SIGNALS AND SYSTEMS. Unit IV. Analysis of DT signals

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1 SIGNALS AND SYSTEMS Unit IV Analysis of DT signals Contents: 4.1 Discrete Time Fourier Transform 4.2 Discrete Fourier Transform 4.3 Z Transform 4.4 Properties of Z Transform 4.5 Relationship between Z Transform and DTFT 4.1 Discrete Time Fourier Transform Given a general aperiodic signal of finite duration, Properties of DTFT Periodicity:

2 Linearity: The DTFT is linear. If and then Time Shifting and Frequency Shifting: If, then, and, Time reversal Let us find the DTFT of x[-n]

3 Convolution Property: Let h[n] be the impulse response of a discrete time LSI system. Then the frequency response of the LSI system is Now and If then Proof

4 now put n-k =m, for fixed k, 4.2 Discrete Fourier Transform (DFT) A periodic discrete time signal x[n] with period N can be represented as a Fourier series: Where Here the summation ranges over any consecutive N integers of x[n], Where N is the period of the discrete time signal x[n]. Here equation (i) is called the Synthesis Equation and equation (ii) is called the Analysis Equation.

5 4.3 Z-transform The response of a linear time-invariant system with impulse response h[n] to a complex exponential input of the form can be represented in the following way: Where In the complex z-plane, we take a circle with unit radius centered at the origin. H(w) is periodic with period with respect to ' w '. When we replace z by, we get periodicity of in the form of a circle. Nature of Region of Convergence Z-transform: The ROC of X(z) of a two sided signal consists of a ring in the z-plane centered about the origin.

6 and depend only on magnitude of z. As in the case of Laplace transform for a right-sided sequence and for a left-sided sequence. If x[n] is two-sided; the ROC will consist of a ring with both and finite and non-zero.

8 4.4 Properties of Z-Transform 1) Linearity R2 If with ROC = R1 and with ROC = then with ROC containing. 2) Differentiation in z domain Now for the discrete variable case, the z - Transform of x[n] is given by On differentiating both sides w.r.t z and then multiplying both sides by -z, we get

9 Provided it exists. This can be thought of as taking the z Transform of nx[n]. Hence we observe that, if Similarly we can proceed further, by differentiating again and again. 3) Time shift If with ROC = R then Because of addition or deletion of the origin or infinity with ROC = R except for the possible the multiplication by for poles will be introduced at z=0, which may cancel corresponding zeroes of X(z) at z=0. In this case the ROC for equals the ROC of X(z) but with the origin deleted. Similarly, if may get deleted. 4) Time scaling The continuous-time concept of time scaling does not directly extend to discrete time. However, the discrete time concept of time expansion i.e. of inserting a number of zeroes between successive values of a discrete time sequence can be defined. The new sequence can be defined as x (k) [n] = x[n / k] if n is a multiple of k = 0 if n is not a multiple of k

10 has k - 1 zeroes inserted between successive values of the original sequence. This is known as up sampling by k. If x[n] X(z) with ROC = R then i.e. x (k) [n] X(z k ) with ROC = R 1/k i.e. z k R X(z k ) = S x[n](z k ) -n, < n < = S x[n]z -nk, < n < Problem1:

11 Determine the z-transform for each of the following sequences. Sketch the pole zero plot, and indicate the region of convergence. Indicate whether or not the Fourier transform of the sequence exist. (a) δ[n + 5] (b) δ[n - 5] (c) (-1) n u[n] (d) (e) (f) (g) (h) Solution: (a) δ[n + 5] The Z-transform of δ[n] is 1, now the Z-transform of δ[n + 5] will be Z 5, by the property that if Z-transform of x[n] is X(Z) then the Z-transform of x[n-m] will be.. The region of convergence in this case is the entire z plane except

12 (b) δ[n - 5] The Z-transform of δ[n] is 1, now the Z-transform of δ[n - 5] will be Z -5, by the property that if Z-transform of x[n] is X(Z) then the Z-transform of x[n-m] will be. The region of convergence in this case is the entire z plane except (c) (-1) n u[n] has the Z-transform hence the sequence (-1) n u[n] will have the Z-transform with the region of convergence (d) property and the above used Hence from combining the shifting property we can get the Z-transform to be as follows. The region of convergence will be note that infinity is not in the ROC

13 (e) If the X(Z) is the Z-transform of x[n] then we shall use the following properties to solve the above problem. 1. Z-transform of x[n-m] will be. The ROC is all Z except 0( if m > 0) or infinity(if m < 0 ) 2. has the Z-transform The ROC is Hence the Z-Transform will be, (in k, we now shift it by m = -1) Hence the final transform will be with region of convergence.

14 (f) now the, hence Z-transform using a procedure similar to the one above will be with a region of convergence note here 0 is not in the ROC (g), hence the Z-transform will be, with region of convergence

15 (h) By the shifting property and the property has the Z-transform, we get the Z-transform to be with the region of convergence Inverse Z Transform We know that there is a one to one correspondence between a sequence x[n] and its ZT which is X[z]. Obtaining the sequence 'x[n]' when 'X[z]' is known is called Inverse Z - Transform. For a ready reference, the ZT and IZT pair is given below. X[z] = Z {x[n]} x[n] = Z -1 {X[z]} Forward Z - Transform Inverse Z - Transform For a discrete variable signal x[n], if its z - Transform is X(z), then the inverse z - Transform of X(z) is given by Where ' C ' is any closed contour which encircles the origin and lies ENTIRELY in the Region of Convergence.

16 4.5 Relationship between Z - Transform and Discrete Time Fourier Transform (DTFT) Similarly, on making the same substitution in the inverse z - Transform of X(z); provided the substitution is valid, that is, z =1lies in the ROC. Hence we conclude that the z -Transform is just an extension of the Discrete Time Fourier Transform. It can be applied to a broader class of signals than the DTFT, that is, there are many discrete variable signals for which the DTFT does not converge but the z-transform does so we can study their properties using the z - Transform.

17 Examples: The z - Transform of this sequence is Also we observe that the DTFT of the sequence does not exist since the summation diverges. This example confirms that in some cases the z - Transform may exist but the DTFT may not. Problem1: Consider the z transform Solution: Problem2: Consider the z transform

18 Solution: There are two poles one at z=1/4 and at z=1/3. The partial fraction expansion, expressed in polynomials in 1/z, is Thus, x[n] is the sum of 2 terms, one with z - transform 1/[1-(1/4z)] and the other with z - transform 2/[1-(1/3z)]. Thus, As the ROC is not mentioned, we get different inverses for different possible ROCs. Pattern when ROC is right sided, i.e. z >1/3 We can identify by inspection,

19 When the ROC is two-sided, i.e. 1/4< z <1/3 When the ROC is left sided, i.e. z <1/4

Chapter Intended Learning Outcomes: (i) Understanding the relationship between transform and the Fourier transform for discrete-time signals

z Transform Chapter Intended Learning Outcomes: (i) Understanding the relationship between transform and the Fourier transform for discrete-time signals (ii) Understanding the characteristics and properties