HW 07 - Revision history

Fonggr: /* Answer 3 */

2008-12-08T21:58:41Z

Answer 3

← Older revision		Revision as of 14:58, 8 December 2008
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	*~~Since~~ the ~~sinc~~ function ~~extends out to +/- infinity in time~~, it is ~~not possible to build~~ a ~~brick wall filter for~~ it.		*The impulse response is a sinc function. The sinc function was "bouncing" around before the impulse function even came along. In other words, the impulse response knew when the impulse would occur, even though it hadn't happened yet (and it knew this, at t=-\infty). This is a non-causal system and is thus unrealizeable. If it had a u[t] stuck on the end of it, then it would become a causal system and would be perfectly legit.

Fonggr: /* Answer 3 */

2008-12-08T21:46:18Z

Answer 3

← Older revision		Revision as of 14:46, 8 December 2008
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	\|		\|
	\|<math>= 2\,f_0\,\mathrm{sinc}(~~j\,~~2\,\pi\,f_0\,(t-t_0))</math>		\|<math>= 2\,f_0\,\mathrm{sinc}(2\,\pi\,f_0\,(t-t_0))</math>
	\|}		\|}

	*Since the sinc function extends out to +/- infinity in time, it is not possible to build a brick wall filter for it.		*Since the sinc function extends out to +/- infinity in time, it is not possible to build a brick wall filter for it.

Fonggr: /* Answer 3 */

2008-12-08T21:45:20Z

Answer 3

← Older revision		Revision as of 14:45, 8 December 2008
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	\|-		\|-
	\|		\|
	\|<math>= e^{-j\,2\,\pi\,f\,t_0}~~\int_{-f_0}^{f_0~~}~~\,e^~~{j\,2\,\pi\,~~f\,~~t}\~~,df~~</math>		\|<math>= \left . \frac{\,e^{j\,2\,\pi\,f\,(t-t_0)}}{j\,2\,\pi\,(t-t_0)}\right \|^{f_0}_{-f_0}</math>
	\|-		\|-
	\|		\|
	\|<math>= e^{-j\,2\,\pi\,f\,t_0}~~\left[\left . \frac{\,~~e^{j\,2\,\pi\,f\,t}}{j\,2\,\pi\,t~~}\right \|^{f_0}_{~~-~~f_0~~}~~\right]~~</math>		\|<math>= \frac{\,e^{j\,2\,\pi\,f_0\,(t-t_0)}-e^{-j\,2\,\pi\,f_0\,(t-t_0)}}{j\,2\,\pi\,(t-t_0)}</math>
	\|-		\|-
	\|		\|
	\|<math>= ~~e^{-j\,2\,\pi\,f\,t_0}\left[~~\frac{\~~,e^{j\,~~2\,\pi\,f_0\,t}-~~e^{-j\,2\,\pi\,f_0\,t}~~}{~~j\,2\,~~\pi\,t}~~\right]~~</math>		\|<math>= \frac{\sin(2\,\pi\,f_0\,(t-t_0))}{\pi\,(t-t_0)}</math>
	\|-		\|-
	\|		\|
	\|<math>= e^{-j\,2\,~~\pi\,f\,t_0~~}~~\left[~~\frac{\sin(~~j\,~~2\,\pi\,f_0\,t)}{\pi\,t}~~\right]~~</math>		\|<math>= \frac{2\,f_0}{2\,f_0}\frac{\sin(2\,\pi\,f_0\,(t-t_0))}{\pi\,(t-t_0)}</math>
	\|-		\|-
	\|		\|
	\|<math>= ~~e^{-j\,2\,\pi\,f\,t_0}\left[\frac{j\,2\,f_0}{j\,2\,f_0}\frac{\sin(j\,2\,\pi\,f_0\,t)}{\pi\,t}\right]</math>~~		\|<math>= 2\,f_0\,\mathrm{sinc}(j\,2\,\pi\,f_0\,(t-t_0))</math>
	\|-
	\|
	~~\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[j\,~~2\,f_0\,\mathrm{sinc}(j\,2\,\pi\,f_0\,t)~~\right]~~</math>
	\|}		\|}

	*Since the sinc function extends out to +/- infinity in time, it is not possible to build a brick wall filter for it.		*Since the sinc function extends out to +/- infinity in time, it is not possible to build a brick wall filter for it.

Fonggr: /* Answer 3 */

2008-12-08T21:20:58Z

Answer 3

← Older revision		Revision as of 14:20, 8 December 2008
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	\|}		\|}

	Since the sinc function extends out to +/- infinity, it is not possible to build a brick wall filter for it.		*Since the sinc function extends out to +/- infinity in time, it is not possible to build a brick wall filter for it.

Fonggr: /* Answer 3 */

2008-12-08T21:19:50Z

Answer 3

← Older revision		Revision as of 14:19, 8 December 2008
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	\|-		\|-
	\|<math>H(t)\,\!</math>		\|<math>H(t)\,\!</math>
	\|<math>= \int_{-\infty}^{\infty}[u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}e^{-j\,2\,\pi\,f\,t}\,dt</math>		\|<math>= \int_{-\infty}^{\infty}[u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}e^{j\,2\,\pi\,f\,t}\,df</math>
	\|-		\|-
	\|		\|
	\|<math>= \int_{-f_0}^{f_0}\,e^{-j\,2\,\pi\,f\,\left(t_0+t\right)}\,dt</math>		\|<math>= \int_{-f_0}^{f_0}\,e^{j\,2\,\pi\,f\,\left(t-t_0\right)}\,df</math>
	\|-		\|-
	\|		\|
	\|<math>= e^{-j\,2\,\pi\,f\,t_0}\int_{-f_0}^{f_0}\,e^{-j\,2\,\pi\,f\,t}\,dt</math>		\|<math>= e^{-j\,2\,\pi\,f\,t_0}\int_{-f_0}^{f_0}\,e^{j\,2\,\pi\,f\,t}\,df</math>
	\|-		\|-
	\|		\|
	\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\left . \frac{\,e^{-j\,2\,\pi\,f\,t}}{-j\,2\,\pi\,f}\right \|^{f_0}_{-f_0}\right]</math>		\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\left . \frac{\,e^{j\,2\,\pi\,f\,t}}{j\,2\,\pi\,t}\right \|^{f_0}_{-f_0}\right]</math>
	\|-		\|-
	\|		\|
	\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\frac{\,e^{-j\,2\,\pi\,f\,f_0}-e^{j\,2\,\pi\,f\,f_0}}{-j\,2\,\pi\,f}\right]</math>		\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\frac{\,e^{j\,2\,\pi\,f_0\,t}-e^{-j\,2\,\pi\,f_0\,t}}{j\,2\,\pi\,t}\right]</math>
			\|-
			\|
			\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\frac{\sin(j\,2\,\pi\,f_0\,t)}{\pi\,t}\right]</math>
			\|-
			\|
			\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\frac{j\,2\,f_0}{j\,2\,f_0}\frac{\sin(j\,2\,\pi\,f_0\,t)}{\pi\,t}\right]</math>
			\|-
			\|
			\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[j\,2\,f_0\,\mathrm{sinc}(j\,2\,\pi\,f_0\,t)\right]</math>
	\|}		\|}

			Since the sinc function extends out to +/- infinity, it is not possible to build a brick wall filter for it.

Fonggr: /* Answer 3 */

2008-12-08T20:59:50Z

Answer 3

← Older revision		Revision as of 13:59, 8 December 2008
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	\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\left . \frac{\,e^{-j\,2\,\pi\,f\,t}}{-j\,2\,\pi\,f}\right \|^{f_0}_{-f_0}\right]</math>		\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\left . \frac{\,e^{-j\,2\,\pi\,f\,t}}{-j\,2\,\pi\,f}\right \|^{f_0}_{-f_0}\right]</math>
			\|-
			\|
			\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\frac{\,e^{-j\,2\,\pi\,f\,f_0}-e^{j\,2\,\pi\,f\,f_0}}{-j\,2\,\pi\,f}\right]</math>
	\|}		\|}

Fonggr: /* Answer 3 */

2008-12-08T20:58:19Z

Answer 3

← Older revision		Revision as of 13:58, 8 December 2008
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	\|<math>= \int_{-~~\infty~~}^{~~\infty~~}~~[u(f+f_0)-u(f-f_0)]~~\,e^{-j\,2\,\pi\,f\,\left(t_0+t\right)}\,dt</math>		\|<math>= \int_{-f_0}^{f_0}\,e^{-j\,2\,\pi\,f\,\left(t_0+t\right)}\,dt</math>
	\|-		\|-
	\|		\|
	\|<math>= \int_{-\~~infty~~}^{\~~infty~~}[~~u(f+f_0)~~-u(f-~~f_0)]~~\,\~~delta~~\~~left(t_0+t~~\right)\~~,dt~~</math>		\|<math>= e^{-j\,2\,\pi\,f\,t_0}\int_{-f_0}^{f_0}\,e^{-j\,2\,\pi\,f\,t}\,dt</math>
			\|-
			\|
			\|<math>= e^{-j\,2\,\pi\,f\,t_0}\left[\left . \frac{\,e^{-j\,2\,\pi\,f\,t}}{-j\,2\,\pi\,f}\right \|^{f_0}_{-f_0}\right]</math>
	\|}		\|}

Fonggr: /* Answer 3 */

2008-12-07T19:40:43Z

Answer 3

← Older revision		Revision as of 12:40, 7 December 2008
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	\|-		\|-
	\|<math>H(t)\,\!</math>		\|<math>H(t)\,\!</math>
	\|<math>= \int_{-\infty}^{\infty}[u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}e^{-j\,2\,\pi\,f\,t}</math>		\|<math>= \int_{-\infty}^{\infty}[u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}e^{-j\,2\,\pi\,f\,t}\,dt</math>
			\|-
			\|
			\|<math>= \int_{-\infty}^{\infty}[u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,\left(t_0+t\right)}\,dt</math>
			\|-
			\|
			\|<math>= \int_{-\infty}^{\infty}[u(f+f_0)-u(f-f_0)]\,\delta\left(t_0+t\right)\,dt</math>
	\|}		\|}

Fonggr: /* Answer 3 */

2008-12-07T19:35:39Z

Answer 3

← Older revision		Revision as of 12:35, 7 December 2008
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	===Answer 3===		===Answer 3===
	*We're interested in taking it back to the time domain		*We're interested in taking it back to the time domain

	~~<math>H(f) = [u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}</math>~~

	{\| border="0" cellpadding="0" cellspacing="0"		{\| border="0" cellpadding="0" cellspacing="0"
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	\|-		\|-
	\|<math>H(t)\,\!</math>		\|<math>H(t)\,\!</math>
	\|<math>= [u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}*</math>		\|<math>= \int_{-\infty}^{\infty}[u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}e^{-j\,2\,\pi\,f\,t}</math>
	\|}		\|}

Fonggr: /* Answer 3 */

2008-12-05T19:11:08Z

Answer 3

← Older revision		Revision as of 12:11, 5 December 2008
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	By investigating its impulse response, argue that an ideal "brick wall" low pass filter having <math>H(f) = [u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}</math> is always going to be an unrealizeable filter, meaing that we won't ever be able to build a system with this ''exact'' response.		By investigating its impulse response, argue that an ideal "brick wall" low pass filter having <math>H(f) = [u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}</math> is always going to be an unrealizeable filter, meaing that we won't ever be able to build a system with this ''exact'' response.
	===Answer 3===		===Answer 3===
			*We're interested in taking it back to the time domain

			<math>H(f) = [u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}</math>

			{\| border="0" cellpadding="0" cellspacing="0"
			\|-
			\|<math>H(f)\,\!</math>
			\|<math>= [u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}</math>
			\|-
			\|<math>H(t)\,\!</math>
			\|<math>= [u(f+f_0)-u(f-f_0)]\,e^{-j\,2\,\pi\,f\,t_0}*</math>
			\|}