Z-TRANSFORM
In today s class Z-transform Unilateral Z-transform Bilateral Z-transform Region of Convergence Inverse Z-transform Power Series method Partial Fraction method Solution of difference equations
Need for transformation? Why do we need to transform our signal from one domain to another? Information available in one domain is not sufficient for complete analysis Looking at a sine wave in time-domain, we cannot really know the frequency content So we have to look into the frequency domain An alternate domain may express the information more comprehensively A pole-zero map easily tells whether a systems is stable or not
Z-transform Digital counterpart for the Laplace transform used for analog signals Mathematically defined as, X (z) x[n] z n n This equation is in general a power series, where z is a complex variable.
Derivation The continuous-time Fourier transform of x(t) is given as, Fxt xt e j 2ft dt And the discrete-time Fourier transform of x[nt] is given as, F xnt D n xnt e j 2fnT The Z-transform of x[n] is given as the Fourier transform of x[n] multiplied by r n nt D n x F r x nt D
Bi-lateral Z-transform D x nt F D r D x nt x nt D x nt D z re D j 2 ft x nt n x nt r n x nt e n n n n x nt re x nt z n j 2 fnt x nt r n j 2 fnt e j 2 ft n
Uni-lateral Z-transform xnt D n xnt z xnt xnt z n n0 X z xnt z n n0 where z -1 would show a delay by one sample time n
Example 1: Find the z-transform of the following finite-length sequence 4 y nt 3. 5 3 2. 5 2 1. 5 1 0.5 0-2 - 1 0 1 2 3 4 5 6 7 ynt 0 0 0 2 0 4 3 2 0 0 ynt 2 n 1T 4 n 3T 3 n 4T 2 n 5T
ynt 0 2 0 4 3 2 0 0 Y z ynt z n n0 Y z 0 n0 2 0 4 3 2 0 0z n Y z 0z 0 2z 1 0z 2 4z 3 3z 4 2z 5 0z 6 0z 7 Y z 2z 1 4z 3 3z 4 2z 5 So, Y(z) would exist on the entire z-plane except the point z=0
Z-transform as Rational Function Often it is convenient to represent Z-transform X(z) as a rational function X (z) P(z) Q(z) Where P(z) and Q(z) are two polynomials The values of z at which X(z) becomes zero (X(z) = 0) are called zeros The values of z at which X(z) becomes infinite (X(z) = called poles ), are
Significance of Poles & Zeros Poles Roots of the denominator Q(z) The point where H(z) becomes infinite The point where H(e jw ) shows a peak value System may become unstable Zeros Roots of the numerator P(z) The point where H(z) becomes zero The point where H(e jw ) shows maximum attenuation
Convergence issues A power series may not necessarily converge The infinite sum may not always be finite The set of values of z for which the z-transform converges is called Region of Convergence (RoC) The convergence of X(z) depends only on z and it converges for n n x [ n ] z
X ( z ) x [ n ] z n n Replacing z re j 2 fn re jw n x [ n ] re jw n n x [ n ] r n e jwn This equation can be segmented into two parts, one for the right-sided (causal) signal and second for the left-sided (non-causal) signal 1 X (z) x[n]r n n x[n]r n n0
X (z) x[n]r n n1 x[n] n n0 r For X(z) to exist in a particular region (for certain values of z), both summations must be finite in that region For the first summation, r should be small enough x[-n]r n converges when summed to infinite terms so that For the second summation, r should be large enough so that x[n]/r n converges when summed to infinite terms
So, there are two circles with radius r L & r R for the sequence x[n] If it is defined as a left-sided sequence (non-causal), then the second summation becomes zero (by definition), and the radius r L should be small enough to make the first summation converge rr RR r L If x[n] is defined as a right-sided sequence (causal), then the first summation becomes zero (by definition), and the radius r R should be large enough to make the second summation converge
Imaginary Part Example 2: Find the z-transform of the following finite-length sequence x[n] 2 [n 2] [n 1] 2 [n] [n 1] 2 [n 2] The z-transform of this sequence is given as, X ( z) 2 z 2 z 2 z 1 2 z 2 it is clear to see that the sequence does not have any poles (denominator is 1), it has 4 zeros 1 0.8 0.6 0.4 0.2 0-0.2 4 It can be observed that X(z) becomes undetermined at z = 0 and z =, so the RoC is entire z-plane except at z = 0 and z = -0.4-0.6-0.8-1 -1-0.5 0 0.5 1 Real Part
Example 3: Find the z-transform of the following right-sided sequence x[n] a n u[n] X (z) a n u(n)z n n a n z n n0 (az 1 ) n n0 1 z X (z) (az 1 ) n 1 n0 1 az z a For convergence we require X (z) Now, X(z) will not exist for z=a & RoC is entire z-plane except z=a However, since the z-plane is a circle so we have to use the following condition (the sequence is right-sided) z a
Example 4: Find the z-transform of the following right-sided sequence X (z) a n u(n 1)z n n 1 a n z n n a n z n n1 x[n] a n u[n 1] 1 a n z n n0 X (z) 1 1 z (a 1 z) n 1 1 n0 1 a z z a For convergence we require X (z) Now, X(z) will not exist for z=a & RoC is entire z-plane except z=a However, since the z-plane is a circle so we have to use the following condition (the sequence is left-sided) z a
Concepts From the two examples we observe that the closed form equations for the z-transform of causal & noncausal signals come out to be same This creates an ambiguity about the existence of their z-transform Therefore, we require complimentary information apart from the closed form equations, i.e. the RoC
Properties of RoC Property 1: The RoC is a ring or disk in the z-plane centred at the origin; i.e., 0 r R z r L Property 2: The RoC cannot contain any poles Property 3: If x[n] is a finite-duration sequence i.e. a sequence that is zero except in a finite interval N 1 n N 2, then the RoC is the entire z-plane except possibly z=0 and z=
Property 4: If x[n] is a right-sided sequence i.e. a sequence that is zero for, n N 1, the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z= Property 5: If x[n] is a left-sided sequence i.e. a sequence that is zero for, n N 2, the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z=0
Z-transform pairs Sequence z-transform RoC ( n ) 1 All z ( n m ) z m All z except 0 (if m>0) or (if m<0) a n u ( n ) 1 1 az 1 z a a n u(n 1) 1 1 az 1 z a na n u(n) (1 az 1 ) 2 az 1 z a
Sequence z-transform RoC 1[cos 0 ]z [cos 0 n]u(n) 1 z 1 1[2 cos ]z 1 z 2 0 [sin 0 n]u(n) [sin 0 ]z 1 z 1 1 [2 cos ]z 1 z 2 0 n [r cos 0 n]u(n) 1[r cos ]z 1 1[2r cos ]z 1 r 2 z 2 0 0 z r n [r sin 0 n]u(n) [r sin 0 ]z 1 1 2 2 z r 1[2r cos 0 ]z r z
Example 5: Find the RoC of x[n] (0.5) n u[n] (0.4) n u[n] Using the properties of z-transform we get X (z) 1 1 0.5z 1 1 1 0.4z 1 z z z(z 0.4) z(z 0.5) z 0.5 z 0.4 (z 0.5)(z 0.4) It is clear that the RoC is given by z 0.4 and z 0.5 So we can conclude that the RoC is z 0.5
Example 6: Find the RoC of x[n] (0.5) n u[n] (0.9) n u[n 1] Using the properties of z-transform we get X (z) 1 1 0.5z 1 1 1 0.9z 1 z z z(z 0.9) z(z 0.5) z 0.5 z 0.9 (z 0.5)(z 0.9) The RoC due to the first part is z 0.5 since it is a right-sided sequence however, the second part is a left-hand sequence, therefore its RoC is z 0.9 So we can conclude that the RoC for X(z) is 0.5 z 0.9
Inverse Z-transform Power Series method Simple Tedious for large n Not accurate Partial Fraction method Complicated More accurate
IZT: Power Series method In this method we divide the numerator of a rational Z-transform by its denominator The basic idea is Given a Z-transform X(z) with its corresponding RoC, we can expand X(z) into a power series of the form X (z) c n z n which converges in the given RoC n
Example 7: Find the Inverse Z-transform of X(z) 1 X(z) 11.5z 1 0.5z 2 RoC z >1 Since RoC is the exterior of the circle, so we expect a right-sided sequence, so we seek an expansion in the negative powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series 1 1 3 z 1 1 z 2 2 2 1 3 z 1 7 z 2 15 z 3 31 z 4... 2 4 8 16 x[n] = [1, 3/2, 7/4, 15/8, 31/16,. ]
Example 8: Find the Inverse Z-transform of X(z) 1 X(z) 11.5z 1 0.5z 2 RoC z <1 Since RoC is the interior of the circle, so we expect a left-sided sequence, so we seek an expansion in the positive powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series 1 1 3 z 1 1 z 2 2 2 2 z 2 6 z 3 14 z 4 30 z 5... x[n] = [., 30, 14, 6, 2, 0, 0]
IZT: Partial Fraction method Steps to follow Eliminate the negative powers of z for the z-transform function X(z) Determine the rational function X(z)/z (assuming it is proper), and apply the partial fraction expansion to the determined rational function X(z)/z using formulae in table (next slide)
Partial fraction(s) and formulas for constant(s) Partial fraction with the first-order real pole: A z p Partial fraction with the first-order complex poles: A (z p) X (z) z p z Az A* z A (z p) X (z) z p z p * z p z, A A* Partial fraction with mth-order real poles: A k A A k 1 1 A z p (z p) 2 (z p) k k k 1 1 d (z p) k (k 1)! dz k 1 X (z) z z p
An example for Simple Real Poles
An example for Multiple Real Poles f n =[9(0.3) n 8(0.2) n +2n(0.2) n ]u(n)
Pulse Transfer Function Pulse transfer function H(z) is defined as the ratio of the Z-transform of the input x[n] to the Z-transform of the output y[n] H z Y z X z
Derivation l y n b i xn i a i y n i i 0 i 1 k Applying Z-transform and moving the terms of y to one side Y z k i1 Y za z i i l i i0 b X zz i Y z Y z Y z1 k l i a i z X z i i1 i0 k i i1 a z i X z l i i0 b z b z i i H z Y z X z 1 l b i z i0 k i1 i a i z i
Example 8: Find the Pulse Transfer function of the difference equation yn 0.1yn 1 0.02 yn 2 2xn xn 1 a 1 0.1 b 0 2 H z 1 a 2 0.02 b 1 1 1 b i z i0 2 i i1 i a z i H z Y z X z l i b i z i0 k i a z i 1 i1 H z 2z 1z 0 1 1 0.1z 1 0.02z 2 1 2 z 1 0.1z 1 0.02z 2