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Unformatted text preview: ECE 421 - Sum 2010 Notes Set 6: Digital Filter Frequency Response 1 INTRODUCTION In a previous set of Notes we designed digital filters using properties of the analog dif- ferential equation and the analog transfer function H s . In a future set of Notes we will take a more direct approach for designing digital filters without having to use analog do- main properties. To use this direct design method effectively we have to understand the digital filter frequency response, which is often called the digital frequency response. This property is analogous to the analog filter frequency response. One of the strong points of digital filters is that they may be designed independently of the Hertz frequency range of the application. A digital filter algorithm would have the same mathematical form regardless of whether it is applied to a speech or audio or video application. The important frequency for a digital filter design is not cycles per second (Hertz frequency f ) or radians per second (radian frequency ). Digital filters are designed to operate in a frequency domain having units of radians per sampling interval, which will be given the symbol in these Notes. In this set of Notes we will first examine some simple geometrical methods for helping us understand the concept and properties of a digital filter frequency response. After that we will study the differences between analog and digital frequency responses. ECE 421 - Sum 2010 Notes Set 6: Digital Filter Frequency Response 2 RADIANS PER SAMPLING INTERVAL Lets examine the frequency unit , called digital frequency, which has the units of radians per sampling interval. Suppose we wanted to design a digital filter algorithm to work in the 100 KHz analog (Hz) frequency range. The input to this filter would be samples of the analog signal x t , which have been sampled by an Analog-to-Digital Converter. As an example, let the analog signal x t have a single Hertz frequency f c = 100 KHz and be given by x t = 5 cos 2 10 5 t 1 Let x t be sampled at t = mT s , where T s is called the sampling interval. This operation produces the digital signal x [ m , where x [ m = x mT s = x t t = mT s 2 Suppose we choose the sampling interval T s = 0.5 sec = 5 10 7 sec. Then equation (2) gives x [ m = 5cos 2 10 5 m 5 10 7 = 5cos : 1 m 3 Equation (3) has the general form x [ m = 5cos c m ; c = : 1 4 Note in (4) that c has the units of radians. We can no longer see that the original signal was a 100 KHz cosine. To see why c is given the units of radians per sampling interval, observe the angle of the cosine in (4) at m = 0: x [ = 5cos c = 5cos 5 Now observe the angle of the cosine in (4) at one sample interval later, that is, at m = 1: x [ 1 = 5cos c 1 = 5cos c = 5cos : 1 6 The phase of the cosine has advanced by c = : 1 radians in the time of one sampling interval. In these notes, the general digital frequency having units of radians per samplinginterval....
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