Frequency Response and Sinusoidal Steady-State Response in Digital Signal Processing

Digital Signal Processing I / 4th Class/ 2020-2021 Dr. Abbas Hussien & Dr. AmmarGhalib Frequency response and Sinusoidal Steady State Response The Frequency domain representation: In continuous linear time invariant (CLTI) system, it was important to know the frequency response of a system ( H(j ) ) , which could be used to find the steady-state response of the system. For discrete linear time invariant (DLTI) system, H(ejW) will be used to find the frequency response of the system. W = Ts rad/sample digital frequency. = 2 f rad/sec. analog frequency. Ts = 1 / fs sec. where, sampling rate = 1 /Ts. Response to a complex exponential sequence: If x(n) = ejnW (4.1) h(k)e jWk k= y(n)= h(k)x(n k) = h(k)ejW(n k)=ejnW k = And (4.2) k= H(ejW)= h(k)e jWk k= Let (4.3) y(n) =e jnnW H (ejW ) (4.4) H(ejW) = HR (ejW) + j HI(ejW) = H(ejW) (ejW) (4.5) (4.6.a) H(ejW) = [ { HR (ejW) }2 + { HI(ejW) }2 ]1/2 (ejW) = tan -1 [HI(ejW) / HR (ejW) ] (4.6.b) Response to a sinusoidal sequence: A 2 j j e jW n o) o + e jW n (e e If x(n) = A cos ( W n + ) = (4.7) o Substituting equation (4.7) into equation (4.4), and rearrange the terms, then: jWo ) ejnWoe j y(n) = 2 Re [0.5 A H (e y(n) = A H (e ] jWon jWo ) cos [ n Wo + + ( e (4.8) ) ] 35

Digital Signal Processing I / 4th Class/ 2020-2021 Dr. Abbas Hussien & Dr. AmmarGhalib Note: the output to a sinusoid is another sinusoid of the same frequency but with different phase and magnitude. Example (1): A discrete time system has a unit sample response h(n) h(n) = 0.5 (n) + (n 1) + 0.5 (n 2) a) Find the system frequency response. Plot magnitude and phase. b) Find the steady-state response of the system to x(n) = 5 cos ( n /4). c) Find the steady-state response of the system to x(n) = 5 cos ( 3 n /4). d) Find the total response to x(n) = u(n) assuming the system is initially at rest. Solution: H(ejW)= h(n) e jWn = 0.5 e-0 + e jW + 0.5e-j2W a ) n= = e jW [ 0.5 e jW +1 + 0.5 e jW ] = e jW ( 1 + cos W ) H(e jW ) = e jW . ( 1 + cos W ) = 1 + cos W (e jW ) = tan-1 (e jW) + tan -1 (1 + cos W ) = W H(ejW) (ejW) 2 - W 2 0 W 2 b )Applying equation (4.8), where, W0 = /4 H ( ejWo ) = H(e j /4 ) = 1 + cos ( / 4 ) = 1.707 ( ejWo) = / 4 Then y(n) = 5 ( 1.707) cos [ (n / 4 ) ( / 4 ) ] = 8.535 cos [ ( n 1) / 4 ] C) H(e j 3 / 4 ) = 1 + cos ( 3 / 4 ) = 0.2928 ( ejWo ) = 3 /4 y(n) = 5 ( 0.2928) cos [ (n /4 ) ( 3 / 4 ) ] = 1.4644 cos [ 3 ( n 1) / 4 ] In part (b) the input signal is amplified, while in part (c) the input signal is attenuated. y(n) 2 d) y(n) = x(n) h(n) 1.5 0.5 . = 0.5 x(n) + x(n 1) + 0.5 x(n 2) = 0.5 u(n) + u(n 1) + 0.5 u(n 2) 0 1 2 3 n 36

Digital Signal Processing I / 4th Class/ 2020-2021 Dr. Abbas Hussien & Dr. AmmarGhalib Note: (t to) f (t) = = f (t to) Properties of frequency response: 1- H(e jW ) is a continuous function in W. 2- H(e jW ) is periodic in W with period2 . 3- H(e jW ) is an even function of W and symmetrical about . 4- (e jW ) is an odd function of W and unti-symmetrical about . Example (2): Find and plot the frequency response of a rectangular window filter if : h(n) = 1 0 n N 1 elsewhere h(n) 1 0 ... 0 1 2 N-1 n Solution: 1 e jWN N 1 H(ejW) = h(k) e jWk= e jWk= 1 e jW k = k =0 1 an+1 1 a n By using ak= a 1 , k =0 )e jWN / 2 ( e jWN / 2 e jWN / 2) e jW / 2 ( e jW / 2 e jW / 2) =e jW ( N 1) / 2sin(WN / 2) jW H(e sin(W /2) H(e jW ) = sin(WN /2) sin(W / 2) sin(WN / 2)} sin(W /2) (e jW ) = W ( N 1) /2 + tan-1{ Assuming N = 5 ,then 37

Digital Signal Processing I / 4th Class/ 2020-2021 Dr. Abbas Hussien & Dr. AmmarGhalib Z-Transform 5.1 Definition of Z.T The z-transform is a very important tool in describing and analyzing digital systems. It also offers the techniques for digital filter design and frequency analysis of digital signals. The z- transform of a causal sequence x(n), designated by X(z) or Z(x(n)), is definedas: Where, z is the complex variable. Here, the summation taken from n = 0 to n = is according to the fact that for most situations, the digital signal x(n) is the causal sequence, that is, x(n) = 0 for n 0. For non-causal system, the summation starts at n = - . Thus, the definition in Equation (5.1) is referred to as a one-sided z-transform or a unilateral transform. The region of convergence is defined based on the particular sequence x(n) being applied. The z-transforms for common sequences are summarized below: 38