) + cos(
3t),We
can use the convolution property,Y (j?) = X(j?) × H(j?),We have H(j?),we need
to determine X(j?),
1 1
X(j?) =?[?( ) +?(? + ) +?( 1) +?(? + 1)?
3
3
+?(
3) +?(? +
3)] (14)
=?[X
a
(j?) + X
b
(j?) + X
c
(j?)],(15)
7
3
Because the system is LTI,we can write the output as the sum of three terms,
Y (j?) = Y
a
(j?) + Y
b
(j?) + Y
c
(j?),(16)
where the?rst term is de?ned as
Y
a
(j?) = H(j?)X
a
(j?) (17)
a? j?
=?X
a
(j?) (18)
a + j?
= e
2j tan
1
a
X
a
(j?) (19)
1
=?e
2j arctan
1
3
) +?e
2j arctan
a
(
a
(? + ),(20)?
3
We now substitue a = 1 into Y
a
(j?),Also,because X
a
(j?) is the sum of two delta
functions,we can substitute? =
1
3
into the arctangent term in the?rst? term of
Y
a
(j?) and? =
1
into the arctangent term for the second? term of Y
a
(j?),This
3
gives,
Y
a
(j?) =?e
j
3
3
+
1
.
3
1
+?e
j
3
This is now recognizable as a cosine function in the time domain with a phase change
of
,Hence,
1?
y
a
(t) = cos t?,?
3
3
The other terms,Y
b
(j?) and Y
c
(j?) can be solved similarly to give cosine functions
with various phase changes,We get,
j
Y
b
(j?) =?e
j
2
( 1) +?e
2
(? + 1),(21)
y
b
(t) = cos t? = sin(t),(22)
2
j
2?
3 3
Y
c
(j?) =?e
j
2?
(
3) +?e?(? +
3),(23)
2?
y
c
(t) = cos
3t?, (24)
3
Summing all the terms,
1? 2?
y(t) = cos t? + sin(t) + cos
3t?,?
3
3 3
These two functions,y(t) and x(t) are plotted together below,They are nonperiodic
functions,Why are they nonperiodic? Because the three periods that sum to make
8
up the function are 3 di?erent irrational numbers and hence we will?nd no rational
number,T,where each term will have an integer number of waveforms in that T,The
period of the?rst term is T
a
= 2?
3,,the period of the second term is T
b
= 2?,and
the period of the last term is T
c
= 2?
1
.
3
Problem 1? Both x(t) and y(t)
x(t) and y(t)
3
2
1
0
1
2
3
x(t)
y(t)
10?8?6?4?2 0 2 4 6 8 10
time (pi*seconds)
Problem 2 O&W 6.28 (a)-(iii) and (v)
Solution,
16
(a)-(iii) To sketch the frequency response of H(j?) =
(j?+2)
4
,we recognize that this is a cascade
of 4 identical?rst order systems,It’s magnitude,H(j?) can be written as,| |
1 1 1 1
|H(j?) = H
1
(j?)
4
= ( )( )( )( )| | | |
0.5j? + 1 0.5j? + 1 0.5j? + 1 0.5j? + 1
|
1 1 1 1
= ( )( )( )( )?
1 + 0.25?
2
1 + 0.25?
2
1 + 0.25?
2
1 + 0.25?
2
Because the system is LTI,the bode plot of the cascade of the 4 identical?rst order
systems is equal to the sum of the bode plots for each?rst order system,The?rst
order system,20 log H
1
(j?) =?10 log(1 + 0.25?
2
) has the plot,| |
0 for 2
(25)?10 log(1 + 0.25?
2
)?
20 log(0.5?) for? ≤ 2
9
Therefore the bode plot of the cascade of the 4?rst order systems has the plot,
H(j?)
0 for 2
(26)| |?
80 log(0.5?) for? ≤ 2
The following graph illustrates the plot,The two straight-line approximations intersect
at? = 2,The solid line is the approximation to H(j?) and the dashed line is the real | |
H(j?),| |
Problem 2? Bode plot of approximate and real |H(j?)|
|H(j
)| and |H
approx
(j
)|
(dB)
0
50
100
150
|H
approx
(j?)|
|H(j?)|
10
1
10
0
10
1
10
2
frequency (?)
We can see from the graph and by calculation that at? = 2,the actual magnitude is
H(j?) =?40 log(1 + 0.25 2
2
) =?40 log(2) =?12 dB,| ←
The phase is next determined and sketched,Again expressing the function as the
cascade of 4 identical?rst order functions we have,
j?H(j?)
e =
e
j tan
1
(0)
e
j tan
1
(0)
e
j tan
1
(0)
e
j tan
1
(0)
e
j tan
1?
2
e
j tan
1?
2
e
j tan
1?
2
e
j tan
1?
2
= e
4j tan
1?
2
,
To plot this we can plot each of the?rst order phase plots and then sum them to get
the?nal fourth order plot,For the?rst order phase plot and making straight line
approximations,we have
H(j?)?
0
4
(log(
2
) + 1)
2
for? √ 0.2
for 0.2 <? < 20
for? → 20
(27)
10
By summing this plot with itself four times,we get the overall phase plot characteristics,
0 for? √ 0.2
H(j?)(log(
2
) + 1) for 0.2 <? < 20 (28)
for? 20?2? →
The approximation is plotted along with the exact function in the graph below,The
exact function,< H(j?) is shown as a dashed line and the approximation is shown as
a solid line,
Problem 2? Bode plot of <H(j?)? approximation and exact function
<H(j
) and <H
approx
(j
) in units of radians/
0.5
0
0.5
1
1.5
2
2.5
<H
approx
(j?)
<H(j?)
10
1
10
0
10
1
10
2
frequency (?)
(a)-(v) To sketch the Bode plots,we again break the function down into its?rst order systems,
nd the bode plot of each?rst order system and then add the two plots together to
get the overall plot,We need to determine the magnitude and phase of the function,
j?
10
1
H(j?) =
1 + j?
The magnitude is found by dividing the magnitude of the numerator with the magni-
tude of the denominator,
0.01?
2
+ 1
H(j?) =| |?
2
+ 1
The logarithm of H(j?) is then,| |
20 log H(j?) = 10 log (0.01?
2
+ 1)? 10 log (?
2
+ 1) (29)| |
11
= 20 log H
1
(j?) + 20 log H
2
(j?)| | | |
To plot 20 log H(j?) we determine the bode plot of each term in Equation 29 and | |
then sum the two plots to get the bode plot for the magnitude of H(j?),| |
20 log H
1
(j?)
10 log 1 = 0 for 10
(30)| |?
20 log 20 log 10 for? ≤ 10
20 log H
2
(j?)
10 log 1 = 0 for 1
(31)| |?
20 log? for? ≤ 1
The bode plots for the two terms are shown below separately,Below the two separate
plots is a plot of the addition of the two plots to give the bode plot for the magnitude
of H(j?),
Prob 2(v)? Bode plot of approximate and exact |H
1
(j?)| Bode plot of approximate and exact |H
2
(j?)|
40
35
30
25
20
15
10
5
0
5
|H
2
(j
)| and |H
2approx
(j
)|
(dB)
5
0
5
10
15
20
25
|H
1
(j
)| and |H
1approx
(j
)|
(dB)
|H
1approx
(j?)|
|H
1
(j?)|
|H
2approx
(j?)|
|H
2
(j?)|
10
1
10
1
10
1
10
0
10
2
10
0
10
2
10
1
frequency (?) frequency (?)
12
Bode plot of approximate and exact |H(j?)|
|H
(
j
)| and |H
(j
)|
(dB)
approx
5
0
5
10
15
20
25
|H
approx
(j?)|
|H
(
j?)|
10
1
10
0
10
1
10
2
frequency (?)
To plot the phase of H(j?),we plot the approximate phase of the numerator of H(j?)
and we plot the approximate phase of the denominator of H(j?) and we add the two
plots together to get the phase plot of H(j?),For the numerator,we have
for? √ 1
H
num
(j?)?
4
(log(
10
)? 1) +
for 1 <? < 100 (32)
2
for? 100
2
→
The approximation is plotted along with the exact function in the graph below,
Problem 2? Bode plot of numerator of <H(j?) (exact and approximate)
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
<H
num
(j
) and <H
num approx
(j
) in units of radians/
<H
num approx
(j?)
<H
num
(j?)
10
1
10
0
10
1
10
2
10
3
frequency (?)
13
For the denominator,we have
0 for? √,1
H
den
(j?)?
4
(log(?) + 1) for,1 <? < 10 (33)
for? 10
2
→
The approximation is plotted along with the exact function in the graph below,
Problem 2? Bode plot of denominator of <H(j?) (exact and approximate)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.1
0.2
<H
den
(j
) and <H
den approx
(j
) in units of radians/
<H
den approx
(j?)
<H
den
(j?)
10
2
10
1
10
0
10
1
10
2
10
3
frequency (?)
The sum of the phases for the numerator and the denominator gives the phase of the
overall function H(j?) and is shown below,
14
Problem 2? Bode plot of <H(j?) (exact and approximate)
10
2
10
1
10
0
10
1
10
2
10
3
0
0.2
0.4
0.6
0.8
1
<H(j
) and <H
approx
(j
) in units of radians/
<H
approx
(j?)
<H(j?)
frequency (?)
Problem 3 O&W 6.32 (b)
Solution,
The system of Figure P6.32 consists of a cascade of a compensator with a a system with
1
the frequency response,H
2
(j?) =
j?+50
,We want the log magnitude of the cascade of the
compensator with H
2
(j?) to have the following specs,
(a) It should have a slope of +20 dB/decade for 0 <? < 10,
(b) It should be between +10 and +30 dB for 10 <? < 100,
(c) It should have a slope of -20 dB/decade for 100 <? < 1000,
(d) It should have a slope of -40 dB/decade for? > 1000,
To get those specs we need to sum the bode plot of H
2
(j?) with the bode plot of the
compensator,C(j?),
The bode plot of the magnitude of H
2
(j?) is as shown below,
20 log H
2
(j?)
20 log 50 for?√ 50
(34)| |?
20 log? for? 50→
The approximation is plotted along with the exact function in the graph below,
15
Bode plot of approximate and exact |H
2
(j?)| for Problem 3
30
35
|H
2
(j
)| and |H
2 approx
(j
)|
(dB)
40
45
50
55
60
65
|H
2approx
(j?)|
|H
2
(j?)|
10
0
10
1
10
2
10
3
frequency (?)
The above bode plot needs to be cascaded with some compensator function,C(j?) to
give us the specs listed above,We will determine C(j?) to be some cascaded combination
of a constant and one or more of the four frequency responses,H
a
(j?) = j?,H
b
(j?) =
j?
,
1
H
c
(j?) =
1+j
k
and H
d
(j?) = (1 + j?ω
k
) with appropriate ω
k
’s selected,
C
k
To meet the?rst and second spec,we need H
a
(j?) = j? and H
c
(j?) =
1+j
k
,By
selecting ω
k
= 0.1,there will be +20 dB/decade for? < 10 due to H
a
(j?) and there will be 0
dB/decade for? > 10 due to the cascade of H
a
(j?) and H
c
(j?),H
a
(j?) will meet the?rst
spec because it is a continuous slope of +20 dB/decade and H
c
(j?) will ‘turn o?’ the +20
dB/decade at an appropriate? and make the slope = 0 for frequencies greater than that?,
For the third spec,we don’t need to add anything since H
2
(j?) will give us -20 dB/decade
for? > 50,However we need to cascade another H
c
(j?) to get -40 dB/decade for? > 1000,
Choosing ω
k
= 0.001 for H
c
(j?) will cause the slope change from -20 dB/decade to -40
dB/decade to occur at? = 1000,
Thus far we have the following frequency response for the compensator,
C
k
j?
C(j?) =
(1 + j0.1?)(1 + j0.001?)
The last part is to determine the constant,C
k
of the compensator,Because 20 log H
2
(j?) =
20 log 50 =?34 dB for? < 50 and because the second spec requires the overall log
magnitude to be between +10 dB and +30 dB for 10 <? < 100 we select the log magnitude
of C(j?) at? = 10 to add to the -34 dB and give +29.9 dB,We choose to be close to the
16
1
upper bound of the spec (close to +30 dB as opposed to somewhere in the middle of +10
dB and +30 dB) because at? = 50,H
2
(j?) begins to decrease at a slope of -20 dB/decade,
Thus,we algebraically determine
34 + 20 log C
k
+ 20 log 10 = 29.9
yields C
k
= 156.3,The?nal compensator frequency response is
156.3j?
C(j?) =,
(1 + j0.1?)(1 + j0.001?)
The?nal frequency response for H(j?) is
156.3j?
H(j?) =,
(1 + j0.1?)(50 + j?)(1 + j0.001?)
The?nal frequency response for H(j?) is plotted as an approximation and with the exact
function in the graph below,
Bode plot of approximate and exact |H(j?)| for Problem 3
|H(j
)| and |H
approx
(j
)|
(dB)
30
20
10
0
10
20
30
40
|H
approx
(j?)|
|H(j?)|
10
1
10
0
10
1
10
2
10
3
10
4
frequency (?)
Problem 4 O&W 6.39 (j)
Solution,
We need to sketch the log magnitude and phase of the the function,
1
H(e
j?
) =,
(1? 0.25e
j?
)(1 + 0.75e
j?
)
17
1
We can write this as a cascade of two frequency responses,H
1
(e
j?
) =
(1?0.25e
j?
)
and
1
,ThenH
2
(e
j?
) =
(1+0.75e
j?
)
20 log H(e
j?
) = 20 log |H
1
(e
j?
) + 20 log H
2
(e
j?
)| | | | |
and
H(e
j?
) =?H
1
(e
j?
) +?H
2
(e
j?
)
We need to determine 20 log |H
1
(e
j?
) and 20 log H
2
(e
j?
), Using complex number manipu-| | |
lation,we get,
20 log H
1
(e
j?
) =?10 log ((1? 0.25 cos?)
2
+ 0.25
2
sin
2
)| |
and
20 log H
2
(e
j?
) =?10 log ((1 + 0.75 cos?)
2
+ 0.75
2
sin
2
)| |
The two bode plots were calculated using Matlab and are shown below separately,Note the
di?erent ordinate axis scales for each frequency response,The magnitude of H
1
(e
j?
) is much
less than the magnitude of H
2
(e
j?
),Note also that because the coe?cent in the denominator
is positive (+0.25) for H
1
(e
j?
) whereas it is negative (?0.75) for H
2
(e
j?
),their peak values
are shifted by? from each other,H
1
(e
j?
) has more low frequency components and H
2
(e
j?
)
has more high frequency components,
20log|H
1
(e
j?
)|? Prob,4 (O&W 6.39(j)) 20log|H
2
(e
j?
)|
3 14
12
2
10
8
1
20log|H
2
(e
j
)|
dB
20log|H
1
(e
j
)| dB
6
0 4
2
1
0
2
2
4
3?6
2?1 0 1 2?2?1 0 1 2
frequency normalized by? (?/?) frequency normalized by? (?/?)
The two functions are shown summed together to give the overall log magnitude of H(e
j?
),
18
20log|H(e
j?
)| for Problem 4 (O&W 6.39(j))
12
10
20log|H(e
j
)|
dB
8
6
4
2
0
2
4
2?1.5?1?0.5 0 0.5 1 1.5 2
frequency normalized by? (?/?)
To determine?H(e
j?
),we?nd and plot?H
1
(e
j?
) and?H
2
(e
j?
) and then sum the two
functions to get?H(e
j?
),
1 0.25 sin?
H
1
(e
j?
) =?
1? 0.25e
j?
=? tan
1
1? 0.25 cos?
and
1 0.75 sin?
H
2
(e
j?
) =? =? tan
1
,
1 + 0.75e
j?
1 + 0.75 cos?
Each of these phases are plot separately and summed together to give the overall phase of
H(e
j?
),
19
<H e
j?
<H e
j?
1 2
0.1 0.3
0.08
0.2
0.06
<H
1
(e
j
normalized by
(radians/
)
0.04
0.02
0
0.02
0.04
<H
2
(e
j
normalized by
(radians/
)
0.1
0
0.1
0.2
<H(e
j
normalized by
(radians/
)
0.06
0.3
0.08
0.1
2?1 0 1
frequency normalized by? (?/?)
0.4
2?2?1 0 1 2
frequency normalized by? (?/?)
<He
j?
0.25
0.2
0.15
0.1
0.05
0
0.05
0.1
0.15
0.2
0.25
2?1.5?1?0.5 0 0.5 1 1.5 2
frequency normalized by? (?/?)
Problem 5 O&W 6.43 (a)
Solution,
We need to transform the low pass?lter,H
lp
(e
j?
) shown in Figure P6.43 of O&W into a
high pass?lter,H
lp
(e
j?
),In this problem we accomplish this by modulating h
lp
[n],Speci?-
20
cally,h
hp
[n] = (? 1)
n
h
lp
[n],This is equivalent to shifting the frequency response by? since
F
from Table 5.1 of O&W we have e
j?
o
n
x[n]?≈ X(e
j(
o
)
) and we know that e
j?n
= (? 1)
n
,
Thus,
j()
H
hp
(e
j?
) = H
lp
e,
The frequency response,H
hp
(e
j?
) is shown below superimposed onto the original H
lp
(e
j?
):
H
j? j?
)
H
j?
)
H
j?
)
2?
lp
(e ),H
hp
(e
2?
lp
(e
hp
(e
Because a discrete signal is periodic with period = 2?,its low frequency components
occur periodically at 0,± 2?,± 4?,etc,Thus,its high frequency components occur halfway
between the low frequency components at ±?,± 3?,etc,The shifted frequency response
corresponds to a high pass?lter since the original signal had its peak magnitude centered
around 0,± 2?,etc,and the new signal is shifted by? from that original signal,
Problem 6 O&W 6.58 (a),(b)
Solution,
21
(a) 1,We need to?nd h
1
[n] as a function of h[n],To accomplish this we investigate
H
1
(e
j?
),From Figure P6.58 of O&W,we can write
j?
)
G(e
j?
)
H(e =
j?
)
,(35)
X(e
j?
)
R(e
j?
)
H(e = (36)
G(e
j?
)
R(e
j?
) = S(e
j?
),(37)
We can manipulate the equations to write,
R(e
j?
) = G(e
j?
)H(e
j?
) (38)
j?
)H(e
j?
)H(e
j?
).= X(e (39)
Thus,
H
1
(e
j?
) =
S(e
j?
)
(40)
j?
)X(e
j?
)H(e
j?
).= H(e (41)
From this function,we recognize by the convolution property,h
1
[n] = h[n] h[?n].←
We can show that the system has zero phase characteristic by noting that since
h[n] is real,H(e
j?
) = H
(e
j?
),Thus,
H
1
(e
j?
) = H(e
j?
)H(e
j?
) = H(e
j?
)H
(e
j?
) = |H(e
j?
)
2
.|
2,Because H
1
(e
j?
)| = |H(e
j?
)
2
,then the magnitude H
1
(e
j?
) = H(e
j?
)
2
and the | | | | |
phase?H
1
(e
j?
) = 0,
(b) 1,We look at the Fourier transforms to determine h
2
[n],Since y[n] = g[n] + r[?n],
Y (e
j?
) = G(e
j?
) + R(e
j?
),From Figure P6.58 of O&W we determine
j?
)X(e
j?
)Y (e
j?
) = G(e
j?
) + R(e
j?
) = H(e
j?
) + H(e
j?
)X(e
Thus,
H
2
(e
j?
) = H(e
j?
) + H(e
j?
)?≈ h
2
[n] = h[n] + h[?n],
To determine if |H
2
(e
j?
) has zero phase characteristic,we manipulate H
2
(e
j?
)| | |
to get it into polar form,We use the property in Table 5.1 that since h[n] is real,
then ≥e{H(e
j?
)} = ≥e{H(e
j?
)} and ∞m{H(e
j?
)} = {H(e
j?
)},Hence,?∞m
|H
2
(e
j?
) = {H(e
j?
)} + ∞m{H(e
j?
)} + ≥e{H(e
j?
)} + ∞m{H(e
j?
)}|(42)| |≥e
j?
)}= 2≥e{H(e (43)
= 2 H(e
j?
)| cos (< H(e
j?
)),(44)|
From this,we can see that |H
2
(e
j?
) has zero phase characteristic,|
2,Since we already put |H
2
(e
j?
) in polar form,we know that |
|H
2
(e
j?
) = 2 H(e
j?
) cos (< H(e
j?
)). (45)| | |
and we know that?H
2
(e
j?
) = 0,
22
Problem 7 O&W 7.22
Solution,
In this problem we need to?gure out a range of values for the sampling period,T,to recover
y(t) completely from y
p
(t),To do this we need to determine the bandwidth of the original
Y (j?) and use the sampling theorem,By the convolution property,Y (j?) = X
1
(j?)X
2
(j?),
The bandwidth of Y (j?) then will be the bandwidth of the smaller of the two bandwidths,
X
1
(j?) or X
2
(j?),Hence,Y (j?) = 0 for? > 1000?,Then,using the sampling theorem,| |
2?
s
= > 2?
m
= 2(1000?),
T
This gives the range of T as 0 < T < 0.001 seconds,
Problem 8 O&W 7.23
Solution,
(a) We need to sketch X
p
(j?) and Y (j?),In the frequency domain,X
p
(j?) =
1
X(j?) ←
2?
P(j?),We need to determine P(j?),Since P(j?) is periodic,we need to use the
periodic Fourier transform formula,That is
P(j?) = 2? a
k
( k?
o
),
k=
1
T
p(t)e
jk?
o
t
Here,?
o
=
2?
=
,We need to determine the a
k
’s using the formula a
k
=,
T T
a
A few are shown below:
1
2?
0
= (?(t)(t))dt = 0
2?
0
1
2?
1 1
t
dt = (1? 1 e
j?
) =
2?
·
j?
(?( )?(?))et t
a
1
=
2?
1
0
2?
1
t
dt = (1? 1 e
j2?
) = 0
2?
·
2j?
(?( )?(?))et t
a
2
=
2?
1
0
2?
1 1
t
dt = (1? 1 e
j3?
) =
2?
·
3j?
(?( )?(?))et t
a
3
=
2?
0
1
Thus,a
k
= 0 for k even and a
k
=
for k odd and
2
2?
P(j?) =?( k ) = (2k + 1)
kodd k=
From this Fourier transform for P(j?),we can sketch X
p
(j?) as copies of X(j?) scaled
1
by
and replicated at intervals of? = (2k + 1)
,for all k,This can be seen in the
gure below,
23
X
p
)
1/?
(j
4?/3?/2?///? 2?/? 3?/? 4?/?
H(j?) is a sum of two ideal unity gain bandpass?lters,Thus,Y (j?) is the part of
X
p
(j?) that is passed through H(j?),This is shown below,
Y?)
1/?
(j
4?/3?/2?///? 2?/? 3?/? 4?/?
(b) To recover
p
(t x
) we need to do two things,First,we need to multiply
with a cosine function,cos
x(t) from x
p
(t)
t,This will shift X
p
(j?) such that one of the copies of
X(j?) is centered around? = 0,Second,we send the shifted signal through a lowpass
lter,R(j?),to eliminate the extra copies of X(j?),To achieve this we have a?lter,
R(
2?
j?) with gain =?,bandwidth and centered around? = 0,This is shown below,
)
R(j
//? 2?/?
The overall system is shown below,
cos(?/?t)
x
p
)
x x(t) (t)
R(j
24
=
(c) To recover x(t) from y(t) we need to run Y (j?) through two parallel?lter systems.
The top parallel path will multiply ( ) by cos ty
t which will shift the demi-replicate of
X(j?) that is centered at? = over to? = 0,The shifted signal then passes through
the lowpass?lter,R(j?) described above in part (b) to eliminate the extra copies.
3?
The bottom parallel path will multiply ( ) by cos ty
t which will shift the demi-replicate
of X(j?) that is centered at? =
3?
over to? = 0,The shifted signal then passes
through the lowpass?lter,R(j?) described above in part (b) to eliminate the extra
copies,Thus,the two halves combine together to form a complete X(j?) and x(t) is
recovered,The overall system is shown below,
+
x
x
t/?)
)
)
x(t) y(t)
cos(
R(j
R(j
cos(3?t/?)
(d) To recover x(t) from x
p
(t) and y(t),X
p
(j?) cannot have any overlap in the copies of
X(j?),Because of this particular p(t),the copies of X(j?) are at? = (2k + 1)
for all
k,Thus,just looking at one interval to make sure the copies of X(j?) don’t overlap,
we have one copy of X(j?) centered at? =
and one copy of X(j?) centered at
3?
,(See Figure of X
p
(j?) above),For no overlap between these copies,
3?
+?
m
<
m
,
which gives
< or?
max
=,
m
m
25
