Modulation Property. From this homework problem, you'll be able to derive a fundamental principle called the Modulation Property. It says that multiplying a signal v(t) by cos(2 pi fc t) in the time domain will in the frequency domain shift the frequency content in v(t) by fc and by -fc.
This might not seem too important, but it allows us to convert an audio signal v(t) into a wireless electromagnetic signal centered at fc for long-range broadcast across the state and country, or a communication signal over the 2.4 GHz band for Wi-Fi. 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.
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. Using placeholder terms will also reveal how the terms interact:
y(t) = (x(t) + A) cos(2 pi fc t) = (x(t) + A) cos(wc t)
x(t) = 3 cos(w1 t + pi/4) + cos(w2 t + pi/2)
where
w1 = 2 pi (1000 Hz) = 2000 pi
w2 = 2 pi (2000 Hz) = 4000 pi
wc = 2 pi (1300 kHz) = 2600 x 103 pi
Per the hint on problem 2.2, substitute x(t) into the equation for y(t)
y(t) = (3 cos(w1 t + pi/4) + cos(w2 t + pi/2) + A) cos(wc t)
and expand terms
y(t) = 3 cos(w1 t + pi/4) cos(wc t) + cos(w2 t + pi/2) cos(wc t) + A cos(wc t)
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 as well as the textbook.
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.
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. The Fourier series analysis computes the average strength of a frequency component over the period.
This homework problem explores the use of a spectrogram to analyze both periodic and non-periodic signals in the time and frequency domains simultaneously. The spectrogram tell us when in time a particular frequency occurs and at what strength.
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. 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. Nonetheless, the Sandstorm video shows a time-frequency representation that indicates when certain notes (frequencies) are played, but does not show the harmonics of the note frequencies.
For part (b), y(t) = cos2(2 pi f0 t) = (1/2) + (1/2) cos(2 pi (2 f0) t). 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. The spectrogram only shows non-negative frequencies, so we should see straight solid lines at 0 Hz and 880 Hz.
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.
Although not asked, we can see what happens when y(t) = cos3(2 pi f0 t). We know from lecture slide 3-9 that y(t) will have frequency components of -3 f0, -f0, -f0 and 3 f0.
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.
For part (a), y(t) = | cos(2 pi f0 t) |. If plot the spectrogram, which only plots non-negative frequency components, we’ll 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., ....
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.
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)).
bevans@ece.utexas.edu
