102_1_Week-10-Recap

# 102_1_Week-10-Recap - 48 NHAN LEVAN WEEK 10 RECAP4 P13 Let...

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48 NHAN LEVAN WEEK 10 RECAP 4 P13. Let sig t := 1 , t 0 , (sig := signum) := - 1 , t < 0 Then F{ sig t } =? SOLS: Please plot the signal 2 U ( t ) - 1 Then you will see that sign t = 2 U ( t ) - 1 F{ sign t } = F{ 2 U ( t ) - 1 } F{ sign t } = 2 π δ ( ω ) + 2 i ω - 2 π δ ( ω ) = 2 i ω Note: Here F{ sig t } is computed in terms of F{ U ( t ) } . Hence one can also find F{ U ( t ) } in terms of F{ sig t } ? For this you only have to note that d dt sig t = 2 δ ( t ) F{ 2 δ ( t ) } = F{ d dt sig t } = i ω F{ sigt } Therefore F{ sig t } = 2 i ω F{ U ( t ) } = 1 2 F{ 1 + sigt } = 1 2 { 2 πδ ( ω ) + 2 i ω } ( HaHa ) 26. UNIFORM SAMPLING We begin with a simple Problem. P14. Consider the pair: ( Ω - Band-Limited) sin Ω t π t , t R | rec ( ω , Ω ) , ω [ - Ω , Ω ] Now let us define the LTI system — denoted LP — whose IRF is (26.1) h ( t ) := sin Ω t π t , t R then LP is clearly non-causal. Hence it is called an “Ideal” Low Pass Filter (or system). The IPOP relation of LP is, by BT, (26.2) x ( t ) [ LP ] y ( t ) = -∞ sin Ω ( t - τ ) π ( t - τ ) x ( τ ) d τ , t R (i) Show that all Ω -Band-Limited signals x Ω ( t ) , that is, X Ω ( i ω ) := F{ x Ω ( t ) } = 0 , ω / [ - Ω , Ω ] applied to LP will reappear “as is” at the OP terminal of LP . In other words the band-limited signals x Ω ( t ) are impervious to LP , i.e., you don’t even know that x Ω ( t ) took a trip through LP !

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• Spring '08
• Levan
• LP, High-pass filter, Band-pass filter, Low-pass filter, NHAN LEVAN

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