HW 08: Difference between revisions
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Frequency: The frequency is doubled |
Frequency: The frequency is doubled |
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Amplitude: |
Amplitude: The amplitude remains the same |
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Phase: |
Phase: Remains the same |
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===Question 2=== |
===Question 2=== |
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*What is the output due to <math>cos(2\pi ft)\,\!</math>? |
*What is the output due to <math>cos(2\pi ft)\,\!</math>? |
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===Answer 2=== |
===Answer 2=== |
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|Input |
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|LTI System |
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|Reason |
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|<math>\Phi(\lambda,t)\,\!</math> |
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|<math> \Longrightarrow </math> |
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|<math>\Psi(\lambda,t)\,\!</math> |
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|Given |
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===Question 3=== |
===Question 3=== |
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If a signal x(t) only has frequency components near DC, <math>\left|X(f)\right| = 0</math> for <math>|f|>f_{max}\,\!</math>, then x(t) is known as a baseband signal. When x(t) is a baseband signal, <math>x(t)\,\cos(2\pi f_0 t)</math> is known as a double sideband (DSB) signal. Sometimes a double sideband signal is used to send information over a radio frequency communications link. The transmitter and receiver are shown below. |
If a signal x(t) only has frequency components near DC, <math>\left|X(f)\right| = 0</math> for <math>|f|>f_{max}\,\!</math>, then x(t) is known as a baseband signal. When x(t) is a baseband signal, <math>x(t)\,\cos(2\pi f_0 t)</math> is known as a double sideband (DSB) signal. Sometimes a double sideband signal is used to send information over a radio frequency communications link. The transmitter and receiver are shown below. |
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|<math>v(t)\,\!</math> |
|<math>v(t)\,\!</math> |
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|<math>=x(t)\,\cos(2\pi f_0 t)</math> |
|<math>= x(t)\,\cos(2\pi f_0 t)</math> |
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|x(t) is the original signal |
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|<math>v(f)\,\!</math> |
|<math>v(f)\,\!</math> |
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|<math>=\int_{-\infty}^{\infty}x(t) |
|<math>= \int_{-\infty}^{\infty}x(t)\cos(2\pi f_0 t)e^{-j2\pi ft}\,dt</math> |
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|v(f) is x(t) after a low pass filter (cutoff frequency = f_max), multiplied by cos(2\pi f_0 t) |
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|<math>= \int_{-\infty}^{\infty}x(t)\frac{e^{j2\pi f_0 t}+e^{-j2\pi f_0 t}}{2}e^{-j2\pi ft}\,dt</math> |
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|<math>= \frac{1}{2}\int_{-\infty}^{\infty}x(t)\left(e^{j2\pi (f_0-f) t}+e^{-j2\pi (f_0+f) t}\right)\,dt</math> |
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|<math>= \frac{1}{2}\left[X(f_0-f)+X(f_0+f)\right]</math> |
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|<math>w(t)\,\!</math> |
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|<math>= v(t)\cdot \cos(2\pi f_0 t)</math> |
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|w(t) is v(t) multiplied by cos(2\pi f_0 t) |
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|<math>= x(t)\,\cos(2\pi f_0 t)^2</math> |
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|<math>= x(t)\,\frac{e^{j2\pi f_0 t}+e^{-j2\pi f_0 t}}{2}\,\frac{e^{j2\pi f_0 t}+e^{-j2\pi f_0 t}}{2}</math> |
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|<math>x(t)\left[\frac{e^{j2\pi (2 f_0) t}}{4}+\frac{e^{j2\pi f_0 t}}{2}+\frac{e^{j2\pi (-2 f_0) t}}{4}\right]</math> |
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|Need help seeing the math |
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*The bandwidth of v(t) is 1/2 that of x(t). The cosine splits the (amplitude? of the) signal up in half and moves it up and down by f_0. |
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*The spectrum of w(t) is 1/2 of x(t) at the center frequency. At +/- 2*f_0, 1/4th of x(t). |
Latest revision as of 17:08, 17 December 2008
Question 1
If the sound track of a movie was played into a high fidelity playback system at twice the correct speed, what happens to a sine wave's frequency, amplitude and phase, relative to what happens at the correct speed? Explain your answers.
Answer 1
Frequency: The frequency is doubled
Amplitude: The amplitude remains the same
Phase: Remains the same
Question 2
Suppose and where is any real function of t. If we have a linear time invariant system where an input of produces an output of .
- How do you find if you are given ?
- What is the output due to ?
Answer 2
Input | LTI System | Output | Reason |
Given |
Question 3
If a signal x(t) only has frequency components near DC, for , then x(t) is known as a baseband signal. When x(t) is a baseband signal, is known as a double sideband (DSB) signal. Sometimes a double sideband signal is used to send information over a radio frequency communications link. The transmitter and receiver are shown below.
- Find the Fourier Transform of the DSB signal, .
- What is the lowest that can be used and still have the communications system work?
- How does the bandwidth of v(t) compare to the bandwidth of x(t)?
- What does the spectrum of w(t) look like and how does it compare to that of x(t)? A graph would be appropriate showing the spectrum of x(t) and that of w(t).
Answer 3
x(t) is the original signal | ||
v(f) is x(t) after a low pass filter (cutoff frequency = f_max), multiplied by cos(2\pi f_0 t) | ||
w(t) is v(t) multiplied by cos(2\pi f_0 t) | ||
Need help seeing the math |
- The bandwidth of v(t) is 1/2 that of x(t). The cosine splits the (amplitude? of the) signal up in half and moves it up and down by f_0.
- The spectrum of w(t) is 1/2 of x(t) at the center frequency. At +/- 2*f_0, 1/4th of x(t).