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: H(s) =
1 s2 − 8s + 16
=
1 . (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 L{tn eat } =
n! (s − a)n+1
where n = 1, 2, 3, . . .. In this case n = 1 and a = 4, such that L{te4t } =
1 . (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 H(s) =
s−3 s−3 = . 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
Real(H(j!))
1 0.5
0
-0.2 10-1
0 -0.5
Imag(H(j!))
Imaginary Axis (seconds-1 )
1.5
-1 -1.5 -3
-2
-1
0
1
2
3
100
101
102
103
101
102
103
0.2 0.1 0 10-1
100
Real Axis (seconds-1 )
Frequency, ! (rad/s)
Figure 1: (a) Pole-zero map. (b) Real and imaginary parts of the frequency response. Impulse Response
Amplitude
0.1
0
-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
3
Real(H(j!))
0.5
2 1 0
0
10-2
10-1
100
10-2
10-1
100
2
Imag(H(j!))
Imaginary Axis (seconds-1 )
Pole-Zero Map
-0.5
-1 -1
1
0 -0.5
0
0.5
1
-1
Frequency, ! (rad/s)
Real Axis (seconds )
Figure 3: (a) Pole-zero map. (b) Real and imaginary parts of the frequency response. Impulse Response 1
Amplitude
0.5
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 = Zin ZL
⇒
Z0 =
p Zin 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 quarterwavelength 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 Zi (β) = Z0
1 + Γe−j2βl 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 1+ Zi (ω) = Z0 1−
ZL −Z0 ZL +Z0 ZL −Z0 ZL +Z0
e e
−jπ −jπ
( (
ω ω0 ω ω0
) ) = Z0
(ZL + Z0 ) + (ZL − Z0 ) e−jπ(ω/ω0 ) . (ZL + Z0 ) − (ZL − Z0 ) e−jπ(ω/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) √ = Zin ZL = Z0 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) Zin i ZL /Zin sin(πωn /2)+cos(πωn /2) − 1 Zi (ωn ) − Zin √ Γin (ωn ) = =q Zi (ωn ) + Zin ZL √ZL /Zin cos(πωn /2)+i sin(πωn /2) +1 Zin 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
Reflection coefficient, |! in|
0
-10
-20
-30
-40
-50 0.5
1
1.5
2
2.5
3
Normalized frequency, ! n
Figure 5: Calculated input-referred reflection coefficient of a quarter-wave transformer for ZL /Zin = 1/2.
6. Solution to Exercise 1.6 Assuming that thepcable loss is relatively small, its characteristic impedance can be approximated as Z0 ≈ L0 /C0 where L0 is the inductance per unit length. Given Z0 = 50 Ω and C0√= 100 pF/m, we get L0 = 250 nH/m. Thus, the wave velocity can be calculated as v = 1/ L0 C0 = 2 × 108 m/s. Finally, the time delay of ℓ = 10 m of cable is td = ℓ/v = 50 ns. From Eqn. (1.23), the attenuation constant of the line can be written as α≈
β 2(QL ||QC )
where β = 2π/λ, QL = ωL0 /R0 , QC = ωC0 /G0 , and R0 and G0 are the cable resistance and conductance per unit length, respectively. Since the dielectric is assumed to be lossless, G0 = 0 and QC → ∞. Thus, the expression for α simplifies to α≈
β 2π R0 R0 = = . 2QL λω L0 vL0
Thus, the resistance per unit length can be written as R0 = (vL0 )α. The given cable has an loss of 5 dB, implying that 20 log10 (e−αℓ ) = −5 dB for ℓ = 10 m. Simplifying, α = 5/86.859 = 0.0576 m−1 . Substituting this value in the expression for R0 , we get R0 = 2.88 Ω/m. 7. Solution to Exercise 1.7
p Since the line is lossless, Z0 = L0 /C0 where Z0√= 50 Ω and C0 = 120 pF/m. Substituting these values, we get L0 = 300 nH/m and v = 1/ L0 C0 = 1.67 × 108 m/s. Thus, the time delay for ℓ = 1 m of line is td = ℓ/v = 6 ns. The reflection coefficients at the source and load ends of the line are Γs = (Zs −Z0 )/(Zs +Z0 ) = 0 and ΓL = (ZL − Z0 )/(ZL + Z0 ) = 1/3, respectively. Thus, the pulse will only reflect once (from the load end) before being completely absorbed (at the source end). The resulting waveforms at the source and load are shown in Fig. 6.
8. Solution to Exercise 1.8 5