HW 08: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
 
(3 intermediate revisions by the same user not shown)
Line 6: Line 6:
Amplitude: The amplitude remains the same
Amplitude: The amplitude remains the same


Phase:
Phase: Remains the same


===Question 2===
===Question 2===
Line 13: Line 13:
*What is the output due to <math>cos(2\pi ft)\,\!</math>?
*What is the output due to <math>cos(2\pi ft)\,\!</math>?
===Answer 2===
===Answer 2===
{| border="1" cellpadding="5" cellspacing="0"
|-
|Input
|LTI System
|Output
|Reason
|-
|<math>\Phi(\lambda,t)\,\!</math>
|<math> \Longrightarrow </math>
|<math>\Psi(\lambda,t)\,\!</math>
|Given
|-
|}

===Question 3===
===Question 3===
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.
Line 23: Line 37:
|-
|-
|<math>v(t)\,\!</math>
|<math>v(t)\,\!</math>
|<math>=x(t)\,\cos(2\pi f_0 t)</math>
|<math>= x(t)\,\cos(2\pi f_0 t)</math>
|x(t) is the original signal
|x(t) is the original signal
|-
|-
|<math>v(f)\,\!</math>
|<math>v(f)\,\!</math>
|<math>=\int_{-\infty}^{\infty}x(t)\cos(2\pi f_0 t)e^{-j2\pi ft}\,dt</math>
|<math>= \int_{-\infty}^{\infty}x(t)\cos(2\pi f_0 t)e^{-j2\pi ft}\,dt</math>
|v(f) is x(t) after a low pass filter (cutoff frequency = f_max), multiplied by cos(2\pi f_0 t)
|v(f) is x(t) after a low pass filter (cutoff frequency = f_max), multiplied by cos(2\pi f_0 t)
|-
|-
|
|
|<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>
|<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>
|-
|-
|
|
|<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>
|<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>
|-
|-
|
|
|<math>=\frac{1}{2}\left[X(f_0-f)+X(f_0+f)\right]</math>
|<math>= \frac{1}{2}\left[X(f_0-f)+X(f_0+f)\right]</math>
|-
|-
|<math>w(t)\,\!</math>
|<math>w(t)\,\!</math>
|<math>=v(t)\cdot \cos(2\pi f_0 t)</math>
|<math>= v(t)\cdot \cos(2\pi f_0 t)</math>
|w(t) is v(t) multiplied by cos(2\pi f_0 t)
|w(t) is v(t) multiplied by cos(2\pi f_0 t)
|-
|-
|
|
|<math>=\int_{-\infty}^{\infty}\frac{1}{2}\left[X(f_0-f)+X(f_0+f)\right]\frac{e^{j2\pi f_0 t}+e^{-j2\pi f_0 t}}{2}e^{j2\pi ft}\,df</math>
|<math>= x(t)\,\cos(2\pi f_0 t)^2</math>
|-
|
|<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>
|-
|
|<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>
|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 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).

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).