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SIGNAL► Signal is a physical quantity that varies with respect to time , space or any other independent variable Eg x(t)= sin t.► the major classifications of the signal are:(i) Discrete time signal(ii) Continuous time signal
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Unit Step &Unit Impulse Discrete time Unit impulse is defined as δ [n]= {0, n≠ 0 {1, n=0 Unit impulse is also known as unit sample. Discrete time unit step signal is defined by U[n]={0,n=0 {1,n>= 0Continuous time unit impulse is defined as δ (t)={1, t=0 {0, t ≠ 0 Continuous time Unit step signal is defined as U(t)={0, t<0 {1, t≥0
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SIGNAL► Periodic Signal & Aperiodic Signal A signal is said to be periodic ,if it exhibits periodicity.i.e., X(t +T)=x(t), for all values of t. Periodic signal has the property that it is unchanged by a time shift of T. A signal that does not satisfy the above periodicity property is called an aperiodic signal► even and odd signal ? A discrete time signal is said to be even when, x[-n]=x[n]. The continuous time signal is said to be even when, x(- t)= x(t) For example,Cosωn is an even signal.
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Energy and power signal► A signal is said to be energy signal if it have finite energy and zero power.► A signal is said to be power signal if it have infinite energy and finite power.► If the above two conditions are not satisfied then the signal is said to be neigther energy nor power signal
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Fourier SeriesThe Fourier series represents a periodic signal in terms offrequency components: p −1 ∞ ikω0n ikω0 t x(n) = ∑ Xk e x( t ) = ∑ Xk e k =0 k = −∞We get the Fourier series coefficients as follows: 1 p −1 1p −ikω0 t Xk = ∑ x(n)e −ikω0n Xk = ∫ x( t )e dt p n =0 p0The complex exponential Fourier coefficients are a sequence ofcomplex numbers representing the frequency component ω0k.
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Fourier series► Fourier series: a complicated waveform analyzed into a number of harmonically related sine and cosine functions► A continuous periodic signal x(t) with a period T0 may be represented by: X(t)=Σ∞k=1 (Ak cos kω t + Bk sin kω t)+ A0► Dirichlet conditions must be placed on x(t) for the series to be valid: the integral of the magnitude of x(t) over a complete period must be finite, and the signal can only have a finite number of discontinuities in any finite interval
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Trigonometric form for Fourier series► If the two fundamental components of a periodic signal areB1cosω0t and C1sinω0t, then their sum is expressed by trigonometric identities:► X(t)= A0 + Σ∞k=1 ( Bk 2+ Ak 2)1/2 (Ck cos kω t- φk) or► X(t)= A0 + Σ∞k=1 ( Bk 2+ Ak 2)1/2 (Ck sin kω t+ φk)
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Fourier Transform► Viewed periodic functions in terms of frequency components (Fourier series) as well as ordinary functions of time► Viewed LTI systems in terms of what they do to frequency components (frequency response)► Viewed LTI systems in terms of what they do to time-domain signals (convolution with impulse response)► View aperiodic functions in terms of frequency components via Fourier transform► Define (continuous-time) Fourier transform and DTFT► Gain insight into the meaning of Fourier transform through comparison with Fourier series
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The Fourier Transform►A transform takes one function (or signal) and turns it into another function (or signal)► Continuous Fourier Transform: ∞ H ( f ) = ∫ h ( t ) e 2πift dt −∞ ∞ h ( t ) = ∫ H ( f ) e −2πift df −∞
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Continuous Time Fourier TransformWe can extend the formula for continuous-time Fourier seriescoefficients for a periodic signal 1p −ikω0 t 1 p/2 Xk = ∫ x( t )e dt = ∫ x( t )e −ikω0 t dt p0 p −p / 2to aperiodic signals as well. The continuous-time Fourierseries is not defined for aperiodic signals, but we call theformula ∞ X(ω) = ∫ x( t )e −iωt dt the (continuous time) −∞ Fourier transform.
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Inverse TransformsIf we have the full sequence of Fourier coefficients for a periodicsignal, we can reconstruct it by multiplying the complexsinusoids of frequency ω0k by the weights Xk and summing: p −1 ∞ ikω0n ikω0 t x(n) = ∑ Xk e x( t ) = ∑ Xk e k =0 k = −∞We can perform a similar reconstruction for aperiodic signals 1 π 1 ∞x(n) = ∫ X(ω)eiωn dω x( t ) = ∫ X(ω)eiωt dω 2π − π 2π − ∞These are called the inverse transforms.
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Fourier Transform of Impulse FunctionsFind the Fourier transform of the Dirac delta function: ∞ ∞ X(ω) = ∫ x( t )e −iωt dt = ∫ δ( t )e −iωt dt = e −iω0 = 1 −∞ −∞Find the DTFT of the Kronecker delta function: ∞ ∞ −iωn −iωn −iω0X( ω) = ∑ x(n)e = ∑ δ(n)e =e =1 n = −∞ n = −∞The delta functions contain all frequencies at equal amplitudes.Roughly speaking, that’s why the system response to an impulseinput is important: it tests the system at all frequencies.
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Laplace Transform► Lapalce transform is a generalization of the Fourier transform in the sense that it allows “complex frequency” whereas Fourier analysis can only handle “real frequency”. Like Fourier transform, Lapalce transform allows us to analyze a “linear circuit” problem, no matter how complicated the circuit is, in the frequency domain in stead of in he time domain.► Mathematically, it produces the benefit of converting a set of differential equations into a corresponding set of algebraic equations, which are much easier to solve. Physically, it produces more insight of the circuit and allows us to know the bandwidth, phase, and transfer characteristics important for circuit analysis and design.► Most importantly, Laplace transform lifts the limit of Fourier analysis to allow us to find both the steady-state and “transient” responses of a linear circuit. Using Fourier transform, one can only deal with he steady state behavior (i.e. circuit response under indefinite sinusoidal excitation).► Using Laplace transform, one can find the response under any types of excitation (e.g. switching on and off at any given time(s), sinusoidal, impulse, square wave excitations, etc.
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Application of Laplace Transform to Circuit Analysis
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system►• A system is an operation that transforms input signal x into output signal y.
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LTI Digital Systems► Linear Time Invariant• Linearity/Superposition:► If a system has an input that can be expressed as a sum of signals, then the response of the system can be expressed as a sum of the individual responses to the respective systems.► LTI
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Time-Invariance &Causality► Ifyou delay the input, response is just a delayed version of original response.► X(n-k) y(n-k)► Causality could also be loosely defined by “there is no output signal as long as there is no input signal” or “output at current time does not depend on future values of the input”.
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Convolution► The input and output signals for LTI systems have special relationship in terms of convolution sum and integrals.► Y(t)=x(t)*h(t) Y[n]=x[n]*h[n]
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Sampling theory► The theory of taking discrete sample values (grid of color pixels) from functions defined over continuous domains (incident radiance defined over the film plane) and then using those samples to reconstruct new functions that are similar to the original (reconstruction).► Sampler: selects sample points on the image plane► Filter: blends multiple samples together
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Sampling theory► For band limited function, we can just increase the sampling rate► • However, few of interesting functions in computer graphics are band limited, in particular, functions with discontinuities.► • It is because the discontinuity always falls between two samples and the samples provides no information of the discontinuity.
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Z-transforms► Fordiscrete-time systems, z-transforms play the same role of Laplace transforms do in continuous-time systems Bilateral Forward z-transform Bilateral Inverse z-transform ∞ 1 ∑ h[ n] z − n 2 π j ∫R H [ z] = h[n] = H [ z ] z − n +1dz n = −∞► Aswith the Laplace transform, we compute forward and inverse z-transforms by use of transforms pairs and properties
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Region of Convergence► Region of the complex ► Four possibilities (z=0 z-plane for which is a special case and forward z-transform may or may not be converges Im{z} included) Im{z} Entire Disk plane Re{z} Re{z} Im{z} Im{z} IntersectionComplement of a disk andof a disk Re{z} complement Re{z} of a disk
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Z-transform Pairs► h[n] = δ[n] ► h[n] = an u[n] ∞ 0 ∑ δ [ n] z = ∑ δ [ n] z − n = 1 ∞ ∑ a u[ n] z −n H [ z] = H [ z] = n −n n = −∞ n =0 n = −∞ Region of convergence: ∞ a ∞ n = ∑ a z = ∑ n −n entire z-plane n =0 n =0 z 1 a = if <1► h[n] ∞= δ[n-1]1 1− a z H [ z ] = ∑ δ [ n − 1] z −n = ∑ δ [ n − 1] z −n = z −1 z n = −∞ n =1 Region of convergence: Region of entire z-plane convergence: |z| > |a| which is h[n-1] ⇔ z-1 H[z] the complement
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Stability► Rule #1: For a causal sequence, poles are inside the unit circle (applies to z-transform functions that are ratios of two polynomials)► Rule #2: More generally, unit circle is included in region of convergence. (In continuous-time, the imaginary axis would be in the region of convergence of the Z 1 a u[ n] ↔ Laplace transform.)a z for z > a n −1 1− This is stable if |a| < 1 by rule #1.
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Inverse z-transform c + j∞ 1 f [ n] = F [ z ] z n −1dz 2πj c∫ j∞ −► Yuk! Using the definition requires a contour integration in the complex z-plane.► Fortunately, we tend to be interested in only a few basic signals (pulse, step, etc.) Virtually all of the signals we’ll see can be built up from these basic signals. For these common signals, the z-transform pairs have been tabulated (see Lathi, Table 5.1)
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Example z 2 + 2z +1 ► Ratio of polynomial z-X [ z] = 3 1 z2 − z + 2 2 domain functions 1 + 2 z −1 + z −2 ► Divide through by theX [ z] = 3 1 1 − z −1 + z − 2 highest power of z 2 2 ► Factor denominator into 1 + 2 z −1 + z −2X [ z] = first-order factors 1 − z (1 − z ) 1 −1 −1 2 ► Use partial fractionX [ z ] = B0 + A1 + A2 decomposition to get 1 −1 1 − z −1 1− z 2 first-order terms
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Example (con’t) 2 1 − 2 3 −1 z − z + 1 z − 2 + 2 z −1 + 1 ► FindB0 by 2 2 z − 2 − 3 z −1 + 2 polynomial division 5 z −1 − 1 − 1 + 5 z −1 X [ z] = 2 + ► Express in terms of 1 −1 1 − z 1 − z −1 ( ) 2 B0 1 + 2 z −1 + z −2 1+ 4 + 4A1 = = = −9 1 − z −1 z −1 = 2 1− 2 1 + 2 z −1 + z − 2 1+ 2 +1 ► Solve for A1 and A2A2 = = =8 1 1 1 − z −1 2 z −1 =1 2
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Example (con’t)► Express X[z] in terms of B0, A1, and A2 9 8 X [ z] = 2 − + 1 −1 1 − z −1 1− z 2► Use table to obtain inverse z-transform n 1 x[ n] = 2 δ [ n] − 9 u[ n] + 8 u[ n] 2► With the unilateral z-transform, or the bilateral z-transform with region of convergence, the inverse z-transform is unique
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Z-transform Properties► Linearity a1 f1 [ n] + a2 f 2 [ n] ⇔ a1 F1 [ z ] + a2 F2 [ z ]► Right shift (delay) f [ n − m] u[ n − m] ⇔ z − m F [ z ] m f [ n − m] u[ n] ⇔ z F [ z ] + z ∑ f [ − n] z − n −m −m n =1
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Z-transform Properties ► Convolution definition ∞ f1 [ n] ∗ f 2 [ n] = ∑ f [ m] f [ n − m] m = −∞ 1 2 ► Take z-transform ∞ Z { f1 [ n] ∗ f 2 [ n]} = Z ∑ f1 [ m] f 2 [ n − m] m = −∞ ► Z-transform definition ∞ ∞ = ∑ ∑ f1 [ m] f 2 [ n − m] z − n ► Interchange summation n = −∞ m = −∞ ∞ ∞ = ∑ f [ m] ∑ f [ n − m] z 1 2 −n ► Substitute r = n - m m = −∞ n = −∞ ∞ ∞ f1 [ m] ∑ f 2 [ r ] z −( r + m ) ► Z-transform definition = ∑ m = −∞ r = −∞ ∞ − m ∞ = ∑ f1 [ m]z ∑ f 2 [ r ] z − r m = −∞ r = −∞ = F1 [ z ] F2 [ z ]
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Introduction► Impulse response h[n] can fully characterize a LTI system, and we can have the output of LTI system as y[ n] = x[ n] ∗ h[ n]► The z-transform of impulse response is called transfer or Y ( z ) = X ( z ) H ( z ). system function H(z).► ( ) Frequency response at H e = H ( z ) jω z =1 is valid if z = ROC includes 1, and Y ( e jω ) = X ( e jω ) H ( e jω )
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5.1 Frequency Response of LIT System jω jω j∠X ( e jω ) jω jω j∠H ( e jω )► Consider X (e ) = X (e ) e and H (e ) = H (e ) e , then magnitude Y ( e jω ) = X ( e jω ) H ( e jω ) phase jω jω jω ∠Y ( e ) = ∠X ( e ) + ∠H ( e )► We will model and analyze LTI systems based on the magnitude and phase responses.
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System Function► General formNof LCCDE M ∑ a y [ n − k ] = ∑ b x[ n − k ] k =0 k k =0 k N M ∑ ak z Y ( z ) = ∑ bk z −k X ( z ) −k k =0 k =0► Compute the z-transform M Y ( z) ∑ bk z −k H ( z) = = k =0 X ( z) N ∑ ak z − k k =0
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System Function: Pole/zero Factorization► Stability requirement can be verified.► Choice of ROC determines causality.► Location of zeros and poles determines the frequency response1 )and phase : c , c ,..., c . M (1 − c z − b0 ∏ k zeros 1 2 M H ( z) = k =1 ∏ (1 − d z ) N a0 −1 poles : d1 , d 2 ,..., d N . k k =1
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Second-order System► Suppose the system function of a LTI system is (1 + z −1 )2 H ( z) = . 1 −1 3 −1 (1 − z )(1 + z ) 2 4► Tofind the difference equation that is satisfied by the input and out of this system (1 + z −1 )2 1 + 2 z −1 + z −2 Y ( z) H ( z) = = = 1 3 1 3 (1 − z −1 )(1 + z −1 ) 1 + z −1 − z −2 X ( z ) 2 4 4 8 1 3 y[n ] + y[n − 1] − y[n − 2] = x[n ] + 2 x[n − 1] + 2 x[n − 2] 4 8► Can we know the impulse response?
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System Function: Stability► Stability of LTI system: ∞ ∑ h[n] < ∞ n = −∞► This condition is identical to the condition ∞ that ∑ h[n ]z −n < ∞ when z = 1. n = −∞ The stability condition is equivalent to the condition that the ROC of H(z) includes the unit circle.
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System Function: Causality► If the system is causal, it follows that h[n] must be a right- sided sequence. The ROC of H(z) must be outside the outermost pole.► If the system is anti-causal, it follows that h[n] must be a left-sided sequence. The ROC of H(z) must be inside the innermost pole. Im Im Im a 1 a 1 Re a b Re Re Right-sided Left-sided Two-sided (causal) (anti-causal) (non-causal)
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Determining the ROC► Consider the LTI system 5 y[n ] − y[n − 1] + y[n − 2] = x[n ] 2► The system1 function is obtained as H ( z) = 5 −1 −2 1− z +z 2 1 = 1 −1 (1 − z )(1 − 2 z −1 ) 2
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System Function: Inverse Systems H i ( z) H ( z)► is an inverse system for , if G ( z ) = H ( z ) H i ( z ) = 1 ⇔ g [ n ] = h[ n ] ∗ hi [ n ] = δ [ n ] 1 1 Hi ( z) = jω ⇔ H i (e ) = H ( z) H ( e jω )► The ROCs of H ( z ) and H i ( z ) overlap. must► Useful for canceling the effects of another system► See the discussion in Sec.5.2.2 regarding ROC
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All-pass System►A system of the form ∗(or cascade of these) −1 z −a H Ap ( Z ) = pole : a = re jθ 1 − az −1 zero : 1 / a* = r −1e jθ − jω ∗ jω e −a − jω 1 − a * e H Ap ( e ) = jω − jω =e − jω 1 − ae 1 − ae H Ap ( e jω ) = 1
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All-pass System: General Form► In general, all pass systems have form Mr z −1 − d k M c ( z −1 − ek )( z −1 − ek ) * H Ap ( z ) = ∏ −1 ∏ −1 * −1 k =1 1 − d k z k =1 (1 − ek z )(1 − ek z ) real poles complex poles Causal/stable: ek , d k < 1
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All-Pass System Example Im Unit circle z-planeM r = 2 and M c = 1 0.8 Re 4 3 0.5 − − 2 3 4pole : re jθ reciprocal & conjugate → zero : r −1e jθ This all - pass system has M = N = 2 M c + M r = 4 poles and zeros.
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Minimum-Phase System► Minimum-phase system: all zeros and all poles are inside the unit circle.► The name minimum-phase comes from a property of the phase response (minimum phase-lag/group-delay).► Minimum-phase systems have some special properties.► When we design a filter, we may have multiple choices to satisfy the certain requirements. Usually, we prefer the minimum phase which is unique.► All systems can be represented as a minimum-phase system and an all-pass system.
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Example y[n] = a1y[n − 1] + a2 y[n − 2] + b0x[n]► Block diagram representation of
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Block Diagram Representation► LTI systems with rational system function can be represented as constant-coefficient difference equation► The implementation of difference equations requires delayed values of the
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Direct Form I N M ∑ ˆ y[n − k ] = ∑ b x[n − k ] a k =0 k ˆ k =0 k► General form of difference equation N M y[n] − ∑ a y[n − k ] = ∑ b x[n − k ] k k k =1 k =0► Alternative equivalent form
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Direct Form I M ∑ bk z −k H( z ) = k =0 N 1 − ∑ ak z −k► Transfer function can be written as k =1 MH( z ) = H2 ( z )H1 ( z ) = 1 b z −k ∑ N k = 0 k −k 1 − ∑ ak z k =1 M −k V ( z ) = H1 ( z ) X( z ) = ∑ bk z X( z ) M k =0 v[n] = ∑ b x[n − k ] k► Direct Form I Represents k =0 N 1 y[n] = ∑ a y[n − k ] + v[n]Y ( z ) = H2 ( z ) V ( z ) = V ( z ) k k =1 N 1 − ∑ ak z −k k =1
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Alternative Representation► Replace order of cascade LTI systems M 1H( z ) = H ( z )H ( z ) = ∑ b z −k 1 2 k 1− N k =0 ∑a z k =1 k −k N w[n] = ∑ a w[n − k ] + x[n] k k =1 1W( z ) = H2 ( z ) X( z ) = X ( z ) M N y[n] = ∑ b w[n − k ] k 1 − ∑ ak z −k k =0 k =1 M Y ( z ) = H1 ( z ) W( z ) = ∑ bk z −k W( z ) k =0
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Alternative Block Diagram► We can change the order of the cascade systems N w[n] = ∑ a w[n − k ] + x[n] k k =1 M y[n] = ∑ b w[n − k ] k k =0
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Direct Form II► No need to store the same data twice in previous system► So we can collapse the delay elements into one chain► This is called Direct Form II or the Canonical Form► Theoretically no difference between Direct Form I and II► Implementation wise Less memory in Direct II Difference when using finite-precision arithmetic
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Signal Flow Graph Representation► Similar to block diagram representation Notational differences►A network of directed branches connected at nodes
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Example► Representation of Direct Form II with signal flow graphs w1 [n] = aw4 [n] + x[n] w2 [n] = w1 [n] w3 [n] = b0w2 [n] + b1w4 [n] w4 [n] = w2 [n − 1] y[n] = w3 [n] w1 [n] = aw1 [n − 1] + x[n] y[n] = b0w1 [n] + b1w1 [n − 1]
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Determination of System Function from Flow Graph w1 [n] = w4 [n] − x[n] w2 [n] = αw1 [n] w3 [n] = w2 [n] + x[n] w4 [n] = w3 [n − 1] y[n] = w2 [n] + w4 [n]W1 ( z ) = W4 ( z ) − X( z ) ( αX( z ) z −1 − 1 ) W2 ( z ) =W2 ( z ) = αW1 ( z ) 1 − αz −1 Y ( z) z −1 − α H( z ) = =W3 ( z ) = W2 ( z ) + X( z ) X( z ) z −1 (1 − α ) X( z ) 1 − αz −1 W4 ( z ) =W4 ( z ) = W3 ( z ) z −1 1 − αz −1 h[n] = αn −1u[n − 1] − αn +1u[n] Y ( z ) = W2 ( z ) + W4 ( z ) Y ( z ) = W2 ( z ) + W4 ( z )
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Basic Structures for IIR Systems: Direct Form I
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Basic Structures for IIR Systems: Direct Form II
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Basic Structures for IIR Systems: Cascade Form► General form for cascade implementation M1 M2 ∏ (1 − f z )∏ (1 − g z )(1 − g z ) k −1 k −1 ∗ k −1 H( z ) = A k =1 N1 k =1 N2 ∏ (1 − c z )∏ (1 − d z )(1 − d z ) k =1 k −1 k =1 k −1 ∗ k −1 M1 b0k + b1k z −1 − b2k z −2 H( z ) = ∏ k =1 1 − a1k z −1 − a2k z −2► More practical form in 2nd order systems
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Example H( z ) = 1 + 2z + z −1 (1 + z −2 = −1 )(1 + z ) −1 1 − 0.75z −1 + 0.125z −2 (1 − 0.5z −1 )(1 − 0.25z ) −1 = (1 + z ) −1 (1 + z ) −1 (1 − 0.5z ) (1 − 0.25z ) −1 −1► Cascade of Direct Form I subsections► Cascade of Direct Form II subsections
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Basic Structures for IIR Systems: Parallel Form► Represent system function using partial fraction expansion H( z ) = NP NP Ak NP ( Bk 1 − ek z −1 ) ∑ Ck z − k k =0 +∑ k =1 1 − ck z −1 +∑ ( k =1 1 − dk z −1 )( 1 − dk z −1 ∗ ) NS NP e0k + e1k z −1H( z ) = ∑C z k −k +∑ k =0 k =1 1 − a1k z −1 − a2k z −2► Or by pairingthe real poles
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Transposed Forms► Linear signal flow graph property: Transposing doesn’t change the input-output relation 1 H( z ) = −1► Transposing:1 − az Reverse directions of all branches Interchange input and output nodes► Example:
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Example Transpose y[n] = a1y[n − 1] + a2y[n − 2] + b0x[n] + b1x[n − 1] + b2x[n − 2]► Both have the same system function or difference equation
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Basic Structures for FIR Systems: Direct Form► Special cases of IIR direct form structures► Transpose of direct form I gives direct form II► Both forms are equal for FIR systems► Tapped delay line
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Basic Structures for FIR Systems: Cascade Form► Obtainedby factoring the polynomial system function MS ∏ (b ) M H( z ) = ∑ h[n]z n=0 −n = k =1 0k + b1k z −1 + b2k z −2
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Structures for Linear-Phase FIR Systems ► Causal FIR system with generalized linear phase are symmetric: h[M − n] = h[n] n = 0,1,..., M (type I or III) h[M − n] = −h[n] n = 0,1,..., M (type II or IV) ► Symmetry means we can half the number of multiplications ► Example: For even M and type I or type III systems: M M / 2 −1 My[n] = ∑ h[k ]x[n − k ] = ∑ h[k ]x[n − k ] + h[M / 2]x[n − M / 2] + ∑ h[k ]x[n − k ] k =0 k =0 k = M / 2 +1 M / 2 −1 M / 2 −1 = ∑ h[k ]x[n − k ] + h[M / 2]x[n − M / 2] + ∑ h[M − k ]x[n − M + k ] k =0 k =0 M / 2 −1 = ∑ h[k ]( x[n − k ] + x[n − M + k ] ) + h[M / 2]x[n − M / 2] k =0
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Structures for Linear-Phase FIR Systems► Structure for even M► Structure for odd M
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