Chapter 10Representation of Discrete-Time SignalsRepresentation of Discrete-Time Signals - ExampleConvolution for Discrete-Time SystemsConvolution for Discrete-Important PropertiesProperties of ConvolutionExampleExample – cont.Example – cont. (Graphical Representation)Slide 10Slide 11Slide 12Remember These Geometric Series:Properties of Discrete-Time LTI SystemsSlide 15Example 1Example 2Example 3Chapter 10Discrete-Time Linear Time-Invariant SystemsSections 10.1-10-3Representation of Discrete-Time Signals•We assume Discrete-Time LTI systems •The signal X[n] can be represented using unit sample function or unit impulse function: [n]•Remember:•Notations: kknkxnx ][][][notesknkxelseknnx],[,0][][1],1[]1[]1[]1[][][0],0[]0[]0[]0[][][10nxnxnnxnxnxnxnnxnxRepresentation of Discrete-Time Signals - Example)]1([]1[][]0[)]1([]1[][][][][101nxnxnxnxnxnxnxConvolution for Discrete-Time Systems•LTI system response can be described using:•For time-invariant: [n-k] h[n-k]•For a linear system: x[k][n-k]x[k]h[n-k]•Remember: •Thus, for LTI:•We call this the convolution sum •Remember:kknkxnx ][][][System[n] h[n]][*][][][][][][][ nhnxknhkxnyknkxnxkk][*][][][][][*][][][][nxnhknxkhnynhnxknhkxnykk][][*][][*][000nnhnnnhnnnh Impulse Response of a SystemConvolution for Discrete-Important Properties•By definition•Remember (due to time-invariance property):•Multiplication][][*][][*][000nnhnnnhnnnh ][][*][][ nhnnhny ][][][][00ngnnngn Properties of Convolution•Commutative •Associative •Distributive]0[]0[]1[]1[]2[]2[....]1[]1[]0[]0[]1[]1[]2[]2[]3[]3[...][*][][][][][*][][][][hnxhnxhnxhnxhnxhnxhnxhnxnxnhknxkhnynhnxknhkxnykkExample•Given the following block diagram–Find the difference equation –Find the impulse response: h[n]; plot h[n]–Is this an FIR (finite impulse response) or IIR system? –Given x[1]=3, x[2]=4.5, x[3]=6, Plot y[n] vs. n–Plot y[n] vs. n using Matlab •Difference equation•To find h[n] we assume x[n]=[n], thus y[n]=h[n]–Thus: h[0]=h[1]=h[2]=1/3•Since h[n] is finite, the system is FIR•In terms of inputs:])2[]1[][(31][  nxnxnxnyFigure 10.3])2[]1[][(31][][  nnnnhnyFigure 10.3FIR system contains finite number of nonzero termsExample – cont.•Given the following block diagram–Find the difference equation –Find the impulse response: h[n]; plot h[n]–Is this an FIR (finite impulse response) or IIR system? –Given x[1]=3, x[2]=4.5, x[3]=6, Plot y[n] vs. n–Plot y[n] vs. n using Matlab •In terms of inputs:•Calculate for n=0, n=1, n=2, n=3, n=4, n=5, n=6–n=0; y[0]=0–n=1; y[n]=1–n=2; y[2]=2.5–n=3; y[2\3]=4.5–n=4; y[4]=3.5–n=5; y[5]=2–n=6; y[6]=0• Figure 10.3]0[]0[]1[]1[]2[]2[... .]1[]1[]0[]0[]1[]1[]2[]2[]3[]3[...][hnxhnxhnxhnxhnxhnxhnxhnxnyTry for different values of nExample – cont. (Graphical Representation)]0[]0[]1[]1[]2[]2[....]1[]1[]0[]0[]1[]1[]2[]2[]3[]3[...][hnxhnxhnxhnxhnxhnxhnxhnxnyh[0]=h[1]=h[2]=1/3 x[1]=3, x[2]=4.5, x[3]=6X[n-k]X[m]Example•Consider the following difference equation:y[n]=ay[n-1]+x[n]–Draw the block diagram of this system –Find the impulse response: h[n]–Is it a causal system?–Is this an IIR or FIR system?Example•Consider the following difference equation:y[n]=ay[n-1]+x[n]; –Draw the block diagram of this system –Find the impulse response: h[n]–Is this an IIR or FIR system? We assume x[n]=[n]y[n]=h[n]=ah[n-1]+[n];y[0]=h[0]=1y[1]=h[1]=ay[2]=h[2]=a^2y[3]=h[3]= a^3h[n]=a^n ; n>=0It is IIR (unbounded)Causal system (current and past)Example•Assume h[n]=0.6^n*u[n] and x[n]=u[n]–Find the expression for y[n]–Plot y[n] –Plot y[n] using Matlab 0];6.01[5.26.0][][6.0][*][][][][10nkuknunxnhknxkhnynnkkkkky[0]=1y[1]=1.6…..y(100)=2.5 Steady State Value is 2.5h[n]x[n]y[n]Remember These Geometric Series:Properties of Discrete-Time LTI Systems•Memory: –A memoryless system is a pure gain system: iff h[n]=K[n]; •K=h[0] = constant & h[n]=0 otherwise•Causality –y[n] has no dependency on future values of x[n]; thus h[n]=0 for n<0 (note h[n] is non-zero only for [n=0]. ][*][][][][][*][][][][nxnhknxkhnynhnxknhkxnykk...]1[]1[][]0[...]0[][][][][0nhxnhxhnxknxkhnyk...]1[]1[][]0[. ..]0[][][][][ nhxnhxhnxknkkxnynk]0[]0[]1[]1[]2[]2[....]1[]1[]0[]0[]1[]1[]2[]2[]3[]3[...][hnxhnxhnxhnxhnxhnxhnxhnxnyNote that if k<0depending on future; Thus h[k] should be zero to remove dependency on the future.Properties of Discrete-Time LTI Systems•Stability –BIBO: |x[n]|< M–Absolutely summable:•Invertibility: –If the input can be determined from output–It has an inverse impulse response –Invertible if there exists: hi[n]*h[n]=[n] kkh |][|Example 1•Assume h[n]= u[n] (1/2)^n–Memoryless? –Casual system? –Stable? –Has memory (dynamic): h[n] is not K[n] (not pure gain); h[n] is non-zero–Is causal: h[n]=0 for n<0–Stable:  k kkkh025.011|21||][|h[n]x[n]y[n]Example 2•Assume h[n]= u[n+1] (1/2)^n–Memoryless? –Casual system? –Stable? –Has memory (dynamic): h[n] is not K[n] (not pure gain)–Is NOT causal: h[n] not 0 for n<0; h[-1]=2–Stable:  k kkkh145.0112|21||][|h[n]x[n]y[n]Example 3•Assume h[n]= u[n] (2)^n–Memoryless? –Casual system? –Stable? –Has memory (dynamic): h[n] is not K[n] (not pure gain)–Is causal: h[n]=0 for n<0–Not Stable:  k