102_1_Week-9-Recap

102_1_Week-9-Recap - 38 NHAN LEVAN WEEK 9 RECAP 25. FT and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
38 NHAN LEVAN WEEK 9 RECAP 25. FT and Linear TI Systems Q A PROBLEMS FEST Q1. What is the meaning of Property-6 — Frequency-Shifting: e ± i ω 0 t f ( t ) | F ( i [ ω ω 0 ] ) ? A1. Consider the “modulated” signal f m ( t ) defined by f m ( t ) := f ( t ) cos ω 0 t, t R where f ( t ) := modulating signal, cos ω 0 t := carrier signal Let F m ( i ω ) denote the FT of f m ( t ) and F ( i ω ) is that of the f ( t ). Then F m ( i ω ) := F{ f ( t ) cos ω 0 t } = 1 2 [ F ( i [ ω - ω 0 ]) + F ( i [ ω + ω 0 ])] Thus the amplitude spectrum of f ( t ) is shifted by ± ω 0 . The modulating signal modulates the amplitude of cos ω 0 t . Hence the name amplitude modulation. ± P1. Show that for a LTI system S : x ( t ) [ S : LT I, h ( t )] y ( t ) , t R You have (25.1) Y ( i ω ) = H ( i ω ) X ( i ω ) , ω R S1 . By BT: (25.2) y ( t ) = ± -∞ h ( t - τ ) x ( τ ) d τ , t R This is also called a Convolution Integral. Taking the FT on both sides we get Y ( i ω ) = ± -∞ e - i ω t { ± -∞ h ( t - τ ) x ( τ ) d τ } dt, t R = ± -∞ x ( τ ) { ± -∞ e - i ω t h ( t - τ ) dt } d τ t - τ := σ Y ( i ω ) = ± -∞ x ( τ ) { ± -∞ e - i ω ( τ + σ ) h ( σ ) d σ } d τ Y ( i ω ) = ± -∞ e - i ωτ x ( τ ) { ± -∞ e - i ωσ h ( σ ) d σ } d τ = H ( i ω ) X ( i ω ) ± Conclusions-1: (i) Property 7: Time-Convolution F{ ± -∞ h ( t - τ ) x ( τ ) d τ } = H ( i ω ) X ( i ω )
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
39 (ii) Application to IPOP of LTI Systems: t - Domain : x ( t ) [ S : LTI , h ( t )] y ( t ) = ± -∞ h ( t - τ ) x ( τ ) d τ , t R f - Domain : X ( i ω ) [ S : LTI , H ( i ω )] Y ( i ω ) = H ( i ω ) X ( i ω ) , ω R (iii) Definitions of FRF: H ( i ω ) := Frequency Response Function (FRF) = F{ h ( t ) } (25.3) := Y ( i ω ) X ( i ω ) = F{ OP } F{ IP } (25.4) ± P2. Find y(t) given that: t R : e i ω 0 t [ S : LT I, h ( t )] y ( t ) =? for some ω 0 R . S2 . BT y ( t ) = ± -∞ h ( t - τ ) e i ω 0 τ d τ t - τ := σ y ( t ) = e i ω 0 t ± -∞ e - i ω 0 σ h ( σ ) d σ = H ( i ω ) | ω = ω 0 e i ω 0 t , t R ± Result: (25.5) t R : e ± i ω 0 t [ S : LTI , h ( t )] y ( t ) = H ( ± i ω 0 ) e ± i ω 0 t , t R Important Remarks: (i) The signal e i ω 0 t , t R , is called a Steady-State Signal (SSS) since it keeps on going for a long long time, i.e., forever, i.e., everlasting! Note that e i ω t U ( t ) is not a SSS. (ii) It follows from (25.6) y ( t ) = H ( i ω 0 ) e i ω 0 t , t R that the SSS e i ω 0 t is very “special” for a L, TI, system S ! Why so? It is because of the fact that e i ω 0 t enters S then it reappears at the OP terminal as “itself” — except for the multiplicative constant H ( i ω 0 )! Such a function (signal) is called an Eigenfunction (Eigensignal) — of S . Moreover, it is clear that the corresponding OP y ( t ) is also a SSS. (iii) Returning to (25.6):
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

102_1_Week-9-Recap - 38 NHAN LEVAN WEEK 9 RECAP 25. FT and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online