phase2.667 sample shift€ 2.667 samples⋅2πrad /cycle16 samples /cycle=π3rad€ 7cosπ8n +π3( )€ 7cosπ8n( )phase is always relativehere, compare peaks of shifted and unshifted cosinesshifted peak occurs before unshifted peak,so lead, so +φpeak occurs between samples, so interpolate€ ˆ ω =π8=2π16= 2πf ⇒ f =116T =1f= 16 samples /cyclephase€ 2.667 samples⋅2πrad /cycle16 samples /cycle=π3rad€ 7cosπ8n +π3( )phase just from plot(unshifted not plotted)here, look at “missing” sample shift topeak of cosineshifted peak occurs before unshifted peak,so lead, so +φ13.33 sample shift16 samples/cycle2.67€ 16 −13.33( )samples⋅2πrad /cycle16 samples /cycle=π3radrelative phase€ 6 samples⋅2πrad /cycle16 samples /cycle=3π4rad€ 7cosπ8n +π3( )phase between two shifted cosineshere, look at “missing” sample shift topeak of cosineshifted peak occurs before unshifted peak,so lead, so +φ10 sample shift€ 16 −10( )samples⋅2πrad /cycle16 samples /cycle=3π4rad6 sample shift€ 7cosπ8n +3π4( )€ 7cosπ8n( )estimate phase of x2relative phase€ 7cosπ8n +π3( )phase between two shifted cosines€ 3π4−π3=5π12rad6 sample shift€ 7cosπ8n +3π4( )€ 7cosπ8n( )Subtract phases, φx2- φx2.67€ 6 − 2.67( )samples ⋅2π16rad /cyclesamples /cycle=5π12radcompare€ 3.33( )samples ⋅2π16rad /cyclesamples /cycle=5π12rad€ 103      samples ⋅2π16rad /cyclesamples /cycle=5π12radrelative phase€ 7cosπ8n +π3( )phase between two shifted cosines3.33 sample shift€ 7cosπ8n +3π4( )€ 7cosπ8n( )€ 3.33( )samples ⋅2π16rad /cyclesamples /cycle=5π12radlook at shift between zero-crossingszero-crossing occurs between samples,so interpolate.easier to interpolate zero-crossing, than peak.represent the systemsolve system response to arbitrary inputSystems€ h[n] = y[n]x[ n ]=δ[ n]⇔ H z( )=bkz−kk=0M∑1− akz−kk=1N∑=Πi=0Mz − zzi( )Πi=0Nz − zpi( )system functionpolynomialimpulse responsesequencepole-zerolocationsfrequency response€ y n[ ]= aly n − l[ ]l=1N∑+ bkx n − k[ ]k=0M∑difference equation € Hω( )= H ejω( )= H z( )z=ejωThe region of convergence mustContain the unit circle for to converge and the systemto be stable. (general)(FIR filter always stable) € Hω( ) € c € cEquivalent ways to represent the systemx[n]y[n]unit delay€ b0€ b1+ +€ a1block diagram€ ⇔ € czx[n]=δ[n]inspection€ z = ejω123456€ h[n] = y[n]x[n ]=δ[n ]impulse responsefrequency response€ y n[ ]= aly n − l[ ]l=1N∑+ bkx n − k[ ]k=0M∑iteration of difference equation € Hω( )= H ejω( )= H z( )z=ejωEquivalent ways to solve for response to arbitrary input1234€ H z( )=Y (z)X (z)=bkz−kk= 0M∑1− akz−kk=1N∑convolve input with impulse response€ y[n] = h[n]* x[n]z-transform of diff.eqn.use frequency response € y[n]x[ n ]= ejˆ ω n= Hˆ ω ( )ejˆ ω nuse z-transforms€ Y (z) = H(z) ⋅ X(z)€ y[n]€ ⇓ IZTa) convolution sumb) synthetic polynomialmultiplication€ y n[ ]= aly n − l[ ]l=1N∑+ bkx n − k[ ]k=0M∑iteration of difference equationEquivalent ways to solve for response to arbitrary input1234€ H z( )=Y (z)X (z)=bkz−kk= 0M∑1− akz−kk=1N∑convolve input with impulse response€ y[n] = h[n]* x[n]Inverse z-transformuse frequency response € y[n]x[ n ]= ejˆ ω n= Hˆ ω ( )ejˆ ω nuse z-transforms€ Y (z) = H(z) ⋅ X(z)€ y[n]€ ⇓ IZTi) numericalii) graphicali) numericalii) graphicala) FIRb) FIR/IIR € Hˆ ω ( )= h[k]ejˆ ω kk= 0M∑= bkejˆ ω kk= 0M∑ € Hω( )= H ejω( )= H z( )z=ejωLTIFIR h[n]=bnIIR h[n] solved iteratively (see 1) € Hω( )= H ejω( )exists?look at poles / roc.roc must contain unit circlea) long divisionb) lookup tablec) partial fractioni) match coeffii) powers of ziii) powers of z-1Initial rest conditionsright sided sequence (causal)€ z > a€ 11− az−1⇔ anu[n]€ z < aleft sided sequence€ 11− az−1⇔ −anu[−n −1]roc€ y[n] = h[n]* x[n]€ ⇓ ZT€ y n[ ]= aly n − l[ ]l=1N∑+ bkx n − k[ ]k=0M∑iteration of difference equationEquivalent ways to solve for response to arbitrary input1234convolve input with impulse response€ y[n] = h[n]* x[n]use frequency response € y[n]x[ n ]= ejˆ ω n= Hˆ ω ( )ejˆ ω nuse z-transforms€ Y (z) = H(z) ⋅ X(z)€ y[n]€ ⇓ IZTone sample at a time(possibly in sequence order)Do you have eqn?one sample at a time(possibly in sequence order)Do you have impulse response?one frequency componentat a time.Do you know freq content of x[n]?GeneralCan you do inverse z-transform?Fourier TransformsCompute spectrum of signals€ X[k] = x[n]e− j 2πk / N( )nn= 0N−1∑DFT€ Xk=2T0x(t)e− j 2πktT0dt0T0∫Periodic in (cont.) timeDiscrete freqDiscrete & periodic timeDiscrete & periodic freq FourierSeriesDTFTDiscrete timePeriodic in (cont.) freq € Hˆ ω ( )= h[k]ejˆ ω kk=0∞∑Discrete Fourier Transform (DFT)Compute spectrum of discrete-time periodic signalsN samples in time domain N complex numbers in frequency domain€ ⇒DFT€ ⇐IDFT€ x[n] =1NX[k]ej 2πk / N( )nk= 0N−1∑€ X[k] = x[n]e− j 2πk / N( )nn= 0N−1∑DFTIDFTanalysissynthesisDFT: sample continuous H(ω) (DTFT) at N evenly spaced frequencies1zPad to get more samples / “bins”.Window data. (DFT assumes periodicity).FFT is an efficient algorithm to compute DFTDFT Convolution€ y[n]* x[n] ⇔ Z(ˆ ω ) = Y (ˆ ω )X(ˆ ω ) ⇔ z[n]sampledomainfrequencydomain€ Y[k]€ Y (ˆ ω )sampled version ofUse DFT to compute Y[k] and X[k]sampledomainDTFTIDTFT€ y[n] ⊗ x[n] ⇔ Z[k] = Y[k]X[k] ⇔ z[n]DFTIDFTcircularconvolutionTo avoid temporal aliasing:if len(x)=N, len(y)=M, then pad so lengths are N+M-1Filter Design0 0.5 1 1.5 2 2.5 300.20.40.60.810 0.5 1 1.5 2 2.5 300.20.40.60.810 0.5 1 1.5 2 2.5 300.20.40.60.81lowpass highpass bandpass€ x n[ ]€ y n[ ]Ideal filter€ H1ˆ ω ( )€ H2ˆ ω ( )€ H3ˆ ω ( )€ ˆ ω € ˆ ω € ˆ ω passband stopbandtransition bandstopband ripplepassband ripplereal filtersFilter DesignFIR vs. IIR Filters (Matlab Help)FIR filters advantages:They can have exactly linear phase.They are always stable.The design methods are generally linear.They can be realized efficiently in hardware.The filter startup transients have finite duration.FIR filter disadvantageMuch higher filter order than IIR filters to achieve a given level of performance. Delay is greater than for an equal performance IIR filter.Filter DesignFIR Filters Design