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Wdigfreqresp6

# Wdigfreqresp6 - ECE 421 Sum 2010 Detailed Solutions Set 6...

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ECE 421 - Sum 2010 Detailed Solutions - Set 6: Digital Filter Frequency Response 1 6-1. Let the sampling interval in a digital system be T s = 0.1 milliseconds. A pure sinusoid x t sin 2 π f c t is sampled by this system. Compute the radian frequency θ c (units = radians/samp. interval) if the Hz frequency f c is given below: (a) f c = 200 Hz, (b) f c = 1 KHz, (c) f c = 5 KHz DETAILED SOLUTION: A sinewave with digital frequency θ c radians per sampling interval has the mathematical form x m sin θ c m a Using the definition of a digital signal x m obtained from an arbitrary analog signal gives x m x mT s x t t mT s b With x t as given in the problem statement, equation b then produces x m sin 2 π f c t t mT s sin 2 π f c m 10 4 c Rewriting the result of c in a slightly different from will simplify our work to follow. x m sin 2 π f c 10 4 m d Therefore, the problems in parts (a), (b), and (c) can now be solved by substituting the appropriate value of f c into d and simplifying. (a) f c 200 Hz Substitute f c 200 into d and simplify. This procedure gives x m sin 2 π 200 10 4 m sin 0 04 π m e Comparing equation a with equation e shows that the radian frequency for part (a) is given by θ c 0 04 π f

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ECE 421 - Sum 2010 Detailed Solutions - Set 6: Digital Filter Frequency Response 2 6-1. (cont.) (b) f c 1 KHz Substitute f c 1000 into d and simplify. This procedure gives x m sin 2 π 10 3 10 4 m sin 0 2 π m g Equation g shows that the radian frequency for part (b) is therefore given by θ c 0 2 π h (c) f c 5 KHz Substitute f c 5000 into d and simplify. This procedure gives x m sin 2 π 5 10 3 10 4 m sin π m i Equation i shows that the radian frequency for part (c) is thus given by θ c π j
ECE 421 - Sum 2010 Detailed Solutions - Set 6: Digital Filter Frequency Response 3 6-2. A digital filter impulse response h m has the analytical form h m a m u m . Compute the digital frequency response H θ if a has the values given below. Also accurately plot H θ for each case. (a) A 0 45, (b) A 0 72, (c) A 0 3 e j 0 1 π , (d) A 0 5 j 0 2 DETAILED SOLUTION: To work this problem, find the filter transfer function H z and then compute the frequency response H θ by evaluating H z on the unit circle. The transfer function H z is the z -transform of the h m given in the problem statement: h m A m u m H z 1 1 Az 1 z z A a Thus, for this problem we have H z z z A b The digital filter frequency response, H θ , is found by evaluating H z on the unit circle: H θ H z z e j θ z z A z e j θ c Performing the evaluation in equation c then gives H θ e j θ e j θ A d Therefore, to find the frequency response requested in parts (a) - (d) of this problem, we simply need to substitute the appropriate value of A into equation d . We can also find a general form for the magnitude H θ which can be used in each of parts (a) - (d). To find the magnitude H θ , recognize that the magnitude of a ratio is the ratio of the magnitudes. Using this property on equation d gives H θ e j θ e j θ A 1 cos θ j sin θ A 1 cos θ A j sin θ e Equation e can be evaluated to produce H θ 1 cos θ A 2 sin θ 2 1 A 2 2 A cos θ cos 2 θ sin

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