University of Illinois at Urbana-Champaign ECE 310: Digital Signal Processing

Transcription

1 University of Illinois at Urbana-Champaign ECE 3: Digital Signal Processing Chandra Radhakrishnan PROBLEM SE 5: SOLUIONS Peter Kairouz Problem o derive x a (t (X a (Ω from X d (ω, we first need to get rid of the repeated frequency component in X d (ω. hus, multiply an ideal LPF on both sides. { Ω < π G a (Ω = else ( π g a (t = sinc t herefore, x a (t = ( ω X a = X d (ωg a (Ω X a (Ω = X d (ωg a (Ω n= x a (t = x(n g a (t ( π x[n]sinc (t n Problem (a he Nyquist sampling rate is given by, Nyquist = ( 3 =.4 sec =.4 ms ( f Nyquist = = 4 khz Nyquist ω max = Nyquist (π(6 = π ω min = Nyquist (π(3 = π 4 he sketch of the frequency response of the discrete-time filter, when sampling at the Nyquist rate is shown in Fig. (c Some aliasing of the input signal is allowed with the condition that the minimum aliasing frequency is greater than the cutoff frequency of the filter. π π max π 6 max 8 max max = 8 sec Also, the maximum frequency of H d, (ω should be less than or equal to π. ω max = max (π(6 π max = sec herefore, = min ( max, max = max = 8 sec

2 he sketch of the frequency response of the discrete-time filter, when sampling at the Nyquist rate is given below (c Some aliasing of the input signal is allowed with the condition that the minimum aliasing frequency is greater than the cutoff frequency of the filter. π π max π 6 max (d 8 max max = 8 sec Also, the maximum frequency of H d, (ω should be less than or equal to π. ω max = max (π(6 π max = sec herefore, = min ( Figure : Figure for Problem max, max = max = 8 sec (d ω max = ω max = 8 (π(6 = π 8 (π(6 = 3 π 3 ω min = ω min = 8 (π(3 = π 8 (π(3 = 3 π he sketch of the frequency response of the discrete-time filter, 3 when sampling at the maximum he sketch of rate the frequency is given below response of the discrete-time filter, when sampling at the maximum rate is shown in Fig. Figure : Figure for Problem (d 3. (a he Nyquist rate is twice the highest frequency component, f s khz. herefore, max = Problem 3 sec. (a he Nyquist rate is twice the highest frequency component, f s khz. herefore, ω = Ω max = sec. π 8 = Ω (c (c Ω = ω = π Ω 65 f π= 65 Hz 8 = Ω Ω = π 65 ω = f Ω = 65 Hz π 8 = Ω Ω = ω = π Ω 5 f π= 5 Hz 8 = Ω Ω = π 5 f = 5 Hz

3 Problem 4 (a he sketches for X d (ω, Y d (ω, and Y a (Ω are given below 4. (a he sketches for X d (ω, Y d (ω, and Y a (Ω are given below Figure 3: Figure for Problem 4(a P a (Ω = e jωt dt ( = e j Ω Ω sinc P a (Ω = e jωt ( dt Y a (Ω = e j Ω Ω sinc ( Y d (Ω = e j Ω Ω sinc he sketch of the magnitude of Y a (Ω is given below: ( Y a (Ω = e j Ω Ω sinc Y d (Ω he sketch of the magnitude of Y a (Ω is given in Fig. 4: he component of Y a (Ω for Ω > π = π is due to the nonideal D/A. he highest amplitude of this unwanted component of Y a (Ω is at Ω = 5π rad/sec and ( ( 5π 5π Y a (5π = Y d sinc = as shown in the figure above. Problem 5 3 3

4 Y a Amplitude X: 5 Y: Ω (x π rad/sec (a (c he component of Y a (Ω for Figure Ω > 4: π Figure = π for Problem is due to4 the nonideal D/A. he highest amplitude of this unwanted component of Y a (Ω is at Ω = 5π rad/sec and ( ( 5π 5π Y a (5π = Y d sinc =.353 X (Ω 4 = P (ΩX d (Ω as shown in the figure above. X (Ω = P (ΩX d (Ω H(Ω = X(Ω X = P(Ω (Ω P (Ω 5. (a For zero-order hold: ( P (Ω = X (Ω e= jωt P (ΩX dt = d (Ω e j Ω Ω sinc X (Ω = P (ΩX d (Ω For first-order hold: H(Ω = X (Ω X (Ω = P (Ω p (t = ( ( t t P (Ω rect rect = ( p t + ( p t + For zero-order hold: ( ( ( P (Ω = e j Ω Ω sinc e j Ω ( Ω = sinc P (Ω = e jωt dt = e j Ω Ω sinc sinc ( Ω ( H(Ω = For first-order hold: e j Ω sinc ( = e j Ω Ω Ω sinc p (t = ( ( t t rect rect = ( p t + ( p t + For an ideal { D/A, Y a (Ω = G ideal (ΩY d (Ω where:, Ω π G ( ( P (Ω = e j Ω Ω ( ideal (Ω =, otherwise For the FOH, to form an ideal D/A, Y a(ω = F sinc e j Ω (ΩP Ω (ΩY d (Ω where F (Ω is an analog filter = sinc that follows the FOH and H(Ω = sinc( { Ω ( e j Ω sinc ( P = e j Ω F (Ω = = (Ω sinc ( Ω, Ω π Ω, otherwise Suppose F (Ω For and Fideal (Ω D/A, are the Y a (Ω analog = G ideal filters (ΩY that d (Ω follow where: the ZOH and FOH, respectively. For reference, the frequency responses of the ZOH and FOH and their subsequent filters { are given in Fig. 5. he cutoff frequency for both F (Ω and F, Ω π (Ω is Ω = π. he magnitudes of these LPFs are G ideal (Ω =, otherwise F = sinc( Ω 4 F = sinc ( Ω 4

5 given below. P P F F pi/ pi/ Ω pi/ pi/ Ω he cutoff frequency for both FFigure (Ω and 5: Figure F (Ω is forω Problem = π. he 5(c magnitudes of these LPFs are F = sinc ( F has a frequency response which is steeper that F. Ωherefore, the FOH (linear interpolation might interpolate y[n] more precisely than a ZOH (piecewise constant interpolation at a cost of a more complicated/expensive analog filter that follows the FOH. F = sinc ( Ω Problem 6 F has a frequency response which is steeper that F. herefore, the FOH (linear (a It is given that interpolation might interpolate y[n] more precisely than a ZOH (piecewise constant interpolation at a cost of a more complicated/expensive analog filter that follows the FOH. y a (t = x a (t + α x a (t τ d + β x a (t τ d Applying the Fourier transform on both sides, we have: Y a (Ω = X a (Ω + αx a (Ωe jωτ d + βx a (Ωe jωτ d H a (Ω = Y a(ω X a (Ω = + αe jωτ d + βe jωτ d aking appropriate sampling period to mean the Nyquist sampling period: = = = 5 µs f max 3 (c With the sampling period = 5 µs there is no aliasing in the system. With an ideal D/C, the digital filter we need is: (d With τ d =, Equation ( above simplifies as: H d (ω = H a ( ω = + αe jωτ d/ + βe jωτ d/ ( = + αe j4ωτ d + βe j8ωτ d 5 H d (ω = + αe jω + βe jω By substituting z = e jω, we obtain the following transfer function: which can be implemented using the block diagram below. H(z = + αz + βz, 5

6 z xn [ ] yn [ ] z Figure 6: Figure for Problem 6(d Problem 7 he Fourier transform of y c (t is sketched in Fig. 7, Figure 7: Figure for Problem 7 Problem 8 (a H(z is not a polynomial in z or z the system is IIR. H(z is not a polynomial in z or z the system is IIR. (c H(z is a polynomial of z the system is an FIR filter. 6