Generalizing the DTFT!

The Transform

Generaliing the DTFT! The forward DTFT is defined by X e jω ( ) = x n e jωn in which n= Ω is discrete-time radian frequency, a real variable. The quantity e jωn is then a complex sinusoid whose magnitude is always one and whose phase can range over all angles. It always lies on the unit circle in the complex plane. If we now replace e jω with a variable that can have any complex value we define the transform X n. ( ) = x n n= The DTFT expresses signals as linear combinations of complex sinusoids. The transform expresses signals as linear combinations of complex exponentials. 12/29/10! M. J. Roberts - All Rights Reserved! 2!

Complex Exponential Excitation! Let the excitation of a discrete-time LTI system be a complex exponential of the form A n where is, in general, complex and of an LTI to a complex exponential A is any constant. Using convolution, the response y n system with impulse response h n excitation x n y[ n] = h n is m= A n = A h[ m] n m = A n =x[ n] m= h [ m ] m The response is the product of the excitation and the transform of h[ n] defined by H( ) = h n n. m= 12/29/10! M. J. Roberts - All Rights Reserved! 3!

The Transfer Function! If an LTI system with impulse response h n x[ n], the transform Y ( ) = y n Y is excited by a signal, ( ) of the response y[ n] is n = ( h[ n] x[ n] ) n = h[ m]x n m n= Y Let q = n m. Then ( ) = h[ m] Y m= n= ( ) = h[ m] m= n= n= m= 12/29/10! M. J. Roberts - All Rights Reserved! 4! x[ n m] n x[ q] ( q+m) = h[ m] m x[ q] q m= q= q= H( ) is the transfer function. Y( ) = H( ) X( ) ( ) =H ( ) =X n

Systems Described by Difference The most common description of a discrete-time system is a difference equation of the general form N a k y[ n k] = b k x[ n k]. k=0 M k=0 It was shown in Chapter 5 that the transfer function for a system of this type is or H( ) = H( ) = M M k=0 N k=0 k=0 N k=0 b k k a k k b k k a k k Equations! = b 0 + b 1 1 + b 2 2 + + b M M a 0 + a 1 1 + a 2 2 + + a N N = N M b 0 M + b 1 M 1 + + b M 1 + b M a 0 N + a 1 N 1 + + a N 1 + a N 12/29/10! M. J. Roberts - All Rights Reserved! 5!

Direct Form II Realiation! Direct Form II realiation of a discrete-time system is similar! in form to Direct Form II realiation of continuous-time systems! A continuous-time system can be realied with integrators,! summing junctions and multipliers! A discrete-time system can be realied with delays, summing! junctions and multipliers! 12/29/10! M. J. Roberts - All Rights Reserved! 6!

Direct Form II Realiation! 12/29/10! M. J. Roberts - All Rights Reserved! 7!

The Inverse Transform! The inversion integral is x[ n] = 1 j2π C X( ) n 1 d. This is a contour integral in the complex plane and is beyond the and X( ) form a "-transform pair". scope of this course. The notation x n x n ( ) indicates that X 12/29/10! M. J. Roberts - All Rights Reserved! 8!

Existence of the Transform! Time Limited Signals! If a discrete-time signal x[ n] is time limited and bounded, the transformation summation n= x[ n] n is finite and the transform of exists for any non-ero x n value of. 12/29/10! M. J. Roberts - All Rights Reserved! 9!

Existence of the Transform! Right- and Left-Sided Signals! A right-sided signal x r [ n] is one for which x r n n < n 0 and a left-sided signal x l for any n > n 0. = 0 for any [ n] is one for which x l [ n] = 0 12/29/10! M. J. Roberts - All Rights Reserved! 10!

Existence of the Transform! Right- and Left-Sided Exponentials! x[ n] = α n u[ n n ] 0, α x[ n] = β n u[ n 0 n], β 12/29/10! M. J. Roberts - All Rights Reserved! 11!

Existence of the Transform! The transform of x[ n] = α n u[ n n ] 0, α is ( ) = α n u n n 0 X n = α 1 n= if the series converges and it converges if > α. The path of integration of the inverse transform must lie in the region of the plane outside a circle of radius α. n=n 0 ( ) n 12/29/10! M. J. Roberts - All Rights Reserved! 12!

Existence of the Transform! The transform of x[ n] = β n u[ n 0 n], β is ( ) = β n n X n 0 = β 1 = β 1 n= n 0 n= ( ) n n= n 0 ( ) n if the series converges and it converges if < β. The path of integration of the inverse transform must lie in the region of the plane inside a circle of radius β. 12/29/10! M. J. Roberts - All Rights Reserved! 13!

Existence of the Transform! 12/29/10! M. J. Roberts - All Rights Reserved! 14!

Some Common Transform Pairs! u n α n u n δ n 1, All 1 = 1, > 1, u n 1 1 1 α = 1, > α, α n u n 1 1 α 1 nu[ n] ( 1) = 1 2 1 1 nα n u[ n] α ( α ) = α 1 2 1 α 1 sin( Ω 0 n)u n cos( Ω 0 n)u n α n sin( Ω 0 n)u n sin Ω 0 2 2 cos Ω 0 α n cos( Ω 0 n)u n cos Ω 0 2 2 cos Ω 0 α sin Ω 0 2 2α cos Ω 0 ( ) 2, > 1, nu n 1 ( ) 2, > α, nα n u n 1 ( ) ( ) + 1, > 1, sin Ω n 0 ( ) ( ) + 1, > 1, cos Ω n 0 ( ), > α, α ( ) n sin Ω + α 2 0 n ( ), > α, α ( ) n cos Ω + α 2 0 n α cos Ω 0 2 2α cos Ω 0 α n α α 1 u[ n n 0 ] u[ n n 1 ], α < < α 1 1, < 1 α = 1, < α 1 α 1 ( 1) = 1 2 1 1 ( ) 2, < 1 α ( α ) = α 1 2 1 α 1 ( )u [ n 1 ] sin Ω 0 2 2 cos Ω 0 ( )u [ n 1 ] = cos Ω 0 2 2 cos Ω 0 α sin Ω ( )u[ n 1] 0 2 2α cos Ω 0 ( )u[ n 1] α cos Ω 0 2 2α cos Ω 0 1 n 0 n 1 ( ) = n 1 n 0 1 + n 1 n 0 2 + + + 1 n 1 1, > 0 ( ) 2, < α ( ) ( ) + 1, < 1 ( ) ( ) + 1, < 1 ( ), < α ( ) + α 2 ( ), < α ( ) + α 2 12/29/10! M. J. Roberts - All Rights Reserved! 15!

-Transform Properties! Given the -transform pairs g n G( ) and h n with ROC's of ROC G and ROC H respectively the following properties apply to the transform. H( ) Linearity α g n + β h[ n] α G ( ) + β H( ) ROC = ROC G ROC H n 0 G( ) Time Shifting g n n 0 ROC = ROC G except perhaps = 0 or Change of Scale in α n g n G( / α ) ROC = α ROC G 12/29/10! M. J. Roberts - All Rights Reserved! 16!

Time Reversal -Transform Properties! g n ( ) G 1 Time Expansion ROC = 1 / ROC G, n / k and integer g n / k 0, otherwise ( ) 1/k ROC = ROC G G( k ) Conjugation g * n ( ) G * * ROC = ROC G -Domain Differentiation n g n d d G( ) ROC = ROC G 12/29/10! M. J. Roberts - All Rights Reserved! 17!

-Transform Properties! Convolution g[ n] h n H( )G( ) First Backward Difference g[ n] g[ n 1] ROC ROC G > 0 ( 1 1 )G ( ) Accumulation Initial Value Theorem Final Value Theorem n m= g m 1 G ( ) ROC ROC G > 1 If g[ n] = 0, n < 0 then g[ 0] = lim If g[ n] = 0, n < 0, lim g n n = lim 1 if lim n g[ n] exists. G( ) ( 1)G ( ) 12/29/10! M. J. Roberts - All Rights Reserved! 18!

-Transform Properties! ( ) all the ( ) must lie in the open For the final-value theorem to apply to a function G finite poles of the function ( 1)G interior of the unit circle of the plane. Notice this does not say that all the poles of G the unit circle. G ( ) must lie in the open interior of ( ) could have a single pole at = 1 and the final-value theorem could still apply. 12/29/10! M. J. Roberts - All Rights Reserved! 19!

The Inverse Transform! Synthetic Division! For rational transforms of the form H( ) = b M M + b M 1 M 1 + + b 1 + b 0 a N N + a N 1 N 1 + + a 1 + a 0 we can always find the inverse transform by synthetic division. For example, ( ) = ( 1.2) ( + 0.7) ( + 0.4) ( 0.2) ( 0.8) ( + 0.5) H, > 0.8 H ( ) = 3 0.1 2 1.04 0.336 3 0.5 2 0.34 + 0.08, > 0.8 12/29/10! M. J. Roberts - All Rights Reserved! 20!

Synthetic Division! 1+ 0.4 1 + 0.5 2 3 0.5 2 0.34 + 0.08 3 0.1 2 1.04 0.336 3 0.5 2 0.34 + 0.08 0.4 2 0.7 0.256 0.4 2 0.2 0.136 0.032 1 0.5 0.12 + 0.032 1 The inverse transform is + 0.4δ [ n 1] + 0.5δ n 2 δ n The Inverse Transform! 1+ 0.4 1 + 0.5 2 12/29/10! M. J. Roberts - All Rights Reserved! 21!

The Inverse Transform! Synthetic Division! We could have done the synthetic division this way. 4.2 30.85 158.613 2 0.08 0.34 0.5 2 + 3 0.336 1.04 0.1 2 + 3 0.336 +1.428+2.1 2 4.2 3 4.2δ [ n] 30.85δ [ n +1] 158.613δ n + 2 2.468 2.2 2 + 5.2 3 2.468 +10.489 2 +15.425 3 30.85 4 12.689 2 10.225 3 + 30.85 4 4.2 30.85 158.613 2 but with the restriction > 0.8 this second form does not converge and is therefore not the inverse transform. 12/29/10! M. J. Roberts - All Rights Reserved! 22!

The Inverse Transform! Synthetic Division! We can always find the inverse transform of a rational function with synthetic division but the result is not in closed form. In most practical cases a closed-form solution is preferred. 12/29/10! M. J. Roberts - All Rights Reserved! 23!

Partial Fraction Expansion! Partial-fraction expansion works for inverse transforms the same way it does for inverse Laplace transforms. But there is a situation that is quite common in inverse transforms which deserves mention. It is very common to have -domain functions in which the number of finite eros equals the number of finite poles (making the expression improper in ) with at least one ero at = 0. H ( ) 2 ( p 1 ) p 2 ( ) = N M 1 ( ) M ( ) ( p N ) ( ) 12/29/10! M. J. Roberts - All Rights Reserved! 24!

Dividing both sides by we get H( ) = N M 1 ( 1 )( 2 ) ( M ) ( p 1 )( p 2 ) ( p N ) and the fraction on the right is now proper in and can be expanded in partial fractions. ( ) H = K 1 p 1 + K 2 p 2 + + p N Then both sides can be multiplied by and the inverse transform can be found. Partial Fraction Expansion! 12/29/10! M. J. Roberts - All Rights Reserved! 25! K N H( ) = K 1 + K 2 + + K N p 1 p 2 p N h[ n] = K 1 p n 1 u[ n] + K 2 p n 2 u[ n] + + K N p n N u[ n]

-Transform Properties! An LTI system has a transfer function ( ) = Y( ) X( ) = 1 / 2 H 2 + 2 / 9, > 2 / 3 Using the time-shifting property of the transform draw a block diagram realiation of the system. Y( ) ( 2 + 2 / 9) = X( ) ( 1 / 2) 2 Y( ) = X( ) ( 1 / 2) X( ) + Y( ) 2 / 9 ( ) = 1 X( ) ( 1 / 2) 2 X( ) + 1 Y( ) 2 / 9 Y ( )Y( ) ( ) 2 Y ( ) 12/29/10! M. J. Roberts - All Rights Reserved! 26!

-Transform Properties! Y( ) = 1 X( ) ( 1 / 2) 2 X( ) + 1 Y( ) ( 2 / 9) 2 Y( ) Using the time-shifting property y n ( )x n 2 = x[ n 1] 1 / 2 + y[ n 1] 2 / 9 ( )y n 2 12/29/10! M. J. Roberts - All Rights Reserved! 27!

Let g n -Transform Properties! G ( ) = pole-ero diagram for G The poles of G 1 ( jπ /4 0.8e ) 0.8e ( + jπ /4 ). Draw a. ( ) and for the transform of e jπn/8 g n ( ) are at = 0.8e ± jπ /4 and its single finite ero is at = 1. Using the change of scale property e jπn/8 g n G( jπ /8 e ) = jπ /8 G e G e ( jπ /8 ) = e jπ /8 1 ( e jπ /8 jπ /4 0.8e ) e jπ /8 0.8e ( ) ( + j 3π /8 ) e jπ /8 jπ /8 e e jπ /8 ( jπ /8 0.8e )e jπ /8 0.8e ( ) e jπ /8 = e jπ /8 ( jπ /8 0.8e )( + j 3π /8 0.8e ) ( + jπ /4 ) 12/29/10! M. J. Roberts - All Rights Reserved! 28!

-Transform Properties! G( jπ /8 e ) has poles at = 0.8e jπ /8 and 0.8e + j 3π /8 and a ero at = e jπ /8. All the finite ero and pole locations have been rotated in the plane by π /8 radians. 12/29/10! M. J. Roberts - All Rights Reserved! 29!

-Transform Properties! Using the accumulation property and u n show that the transform of n u[ n] is nu n u n 1 n m=0 nu n = u[ m 1] n m=0 = u[ m 1] ( 1), > 1. 2 1 1 = 1 1, > 1 1 1 1 = 1, > 1 ( 1), > 1 2 12/29/10! M. J. Roberts - All Rights Reserved! 30!

Inverse Transform Example! Find the inverse transform of X( ) = 0.5 + 2, 0.5 < < 2 Right-sided signals have ROC s that are outside a circle and left-sided signals have ROC s that are inside a circle. Using We get α n u n α n u n 1 ( 0.5) n u[ n] + ( 2) n u[ n 1] α = 1, > α 1 α 1 α = 1, < α 1 α 1 X( ) = 0.5 + 2, 0.5 < < 2 12/29/10! M. J. Roberts - All Rights Reserved! 31!

Inverse Transform Example! Find the inverse transform of X( ) = 0.5 + 2, > 2 In this case, both signals are right sided. Then using We get α n u n ( 0.5) n 2 ( ) n u n α = 1, > α 1 α 1 X( ) = 0.5 + 2, > 2 12/29/10! M. J. Roberts - All Rights Reserved! 32!

Inverse Transform Example! Find the inverse transform of X( ) = 0.5 + 2, < 0.5 In this case, both signals are left sided. Then using We get 0.5 α n u n 1 ( ) n ( 2) n u n 1 α = 1, < α 1 α 1 X( ) = 0.5 + 2, < 0.5 12/29/10! M. J. Roberts - All Rights Reserved! 33!

The Unilateral Transform! Just as it was convenient to define a unilateral Laplace transform it is convenient for analogous reasons to define a unilateral transform ( ) = x n X n=0 n 12/29/10! M. J. Roberts - All Rights Reserved! 34!

Properties of the Unilateral Transform" If two causal discrete-time signals form these transform pairs, g n ( ) and h[ n] G hold for the unilateral transform. Time Shifting Delay: g n n 0 Advance: g n + n 0 Accumulation: n m=0 g m H ( ) then the following properties n 0 G( ), n 0 0 n 0 G 1 G ( ) n 0 1 m=0 ( ) g m m, n > 0 0 12/29/10! M. J. Roberts - All Rights Reserved! 35!

Solving Difference Equations! The unilateral transform is well suited to solving difference equations with initial conditions. For example, y[ n + 2] 3 2 y [ n +1 ] + 1 2 y [ n ] = ( 1 / 4) n, for n 0 = 10 and y[ 1] = 4 y 0 transforming both sides, 2 Y( ) y[ 0] 1 y[ 1] 3 2 Y ( ) y [ 0 ] the initial conditions are called for systematically. + 1 2 Y( ) = 1/ 4 12/29/10! M. J. Roberts - All Rights Reserved! 36!

Solving Difference Equations! Applying initial conditions and solving, and Y( ) = 16 / 3 1/ 4 + 4 1/ 2 + 2 / 3 1 y[ n] = 16 n 1 3 4 + 4 1 n 2 + 2 u[ n] 3 This solution satisfies the difference equation and the initial conditions. 12/29/10! M. J. Roberts - All Rights Reserved! 37!

Pole-ero Diagrams and Frequency Response! For a stable system, the response to a sinusoid applied at! time t = 0 approaches the response to a true sinusoid (applied! for all time).! 12/29/10! M. J. Roberts - All Rights Reserved! 38!

Homework 4. Due date: 09/23/2016

Let the transfer function of a system be H( ) = p 1 = 2 / 2 + 5 /16 = 1+ j2 4 H( e jω ) = Pole-ero Diagrams and, p 2 = Frequency Response! 1 j2 4 e jω e jω p 1 e jω p 2 ( p 1 )( p 2 ) 12/29/10! M. J. Roberts - All Rights Reserved! 39!

Pole-ero Diagrams and Frequency Response! 12/29/10! M. J. Roberts - All Rights Reserved! 40!

Transform Method Comparison! A system with transfer function H( ) = ( 0.3) + 0.8 is excited by a unit sequence. Find the total response. Using -transform methods, Y( ) = y n Y( ) = H( ) X( ) = ( 0.3) + 0.8 ( 0.3) + 0.8 ( ) 2 ( )( 1) = 0.1169 ( ), > 0.8 1, > 1 0.3 + 0.3232 + 0.8 + 0.7937 1 = 0.1169( 0.3) n 1 + 0.3232( 0.8) n 1 + 0.7937 u n 1, > 1 12/29/10! M. J. Roberts - All Rights Reserved! 41!

Using the DTFT, H( e jω ) = Y( e jω ) = H( e jω ) X( e jω ) = Y( e jω ) = Transform Method Comparison! e j 2Ω ( e jω 0.3) e jω + 0.8 e jω ( e jω 0.3) e jω + 0.8 e jω ( ) ( ) 1 ( e jω 0.3) e jω + 0.8 ( ) e jω 1 Y( e jω ) = 0.1169 e jω 0.3 + 0.3232 e jω + 0.8 + 0.7937 e jω 1 + 1 e + πδ ( Ω) jω 2π DTFT of a Unit Sequence jω ( ) + π e ( e jω 0.3) ( e jω + 0.8) δ Ω 2π π 1 0.3 ( )( 1+ 0.8) δ ( Ω) 2π ( ) 12/29/10! M. J. Roberts - All Rights Reserved! 42!

Transform Method Comparison! Using the equivalence property of the impulse and the periodicity of both δ 2π ( Ω) and e jω Y( e jω ) = 0.1169e jω 1 0.3e + 0.3232e jω jω 1+ 0.8e + 0.7937e jω + 2.4933δ jω 1 e jω 2π ( Ω) Then, manipulating this expression into a form for which the inverse DTFT is direct Y( e jω ) = 0.1169e jω 1 0.3e jω + 0.3232e jω 1+ 0.8e + 0.7937 e jω jω 1 e + πδ jω 2π ( ) + 2.4933δ 2π ( Ω) 0.7937πδ 2π Ω =0 ( Ω) 12/29/10! M. J. Roberts - All Rights Reserved! 43!

Transform Method Comparison! Y( e jω ) = 0.1169e jω 1 0.3e jω Finding the inverse DTFT, + 0.3232e jω 1+ 0.8e + 0.7937 e jω jω 1 e + πδ jω 2π = 0.1169( 0.3) n 1 + 0.3232( 0.8) n 1 + 0.7937 ( Ω) y n u n 1 The result is the same as the result using the transform, but the effort and the probability of error are considerably greater. 12/29/10! M. J. Roberts - All Rights Reserved! 44!

System Response to a Sinusoid! A system with transfer function H( ) = is excited by the sinusoid x n 0.9, > 0.9 = cos 2πn /12 ( ). Find the response. The transform of a true sinusoid does not appear in the table of transforms. The transform of a causal sinusoid of the form x n ( )u n = cos 2πn /12 does appear. We can use the DTFT to find the response to the true sinusoid and the result is y n ( ). = 1.995cos 2πn /12 1.115 12/29/10! M. J. Roberts - All Rights Reserved! 45!

System Response to a Sinusoid! Using the transform we can find the response of the system to a causal sinusoid x[ n] = cos( 2πn /12)u[ n] and the response is y[ n] = 0.1217( 0.9) n u[ n]+1.995cos( 2πn /12 1.115)u[ n] Notice that the response consists of two parts, a transient response 0.1217( 0.9) n u n ( )u n and a forced response 1.995 cos 2πn /12 1.115 that, except for the unit sequence factor, is exactly the same as the forced response we found using the DTFT. 12/29/10! M. J. Roberts - All Rights Reserved! 46!

System Response to a Sinusoid! This type of analysis is very common. We can generalie it to say that if a system has a transfer function H( ) = N( ) D( ) causal cosine excitation cos( Ω 0 n)u[ n] is y[ n] = 1 N 1 ( ) D( ) + H( p 1 ) cos Ω 0 n + H( p 1 ) Natural or Transient Response that the response to a ( )u[ n] Forced Response where p 1 = e jω 0. This consists of a natural or transient response and a forced response. If the system is stable the transient response dies away with time leaving only the forced response which, except for the u n factor is the same as the forced response to a true cosine. So we can use the transform to find the response to a true sinusoid. 12/29/10! M. J. Roberts - All Rights Reserved! 47!