Cpe340 Hw1 Sol San
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Let R1 = 1.5 kΩ, R2 = 180 Ω, and C = 0.05 µF Since C is parallel to R1, then the impedance Z1 here is The impedance at x(t) is a combination of impedances Z1 and R2, so it simply is Z1 + R2 The impedance at y(t) is, of course, R2 The transfer function h(t) as well as its frequency response h(ω) can be determined from voltage divider so that it is
Substituting C, R1, and R2; we get h(ω) as
% CPE 340 Principle of Communications, CPE, KMUTT % S1/2009 % Instructor : Asst. Prof. Peerapon S., Ph.D. % TA : San R. % Homework 1, Question 3 % % This M-file is provided by San R. % f = [1:1e5]; w = 2*pi*f; C = 0.05e-6; R1 = 1.5e3; R2 = 180; norm = R2 + j*w*C*R1*R2; denorm = R1 + norm; H = norm./denorm; mag = abs(H); phase = angle(H); figure; plot(f, mag); title('Magnitude Response'); figure; plot(f, phase); title('Phase Response');
Magnitude Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
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10 x 10
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Phase Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
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10 x 10
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4. Consider a signal x(t) in Fig. 3 below. Let A = 2 volts, τ = 0.5 msec, T = 1 msec. x(t) A
0 τ T
Figure 3: (a) Determine the complex Fourier coefficients cn of x(t). Show every step of your calculation. Soln. cn = = = = = = =
1 T
Z
A T
Z
x(t)e−j2nπf0 t dt
T τ /2
e−j2nπf0 t dt
−τ /2
τ /2 A e−j2nπf0 t −(j2nπf0 )T −τ /2 A e−jnπf0 τ − ejnπf0 τ −j2nπ A sin nπf0 τ nπ . sin πx Af0 τ sinc nf0 τ ; sinc x = πx sinc (n/2)
(b) What is the Fourier transform of x(t) ? Soln. The Fourier transform of x(t) is simply an impulse train spaced by f0 and weighted by cn . X(f ) = =
∞ X n=−∞ ∞ X
cn δ(f − nf0 ) sinc(n/2)δ(f − nf0 )
n=−∞
(c) Suppose x(t) is passed through a filter in Fig. 4 below. Determine the magnitude of the complex Fourier coefficients for y(t) and the power of y(t) within the 3-dB bandwidth of the filter. Soln. Let Y (ω) ↔ y(t), ω = 2πf . The transfer function of the filter is given by H(ω) =
1/jωC 1 1 − jωRC = = R + 1/jωC 1 + jωRC 1 + (ωRC)2
with |H(ω)| =
1 p 1 + (ωRC)2
R = 170 C = 0.2 µF
x(t)
y(t)
Figure 4: Therefore, the magnitude of the complex Fourier coefficients for y(t) is |Yn | = |H(f )||cn | = |H(nf0 )||cn | 1 = p |sinc (n/2)|; f0 = 1 kHz, R = 170 Ω, C = 0.2 µF 1 + (2πnf0 RC)2 1 = √ |sinc (n/2)| 1 + 0.0456n2
The 3-dB bandwidth of the filter is 1/2πRC = 4.68 kHz. Therefore, the frequency components up to the fourth harmonic stay within the 3-dB bandwidth. The normalized signal power in such bandwidth range is P
= Y0 + 2
4 X
|Yn |2
n=1
= 1 + 2 ∗ (0.6222 + 0 + 0.1782 + 0) = 1.84 Watts
5. A unity-amplitude periodic square wave with 50% duty cycle is passed through an RC lowpass filter whose 3-dB cutoff frequency is 1,500 Hz. Using a computer, find and plot the output signal if the input square wave has a frequency of 300 Hz, 500 Hz, and 1,000 Hz. Soln. The input square wave can be represented by the Fourier series: x(t) =
∞ X
cn ejnω0 t
n=−∞
where cn =
2 sin(nπ/2) ,
n 6= 0
0,
n=0
nπ
From (2-140) in the Couch text, the output waveform is then given by y(t) = =
∞ X
H(nf0 )cn ejnω0 t
n=−∞ ∞ X
dn ejnω0 t , dn , H(nf0 )cn , H(f ) =
n=−∞
1 1 + j(f /fc )
where fc = 1,500 Hz for the RC low-pass filter. Representing y(t) in the trigonometric form to easily plot, y(t) = D0 +
∞ X
Dn cos(nω0 t + φn )
n=1
where D0 = 0, since c0 = 0 Dn = 2|dn | = 2|H(nf0 )cn |, n > 0 1 4 r = 2 , n odd nπ 0 1 + nf fc 1 − sin(nπ/2) −1 nf0 φn = − tan +π , n odd fc 2
Frequency = 300 Hz 1.5
1
0.5
0
-0.5
-1
-1.5
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Frequency = 500 Hz 1.5
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-0.5
-1
-1.5
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Frequency = 1000 Hz 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1
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Substituting C, R1, and R2; we get h(ω) as
% CPE 340 Principle of Communications, CPE, KMUTT % S1/2009 % Instructor : Asst. Prof. Peerapon S., Ph.D. % TA : San R. % Homework 1, Question 3 % % This M-file is provided by San R. % f = [1:1e5]; w = 2*pi*f; C = 0.05e-6; R1 = 1.5e3; R2 = 180; norm = R2 + j*w*C*R1*R2; denorm = R1 + norm; H = norm./denorm; mag = abs(H); phase = angle(H); figure; plot(f, mag); title('Magnitude Response'); figure; plot(f, phase); title('Phase Response');
Magnitude Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
0
1
2
3
4
5
6
7
8
9
10 x 10
4
Phase Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
0
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9
10 x 10
4
4. Consider a signal x(t) in Fig. 3 below. Let A = 2 volts, τ = 0.5 msec, T = 1 msec. x(t) A
0 τ T
Figure 3: (a) Determine the complex Fourier coefficients cn of x(t). Show every step of your calculation. Soln. cn = = = = = = =
1 T
Z
A T
Z
x(t)e−j2nπf0 t dt
T τ /2
e−j2nπf0 t dt
−τ /2
τ /2 A e−j2nπf0 t −(j2nπf0 )T −τ /2 A e−jnπf0 τ − ejnπf0 τ −j2nπ A sin nπf0 τ nπ . sin πx Af0 τ sinc nf0 τ ; sinc x = πx sinc (n/2)
(b) What is the Fourier transform of x(t) ? Soln. The Fourier transform of x(t) is simply an impulse train spaced by f0 and weighted by cn . X(f ) = =
∞ X n=−∞ ∞ X
cn δ(f − nf0 ) sinc(n/2)δ(f − nf0 )
n=−∞
(c) Suppose x(t) is passed through a filter in Fig. 4 below. Determine the magnitude of the complex Fourier coefficients for y(t) and the power of y(t) within the 3-dB bandwidth of the filter. Soln. Let Y (ω) ↔ y(t), ω = 2πf . The transfer function of the filter is given by H(ω) =
1/jωC 1 1 − jωRC = = R + 1/jωC 1 + jωRC 1 + (ωRC)2
with |H(ω)| =
1 p 1 + (ωRC)2
R = 170 C = 0.2 µF
x(t)
y(t)
Figure 4: Therefore, the magnitude of the complex Fourier coefficients for y(t) is |Yn | = |H(f )||cn | = |H(nf0 )||cn | 1 = p |sinc (n/2)|; f0 = 1 kHz, R = 170 Ω, C = 0.2 µF 1 + (2πnf0 RC)2 1 = √ |sinc (n/2)| 1 + 0.0456n2
The 3-dB bandwidth of the filter is 1/2πRC = 4.68 kHz. Therefore, the frequency components up to the fourth harmonic stay within the 3-dB bandwidth. The normalized signal power in such bandwidth range is P
= Y0 + 2
4 X
|Yn |2
n=1
= 1 + 2 ∗ (0.6222 + 0 + 0.1782 + 0) = 1.84 Watts
5. A unity-amplitude periodic square wave with 50% duty cycle is passed through an RC lowpass filter whose 3-dB cutoff frequency is 1,500 Hz. Using a computer, find and plot the output signal if the input square wave has a frequency of 300 Hz, 500 Hz, and 1,000 Hz. Soln. The input square wave can be represented by the Fourier series: x(t) =
∞ X
cn ejnω0 t
n=−∞
where cn =
2 sin(nπ/2) ,
n 6= 0
0,
n=0
nπ
From (2-140) in the Couch text, the output waveform is then given by y(t) = =
∞ X
H(nf0 )cn ejnω0 t
n=−∞ ∞ X
dn ejnω0 t , dn , H(nf0 )cn , H(f ) =
n=−∞
1 1 + j(f /fc )
where fc = 1,500 Hz for the RC low-pass filter. Representing y(t) in the trigonometric form to easily plot, y(t) = D0 +
∞ X
Dn cos(nω0 t + φn )
n=1
where D0 = 0, since c0 = 0 Dn = 2|dn | = 2|H(nf0 )cn |, n > 0 1 4 r = 2 , n odd nπ 0 1 + nf fc 1 − sin(nπ/2) −1 nf0 φn = − tan +π , n odd fc 2
Frequency = 300 Hz 1.5
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-1
-1.5
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Frequency = 500 Hz 1.5
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-1
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Frequency = 1000 Hz 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1
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