Homework 7 Solution

# Homework 7 Solution - EE 4541 Digital Signal Processing...

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Unformatted text preview: EE 4541 Digital Signal Processing Fall 2007 Problem Set 7 Due: October 25, 2007 1. In the system shown below, ( )( h[n] ) y[n] H ( z ) = 1 − e jπ / 4 z −1 1 − e − jπ / 4 z −1 Determine the output signal y a (t ) . x a (t ) C/D x[n] D/C y a (t ) T=1 /200 s T=1 /200 s 3e − j 2π / 3 X a ( jF ) 3e j 2π / 3 2e jπ / 4 2e − jπ / 4 ‐225 ‐150 0 150 225 F (Hz) Answer: The digital filter is a notch FIR filter with zeros at /4. The two sinusoidal input signals at frequencies 150 and 225 Hz will alias at 50 and 25 Hz, respectively, because they are both larger than Fs/2 = 100 Hz. The resulting normalized frequencies are /2 and /4 2 / ). Therefore, one of the two frequencies will be notched by the digital filter. The ( other frequency, /2, will pass with amplitude modification and phase shift determined by / the frequency response of the digital filter at /2, √2 / . The output is a / . sinusoid with a frequency of 50 Hz, amplitude of √2, and phase of EE 4541 Digital Signal Processing Fall 2007 Problem Set 7 Due: October 25, 2007 2. Consider the single-pole lowpass filter and its impulse response: L 1 H a ( s ) = ⇔ ha (t ) = e −αt u (t ) s +α a. What is the dc gain? What is the 3dB corner frequency for this filter? At what time does the analog impulse response decay to 1/e of its initial value? At what frequency does the frequency response go to zero? b. Define h1 [n] = Tha (nT ) , find the corresponding transfer function H 1 ( z ) . What is the dc gain? What is the 3dB frequency? How many samples are needed before the impulse response decays to 1/e of its initial value? At what frequency does the frequency response go to zero? 2 z −1 c. Consider the transformation, s = , find the corresponding transfer T z +1 function H 2 ( z ) = H a ( s) s = 2 z −1 . What is the dc gain? What is the 3dB frequency? T z +1 How many samples are needed before the impulse response decays to 1/e of its initial value? At what frequency does the frequency response go to zero? d. Bonus (20 points): Can you explain and differences between the results obtained the two methods? Answer: a) DC gain = 1/α and Ωc = α. Impulse response decays to 1/e at t = 1/α. The filter has a zero at infinity. b) Sampling results in a transformation from the s‐plane to the z‐plane . The analog filter has a pole at s = ‐ α. This means that the digital filter will have a pole at z− T h1 [n] = Tha (nT ) = Te −αTn μ[n] ↔ H 1 ( z ) = 1 − e −αT z −1 T ⇒ H 1 (1) = = DC Gain 1 − e −αT T T H 1 ( e jω ) = = 1 − e −αT e − jω 1 − e −αT cos ω + je −αT sin ω ⎛ e −αT − e αT 2 ⎝ ω c = cos −1 ⎜ ⎜ ⎞ ⎟ ⎟ ⎠ Needs 1/(αΤ ) samples to decay to 1/e and has no zero on the unit circle. c) EE 4541 Digital Signal Processing Fall 2007 Problem Set 7 Due: October 25, 2007 T / 2 (1 + z ) ~ 1 − αT / 2 ~ α = 1 + αT / 2 1 + αT / 2 z − α ⇒ H 2 (1) = 1 / α = DC Gain Same as the analog prototype H 2 (z) = ( ) T / 2 1 + e jω T /2 H 2 (e ) = jω ~ ⇒ M 2 (ω ) = 1 + αT / 2 1 + αT / 2 e − α ~2 ⎞ −1 ⎛ α ⎜ ω c = cos ⎜ ~⎟ 1 + 2α ⎟ ⎝ ⎠ jω 2 + 2 cos ω ~ cos ω + α 2 ~ 1 − 2α ~ Needs approximately log α samples to decay to 1/e and has a zero at ω = π. d) The first design approach preserves the impulse response, but the frequency response is aliased and it deviates from the (desired) analog response. On the other hand, the second design approach preserves the magnitude response at DC without aliasing. However, the frequency scaling is no longer linear as with the first design. This can be seen from the following 21 1 1 1 /2 /2 1 1 Ω, Ω /2 Ω /2 1 Ω This shows that the transformation used in the second approach transforms the whole jΩ axis onto (i.e. one‐to‐one) the unit circle in the z‐plane. It also transforms the analog zero at infinity to a digital zero at ω = π. From the definition of the second digital filter given above, H 2 ( z ) = H a ( s) s= 2 z −1 T z +1 ( ) ⇒ H 2 e j 2 tan (ΩT / 2 ) = H a ( jΩ) That is, the frequency response of the digital filter is the same as that of the (desired) ω = 2 tan(ΩT / 2) analog prototype except for a frequency warping . Note that both designs approach each other if αT << 1. Incidentally, sampling results in a transformation from the s‐plane to the z‐plane . The analog filter has a pole at s = ‐ α. This means that the digital filter will have a pole at as can be seen above. This transformation transforms the jΩ axis into the unit circle (on the z‐plane) linearly, but aliasing may occur. EE 4541 Digital Signal Processing Fall 2007 Problem Set 7 Due: October 25, 2007 In practice, the transformation used in the second design approach, called bilinear transformation, is used in designing (IIR) digital filters from analog prototypes. The design procedure accounts for the frequency warping so that the resulting filter meets the magnitude response criteria at the specified corner frequencies of the desired filter. 3. An analog signal of the form x a (t ) = a(t ) cos(2000π t ) is bandlimited to the range 900 ≤ F ≤ 1100 Hz. It is used as an input to the system used in the figure below. a) Determine and sketch the spectra for the sequences x[n] and w[n] . Assume X ( jΩ) to have a general shape within the frequency band where it has non‐zero values. b) We wish to design a lowpass filter so that v[ n] ≈ a[ n] . Suggest appropriate values for ωp ( ) and ωs such that ωp equals the highest frequency in A e jω and ωs equals the lowest frequency of the nearest image resulting from the demodulation. c) Bonus (20 points): Determine the sampling rate of the A/D converter that would allow us to eliminate the frequency conversion. Answer: a) Sampling will result in scaling the spectrum of the analog signal according to the relationship ω = ΩT = Ω/Fs . Therefore, frequency 1000 Hz will appear at ω = 2π X 1000/2500 = 0.8 π. Similarly, frequencies at 900 and 1100 Hz will appear at 0.72 π and 0.88 π, respectively. Multiplying the modulated signal by the sinusoidal term with the same frequency as the carrier, ⎛ 1 + cos(1.6πn ) ⎞ ⎛ 1 + cos(0.4πn ) ⎞ w[ n] = a[ n] cos(0.8πn ) cos(0.8πn ) = a[n]⎜ ⎟ = a[n]⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ EE 4541 Digital Signal Processing Fall 2007 Problem Set 7 Due: October 25, 2007 ( ) X e jω Fs/2 −π ω = 0.8 π π 0 ω ( ) W e jω Fs/2 ω = 0.4 π Fs/4 −π π 0 ω . ( ) b) The highest frequency component of A e jω is 0.08π. This gives a passband edge frequency of ωp=0.08π. The stopband edge frequency is ω s = 0.4π − 0.08π = 0.32π . c) A simple way to do this is by choosing Fs such that 2π1000/Fs = 2πk. For example, k = 1 achieves the desired result, thanks to aliasing! Note that the bandwidth of the signal (200 Hz) is narrow compared to the carrier. The value of k must be chosen to avoid aliasing in a(t) itself. You can learn more about sampling of bandpass signals by reading Section 6.4 in P&M. We will not cover this in EE 4541, but this is an area of significant practical interest in communications, radar, ultrasonic imaging, etc. due to the extensive use of modulated waveforms in these applications. ˆ x a (t ) x[n ] w[n] v[n] a (t ) A/D X H e jω ( ) cos(0.8πn) D/A EE 4541 Digital Signal Processing Fall 2007 Problem Set 7 Due: October 25, 2007 Fs = 1 = 2500 Ts Fs = 1 = 2500 Ts ...
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