M. J. Roberts - 8/16/04
Chapter 6 - Fourier Transform Analysis of
Signals and Systems
Selected Solutions
(In this solution manual, the symbol, ⊗, is used for periodic convolution because the
preferred symbol which appears in the text is not in the font selection of the word processor
used to create this manual.)
1. A system has an impulse response,
h LP ( t) = 3e −10 t u( t) ,
and another system has an impulse response,
h HP ( t) = δ ( t) − 3e −10 t u( t) .
(a)
Sketch the magnitude and phase of the transfer function of these two systems in a
parallel connection.
3
3
H LP ( jω ) =
, H HP ( jω ) = 1 −
jω + 10
jω + 10
H P ( jω ) =
3
3
+ 1−
=1
jω + 10
jω + 10
|HP(jω)|
1
-40
40
ω
40
ω
Phase of HP(jω)
π
-40
-π
(b)
Sketch the magnitude and phase of the transfer function of these two systems in a
cascade connection.
2. Below are some pairs of signals, x( t) and y( t) . In each case decide whether or not y( t)
is a distorted version of x( t) .
Solutions 6-1
M. J. Roberts - 8/16/04
(a)
(b)
x(t)
x(t)
2
2
2
t
2
-2
t
-2
y(t)
y(t)
2
2
2
t
2
-2
t
-2
(a)
Inverted and amplified, undistorted
(b)
Time shifted, undistorted
(c)
(d)
x(t)
x(t)
2
1
2
t
-2
2
-1
y(t)
y(t)
2
1
2
t
-2
2
t
-1
(c)
Clipped at a negative value, distorted
(d)
Time compressed, distorted
(e)
(f)
x(t)
x(t)
2
2
2
t
-2
2
t
-2
y(t)
y(t)
2
2
2
-2
t
t
2
-2
Solutions 6-2
t
M. J. Roberts - 8/16/04
(e)
Constant added, distorted
(f)
Log-amplified, distorted
3. Classify each of these transfer functions as having a lowpass, highpass, bandpass or
bandstop frequency response.
(a)
(b)
|H( f )|
|H(F)|
1
1
-10
10
f
-2
(c)
(d)
|H(jΩ)|
|H(jω)|
1
Lowpass
(b)
4π
Ω
Bandpass
-100
(c)
(e)
(f)
|H(F)|
|H(jΩ)|
(f)
ω
(d)
Bandpass
1
-2
Highpass
100
Lowpass
1
(e)
F
1
-4 π
(a)
2
2
F
-4 π
4π
Ω
Bandstop
4. Classify each of these transfer functions as having a lowpass, highpass, bandpass or
bandstop frequency response.
(a)
(c)
 f − 100 
H( f ) = 1 − rect 
(b)
H( F ) = rect (10 F ) ∗ comb( F )

 10 

π 
π  




 Ω
H( jΩ) = rect  20π  Ω −   + rect  20π  Ω +    ∗ comb 


 2π 


4 
4  

Bandpass
5. A system has an impulse response,
 t − 0.01
h( t) = 10 rect 
 .
 0.02 
Solutions 6-3
M. J. Roberts - 8/16/04
What is its null bandwidth?
6. A system has an impulse response,
n
 7
h[ n ] =   u[ n ] .
 8
What is its half-power DT-frequency bandwidth?
Using
F
α n u [ n ] ←
→
1
1 − α e− jΩ
the transfer function is
8
1
.
=
7 − jΩ 8 − 7e − jΩ
1− e
8
This is a DT lowpass filter. Its maximum transfer function magnitude occurs at Ω = 0 . The
-3 dB point must be the first frequency at which the square of the magnitude of the transfer
function is one-half of its maximum value (the “half-power” bandwidth).
H( jΩ) =
The low-frequency gain is
H(0) = 8
The -3 dB point occurs where
H( jΩ−3 dB )
Solving,
2
82
=
= 32 .
2
Ωhp = 0.1337 ± 2 nπ
So the -3 dB DT-frequency bandwidth in radians is 0.1337. In cycles it is 0.0213. (Notice
that the bandwidths are not in radians/s or in Hz. This is because they are DT bandwidths,
not CT bandwidths.)
7. Determine whether or not the CT systems with these transfer functions are causal.
(a)
H( f ) = sinc( f )
(c)
H( jω ) = rect (ω )
(d)
H( jω ) = rect (ω )e − jω
(e)
H( f ) = A
(f)
H( f ) = Ae j 2πf
(b)
h( t) =
H( f ) = sinc( f )e − jπf
1
 t − 1
sinc
 Not Causal
 2π 
2π
h( t) = Aδ ( t + 1)
Solutions 6-4
Not Causal
M. J. Roberts - 8/16/04
8. Determine whether or not the DT systems with these transfer functions are causal.
(a)
H( F ) =
sin( 7πF )
sin(πF )
(c)
H( F ) =
sin( 3πF ) − j 2πF
e
sin(πF )
(d)
H( F ) = rect (10 F ) ∗ comb( F )
H( F ) =
(b)
sin( 7πF ) − j 2πF
e
sin(πF )
h[ n ] = rect1[ n − 1]
Causal
9. Find and sketch the frequency response of each of these circuits given the indicated
excitation and response.
(a)
Excitation, v i ( t) - Response, v L ( t)
R = 10 Ω C = 1 µF
+
+
vi (t)
vL(t)
L = 1 mH
-
-
Using voltage-division principles,
VL ( jω )
Z L ( jω )
jω L
−ω 2 LC
H ( jω ) =
=
=
=
Vi ( jω ) Z L ( jω ) + ZC ( jω ) + Z R ( jω ) jω L + 1 + R 1 − ω 2 LC + jω RC
jω C
|H( jω )|
3
ω
-150000
150000
Phase of H( jω )
π
ω
-150000
150000
-π
(b)
Excitation, v i ( t) - Response, iC ( t)
Solutions 6-5
M. J. Roberts - 8/16/04
R = 1 kΩ
+
iC(t)
C = 1 µF
vi (t)
-
In this case the transfer function is the reciprocal of the input impedance.
(c)
Excitation, v i ( t) - Response, v R ( t)
+
vR(t)
-
R = 1 kΩ
+
vi (t)
C = 1 µF
L = 1 mH
(d)
Excitation, ii ( t) - Response, v R ( t)
+
ii (t)
R = 100 Ω
vR(t)
L = 1 mH
C = 1 µF
Divide the excitation current between the two branches and multiply the
current in the right branch by R to get the response voltage. Then solve for
the ratio of the response voltage to the excitation current.
10. Classify each of these transfer functions as having a lowpass, highpass, bandpass or
bandstop frequency response.
1
1 + jf
(a)
H( f ) =
(c)
H( jω ) = −
(b)
j10ω
100 − ω 2 + j10ω
Solutions 6-6
H( f ) =
jf
1 + jf
M. J. Roberts - 8/16/04
(d)
H( F ) =
sin( 3πF )
sin(πF )
This case is not as “pure” as the previous ones. It is generally lowpass because the
transfer function magnitude at lower frequencies is generally greater than at high
frequencies. But there are nulls in the transfer function that make it look somewhat
like a bandstop filter or a multiple bandstop filter.
(e)
H( jΩ) = j[sin(Ω) + sin(2Ω)]
This case is also not perfectly clear. The response at zero frequency is zero and the
response at Ω = π is also zero. These criteria fit a bandpass filter. But the response
2π
is also zero at Ω =
. So it might again look like a bandstop in some ways.
3
11. Plot the magnitude frequency responses, both on a linear-magnitude and on a logmagnitude scale, of the systems with these transfer functions, over the frequency range
specified.
20
(a)
H( f ) =
, −100 < f < 100
2 2
20 − 4π f + j 42πf
(b)
2 × 10 5
H( jω ) =
, −500 < ω < 500
(100 + jω )(1700 − ω 2 + j 20ω )
12. Draw asymptotic and exact magnitude and phase Bode diagrams for the frequency
responses of the following circuits and systems.
(a)
An RC lowpass filter with R = 1 MΩ and C = 0.1 µF .
1
1
1
1
jωC
H( jω ) =
=
=
=
6
−7
1
+ R jωRC + 1 jω10 10 + 1 j 0.1ω + 1
jωC
Solutions 6-7
M. J. Roberts - 8/16/04
0
|H(jω)|dB
-10
-20
-30
-40
-50
-1
10
0
1
10
2
10
3
10
ω
10
Phase of H(jω)
0
-0.5
-1
-1.5
-2
-1
10
0
1
10
2
10
3
10
ω
10
(b)
R = 10 Ω C = 1 µF
+
+
vi (t)
L = 1 mH
-
vL(t)
-
Vo ( f )
, of these active filters and identify them as
Vi ( f )
lowpass, highpass, bandpass or bandstop.
13. Find the transfer functions, H( f ) =
(a)
R1
C1
C2
K
+
vi (t)
-
vx(t)
+
vo(t)
R2
-
The triangle with the “ K ” inside is an ideal voltage amplifier of gain K (not an
operational amplifier). It is, in circuit theory parlance, a “voltage-dependent voltage source”.
“Ideal” means its input impedance is infinite so no input current flows and its output
impedance is zero so the output voltage is independent of the output current (and therefore
any load connected to the output).
Solutions 6-8
M. J. Roberts - 8/16/04
Writing Kirchhoff’s current law at the v X ( t ) node and then writing the
relationship between Vx ( f ) and Vo ( f ) using voltage division and the voltage amplifier gain,
K,




1
Vx ( f ) j 2πfC1 +
+ G1 − Vi ( f ) j 2πfC1 − Vo ( f )G1 = 0
1


+ R2
j 2πfC2


Vx ( f )
( G1 is
R2
1
+ R2
j 2πfC2
K = Vo ( f )
1
. That is, it is the conductance of the resistor, R1 .)
R1


j 2πfC2
Vx ( f ) j 2πfC1 +
+ G1 − Vi ( f ) j 2πfC1 − Vo ( f )G1 = 0
1 + j 2πfR2C2


Vx ( f )
j 2πfR2C2
K = Vo ( f )
1 + j 2πfR2C2
Writing the two equations as one matrix equation,


j 2πfC2
+ G1
 j 2πfC1 +
1 + j 2πfR2C2



j 2πfKR2C2

Vx ( f )   j 2πfC1 
V ( f )  =  0  Vi ( f )

−(1 + j 2πfR2C2 )  o  
−G1
Solving by Cramer’s rule with the excitation voltage as a forcing function,


j 2πfC2
∆ = − j 2πfC1 +
+ G1 (1 + j 2πfR2C2 ) + j 2πfKG1R2C2
1 + j 2πfR2C2


∆ = (2πf ) R2C1C2 − j 2πf [C1 + C2 + G1R2C2 (1 − K )] − G1
2


j 2πfC2
1  j 2πfC1 +
+ G1
Vo ( f ) = 
1 + j 2πfR2C2

∆
j 2πfKR2C2
j 2πfC1
Vi ( f )
0
V (f)
(2πf ) KR2C1C2
H( f ) = o
=
2
Vi ( f ) (2πf ) R2C1C2 − j 2πf [C1 + C2 + G1R2C2 (1 − K )] − G1
2
H( f ) =
(2πf ) 2 KR1R2C1C2
(2πf ) 2 R1R2C1C2 − j 2πf (R1(C1 + C2 ) + R2C2 (1 − K )) − 1
Highpass
Solutions 6-9
M. J. Roberts - 8/16/04
The circuit is a highpass filter because at low frequencies the response
approaches zero and at high frequencies it approaches K. This can be seen
both in the transfer function formula and in the physical nature of the circuit
connection itself. At zero frequency no current flows through the capacitors
and therefore not current flows through resistor, R2 . Therefore the voltage at
the input of the amplifier is zero making the response voltage zero. At high
frequencies the capacitor impedances become practically zero, making the
excitation voltage and the voltage at the amplifier input equal. Therefore the
transfer function must be K at high frequencies.
(b)
C1
R1
R2
K
+
vx(t)
vi (t)
+
vo(t)
C2
-
-
Similar to (a)
14. Show that this system has a highpass frequency response.
y(t)
x(t)
∫
Write the differential equation from the block diagram. You should get
d
d
y( t) + y( t) = x( t)
dt
dt
Fourier transform both sides and solve for the ratio of Y to X.
15. Draw the block diagram of a system with a bandpass frequency response using two
integrators as functional blocks. Then find its transfer function and verify that it has a
bandpass frequency response.
Lowpass cascaded with highpass. Find the transfer function of both stages (lowpass and
highpass) and multiply the transfer functions.
Solutions 6-10
M. J. Roberts - 8/16/04
Y( jΩ)
, and sketch the frequency response of each
X( jΩ)
of these DT filters over the range, −4π < Ω < 4π .
16. Find the transfer function, H( jΩ) =
(a)
x[n]
y[n]
D
y[ n ] = x[ n ] − x[ n − 1]
H ( jΩ ) = 1 − e − j Ω
(b)
Similar to (a)
(c)
y[n]
x[n]
D
D
1 − e − jΩ
H( jΩ) =
1 + e − jΩ
(d)
x[n]
D
y[n]
D
Let z be the output of the left-hand summer. Then
y[ n ] = x[ n ] − x[ n − 1] + z[ n ]
z[ n ] = x[ n ] − z[ n − 1]
Take the DTFT of both equations, eliminate Z and solve for the ratio of Y to X.
H ( jΩ ) =
2 − e− j 2 Ω
1 + e− jΩ
Solutions 6-11
M. J. Roberts - 8/16/04
17. Find the minimum stop band attenuation of a moving-average filter with N = 3. Define
1
the stop band as the frequency region, Fc < F < , where Fc is the DT frequency of the
2
first null in the frequency response.
From the text, for a moving-average filter
H( F ) =
e − jπNF sin(π ( N + 1) F )
N +1
sin(πF )
The first null in the frequency response occurs at
π ( N + 1) F = π ⇒ F =
1
1
= .
N +1 4
The phrase, “minimum stop band attenuation” refers to the point in the stop band at which
the reduction in magnitude is the smallest. That is, the point in the stop band in which the
transfer function is the largest. The biggest magnitude response after the null frequency is at
the next maximum of H( F ) which occurs at
π ( N + 1) F =
3π
3
3
⇒F=
= .
2
2( N + 1) 8
18. In the system below, x t ( t) = sinc( t) , f c = 10 and the cutoff frequency of the lowpass filter
is 1 Hz. Plot the signals, x t ( t) , y t ( t) , y d ( t) and y f ( t) and the magnitudes and phases
of their CTFT’s.
x t(t)
yt (t) = x r(t)
cos(2πfct)
x t ( t) = sinc( t)
yd (t)
LPF
cos(2πfct)
X t ( f ) = rect ( f )
Solutions 6-12
yf (t)
M. J. Roberts - 8/16/04
Modulation
|Xt( f )|
xt(t)
1
1
-2
2
f
Phase of Xt( f )
-4
4
π
t
-2
-0.5
2
f
-π
y t ( t) = sinc( t) cos(20πt)
Fourier transforming,
1
[rect( f − 10) + rect( f + 10)]
2
Yt ( f ) =
Modulated Carrier
|Yt( f )|
yt(t)
0.5
1
-10
-4
4
10
f
Phase of Yt( f )
t
π
-10
-1
y d ( t) = sinc( t) cos2 (20πt)
10
f
-π
Yd ( f ) =
1
[rect( f − 20) + 2 rect( f ) + rect( f + 20)]
4
Solutions 6-13
M. J. Roberts - 8/16/04
Demodulated Carrier
|Yd( f )|
yd(t)
0.5
1
-20
20
f
Phase of Yd( f )
-4
4
π
t
-20
20
-0.5
1
y f ( t) = sinc( t)
2
f
-π
2 rect ( f )
4
Demodulated and Filtered Carrier
Yf ( f ) =
|Xf( f )|
xf(t)
0.5
0.5
-2
2
f
Phase of Xf( f )
-4
4
π
t
-2
-0.25
2
f
-π
19. In the system below, x t ( t) = sinc(10 t) ∗ comb( t) , m = 1, f c = 100 and the cutoff
frequency of the lowpass filter is 10 Hz. Plot the signals, x t ( t) , y t ( t) , y d ( t) and y f ( t)
and the magnitudes and phases of their CTFT’s.
x t(t)
yt (t) = x r(t)
m
1
cos(2πfct)
yd (t)
LPF
yf (t)
cos(2πfct)
Similar to Exercise 18.
20. An RC lowpass filter with a time constant of 16 ms is excited by a DSBSC signal,
Solutions 6-14
M. J. Roberts - 8/16/04
x( t) = sin(2πt) cos(20πt) .
Find the phase and group delays at a the carrier frequency.
The transfer function of the RC lowpass filter is
H( jω ) =
A
A
=
.
1 + jωτ 1 + j 0.016ω
The phase of the transfer function is
φ ( jω ) = − tan −1 (ωτ ) = − tan −1 (0.016ω ) .
The carrier frequency is 10 Hz. Therefore
φ ( j20π ) = − tan −1 (0.016 × 20π ) = 0.788
and the phase delay is −
φ ( j 20π ) 0.788
=
= 0.01254 or 12.54 ms . The derivative of the
ωc
20π
phase shift function is
τ
d
φ ( jω )) = −
(
2 .
dω
1 + (ωτ )
Evaluating this derivative at the carrier frequency we get
τ
0.016
 d

φ ( jω )) 
−
=
(
2 =
2 = 7.95 ms .
1 + (0.016 × 20π )
 dω
ω =ω c 1 + (ω cτ )
21. A pulse train,
p( t) = rect (100 t) ∗ 10 comb(10 t)
is modulated by a signal,
x( t) = sin( 4πt) .
Plot the response of the modulator, y( t) , and the CTFT’s of the excitation and response.
Similar to Exercise 18.
∞
n 
 
y( t) = sin( 4πt) ∑ rect 100 t −  
  10  
n =−∞
Y( f ) =
∞

j  ∞
 k
 k
2
10
sinc
δ
f
+
−
k
−
sinc  δ ( f − 2 − 10 k ) 
(
)


∑
∑

 10 
 10 
20 k =−∞
k =−∞

22. In the system below, let the excitation be
Solutions 6-15
M. J. Roberts - 8/16/04
x( t) = rect (1000 t) ∗ 250 comb(250 t)
and let the filter be ideal, with unity passband gain. Plot the signal power of the response,
y( t) , of this system versus the sweep frequency, f c , over the range, 0 < f c < 2000 for a LPF
bandwidth of
and
(a)
5 Hz
(b)
50 Hz
(c)
500 Hz.
Multiplier
xsh(t)
x(t)
LPF
y(t)
cos(2πfc t)
x sh ( t) = [rect (1000 t) ∗ 250 comb(250 t)] cos(2πf c t)
∞
n 


x sh ( t) = cos(2πf c t) ∑ rect 1000 t −

 250  

n =−∞
 1
 f 
 f  1
X sh ( f ) = 
sinc
 ∗ δ ( f − fc ) + δ ( f + fc )
 comb
 1000 
 250   2
1000
[
]
∞

1
 f 
X sh ( f ) = sinc
 ∑ δ ( f − 250 k )  ∗ δ ( f − f c ) + δ ( f + f c )
 1000  k =−∞
8

[
X sh ( f ) =
]
1 ∞
 k
sinc  δ ( f − f c − 250 k ) + δ ( f + f c − 250 k )
∑
 4
8 k =−∞
[
]
1 ∞

 k
 f 
Y( f ) =  ∑ sinc  δ ( f − f c − 250 k ) + δ ( f + f c − 250 k )  rect  
 4
 2B 
 8 k =−∞

[
]
where “B” is the bandwidth of the LPF.
∞


 f 
 k
rect
sinc  δ ( f − f c − 250 k ) 


∑

 2 B  k =−∞
 4
1

Y( f ) = 

∞
8
 f 
 k
+ rect   ∑ sinc δ ( f + f c − 250 k )


 2 B  k =−∞
 4
1 
 k
Y( f ) =  ∑ sinc  δ ( f − f c − 250 k ) +
 4
8  f c + 250 k < B

 k
sinc  δ ( f + f c − 250 k )
 4

f c − 250 k < B
∑
Solutions 6-16
M. J. Roberts - 8/16/04
Py =
1 
2 k 
 ∑ sinc   +
64  f c + 250 k < B
4
 k  
sinc 2   
 4 
f c − 250 k < B

∑
Signal Power
0.1
2000
f
c
Signal Power
0.1
2000
fc
2000
fc
Signal Power
0.1
23. A signal, x(t) is described by
x( t) = 500 rect (1000 t) ∗ comb(500 t)
(a)
If x(t), is the excitation of an ideal lowpass filter with a cutoff frequency of
3 kHz, plot the excitation , x(t) and the response, y(t) on the same scale and
compare.
Fourier transform the excitation, to yield X. Write the transfer function, H,
of the ideal filter as a rectangle function. Form the transform of the response,
Y, from X times H. Recognize it as a finite summation of impulses. Inverse
transform the impulses in pairs to form y, and graph y.
y(t)
1
-2 ms
2 ms
t
This looks like the partial sums in the discussion of convergence of the CTFS
because, mathematically, the same thing is happening.
(b)
Similar to (a)
24. Determine whether or not the CT systems with these transfer functions are causal.
The test for causality is that a causal system has an impulse response that is zero fall
time, t < 0 .
(a)
H( jω ) =
2
jω
(b)
H( jω ) =
Solutions 6-17
10
6 + j 4ω
M. J. Roberts - 8/16/04
(c)
H( jω ) =
4
25 − ω + j 6ω
2
=
4
( jω + 3) 2 + 16
F
→
Using e− at sin (ω 0t ) u ( t ) ←
ω0
( jω + a )2 + ω 02
h( t) = e −3 t sin( 4 t) u( t)
(d)
H( jω ) =
(f)
H( jω ) =
4
25 − ω + j 6ω
2
Causal
e jω
jω + 9
45 − ω 2 + j 6ω
(e)
H( jω ) =
(g)
H( jω ) =
4
25 − ω 2 + j 6ω
e − jω
49
49 + ω 2
25. Determine whether or not the DT systems with these transfer functions are causal.
(a)
H( F ) = [rect (10 F ) ∗ comb( F )]e − j 20πF
(b)
H( F ) = j sin(2πF )
(d)
H( jΩ) =
H( F ) = 1 − e − j 4 πF
(c)
8e jΩ
8 − 5e − jΩ
Similar to Exercise 24, except for discrete time.
26. Find and sketch the frequency response of each of these circuits given the indicated
excitation and response.
(a)
Excitation, v i ( t) - Response, vC 2 ( t)
R1 = 1 kΩ
R 2 = 10 kΩ
+
vi (t)
+
C1 = 1 µF
C2 = 0.1 µF
-
vC2(t)
-
The transfer function can be found in multiple ways. One way is to think of
this circuit as two voltage dividers. The first voltage division is from the
excitation, vi ( t ) , to the voltage across the first capacitor. The second voltage
division is from that voltage to the response voltage, vC 2 ( t ) .
Solutions 6-18
M. J. Roberts - 8/16/04
H ( jω ) =
1
Zπ ( jω )
1
jω C2
=
1
R1 + Zπ ( jω ) jω R2C2 + 1
R2 +
jω C2
VC 2 ( jω )
Zπ ( jω )
=
Vi ( jω )
R1 + Zπ ( jω )
first volltage division
second voltage division
1 
1 
R2 +

jω C1 
jω C2 
Zπ ( jω ) =
1
1
+ R2 +
jω C1
jω C2
Substitute Zπ ( jω ) into the expression for the transfer function and simplify.
H( jω ) =
(b)
1
1 − ω R1R2C1C2 + jω (C1 + C2 ) R1 + R2C2
[
2
]
Excitation, v i ( t) - Response, iC1 ( t)
R1 = 1 kΩ
R 2 = 10 kΩ
+
iC1(t)
vi (t)
C1 = 1 µF
C2 = 0.1 µF
Think of the transfer function as the transfer function from the excitation to
the current in R1 times the transfer function from the current in R1 to iC1 ( t ) .
1
R2 +
I ( jω )
IR1 ( jω )
1
jω R2C2 + 1
jω C2
H ( jω ) = C1
=
=
1
1
Vi ( jω )
Vi ( jω )
Zi ( jω ) jω R C + 1 + C2
+ R2 +
2 2
jω C2 input impedance
C1
first transsfer function jω C1
second transfer function
Zi ( jω ) = R1 + Zπ ( jω ) = R1 +
second transfer function
jωR2C2 + 1
jω (C1 + C2 ) − ω 2 R2C1C2
Combine expressions and simplify to yield
H( jω ) =
(c)
jωC1 ( jωR2C2 + 1)
1 − ω R1R2C1C2 + jω R1 (C1 + C2 ) + R2C2
[
2
Excitation, v i ( t) - Response, v R 2 ( t)
Solutions 6-19
]
M. J. Roberts - 8/16/04
C1 = 1 µF
C2 = 1 µF
+
+
vi (t)
R 1 = 10 kΩ
vR2(t)
R 2 = 10 kΩ
-
-
Similar to (a).
(d)
Excitation, ii ( t) - Response, v R1 ( t)
C1 = 1 µF
R 2 = 10 kΩ
i i(t)
+
C2 = 1 µF
vR1(t)
R 1 = 10 kΩ
-
(e)
Excitation, v i ( t) - Response, v RL ( t)
R 2 = 10 kΩ
+
+
C1 = 1 µF
vi (t)
R 1 = 10 kΩ
C2 = 1 µF
vRL(t)
RL = 1 kΩ
-
-
Write two nodal equations and solve for the transfer function.
Summing currents to zero at the middle node and the right-hand node,
VR1 ( jω )[ jωC1 + jωC2 + G1 ] − Vi ( jω ) jωC1 − VRL ( jω ) jωC2 = 0
VRL ( jω )[ jωC2 + GL + G2 ] − Vi ( jω )G2 − VR1 ( jω ) jωC2 = 0
Solutions 6-20
M. J. Roberts - 8/16/04
Solve for the transfer function,
−ω 2 R1R2C1C2 + jωR1 (C1 + C2 ) + 1
H( jω ) =

 R + R2

R 
−ω 2 R1R2C1C2 + jω (C1 + C2 )1 + 2  R1 + R2C2  + L
RL
 RL 


|H( jω )|
1
ω
-25000
25000
Phase of H( jω )
π
ω
-25000
25000
-π
27. Find and sketch versus frequency the magnitude and phase of the input impedance,
V(f)
V (f)
Z in ( f ) = i
and transfer function, H( f ) = o
, for each of these filters.
Ii ( f )
Vi ( f )
+
i i(t) 1 µF
vi (t)
(a)
-
+
vo(t)
-
i i(t) 100 Ω 50 mH
vi (t)
(b)
1 kΩ
+
-
10 nF
+
vo(t)
-
28. The signal, x(t), in Exercise 23 is the excitation of an RC lowpass filter with R = 1kΩ
and C = 0.3 µF. Sketch the excitation and response voltages versus time on the same
scale.
From Exercise 23,
x( t) = 500 rect (1000 t) ∗ comb(500 t)
Solutions 6-21
M. J. Roberts - 8/16/04
1 ∞
 n
X( f ) = ∑ sinc  δ ( f − 500 n )
 2
2 n =−∞
The transfer function is
H( f ) =
1
j 2πfRC + 1
Therefore the output is
∞
1
1
 n
Y( f ) =
sinc  δ ( f − 500 n )
∑
 2
2 j 2πfRC + 1 n =−∞
1 ∞
1
 n
δ ( f − 500 n )
Y( f ) = ∑ sinc 
 2  j1000πnRC + 1
2 n =−∞
Converting to the time domain,
 n
sinc 
∞
 2
1
y( t) = ∑
e j1000πnt
2 n =−∞ j1000πnRC + 1
or
∞

e j1000πnt
e − j1000πnt
1
 n 
1
y( t) =  + ∑ sinc  
+
 2   j1000πnRC + 1 − j1000πnRC + 1 
2  n =1
− j1000πnt
∞
− e j1000πnt ) + e j1000πnt + e − j1000πnt 
1 
 n   j1000πnRC (e
y( t) = 1 + ∑ sinc  

2
 2 
2  n =1
nRC
1000
1
π
(
)
+



∞
1 
 n   2000πnRC sin(1000πnt) + 2 cos(1000πnt) 
y( t) = 1 + ∑ sinc  

 2 
2  n =1
(1000πnRC) 2 + 1

1
-2 ms
2 ms
t
29. Draw asymptotic and exact magnitude and phase Bode diagrams for the frequency
responses of the following circuits and systems.
(a)
R1 = 1 kΩ
R 2 = 10 kΩ
+
vi (t)
+
C1 = 1 µF
C2 = 0.1 µF
-
vC2(t)
-
From Exercise 26(b)
Solutions 6-22
M. J. Roberts - 8/16/04
H( jω ) =
1
1 − ω R1R2C1C2 + jω (C1 + C2 ) R1 + R2C2
[
2
H( jω ) =
]
1
1 − 10 ω + j 2.1 × 10 −3 ω
−6
2
0
-20
|H(jω)|dB
-40
-60
-80
-100
-120
-140
1
10
2
3
10
10
4
ω
5
10
6
10
10
Phase of H(jω)
0
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
1
10
2
3
10
10
4
ω
5
10
6
10
10
(b)
X(jω)
(c)
10
jω+10
jω
jω+10
Y(jω)
A system whose transfer function is H( jω ) =
H( jω ) =
j 20ω
=
( jω + 10 − j 99.5)( jω + 10 + j 99.5)
j 20ω
10, 000 − ω 2 + j 20ω
j 20ω

ω
ω2 
−
10000 1 + j
500 10000 

30. Find the transfer function for the following circuit. What function does it perform?
if(t)
Rf
i i(t) C i
+
vi (t)
vx(t)
+
vo(t)
-
-
31. Design an active highpass filter using an ideal operational amplifier, two resistors and
one capacitor and derive its transfer function to verify that it is high pass.
Solutions 6-23
M. J. Roberts - 8/16/04
Use an inverting amplifier configuration. Let the feedback impedance be a simple
resistor. Choose an input impedance that is high at low frequencies and approaches a
Z f ( jω )
constant at high frequencies so that the transfer function, −
approaches zero at
Zi ( jω )
low frequencies and approaches a constant at high frequencies.
Vo ( f )
, of these active filters and identify them as
Vi ( f )
lowpass, highpass, bandpass or bandstop.
32. Find the transfer functions, H( f ) =
(a)
C4
R1
R5
vx(t)
+
C3
+
vi (t)
-
vo(t)
R2
-
Sum currents to zero at node, vx ( t ) , and at the input node of the operational
amplifier, which must be at zero volts because the ideal operational amplifier
gain is infinite. Remember the input impedance of the operational amplifier
is infinite so no current flows into its input terminals.
Vx ( f )(G1 + G2 + j 2πfC3 + j 2πfC4 ) − Vi ( f )G1 − Vo ( f ) j 2πfC4 = 0
− Vx ( f ) j 2πfC3 − Vo ( f )G5 = 0
Solve for the transfer function.
H( f ) =
j 2πfR5C3


(2πf ) 2 R1R5C3C4 − j 2πfR1(C4 + C3 ) − 1 +
What kind of filter is this?
(b)
Solutions 6-24
R1 

R2 
M. J. Roberts - 8/16/04
C4
C1
vx(t)
+
R5
C3
+
vi (t)
vo(t)
R2
-
-
Similar to (a).
(c)
R4
R1
C5
R3
vx(t)
+
+
vi (t)
vo(t)
C2
-
-
Similar to (a)
33. When music is recorded on analog magnetic tape and later played back, a high-frequency
noise component, called tape “hiss” is added to the music. For purposes of analysis
assume that the spectrum of the music is flat at –30 dB across the audio spectrum from
20 Hz to 20 kHz. Also assume that the spectrum of the signal played back on the tape
deck has an added component making the playback signal have a Bode diagram as
illustrated in Figure E33.
-24 dB
-30 dB
f
200
2 kHz
20 kHz
6 kHz 12 kHz
Figure E33 Bode diagram of playback signal
The extra high-frequency noise could be attenuated by a lowpass filter but that would also
attenuate the high-frequency components of the music, reducing its fidelity. One solution to
the problem is to “pre-emphasize” the high-frequency part of the music during the
recording process so that when the lowpass filter is applied to the playback the net effect on
the music is zero but the “hiss” has been attenuated. Design an active filter which could be
used during the recording process to do the pre-emphasis.
Solutions 6-25
M. J. Roberts - 8/16/04
The pre-emphasis active filter should have a low-frequency gain of one and a highfrequency gain of 6 dB and should transition between those gains between 6 and 12 kHz
with a slope (asymptotic slope) of 6 dB/octave. So the pre-emphasis filter should have one
real zero to create the first corner and one real pole to create the second corner. If we use the
inverting amplifier configuration with an op-amp and make the feedback impedance a
resistor then the source impedance can create the zero and pole at the required locations. The
design idea is to make the source impedance have the required low-frequency value and then
to make its high-frequency value lower, increasing the overall gain at high frequencies, as
illustrated in Figure S33.
Rf
R
s1
R s2
Cs
+
vi (t)
-
+
vo(t)
Figure S33 Pre-emphasis filter
The source impedance is

1 
1
Rs1  Rs2 +
j 2πf +

j 2πfCs 

j 2πfCsRs1Rs2 + Rs1
R R
CsRs2
= s1 s2
Zs ( f ) =
=
1
1
j 2πfCs ( Rs1 + Rs2 ) + 1 Rs1 + Rs2 j 2πf +
Rs1 + Rs2 +
j 2πfCs
Cs ( Rs1 + Rs2 )
The numerator provides the zero of the source impedance (the pole of the overall gain) and
the denominator provides the pole of the source impedance (the zero of the overall gain). So
we want the gain-pole location to be set by
1
= 2π × 12000
CsRs2
and the gain-zero location to be set by
1
= 2π × 6000 .
Cs ( Rs1 + Rs2 )
There is no unique solution so let’s arbitrarily set Rs2 = 10 kΩ. Then it follows that
Cs = 1.33 nF and Rs1 = 10 kΩ. The overall gain is
Solutions 6-26
M. J. Roberts - 8/16/04
Rf
H( f ) = −
Rs1Rs2
Rs1 + Rs2
1
j 2πf +
CsRs2
1
j 2πf +
Cs ( Rs1 + Rs2 )
= −Rf
Rs1 + Rs2
Rs1Rs2
1
Cs ( Rs1 + Rs2 )
1
j 2πf +
CsRs2
j 2πf +
At low frequencies,
H( f ) = −
Rf
.
Rs1
To make the low-frequency gain one, set R f = Rs1 = 10 kΩ .
34. One problem with causal CT filters is that the response of the filter always lags the
excitation. This problem cannot be eliminated if the filtering is done in real time but if
the signal is recorded for later “off-line” filtering one simple way of eliminating the lag
effect is to filter the signal, record the response and then filter that recorded response
with the same filter but playing the signal back through the system backward. Suppose
the filter is a single-pole filter with a transfer function of the form,
H( jω ) =
1
ω
1+ j
ωc
,
where ω c is the cutoff frequency (half-power frequency) of the filter.
(a)
What is the effective transfer function of the entire process of filtering the signal
forward, then backward?
(b)
What is the effective impulse response?
(a)
The impulse response of the filter is h( t) = F −1[H( jω )] . The response of the filter
on the first pass through the filter forward is y1 ( t) = x( t) ∗ h( t) in the time domain or
Y1 ( jω ) = X( jω ) H( jω ) in the frequency domain. The response of the filter on the second
pass through the filter backward is y 2 ( t) = y1 (− t) ∗ h( t) in the time domain or
Y2 ( jω ) = Y1* ( jω ) H( jω ) in the frequency domain. (This uses the CTFT property,
F
g ( −t ) ←
→ G * ( jω ) .) The final signal is y 2 (− t) in the time domain and Y2* ( jω ) in the
frequency domain. Therefore the final signal in the frequency domain is
[
]
Y2* ( jω ) = Y1* ( jω ) H( jω ) = Y1 ( jω ) H* ( jω ) = X( jω ) H( jω ) H* ( jω )
*
and in the time domain this is y 2 ( t) = x( t) ∗ h( t) ∗ h(− t) . So the effective transfer function is
Solutions 6-27
M. J. Roberts - 8/16/04
1
H( jω ) H* ( jω ) =
F
and, using e − a t ←
→
1+ j
1
ω
ω
1− j
ωc
ωc
=
1
ω
1+  
 ωc 
2
2a
, the effective impulse response is
a + ω2
2
h( t) ∗ h(− t) = ω c e −ω c t u( t) ∗ ω c eω c t u(− t) =
ω c −ω c t
.
e
2
Just as there is no phase shift in the frequency domain, the effective impulse response,
h(− t) ∗ h( t) is symmetrical about t = 0 which also means it creates no time delay when it is
convolved with the excitation.
35. Repeat Exercise 18 but with the second cos(2πf c t) replaced by sin(2πf c t) .
Similar to Exercise 18.
36. In the system below, x t ( t) = sinc( t) , f c = 10 and the cutoff frequency of the lowpass filter
is 1 Hz. Plot the signals, x t ( t) , y t ( t) , y d ( t) and y f ( t) and the magnitudes and phases
of their CTFT’s.
This is a single sideband system. Analysis is similar to several previous communication
system exercises. The biggest difference is the addition of the filter which removes a
sideband before transmission.
yt (t) = x r(t)
|H( f )|
x t(t)
fm
-fc
fm
fc
yd (t)
f
LPF
cos(2πfct)
cos(2πfct)
Yt ( f ) =
1
 f + 10.25  
 f − 10.25 
rect 

 + rect 

 0.5  


2
0.5
1
 t
y t ( t) = sinc  cos(20.5πt)
 2
2
1
 t
y d ( t) = sinc  cos(20.5πt) cos(20πt)
 2
2
Yd ( f ) =
1 rect (2( f − 20.25)) + rect (2( f + 0.25)) 


4 + rect (2( f − 0.25)) + rect (2( f + 20.25)) 


Solutions 6-28
yf (t)
M. J. Roberts - 8/16/04
Demodulated Carrier
|Yd( f )|
yd(t)
0.5
0.5
-20
20
f
Phase of Yd( f )
-4
4
π
t
-20
20
-0.25
f
-π
Demodulated and Filtered Carrier
|Yf( f )|
yf(t)
0.25
0.25
-2
2
f
Phase of Yf( f )
-4
4
π
t
-2
-0.125
2
f
-π
37. A quadrature modulator modulates a sine carrier, sin(20πt) , with a signal, x1 (t ) = sinc( t) ,
and a cosine carrier, cos(20πt) , with a signal, x 2 (t ) = rect ( t) . The quadrature
π

demodulator has a phase error making its local oscillators be sin 20πt −  and

6
π

cos 20πt −  . Plot the two demodulated and filtered signals, x1 f (t ) and x 2 f (t ) .

6
y( t) = sinc( t) sin(20πt) + rect ( t) cos(20πt)
π


X1d ( f ) = F  sinc ( t ) sin ( 20π t ) + rect ( t ) cos ( 20π t )  sin  20π t −  

6

Solutions 6-29
M. J. Roberts - 8/16/04
j


∗
δ ( f + 10) − δ ( f − 10)] 
rect
f
(
)
[
πf

−j
j
2
X1d ( f ) = 
 ∗ [δ ( f + 10) − δ ( f − 10)]e 60
+ sinc( f ) ∗ 1 [δ ( f − 10) + δ ( f + 10)] 2


2
πf
−j
j  j[rect ( f + 10) − rect ( f − 10)] 


∗
+
f
f
e
10
10
X1d ( f ) =
) − δ ( − )] 60
[δ (
4 +[sinc( f − 10) + sinc( f + 10)]


πf
πf
 

 
−j
−j
60
− δ ( f − 10)e 60   
 rect ( f + 10) ∗ δ ( f + 10)e

 
 
 j
πf
πf  
 − rect ( f − 10) ∗ δ ( f + 10)e − j 60 − δ ( f − 10)e − j 60  

 


j  

X1d ( f ) = 
πf
πf
4 



−j
−j
 sinc( f − 10) ∗ δ ( f + 10)e 60 − δ ( f − 10)e 60  

 
 
+ 
πf
πf  


−j
−j
 + sinc( f + 10) ∗ δ ( f + 10)e 60 − δ ( f − 10)e 60 
 


π
π
 
−j   
j
 rect ( f + 10) ∗ δ ( f + 10)e 6 − rect ( f + 10) ∗ δ ( f − 10)e 6   
  
 
 j
π
π  
 − rect ( f − 10) ∗ δ ( f + 10)e j 6 − rect ( f − 10) ∗ δ ( f − 10)e − j 6  

 

j   

X1d ( f ) = 
π
π
4  
j
− j  
 sinc( f − 10) ∗ δ ( f + 10)e 6 − sinc( f − 10) ∗ δ ( f − 10)e 6  
 
+ 
  
π
π 
j
−j 
 + sinc( f + 10) ∗ δ ( f + 10)e 6 − sinc( f + 10) ∗ δ ( f − 10)e 6 
  

π
π
π
π
 
j
−j
j
−j  
6
6
6
6
j
rect
f
+
e
rect
f
e
rect
f
e
rect
f
e
−
−
+
−
20
20
(
)
(
)
(
)
(
)
 
 
j 
 
X1d ( f ) = 

π
π
π
π
4 
j
−j
j
−j 
+ sinc( f )e 6 − sinc( f − 20)e 6 + sinc( f + 20)e 6 − sinc( f )e 6 
 

π
π
 
j
−j  
π 
6
j
rect
f
+
e
cos
rect
f
rect
f
e
−
+
−
20
2
20
(
)
( )
(
) 6 
 
 
 6
j 
 
X1d ( f ) = 
π
π 
4 
−j
j 
+  j 2 sin π  sinc( f ) − sinc( f − 20)e 6 + sinc( f + 20)e 6 
 6
 

Solutions 6-30
M. J. Roberts - 8/16/04
Now apply the lowpass filter,
X1 f ( f ) =
j

π 
π 
− j 2 cos  rect ( f ) + j 2 sin  sinc( f ) 





4
6
6

X1 f ( f ) =
x1 f ( t) =
1  π 

π 
cos  rect ( f ) − sin  sinc( f ) 

 6
2   6

1  π 

π 
cos  sinc( t) − sin  rect ( t) 

 6
2   6

Similarly,
x 2 f ( t) =
1  π 

π 
cos  rect ( t) + sin  sinc( t) 

 6
2   6

Quadrature Demodulated and Filtered Signal
x1f(t)
1
-4
4
-0.25
t
In-phase Demodulated and Filtered Signal
x2f(t)
1
-4
4
-0.25
t
38. A pulse train,
p( t) =
is modulated by a signal,
1
 t
rect   ∗ 4 comb( 4 t)
 w
w
x( t) = sinc( t) .
Plot the response of the modulator, y( t) , and the CTFT’s of the excitation and response for
and
(a)
w = 10 ms
(b)
w = 1 ms .
This is similar to previous modulation exercises.
Solutions 6-31
M. J. Roberts - 8/16/04
n

t
−
1

4
y( t) = sinc( t) ∑ rect 

w
n =−∞
 w 


∞
∞
Y( f ) = 4 ∑ sinc( 4 wk ) rect ( f − 4 k )
k =−∞
PAM Modulator Response (a)
|Y( f )|
x(t) and y(t)
4
100
-200
200
f
Phase of Y( f )
-3
3
π
t
-200
-50
200
f
-π
Because of the scale, it is difficult to see what is really happening in the magnitude
plot of the transform of the response. It consists of a large number of closely-spaced
impulses.
1
 n
and the cutoff DT frequency of the
39. In the system below, x t [ n ] = sinc  , Fc =
 20 
4
1
lowpass filter is
. Plot the signals, x t [ n ] , y t [ n ] , y d [ n ] and y f [ n ] and the magnitudes
20
and phases of their DTFT’s.
x t[n]
yt [n] = xr[n]
cos(2πFc n)
yd [n]
LPF
yf [n]
cos(2πFc n)
This is a DT modulation system. The analysis is very similar to that used for CT
modulation systems.
40. Repeat Exercise 22 but with an excitation,
x( t) = rect (1000 t) ∗ 20 comb(20 t) .
x sh ( t) = [rect (1000 t) ∗ 20 comb(20 t)] cos(2πf c t)
Solutions 6-32
M. J. Roberts - 8/16/04
∞
n 


x sh ( t) = cos(2πf c t) ∑ rect 1000 t −  


20  
n =−∞
 1
 f 
 f  1
X sh ( f ) = 
sinc
 comb   ∗ δ ( f − f c ) + δ ( f + f c )
 1000 
 20   2
1000
[
X sh ( f ) =
]
∞

1
 f 
sinc
δ ( f − 20 k )  ∗ δ ( f − f c ) + δ ( f + f c )


∑

 1000  k =−∞
8

[
]
1 ∞
 k
X sh ( f ) = ∑ sinc  δ ( f − f c − 20 k ) + δ ( f + f c − 20 k )
 4
8 k =−∞
[
]
1 ∞

 k
 f 
Y( f ) =  ∑ sinc  δ ( f − f c − 20 k ) + δ ( f + f c − 20 k )  rect  
 4
 2B 
 8 k =−∞

[
]
∞


 f 
 k
rect
sinc  δ ( f − f c − 20 k ) 


∑

 2 B  k =−∞
 4
1

Y( f ) = 

∞
8
 f 
 k
+ rect   ∑ sinc δ ( f + f c − 20 k )


 2 B  k =−∞
 4
1 
 k
Y( f ) =  ∑ sinc  δ ( f − f c − 20 k ) +
 4
8  f c + 20 k < B
Py =
1 
2 k 
 ∑ sinc   +
64  f c + 20 k < B
4

 k
sinc  δ ( f + f c − 20 k )
 4

f c − 20 k < B
∑
 k  
sinc 2   
 4 
f c − 20 k < B

∑
Signal Power
0.1
2000
fc
2000
fc
2000
fc
Signal Power
0.1
Signal Power
0.1
41. The diffraction of light can be approximately described through the use of the Fourier
transform. Consider an opaque screen with a small slit being illuminated from the left by
a normally-incident uniform plane light wave (Figure E41).
Solutions 6-33
M. J. Roberts - 8/16/04
Diffracting
Screen
Viewing
Screen
x1
Propagation
Direction
x0
Optical Axis
Slit
Wavefronts
z
Figure E41 One-dimensional diffraction of light through a slit
πx
is a good approximation for any x1 in the slit, then the electric field strength of
λ
the light striking the viewing screen can be accurately described by
If z >>
2
1
j
2πz
π
2π
−j
x 0 x1
e λ j λz x 02 ∞
e
E 0 ( x0 ) = K
∫ E1 ( x1 ) e λz dx1
jλz
−∞
where E1 is field strength at the diffracting screen, E 0 is field strength at the viewing screen,
K is a constant of proportionality and λ is the wavelength of the light. The integral is a
Fourier transform with different notation. The field strength at the viewing screen can be
written as
j
2π z
λ
π
j x02
e
λz
E 0 ( x0 ) = K
e
F  E1 ( t )  f → x0 .
jλ z
λz
The intensity, I 0 ( x 0 ) , of the light at the viewing screen is the square of the magnitude of the
field strength,
2
I( x 0 ) = E 0 ( x 0 ) .
(a)
Plot the intensity of light at the viewing screen if the slit width is 1 mm, the
wavelength of light is 500 nm, the distance, z, is 100 m, the constant of proportionality is
V
10 −3 and the electric field strength at the diffraction screen is 1 .
m
The electric field exiting the diffraction slit is
 x 
E ( x1 ) = rect  1 
 0.001
Finding the screen electric field using the formula above,
Solutions 6-34
M. J. Roberts - 8/16/04
j
2πz
λ
π
j x 02
e
 x0 
λz
e
E 0 ( x0 ) =
sinc

 5 × 10 −2 
j 50
The intensity is the square of the magnitude of the electric field,
 x0 
sinc 2 

 5 × 10 −2 
x
I0 ( 0 ) =
= 4 × 10 −4 sinc 2 (20 x 0 )
2500
I(x )
0
0.0004
-0.2
0.2
x
0
(b)
Now let the slit be replaced by two slits each 0.1 mm in width, separated by 1 mm
(center-to-center) and centered on the optical axis. Plot the intensity of light at the viewing
screen if the other parameters are the same as in part (a).
Similar to (a)
I(x )
0
1.6e-05
-1
x0
1
42. In Figure 42-1 is a circuit diagram of a half-wave rectifier followed by a capacitor to
smooth the response voltage. Model the diode as ideal and let the excitation be a cosine
at 60 Hz with an amplitude of 120 2 volts. Let the RC time constant be 0.1 seconds.
Then the response voltage will look as illustrated in Figure E42-2 . Find and plot the
magnitude of the CTFT of the response voltage.
+
v i (t)
+
R
C
-
vo (t)
-
Figure E42-1 A half-wave rectifier with a capacitive smoothing filter
Solutions 6-35
M. J. Roberts - 8/16/04
175
vo(t)
vi (t)
0.05
t
-175
Figure E42-2 Excitation and response voltages
The response voltage has two parts, the exponential decay time and the cosinusoidal charging
time. The dividing time, td , between these two parts is set by the intersection of the cosine
and the exponential decay. The peak of the cosine is 120 2 . The decay time constant is 0.1
seconds. Therefore the dividing time is the solution of
120 2 cos(120πtd ) = 120 2e
−
td
0.1
or
cos(120πtd ) = e
−
td
0.1
This is a transcendental equation best solved numerically. This equation is simple enough
that a trial-and-error method converges very quickly to a solution. That solution is
td = 15.23906 ms .
Therefore the description of the response voltage over one period is
 − 0t.1
, 0 < t < 15.23906 ms
e
v o1 ( t) = 120 2 
2
cos(120πt) , 15.23906 < t < 16 ms
3

or
 − 0t.1
 t − 0.00761953
 t − 0.01595286  
v o1 ( t) = 120 2 e rect 
 + cos(120πt) rect 

 0.01523906 
 0.001427607  

The CTFT of the response is the CTFT of this voltage convolved with a comb to make it
periodically repeat. The CTFT of one period is

0.01523906 − t − j 2πft
dt

 ∫ e 0.1e
0


Vo1 ( f ) = 120 2

 1
+ [δ ( f − 60) + δ ( f + 60)] ∗ 0.001427607 sinc(0.001427607 f )e − j 2πf ( 0.01595286) 

 2
Solutions 6-36
M. J. Roberts - 8/16/04
0.01523906

 − t

j
ft
2
−
π

 e 0.1e


 −10 − j 2πf 


 0


 1
− j 2π ( f − 60)( 0.01595286) 
Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e


 2
 1
− j 2π ( f + 60)( 0.01595286) 

+ 2 0.001427607 sinc(0.001427607( f + 60))e







 − 0.01523906 − j 2πf 0.01523906 
0.1


e
1

 e
−





−10 − j 2πf
 −10 − j 2πf 





 1
− j 2π ( f − 60)( 0.01595286) 
Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e


 2
 1
− j 2π ( f + 60)( 0.01595286) 

+ 2 0.001427607 sinc(0.001427607( f + 60))e




0.01523906
−


0.1
e − j 2πf 0.01523906

1 − e


10 + j 2πf


 1
− j 2π ( f − 60)( 0.01595286) 
Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e

 2

 1
− j 2π ( f + 60)( 0.01595286) 
+ 2 0.001427607 sinc(0.001427607( f + 60))e





The CTFT of the actual periodic response is the product of this CTFT with the CTFT of
 f 
60 comb(60 t) which is comb  . Therefore
 60 
0.01523906
−


0.1
1
−
e
e − j 2πf 0.01523906




10 + j 2πf


 1
 f 
− j 2π ( f − 60)( 0.01595286) 
Vo ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e
 comb 
60
 2

 1
− j 2π ( f + 60)( 0.01595286) 
+ 2 0.001427607 sinc(0.001427607( f + 60))e





Solutions 6-37
M. J. Roberts - 8/16/04
|V ( f )|
o
dB
50
-600
600
f
Phase of V ( f )
o
π
-600
600
f
-π
43. Create a discrete-space image consisting of 96 by 96 pixels. Let the image be a
“checkerboard” consisting of 8 by 8 alternating black-and-white squares.
(a) Filter the image row-by-row and then column-by-column with a DT filter whose
impulse response is
n
h[ n ] = 0.2(0.8) u[ n ]
and display the image on the screen using the imagesc command in MATLAB.
After defining the checkerboard we can filter it by convolving it with the impulse
response using the MATLAB conv function.
Notice how this lowpass spatial filter blurs the edges. A lowpass filter does not
allow any fast transitions to occur.
(b) Filter the image row-by-row and then column-by-column with a DT filter whose
impulse response is
n
h[ n ] = δ [ n ] − 0.2(0.8) u[ n ]
Solutions 6-38
M. J. Roberts - 8/16/04
and display the image on the screen using the imagesc command in MATLAB.
Notice how this highpass spatial filter empahsize the edges and de-emphasizes the
constant regions between the edges. A highpass filter does allow fast transitions to
occur.
 f
44. In the system of Figure E0 let the CTFT of the the excitation be X( f ) = tri  . This
 fc 
system is sometimes called a scrambler because it moves the frequency components of a
signal to new locations making it unintelligible.
(a) Using only an analog multiplier and an ideal filter, design a “descrambler” which
would recover the original signal.
(b) Sketch the magnitude spectrum of each of the signals in the scrambler-descrambler
system.
Multiplier
ys (t)
x(t)
cos(2πfct)
Figure E0 A “scrambler”
No help here. Left as a challenge for the student.
45. Electronic amplifiers that handle very-low-frequency signals are difficult to design
because thermal drifts of offset voltages cannot be distinguished from the signals. For
this reason a popular technique for designing low-frequency amplifiers is the so called
“chopper-stabilized” amplifier (Figure E45).
Solutions 6-39
M. J. Roberts - 8/16/04
Typical Amplifier
+
vi (t)
-
+
vo(t)
Chopper-Stabilized Amplifier
+
vi (t)
-
BPF
LPF
+
vo(t)
-
Figure E45 A chopper-stabilized amplifier
A chopper-stabilized amplifier “chops” the excitation signal by switching it on and off
periodically. This action is equivalent to a pulse amplitude modulation in which the pulse
train being modulated by the excitation is a 50% duty-cycle square wave which alternates
between zero and one. Then the “chopped” signal is bandpass filtered to remove any slow
thermal drift signals from the first amplifier. Then the amplified signal is “chopped” again
at exactly the same rate and in phase with the chopping signal used at the input of the first
amplifier. Then this signal may be further amplified. The last step is to lowpass filter the
signal out of the last amplifier to recover an amplified version of the original signal. (This is
a simplified model but it illustrates the essential features of a chopper-stabilized amplifier.)
Let the following be the parameters of the chopper-stabilized amplifier:
Chopping frequency
Gain of the first amplifier
Bandpass filter
Gain of the second amplifier
Lowpass filter
500 Hz
100 V/V
Unity-gain, ideal, zero-phase. Passband 250 < f < 750
10 V/V
Unity-gain, ideal, zero-phase. Bandwidth 100 Hz
Let the excitation signal have a 100 Hz bandwidth. What is the effective DC gain of this
chopper-stabilized amplifier?
Let the excitation be v i ( t) . Then the signal after the first amplifier is
v1 ( t) = 100 v i ( t)[rect (1000 t) ∗ 500 comb(500 t)]
or, in the frequency domain,
V1 ( f ) = 50
∞
 m
∑ sinc 2  V ( f − 500m)
m =−∞
i
The response signal from the bandpass filter is that part of the spectrum lying between 250
and 750 Hz which is
 1
Vbpf ( f ) = 50 sinc  [Vi ( f − 500) + Vi ( f + 500)] .
 2
Solutions 6-40
M. J. Roberts - 8/16/04
The response of the second amplifier is the response of the BPF pulse-amplitude modulated
by the synchronous switching. In the frequency domain,
∞
 1
 m
V2 ( f ) = 25 sinc  ∑ sinc  [Vi ( f − 500 m − 500) + Vi ( f − 500 m + 500)]
 2  m =−∞
 2
If the excitation is dc, then Vi ( f ) = Aδ ( f ) and
∞
 1
 m
V2 ( f ) = 25 A sinc  ∑ sinc  [δ ( f − 500 m − 500) + δ ( f − 500 m + 500)] .
 2  m =−∞
 2
The response of the lowpass filter is that part of the signal below 100 Hz, which is

 1
 1 
 1
V2 ( f ) = 25 A sinc  sinc  [δ ( f ) + δ ( f )] = 50 A sinc 2   δ ( f )
 2
 2 
 2

and the effective dc gain is
 1
50 A sinc 2  
 2
DC Gain =
= 20.264 .
A
46. A common problem in over-the-air television signal transmission is “multipath”
distortion of the received signal due to the transmitted signal bouncing off structures.
Typically a strong “main” signal arrives at some time and a weaker “ghost” signal
arrives later. So if the transmitted signal is x t ( t) , the received signal is
(
x r ( t) = K m x t ( t − tm ) + K g x t t − tg
)
where K m >> K g and tg > tm .
(a)
What is the transfer function of this communication channel?
(b)
What would be the transfer function of an “equalization” system that would
compensate for the effects of multipath?
The impulse response of the system is
(
)
h( t) = K mδ ( t − tm ) + K gδ t − tg .
Therefore its transfer function is
H( f ) = K m e − j 2πft m + K g e
− j 2πft g
.
An equalization system would then have a transfer function of the form,
Ae − j 2πft 0
H eq ( f ) =
− j 2πft g
K m e − j 2πft m + K g e
Solutions 6-41