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UK EE 422G - 8-4B Steady-State Frequency Response of a Linear Discrete-Time System

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EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-328-4B Steady-State Frequency Response of a Linear Discrete-Time System In this section, we study various properties of the discrete-time Fourier Transform. As we stated earlier, it is the same as the continuous-time Fourier Transform of the sampled signal∑∞=−=0)()()(nsnTtnTxtxδ: We have also claimed that it can be computing by evaluating the Z-transform around the unit circle (provided that the unit circle is inside the ROC) Definition of DTFT: ∑∞=−===0)()()(nnTjezTjenTxzXeXTjωωω Note that we use X(ejωT) to indicate the substitution z=ejωT. Let’s first show that these two definitions are equivalent. Start with the continuous-time Fourier Transform of the sampled signal: ∑∑∫∫∑∞=−∞=∞∞−−∞∞−−∞===−=−=000)()()()()()()(nTjnTjntjtjnseXenTxdtnTtenTxdtenTtnTxjXωωωωδδω Tπ2 Tπ4 Tπ2− Tπ4− Tπ2 Tπ4 Tπ2− Tπ4− Xs(jω) 0EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-33The primarily reason to study Fourier Transform in continuous-time linear system is that if the input to a continuous-time linear system is a complex sinusoid of frequency ω0, the output is also a complex sinusoid of frequency ω0 with a phase shift and a gain governed by H(jω0). This form of analysis, as you know, is called the steady state analysis (steady state as complex sinusoid is not transient). It would be nice if the DTFT can do the same for Discrete-time linear system or And indeed it is true: )()()()()()(0000000)(0TjnTjmmTjnTjmTmnjmeHemTheemThemThmTnTxnTyωωωωω===−=∑∑∑∞=−∞=−∞= Before we go on, let’s introduce a common representation of the DTFT based on normalized frequency. Tπ4 Tπ2− Tπ4− X(ejωT) -ωr Tπ2ωr 0 πω2Tr 0.5 r X(ej2πr) ω X(ejωT) = X(ej2πr) πω2Tr = maps ω∈[0,π/T] to r∈[0,0.5] r = normalized frequency TπDiscrete-Time linear system, H(⋅) nTjenTx0)(ω= ?)()(00nTjTjeeHnTyωω= Continuous-Time linear system, H(⋅) tjetx0)(ω= tjejHty0)()(0ωω= ω0 ω 1 X(jω) H(jω) H(jω0) Y(jω) • = ω0EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-34Remarks 1. Why called normalized frequency? Given the sampling frequency Tfs/1= and the frequencyπω2=f, r can be more succinctly represented as sffTr ==πω2 When using normalized frequency r, the DTFT is written as )(2 rjeHπ. 2. Why ignore the negative frequency? For real-valued h(nT), we can deduce the negative frequency from the positive frequency: [ ])()(&)()()()(*)()()(**00)(TjTjTjTjTjnnTjnnTjTjeHeHeHeHeHconjugateenThenTheHωωωωωωωω−∠=∠=⇒====−−∞=−∞=−∑∑ Thus, it is sufficient to show only the positive part only. The followings are a list of DTFT properties and common transform pairs. They can be easily deduced from the Z-transform tables.EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-35 Computing DTFT Just like the continuous-time counterpart, it is most common to show the DTFT in terms of its amplitude response )(TjeHω and phase response )(TjeHω∠ . However, unlike continuous-time where we can draw asymptotic approximation (Bode plot) by letting ω→∞, we can’t do that for DTFT as it is periodic. Even though there are geometric techniques to draw the frequency responses based on the locations of poles and zeros, they are beyond the scope of this course. Here we settle with plotting the response using Matlab.EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-36Example 1: A simple delay 1)(−= zzH To get the magnitude and phase response, we can use the matlab routine freqz. In freqz, the Z-transform is specified by the denominator and numerator coefficients: nmznazaazmbzbbzH−−−−++++++++=)1(...)2()1()1(...)2()1()(11 >> [h,w] = freqz([0 1],[1]); % Assume T=1 >> plot(w,abs(h)) % Gives Amplitude Response 0 0.5 1 1.5 2 2.5 3 3.511111111 >> % Gives Phase Response; Unwrap removes phase jumps >> plot(w,unwrap(angle(h))) 0 0.5 1 1.5 2 2.5 3 3.5-3.5-3-2.5-2-1.5-1-0.50 Notice that the amplitude is flat (gain =1) and the phase is linear. A system with response like this is called distortion-less as they essentially keep the input intact. The negative slope of the phase, called the group delay, measures the delay of the frequency component. In this case, TTddeHddTj=−−=∠− )()(ωωωω, i.e. One sample for all frequencies.EE 422G Notes: Chapter 8 Instructor: Cheung Page 5-37Example 2: General Linear Phase System 43211.02.04.02.01.0)(−−−−++++= zzzzzH >> [h,w] = freqz([0.1 0.2 0.4 0.2 0.1],[1]); >> plot(w,abs(h) >> plot(w,unwrap(angle(h))) The amplitude and phase responses are 0 0.5 1 1.5 2 2.5 3 3.50.10.20.30.40.50.60.70.80.910 0.5 1 1.5 2 2.5 3 3.5-7-6-5-4-3-2-10 Notice that the phase is linear with group delay = 2T (2 samples) for all frequencies. It can be shown that the impulse response must be symmetric about the middle sample if it has linear phase. Indeed, it is true for our filter: Linear-phase filter is very important in audio and image applications because 1. In audio, the perception of chords requires different frequencies to register at the same time instance. A non-linear phase filter delays those frequencies by different amount making the chord perception dispersed. 2. In image, color edges require spatial cycles to locate precisely at a particular location. A non-linear phase filter distorts the edges by fattening them. Can a filter have zero delay (phase)? Yes, but such a filter MUST BE ACAUSAL! For example: 0 0 Amplitude Response Phase ResponseEE 422G Notes: Chapter 8 Instructor: Cheung Page 5-38Due to the symmetry requirement, a casual IIR filter can never be linear phase. See the following example. Example 3: a simple zero 211042.063.01.25.01)(−−−+−+=zzzzY >> [h,w] = freqz([1 0.5],[2.1 -0.63 0.042]); >> plot(w,abs(h) >> plot(w,unwrap(angle(h))) 0 0.5 1 1.5 2 2.5 3 3.50.10.20.30.40.50.60.70.80.910 0.5 1 1.5 2 2.5 3


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UK EE 422G - 8-4B Steady-State Frequency Response of a Linear Discrete-Time System

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