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F stable the system is bibo 211 for an ltic system an

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Unformatted text preview: 11 For an LTIC system, an input x(t) produces an output y(t) as shown in Fig. P2.11. Sketch the outputs for the following set of inputs: (i) 5x(t) (ii) 0.5x(t-1)+0.5x(t+1) (iii) x(t+1)-x(t-1) (i) (ii) (iii) Using linearity property 5x(t) → 5y(t) Using linearity property 0.5x(t-1)+ 0.5x(t+1) → 0.5y(t-1)+ 0.5y(t+1) Using linearity property x(t-1)-x(t+1) → y(t-1)-y(t+1) 2.18 The output h[k] of a DT LTI system in response to a unit impulse function δ[k] is shown in Fig. P2.18. Find the output for the following set of inputs: EE 341 Fall 2012 (i) x[k]=δ[k+1]+δ[k]+δ[k-1] (ii) x[k]= (iii) x[k]=u[k] (i) [ −4 ] x[k]=δ[k+1]+δ[k]+δ[k-1] The impulse response is given by δ[k] → h[k]=δ[k+1]-2δ[k]+δ[k-1] So the response to input x[k] is given by δ[k+1] → h[k+1]=δ[k+2]-2δ[k+1]+δ[k] δ[k] → h[k]=δ[k+1]-2δ[k]+δ[k-1] δ[k-1] → h[k-1]=δ[k]-2δ[k-1]+δ[k-2] So the total response is: δ[k+1]-2δ[k]+δ[k-1] → δ[k+2]-δ[k+1]-δ[k-1]+ ]+δ[k-2] (ii) x[k]= [ −4 ] The response will...
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This document was uploaded on 03/06/2014 for the course EE 341 at NMT.

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