96下信號與系çµ&plusmn

96下信號與系çµ&plusmn

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Unformatted text preview: Signals and Systems, Final Exam Solutions (Draft) Spring 2008, Edited by bypeng 1. [4] What is the group delay ( ) τ ω for a system with frequency response ( ) H j ω ? Explain what that means. Solution: The group delay ( ) τ ω is given by { } ( ) ( ) d H j d τ ω ω ω = - s This implies that the effective common time delay experienced by the small band or group of frequencies center at ω ω = is the negative of the slope of the phase at that frequency. 2. [4] A linear, time-invariant continuous-time system ( ) H j ω is distortionless within a signal band, c ω ω < , if for any input signal ( ) x t with ( ) X j ω = , c ω ω ≥ , the output is of the form ( ) ( ) y t kx t t =- for some fixed k and t . What is the condition for the frequency response of this system, ( ) H j ω , to be distortionless within a signal band, c ω ω < ? Solution: ( ) ( ) y t kx t t =- , so ( ) ( ) ( ) ( ) j t Y j ke X j H j X j ω ω ω ω ω- = = . The conclusion is ( ) j t H j ke ω ω- = for c ω ω < . 3. [8] ( ) c x t is a continuous-time signal, [ ] ( ) d c x n x nT = , and ( ) ( ) ( ) p c n x t x nT t nT δ ∞ =-∞ =- ∑ . The discrete-time Fourier transform of [ ] d x n is ( ) j d X e Ω , while the continuous-time Fourier transform of ( ) p x t is ( ) p X j ω . Find the relationship between ( ) j d X e Ω and ( ) p X j ω . Solution: ( ) ( ) ( ) p c n x t x nT t nT δ ∞ =-∞ =- ∑ ⇒ ( ) ( ) j nT p c n X j x nT e ω ω ∞- =-∞ = ∑ ; [ ] ( ) d c x n x nT = ⇒ ( ) [ ] ( ) ( ) j nT j j n j n T d d c c p n n n X e x n e x nT e x nT e X j T Ω ∞ ∞ ∞- Ω- Ω- Ω =-∞ =-∞ =-∞ Ω = = = = ∑ ∑ ∑ 4. [12] A signal ( ) c x t is echoed, so the actually received signal is ( ) ( ) ( ) c c c y t x t ax t T = +- An echo canceller is shown in Figure 4, with a goal to have ( ) ( ) c c z t x t = . Figure 4 (a) [8] Assume the sampling theorem is satisfied, the interpolation is perfect so ( ) [ ] c d z nT z n = , and T n T = , n is an integer. Find the difference equation relating [ ] d y n and [ ] d z n and the frequency response ( ) j d H e Ω . (b) [4] What is the continuous-time frequency response ( ) c H j ω between ( ) c y t and ( ) c z t ? Solution: (a) [ ] ( ) ( ) ( ) ( ) (( ) ) [ ] [ ] d c c c c c d d y n y nT x nT ax nT T x nT ax n n T x n ax n n = = +- = +- = +- . Since we want ( ) ( ) c c z t x t = , we need [ ] [ ] d d z n x n = , so [ ] [ ] [ ] d d d z n az n n y n +- = , and ( ) (1 ) ( ) j n j j d d Y e ae Z e- Ω Ω Ω = + ⇒ ( ) 1 ( ) ( ) 1 j j d d j n j d Z e H e Y e ae Ω Ω- Ω Ω = = + (b) Since we want ( )...
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96下信號與系çµ&plusmn

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