bevans@ece.utexas.edu
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Modulation Property. From this homework problem, you'll be able to derive a fundamental principle called the Modulation Property. D 4 It says that multiplying a signal v(t) by cos(2 pi fc t) in the time E 4 I 4 It says that multiplying a signal v(t) by cos(2 pi fc t) in the time E 4 domain will in the frequency domain shift the frequency content in D 4 v(t) by fc and by -fc. E 4 I 4 v(t) by fc and by -fc. E 4 I 2
E 2 This might not seem too important, but it allows us to convert an D 4 audio signal v(t) into a wireless electromagnetic signal centered at fc E 4 I 4 audio signal v(t) into a wireless electromagnetic signal centered at fc E 4 for long-range broadcast across the state and country, or a communication signal over the 2.4 GHz band for Wi-Fi. I 2 Example carrier frequencies: AM radio about 1 MHz, FM radio about 100 MHz, and Wi-Fi 2.4-2.499 GHz. (There are other Wi-Fi bands.) Audio signals have content between 20 Hz and 20 kHz, which are low frequencies compared to the aforementioned carrier frequencies. Audio signals don't propagate very far; e.g. sound from outdoor concert venues might propagate 1-2 km. By converting the audio signal to an AM radio signal, we can broadcast it for thousands of kilometers. E 2
Visualization of the modulation property in the frequency domain
Logistics: Because there are so many terms in this problem, I'd recommend creating placeholder terms when working the math. D 3 Using placeholder terms will also see how the terms interact: E 3 I 3 Using placeholder terms will also reveal how the terms interact: E 3
D 4 y(t) = (x(t) + A) cos(2 pi fc t) = (x(t) + A) cos(wc t) E 4 I 4 y(t) = (x(t) + A) cos(2 pi fc t) = (x(t) + A) cos(wc t) E 4
D 4 x(t) = 3 cos(w1 t + pi/4) + cos(w2 t + pi/2) E 4 I 4 D 10 x(t) = 3 cos(w1 t + pi/4) + cos(w2 t + pi/2) E 10 I 10 x(t) = 2 cos(w1 t + pi/2) E 10 E 4
where
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w1 = 2 pi (1000 Hz) = 2000 pi
w2 = 2 pi (2000 Hz) = 4000 pi
wc = 2 pi (1300 kHz) = 2600 x 10^3 pi
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w1 = 2 pi (1000 Hz) = 2000 pi
w2 = 2 pi (2000 Hz) = 4000 pi
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w1 = 2 pi (660 Hz) = 1320 pi
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wc = 2 pi (1300 kHz) = 2600 x 103 pi
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Per the hint on problem 2.2, substitute x(t) into the equation for y(t)
D 4 y(t) = (3 cos(w1 t + pi/4) + cos(w2 t + pi/2) + A) cos(wc t) E 4 I 4 D 10 y(t) = (3 cos(w1 t + pi/4) + cos(w2 t + pi/2) + A) cos(wc t) E 10 I 10 y(t) = (2 cos(w1 t + pi/2) + A) cos(wc t) E 10 E 4
and expand terms
D 4 y(t) = 3 cos(w1 t + pi/4) cos(wc t) + cos(w2 t + pi/2) cos(wc t) + A cos(wc t) E 4 I 4 D 10 y(t) = 3 cos(w1 t + pi/4) cos(wc t) + cos(w2 t + pi/2) cos(wc t) + A cos(wc t) E 10 I 10 y(t) = 2 cos(w1 t + pi/2) cos(wc t) + A cos(wc t) E 10 E 4
D 10 The first and second terms involve beat frequencies, and you can reuse what you've learned about beat frequencies from the in-lecture work on Tuesday E 10 I 10 The first term involves beat frequencies, and you can reuse what you'll learned about beat frequencies from the in-lecture work on Tuesday, Sep. 5, 2023, E 10 as well as the textbook.
D 10 This problem is similar to problem 2.2 in fall 2018. Problem 2.2 in fall 2018 gives the spectrum for amplitude modulation and asks to find the time-domain expression. It's the dual of this semester's problem. E 10 I 10 This problem is a simplified version of D 11 to homework problem 2.2 in fall 2021. E 11 I 11 D 15 homework problem 2.2 in fall 2021. E 15 I 15 homework problem 2.2 in fall 2021. E 15 I 13
Solution on the marker board E 13 E 11 E 10 I 3
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D 5 This problem explores the use of a spectrogram to analyze both periodic and non-periodic signals simultaneously in the time and frequency domains. E 5 I 5 Consider the Fourier series analysis of a square wave on lecture slide 3-11. The Fourier series analysis says that the square wave contains a frequency at 25 Hz with strength -j/pi. However, the square wave has an amplitude of 0 from 20ms to 40ms, and hence, no frequency components are present. D 9 The Fourier series analysis computes the strength of a frequency component over the period, and the strength is averaged over one period. E 9 I 9 The Fourier series analysis computes the average strength of a frequency component over the period. E 9
This homework problem explores the use of a spectrogram to analyze both periodic and non-periodic signals in the time and frequency domains simultaneously. E 5 The spectrogram tell us when in time a particular frequency occurs and at what strength. I 5
In the video of "Sandstorm" by Darude (Synthesia version), the keyboard notes (frequencies) are in the horizontal direction and time is in the vertical direction. D 6 This is a time-frequency representation that indicates when certain notes (frequencies) are played. Note that the time-frequency analysis does not show the harmonics of the note frequencies. E 6 I 6 The frequencies have octave spacing; e.g., C3 is the "C" note in the third octave (C3) and has twice the frequency of C2, C2 has twice the frequency of C1, etc. A spectrogram, on the other hand, has uniform spacing of the frequencies. D 8 Nontheless, the Sandstorm video shows a time-frequency representation that E 8 I 8 Nonetheless, the Sandstorm video shows a time-frequency representation that E 8 indicates when certain notes (frequencies) are played, but does not show the harmonics of the note frequencies. I 7
D 10 For part (b), y(t) = cos2(2 pi f0 t) = (1/2) + (1/2) cos(2 pi (2 f0) t). E 10 I 10 For part (a), y(t) = cos2(2 pi f0 t) = (1/2) + (1/2) cos(2 pi (2 f0) t). E 10 In this case, y(t) has frequency components at -2 f0, 0, and 2 f0, as we discussed in lecture. We can see this in spectrogram plot by applying the spectrogram command to y(t) using the code from problem 2.3. D 10 The spectrogram only shows non-negative frequencies, so we should see straight E 10 I 10 By default, the spectrogram only shows non-negative frequencies, so we should see straight E 10 solid lines at 0 Hz and 880 Hz.
D 10 For part (c), y(t) = cos4(2 pi f0 t). As we did on lecture slide 3-9, we can expand cos(theta) using the inverse Euler formula cos(theta) = ( exp(j theta) + exp(-j theta) ) / 2 where theta = 2 pi f0 t and then take ( exp(j theta) + exp(-j theta) ) / 2 to the fourth power. In this case, y(t) will have frequency components of -4 f0, -2 f0, 0, 2 f0 and 4 f0. We can see this in spectrogram plot by applying the spectrogram command to y(t) using the code from problem 2.3. The spectrogram only shows non-negative frequencies, so we should see straight solid lines at 0 Hz, 880 Hz and 1760 Hz.
E 10 Although not asked, we can see what happens when y(t) = cos3(2 pi f0 t). D 10 We know from lecture slide 3-9 that y(t) will have frequency components of -3 f0, -f0, -f0 and 3 f0. E 10 I 10 We know from lecture slide 3-9 that y(t) will have frequency components of -3 f0, -f0, -f0 and 3 f0. E 10
Perhaps now a pattern has emerged. y(t) = cosn(2 pi f0 t) will have odd harmonics up to the nth harmonic if n is odd, and even harmonics up to the nth harmonic if n is even including a zero-frequency component.
D 10 For part (a), y(t) = | cos(2 pi f0 t) |. E 10 I 10 For part (b), y(t) = exp(x(t)). We can use a series expansion to analyze the resulting frequency components. E 10 D 12 If plot the spectrogram, which only plots non-negative frequency components, we’ll E 12 I 12 If plot the spectrogram, which only plots non-negative frequency components, we'll E 12 D 10 see frequency components at 0 Hz, 880 Hz, 1760 Hz, 2640 Hz, etc. The full set of frequency components is the infinite set of even harmonics: ..., -4 f0, -2 f0, 0, 2 f0, 4 f0., .... E 10 I 10 see frequency components at 0 Hz, 440 Hz, 880 Hz, 1320 Hz, 1760 Hz, etc. The full set of frequency components is the infinite set of harmonics: ..., -2 f0, -f0, 0, f0, 2 f0., .... E 10
Please keep in mind that we're sampling the continuous-time signal at a sampling rate of fs = 8000 Hz. Recalling the sampling theorem fs > 2 fmax, we can divide both sides of the inequality by 2 to obtain fmax < (1/2) fs. By sampling at a rate fs, we can only capture continuous-time frequencies up to (1/2) fs, or 4000 Hz. We won't be able to see harmonics at or above 4000 Hz in the spectrogram plot. D 10
Is there a method to represent a transcendental function as a sum of x, x2, x3 etc.? If so, we can reuse the above observations to explain what the spectrogram is showing.
For part (d), y(t) = cos( pi cos(2 pi f0 t)). Again, the relationship between output y(t) and x(t) is through a transcendental function y(t) = cos(pi x(t)). E 10 E 7 E 6 E 5
bevans@ece.utexas.edu
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