Do you have tutor for Discrete signal processing? I need help with only the #6 part of the attachment Document Preview: Page 1 of 9 SIGNALS AND SYSTEMS II LABORATORY 3: The Z Transform, the DTFT, and Digital Filters INTRODUCTION The Z transform pairs that one encounters when solving difference equations involve discrete-time signals, which are geometric (or exponential) in the time domain and rational in the frequency domain. MATLAB provides tools for dealing with this class of signals. Our goals in this lab are to i. gain experience with the MATLAB tools ii. experiment with the properties of the Z transform and the Discrete Time Fourier Transform iii. develop some familiarity with filters, including the classical Butterworth and Chebychev lowpass and bandpass filters, all-pass filters, and comb filters. THE Z TRANSFORM AND THE DTFT The Z transform provides a frequency domain version of a discrete-time signal. Discrete time signals are sequences, and the Z transform is defined by (1) { } S8 -8 ?????? = -k hk H(z) hk z Z transform . Consider, for example, the elementary Z transform pair (2) 1 1 ( ) 1 = ??? = az h a u Z H z k k k where uk is the unit step function. The time domain sequence hk and the frequency function H(z) are alternate ways of describing the same signal. In the time domain, hk is exponential. In the frequency domain, H(z) is rational or, by definition, the ratio of two polynomials. For discrete-time applications, we will use the representation (3) ( )( ) ( ) A zH z = z? B z , where m B z b b z bmz= + -1 +K+ ( ) 0 1 , n A z a z a z an z= + + -2 +K+ 2 1 ( ) 1 1 , and ? is an integer. This numbering of the coefficients is not standard for MATLAB, if we are talking about polynomials, but it is consistent with the way the row vectors a and b are used in the filter function. The indices will be off by one, of course. The representation is unique if we demand that the end coefficients b0 ,bm , and an are not zero. The poles of H(z) are the roots of the denominator polynomial A(z) . At a pole, H(z) becomes infinite. The zeros of H(z) are the roots of the numerator Attachments: Lab-Notes-3.pdf; Do you have tutor for Discrete signal processing? I need help with only the #6 part of the attachment Document Preview: Page 1 of 9 SIGNALS AND SYSTEMS II LABORATORY 3: The Z Transform, the DTFT, and Digital Filters INTRODUCTION The Z transform pairs that one encounters when solving difference equations involve discrete-time signals, which are geometric (or exponential) in the time domain and rational in the frequency domain. MATLAB provides tools for dealing with this class of signals. Our goals in this lab are to i. gain experience with the MATLAB tools ii. experiment with the properties of the Z transform and the Discrete Time Fourier Transform iii. develop some familiarity with filters, including the classical Butterworth and Chebychev lowpass and bandpass filters, all-pass filters, and comb filters. THE Z TRANSFORM AND THE DTFT The Z transform provides a frequency domain version of a discrete-time signal. Discrete time signals are sequences, and the Z transform is defined by (1) { } S8 -8 ?????? = -k hk H(z) hk z Z transform . Consider, for example, the elementary Z transform pair (2) 1 1 ( ) 1 = ??? = az h a u Z H z k k k where uk is the unit step function. The time domain sequence hk and the frequency function H(z) are alternate ways of describing the same signal. In the time domain, hk is exponential. In the frequency domain, H(z) is rational or, by definition, the ratio of two polynomials. For discrete-time applications, we will use the representation (3) ( )( ) ( ) A zH z = z? B z , where m B z b b z bmz= + -1 +K+ ( ) 0 1 , n A z a z a z an z= + + -2 +K+ 2 1 ( ) 1 1 , and ? is an integer. This numbering of the coefficients is not standard for MATLAB, if we are talking about polynomials, but it is consistent with the way the row vectors a and b are used in the filter function. The indices will be off by one, of course. The representation is unique if we demand that the end coefficients b0 ,bm , and an are not zero. The poles of H(z) are the roots of the denominator polynomial A(z) . At a pole, H(z) becomes infinite. The zeros of H(z) are the roots of the numerator Attachments: Lab-Notes-3.pdf

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