HW 08 - Revision history

Fonggr: /* Answer 1 */

2008-12-18T00:08:42Z

Answer 1

← Older revision		Revision as of 17:08, 17 December 2008
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	Amplitude: The amplitude remains the same		Amplitude: The amplitude remains the same

	Phase:		Phase: Remains the same

	===Question 2===		===Question 2===

Fonggr: /* Answer 2 */

2008-12-09T09:29:18Z

Answer 2

@@ Line 13: / Line 13: @@
 *What is the output due to <math>cos(2\pi ft)\,\!</math>?
 ===Answer 2===
 ===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.

Fonggr: /* Answer 3 */

2008-12-09T09:20:25Z

Answer 3

← Older revision		Revision as of 02:20, 9 December 2008
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	\|-		\|-
	\|<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 +/- 2f_0, 1/4th of x(t).		The spectrum of w(t) is 1/2 of x(t) at the center frequency. At +/- 2f_0, 1/4th of x(t).

Fonggr: /* Answer 3 */

2008-12-09T08:55:46Z

Answer 3

← Older revision		Revision as of 01:55, 9 December 2008
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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.		*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 +/- 2f_0, 1/4th of x(t).

Fonggr: /* Answer 3 */

2008-12-09T08:44:23Z

Answer 3

← Older revision		Revision as of 01:44, 9 December 2008
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	\|<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>=\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>
	\|}		\|}

			*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.

Fonggr: /* Answer 3 */

2008-12-09T08:31:18Z

Answer 3

← Older revision		Revision as of 01:31, 9 December 2008
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	\|-		\|-
	\|		\|
	\|<math>=\frac{1}{2}~~x(t)~~\left[~~\delta~~(f_0-f)+~~\delta~~(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>
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	\|-		\|-
	\|		\|
	\|<math>=\int_{-\infty}^{\infty}\frac{1}{2}~~x(t)~~\left[~~\delta~~(f_0-f)+~~\delta~~(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>=\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>
	\|}		\|}

Fonggr: /* Answer 1 */

2008-12-09T08:26:47Z

Answer 1

← Older revision		Revision as of 01:26, 9 December 2008
Line 4:		Line 4:
	Frequency: The frequency is doubled		Frequency: The frequency is doubled

	Amplitude:		Amplitude: The amplitude remains the same

	Phase:		Phase:

Fonggr: /* Answer 3 */

2008-12-07T22:32:50Z

Answer 3

← Older revision		Revision as of 15:32, 7 December 2008
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	\|<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
	\|-		\|-
	\|<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)
	\|-		\|-
	\|		\|
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	\|		\|
	\|<math>=\frac{1}{2}x(t)\left[\delta(f_0-f)+\delta(f_0+f)\right]</math>		\|<math>=\frac{1}{2}x(t)\left[\delta(f_0-f)+\delta(f_0+f)\right]</math>
			\|-
			\|<math>w(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)
			\|-
			\|
			\|<math>=\int_{-\infty}^{\infty}\frac{1}{2}x(t)\left[\delta(f_0-f)+\delta(f_0+f)\right]\frac{e^{j2\pi f_0 t}+e^{-j2\pi f_0 t}}{2}e^{j2\pi ft}\,df</math>
	\|}		\|}

Fonggr: /* Answer 3 */

2008-12-07T21:57:34Z

Answer 3

← Older revision		Revision as of 14:57, 7 December 2008
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	\|-		\|-
	\|<math>v(f)\,\!</math>		\|<math>v(f)\,\!</math>
	\|<math>=\int_{-\infty}^{\infty}x(t)\,\cos(2\pi f_0 t)e^{-j2\pi ft}</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)\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}x(t)\left[\delta(f_0-f)+\delta(f_0+f)\right]</math>
	\|}		\|}

Fonggr at 21:43, 7 December 2008

2008-12-07T21:43:07Z

← Older revision		Revision as of 14:43, 7 December 2008
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	*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).		*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===		===Answer 3===
			{\| border="0" cellpadding="0" cellspacing="0"
			\|-
			\|<math>v(t)\,\!</math>
			\|<math>=x(t)\,\cos(2\pi f_0 t)</math>
			\|-
			\|<math>v(f)\,\!</math>
			\|<math>=\int_{-\infty}^{\infty}x(t)\,\cos(2\pi f_0 t)e^{-j2\pi ft}</math>
			\|}