final-sol

final-sol - MAE143 A Signals and Systems Winter 11 Final...

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Unformatted text preview: MAE143 A - Signals and Systems - Winter 11 Final Instructions (i) This exam is open book. You may use whatever written materials you choose, including your class notes and textbook. You may use a hand calculator with no communication capabilities (ii) You have 3 hours (iii) Do not forget to write your name, student number, and instructor 1. Filtering and sampling. The continuous-time signal x ( t ) = 100 sinc 2 (100 t ) goes through an ADC block that samples signals at a frequency f s = 190 Hz. Answer the following questions (a) (1 point) Compute the Fourier transform of x ( t ) . What is its bandwidth? (b) (1 point) Would you be able to reconstruct the original signal out of the samples taken by the ADC block? Why? (c) (2 points) Consider a system whose impulse response is given by h ( t ) = 150 sinc(150 t ) What kind of filter is this? What is its cutoff frequency in Hz? Plot the magnitude and phase of the transfer function. (d) Suppose the signal x ( t ) passes first through the system in (c) to produce the signal y ( t ) and then goes through the ADC block to produce the samples, see Figure 1. h(t) ADC x(t) samples y(t) Figure 1: Block diagram for question 3, part (d). i. (1 point) What is the bandwidth of y ( t ) ? Would you be able to reconstruct the signal y ( t ) out of the samples taken by the ADC block? Justify your answer. ii. (1 point) Why would one call the system in (c) an antialiasing filter? (e) (1 point (bonus)) With knowledge of y ( t ) , would you be able to recover the original sig- nal x ( t ) ? Why? Solution: (a) Using the table of basic transforms and properties (time scaling), the Fourier transform of x ( t ) is X ( f ) = tri f 100 ( + .5 point ) The bandwidth of this signal is therefore f B = 100 Hz. (+ .5 point) (b) No, because the sampling frequency is below the Nyquist rate, f s = 190 < 2 f B = 200 . (+ 1 point) (c) We compute the transfer function of the system as H ( f ) = rect f 150 ( + 1 point ) Therefore, this is an ideal lowpass filter with cutoff frequency f c = 75 Hz. (+ .5 point) The phase plot of the filter is trivial (identically zero). The magnitude plot looks like-100-50 50 100 0.2 0.4 0.6 0.8 1 (+ .5 point) (d.i) After the signal x ( t ) goes through the ideal filter, all frequencies above the cutoff frequency ( 75 Hz) get cut. Therefore, the bandwidth of y ( t ) is 75 Hz. (+ .5 point) Since the sampling frequency of the ADC block is 190 > 150 = 2...
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final-sol - MAE143 A Signals and Systems Winter 11 Final...

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