Solutions Manual for Sensors, Circuits, and Systems for
Scientific Instruments: Fundamentals and Front-Ends
Soumyajit Mandal
1 Chapter 1
1. Solution to Exercise 1.1
(a) To find the poles, we need to factorize the denominator of the transfer function:
1 1
H(s) = = .
s2 − 8s + 16 (s − 4)2
Thus the pole locations are p1 = p2 = 4.
(b) Recall that H(s) is the Laplace transform of the impulse response, h(t). As a result,
h(t) is the inverse Laplace transform (ILT) of H(s):
h(t) = L−1 {H(s)}.
From a table of Laplace transforms, we find that
n!
L{tn eat } =
(s − a)n+1
where n = 1, 2, 3, . . .. In this case n = 1 and a = 4, such that
1
L{te4t } = .
(s − 4)2
Thus, h(t) = te4t . This function grows exponentially with time, so the system is not
BIBO stable.
2. Solution to Exercise 1.2
(a) The number of poles, P , and zeros, Z, equals the order of the denominator and numerator
polynomials, respectively. In this case, P = 3 and Z = 1.
(b) Recall that the pole and zero locations are the roots of the denominator and numerator
polynomials, respectively. Factorizing the denominator (for example, by using the Factor
function in Mathematica) yields
s−3 s−3
H(s) = = .
s3 + 7s2 + 16s + 10 (s + 1)(s2 + 6s + 10)
Further using the quadratic formula to find the roots of the second term, the denominator
can be written as (s + 1)(s + 3 − i)(s + 3 + i). Thus, the pole locations are p1 = −1,
p2 = −3 + i, and p3 = −3 − i.
Finally, the root of the numerator is 3, so the zero is located at z = 3. The resulting
pole-zero map is shown in Fig. 1(a).
All three poles are located in the left half-plane (LHP), so the system is BIBO stable.
1
, Pole-Zero Map
0.2
1.5
Real(H(j!))
Imaginary Axis (seconds-1 )
1 0
0.5 -0.2
0 10-1 100 101 102 103
0.2
-0.5
Imag(H(j!))
0.1
-1
0
-1.5
-3 -2 -1 0 1 2 3
10-1 100 101 102 103
Real Axis (seconds-1 ) Frequency, ! (rad/s)
Figure 1: (a) Pole-zero map. (b) Real and imaginary parts of the frequency response.
Impulse Response
0.1
0
Amplitude
-0.1
-0.2
0 1 2 3 4 5 6 7 8
Time (seconds)
Figure 2: Computed impulse response.
(c) Using the MATLAB freqs function yields the real and imaginary Bode plots shown in
Fig. 1(b).
(d) Using the MATLAB impulse function yields the impulse response shown in Fig. 2.
3. Solution to Exercise 1.3
(a) Using the same reasoning as for the previous problem, this transfer function has P = 3
poles and Z = 1 zeros.
(b) Using the MATLAB pzmap function yields the pole-zero map shown in Fig. 3(a). All
the poles are located within the unit circle, so the system is BIBO stable.
(c) Using the MATLAB freqz function yields the real and imaginary Bode plots shown in
Fig. 3(b). Note that the sample period was set to 1 sec by default.
(d) Using the MATLAB impulse function with the default sample period of 1 sec yields the
impulse response shown in Fig. 4.
2
, Pole-Zero Map 3
Real(H(j!))
2
Imaginary Axis (seconds-1 )
0.5 1
0
0 10-2 10-1 100
2
Imag(H(j!))
-0.5
1
-1 0
-1 -0.5 0 0.5 1
10-2 10-1 100
-1
Real Axis (seconds ) Frequency, ! (rad/s)
Figure 3: (a) Pole-zero map. (b) Real and imaginary parts of the frequency response.
Impulse Response
1
0.5
Amplitude
0
-0.5
-1
-1.5
0 5 10 15 20 25
Time (seconds)
Figure 4: Computed impulse response.
4. Solution to Exercise 1.4
Quarter-wavelength transformer: Noting that the reflection coefficient is defined as Γ =
(ZL − Z0 )/(ZL + Z0 ), the input impedance of a quarter-wavelength transmission line can be
written as
1−Γ (ZL + Z0 ) − (ZL − Z0 ) Z02
Zin (λ/4) = Z0 = Z0 . (1)
1+Γ (ZL + Z0 ) + (ZL − Z0 ) ZL
When used as an impedance transformer between Zin and ZL , the line must ensure that
Zin (λ/4) = Zin , the source impedance. In other words, we need
Z02 p
= Zin ⇒ Z0 = Zin ZL .
ZL
Note that this value is the geometric mean of the source and load impedance. In our case,
the required value of Z0 for an impedance match in thus given by
√
Z0 = 300 Ω · 50 Ω ∼= 122.5 Ω.
3
, 5. Solution to Exercise 1.5
(a) Based on the analysis in the previous problem, we need
p √
Z0 = Zin ZL = 100 × 50 Ω ∼ = 70.7 Ω.
(b) The electromagnetic wavelength at the design frequency is given by λ0 = v/f0 where
f0 = ω0 /2π = 1 GHz and v = 2 × 108 m/s, resulting in λ0 = 20 cm. Thus, the quarter-
wavelength transformer has a length of λ0 /4 = 5 cm.
(c) For simplicity, we assume that the transmission line is lossless. As discussed in the
previous problem, its input impedance is then given by
1 + Γe−j2βl
Zi (β) = Z0
1 − Γe−j2βl
√
where Z0 = Zin ZL , Γ = (ZL −Z0 )/(ZL +Z0 ), β = 2π/λ, and l = λ0 /4 where λ0 = 5 cm.
This quantity is frequency-dependent since β = 2π/λ = ω/v, thus limiting the overall
impedance-matching bandwidth.
Substituting known quantities, we get
( )
ZL −Z0 −jπ ω
1+ ZL +Z0 e ω0
(ZL + Z0 ) + (ZL − Z0 ) e−jπ(ω/ω0 )
Zi (ω) = Z0 ( ) = Z0 .
ZL −Z0 −jπ ω
(ZL + Z0 ) − (ZL − Z0 ) e−jπ(ω/ω0 )
1− ZL +Z0 e ω0
To simplify the notation, let us denote the normalized frequency as ωn = ω/ω0 . After
some more algebra, we get
ZL ejπωn /2 + e−jπωn /2 + Z0 ejπωn /2 − e−jπωn /2
Zin (ωn ) = Z0
ZL ejπωn /2 − e−jπωn /2 + Z0 ejπωn /2 + e−jπωn /2
√
ZL cos (πωn /2) + iZ0 sin (πωn /2) p ZL cos (πωn /2) + i Zin ZL sin (πωn /2)
= Z0 = Zin ZL √
iZL sin (πωn /2) + Z0 cos (πωn /2) iZL sin (πωn /2) + Zin ZL cos (πωn /2)
p
p ZL /Zin cos (πωn /2) + i sin (πωn /2)
= Zin ZL p .
i ZL /Zin sin (πωn /2) + cos (πωn /2)
Finally, the reflection coefficient seen from the source is given by
q √
ZL √ZL /Zin cos(πωn /2)+i sin(πωn /2)
Zi (ωn ) − Zin Zin i ZL /Zin sin(πωn /2)+cos(πωn /2) − 1
Γin (ωn ) = =q √
Zi (ωn ) + Zin ZL √ZL /Zin cos(πωn /2)+i sin(πωn /2)
Zin +1
i ZL /Zin sin(πωn /2)+cos(πωn /2)
[(ZL /Zin ) − 1] cos (πωn /2)
= p .
[(ZL /Zin ) + 1] cos (πωn /2) + 2i ZL /Zin sin (πωn /2)
Plotting the magnitude of Γin (ωn ) for ZL /Zin = 50/100 = 1/2 yields Fig. 5. The
reflection coefficient is < −10 dB over the range 0.3 < ωn < 1.7, which corresponds to a
bandwidth of 1.4 GHz. Also note that Γin (ωn ) is periodic in frequency with a period of
ωn = 2, as expected for a network based on transmission lines.
4
Scientific Instruments: Fundamentals and Front-Ends
Soumyajit Mandal
1 Chapter 1
1. Solution to Exercise 1.1
(a) To find the poles, we need to factorize the denominator of the transfer function:
1 1
H(s) = = .
s2 − 8s + 16 (s − 4)2
Thus the pole locations are p1 = p2 = 4.
(b) Recall that H(s) is the Laplace transform of the impulse response, h(t). As a result,
h(t) is the inverse Laplace transform (ILT) of H(s):
h(t) = L−1 {H(s)}.
From a table of Laplace transforms, we find that
n!
L{tn eat } =
(s − a)n+1
where n = 1, 2, 3, . . .. In this case n = 1 and a = 4, such that
1
L{te4t } = .
(s − 4)2
Thus, h(t) = te4t . This function grows exponentially with time, so the system is not
BIBO stable.
2. Solution to Exercise 1.2
(a) The number of poles, P , and zeros, Z, equals the order of the denominator and numerator
polynomials, respectively. In this case, P = 3 and Z = 1.
(b) Recall that the pole and zero locations are the roots of the denominator and numerator
polynomials, respectively. Factorizing the denominator (for example, by using the Factor
function in Mathematica) yields
s−3 s−3
H(s) = = .
s3 + 7s2 + 16s + 10 (s + 1)(s2 + 6s + 10)
Further using the quadratic formula to find the roots of the second term, the denominator
can be written as (s + 1)(s + 3 − i)(s + 3 + i). Thus, the pole locations are p1 = −1,
p2 = −3 + i, and p3 = −3 − i.
Finally, the root of the numerator is 3, so the zero is located at z = 3. The resulting
pole-zero map is shown in Fig. 1(a).
All three poles are located in the left half-plane (LHP), so the system is BIBO stable.
1
, Pole-Zero Map
0.2
1.5
Real(H(j!))
Imaginary Axis (seconds-1 )
1 0
0.5 -0.2
0 10-1 100 101 102 103
0.2
-0.5
Imag(H(j!))
0.1
-1
0
-1.5
-3 -2 -1 0 1 2 3
10-1 100 101 102 103
Real Axis (seconds-1 ) Frequency, ! (rad/s)
Figure 1: (a) Pole-zero map. (b) Real and imaginary parts of the frequency response.
Impulse Response
0.1
0
Amplitude
-0.1
-0.2
0 1 2 3 4 5 6 7 8
Time (seconds)
Figure 2: Computed impulse response.
(c) Using the MATLAB freqs function yields the real and imaginary Bode plots shown in
Fig. 1(b).
(d) Using the MATLAB impulse function yields the impulse response shown in Fig. 2.
3. Solution to Exercise 1.3
(a) Using the same reasoning as for the previous problem, this transfer function has P = 3
poles and Z = 1 zeros.
(b) Using the MATLAB pzmap function yields the pole-zero map shown in Fig. 3(a). All
the poles are located within the unit circle, so the system is BIBO stable.
(c) Using the MATLAB freqz function yields the real and imaginary Bode plots shown in
Fig. 3(b). Note that the sample period was set to 1 sec by default.
(d) Using the MATLAB impulse function with the default sample period of 1 sec yields the
impulse response shown in Fig. 4.
2
, Pole-Zero Map 3
Real(H(j!))
2
Imaginary Axis (seconds-1 )
0.5 1
0
0 10-2 10-1 100
2
Imag(H(j!))
-0.5
1
-1 0
-1 -0.5 0 0.5 1
10-2 10-1 100
-1
Real Axis (seconds ) Frequency, ! (rad/s)
Figure 3: (a) Pole-zero map. (b) Real and imaginary parts of the frequency response.
Impulse Response
1
0.5
Amplitude
0
-0.5
-1
-1.5
0 5 10 15 20 25
Time (seconds)
Figure 4: Computed impulse response.
4. Solution to Exercise 1.4
Quarter-wavelength transformer: Noting that the reflection coefficient is defined as Γ =
(ZL − Z0 )/(ZL + Z0 ), the input impedance of a quarter-wavelength transmission line can be
written as
1−Γ (ZL + Z0 ) − (ZL − Z0 ) Z02
Zin (λ/4) = Z0 = Z0 . (1)
1+Γ (ZL + Z0 ) + (ZL − Z0 ) ZL
When used as an impedance transformer between Zin and ZL , the line must ensure that
Zin (λ/4) = Zin , the source impedance. In other words, we need
Z02 p
= Zin ⇒ Z0 = Zin ZL .
ZL
Note that this value is the geometric mean of the source and load impedance. In our case,
the required value of Z0 for an impedance match in thus given by
√
Z0 = 300 Ω · 50 Ω ∼= 122.5 Ω.
3
, 5. Solution to Exercise 1.5
(a) Based on the analysis in the previous problem, we need
p √
Z0 = Zin ZL = 100 × 50 Ω ∼ = 70.7 Ω.
(b) The electromagnetic wavelength at the design frequency is given by λ0 = v/f0 where
f0 = ω0 /2π = 1 GHz and v = 2 × 108 m/s, resulting in λ0 = 20 cm. Thus, the quarter-
wavelength transformer has a length of λ0 /4 = 5 cm.
(c) For simplicity, we assume that the transmission line is lossless. As discussed in the
previous problem, its input impedance is then given by
1 + Γe−j2βl
Zi (β) = Z0
1 − Γe−j2βl
√
where Z0 = Zin ZL , Γ = (ZL −Z0 )/(ZL +Z0 ), β = 2π/λ, and l = λ0 /4 where λ0 = 5 cm.
This quantity is frequency-dependent since β = 2π/λ = ω/v, thus limiting the overall
impedance-matching bandwidth.
Substituting known quantities, we get
( )
ZL −Z0 −jπ ω
1+ ZL +Z0 e ω0
(ZL + Z0 ) + (ZL − Z0 ) e−jπ(ω/ω0 )
Zi (ω) = Z0 ( ) = Z0 .
ZL −Z0 −jπ ω
(ZL + Z0 ) − (ZL − Z0 ) e−jπ(ω/ω0 )
1− ZL +Z0 e ω0
To simplify the notation, let us denote the normalized frequency as ωn = ω/ω0 . After
some more algebra, we get
ZL ejπωn /2 + e−jπωn /2 + Z0 ejπωn /2 − e−jπωn /2
Zin (ωn ) = Z0
ZL ejπωn /2 − e−jπωn /2 + Z0 ejπωn /2 + e−jπωn /2
√
ZL cos (πωn /2) + iZ0 sin (πωn /2) p ZL cos (πωn /2) + i Zin ZL sin (πωn /2)
= Z0 = Zin ZL √
iZL sin (πωn /2) + Z0 cos (πωn /2) iZL sin (πωn /2) + Zin ZL cos (πωn /2)
p
p ZL /Zin cos (πωn /2) + i sin (πωn /2)
= Zin ZL p .
i ZL /Zin sin (πωn /2) + cos (πωn /2)
Finally, the reflection coefficient seen from the source is given by
q √
ZL √ZL /Zin cos(πωn /2)+i sin(πωn /2)
Zi (ωn ) − Zin Zin i ZL /Zin sin(πωn /2)+cos(πωn /2) − 1
Γin (ωn ) = =q √
Zi (ωn ) + Zin ZL √ZL /Zin cos(πωn /2)+i sin(πωn /2)
Zin +1
i ZL /Zin sin(πωn /2)+cos(πωn /2)
[(ZL /Zin ) − 1] cos (πωn /2)
= p .
[(ZL /Zin ) + 1] cos (πωn /2) + 2i ZL /Zin sin (πωn /2)
Plotting the magnitude of Γin (ωn ) for ZL /Zin = 50/100 = 1/2 yields Fig. 5. The
reflection coefficient is < −10 dB over the range 0.3 < ωn < 1.7, which corresponds to a
bandwidth of 1.4 GHz. Also note that Γin (ωn ) is periodic in frequency with a period of
ωn = 2, as expected for a network based on transmission lines.
4
