Profs. Kamalabadi, Katselis, : 5 pm, March 25, 2022
UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering
ECE 310 Digital Signal Processing – Spring 2022
Homework 8
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1. The continuous-time signal xa(t) = sin (10πt) + cos (20πt) is sampled with a sampling period T to obtain the discrete-time signal x[n] = sin π n + cos 2π n
a) Determine a choice for T consistent with this information.
b) Is your choice for T in part (a) unique? If so, explain why. If not, specify another choice of T
consistent with the information given.
a) For information consistency we want x[n] = xa(nT). As a result we have: x[n] = sin( π n) + cos( 2π n) = sin(10πnT ) + cos(20πnT )
which leads to T = 1 50
b) No the choice is not unique. Another choice can be T = 11: 50
sin(10πn11) + cos(20πn11) = sin(11πn) + cos(22πn) = sin(πn) + cos(2πn) 50 50 5 5 5 5
2. The input to an D/A converter is {x[n]} = {−1,0,0,3} with sampling interval T. Determine the ↑
output of the D/A converter if the D/A converter is (a) an ZOH, and (b) an ideal D/A.
a) A ZOH with period T holds the value of a given index for time T. Therefore the signal is given
xc(t) = x[n]ga(t − nT ) n=−∞
andforaZOH,ga(t−nT)=rect(
xc(t)=x[−1]rect(t−T2 +T)+x[2]rect(t−T2 −2T)
=−rect(t−T2 +T)+3rect(t−T2 −2T) TT
2 )suchthat
b) For an ideal D/A, ga(t − nT) = sinc(π(t−nT)) (Here it’s the unnormalized sinc function) T
sin( π(t+T ) ) xc = x[−1] T
sin( π(t−2T ) ) T
sin( π(t+T ) ) sin( π(t−2T ) )
=−T+3T π(t+T ) π(t−2T )
3. Consider the following system with uniform sampling
xn yn xa(t) Hd(ω)
D/A ya(t) (ideal)
The discrete-time system Hd(ω) is an ideal low-pass filter with cutoff frequency π8 .
(a) If xa(t) is bandlimited to 5 kHz, what is the maximum value of T that will avoid aliasing in the
A/D converter?
(b) If T1 = 10 kHz, determine the maximum bandwidth of xa(t) allowed such that the overall system
from xa(t) to ya(t) behaves as an LTI system?
a) The minimum sampling frequency Fmin = 2Fxa = 10000Hz. Tmax = 1 = 1 sec =
Fmin 10000
b) To make the whole system behave as an LTI system, there should not be any aliasing in the
[ −π , π ] interval of the digital signals. Let’s assume the maximum bandwidth is B. −2πBT +2π ≥
π, which will result as B ≤ 15 1 = 150kHz = 9.375kHz.
4. For the digital system in problem 3, assume T = 0.5 msec, with
@ @ @ @
X(Ω) 1 H(ω) ad
1000π B −π 4 π
(a) Sketch Xd(ω), Yd(ω), and Ya(Ω).
(b) Suppose the ideal D/A is now replaced by a zero-order hold, using the pulse
1, 0≤t≤T ga(t) = 0, else.
Sketch Ya(Ω) for |Ω| ≤ 8000π. Find the amplitude of the largest unwanted (out of the band |Ω| ≤ Tπ ) component of Ya(Ω), due to the nonideal D/A.
b) The Ga(Ω) for ZOH is Te−jT/2sinc(ΩT ) with sampling period T (unnormalized sinc function).
The D/A result is shown below. The largest unwanted component would be at 3000π and −3000π with amplitude around 0.2347.
5. A speech signal xa(t) is assumed to be bandlimited to 12kHz. It is desired to filter this signal with a bandpass filter that will pass the frequencies between 300Hz and 6kHz by using a digital filter Hd(ω) sandwiched between an A/D and an ideal D/A.
(a) Determine the Nyquist sampling rate for the input signal.
(b) Sketch the frequency response Hd,1(ω) for the necessary discrete-time filter, when sampling at the Nyquist rate.
(c) Find the largest sampling period T for which the A/D, digital filter response (Hd,2(ω)), and D/A can perform the desired filtering function. (Hint: some amount of aliasing may be permissible during A/D conversion for this part.)
(d) For the system using T from part (c), sketch the necessary Hd,2(ω).
a) Fnyquist = 2Fxa = 24kHz
b) 300 Hz in CTFT will be mapped to 2π0.3 = π rad/sample in DTFT. 6 kHz in CTFT will be
mapped to 2π6 = π rad/sample in DTFT. The frequency response of Hd,1(ω) can be plotted as 24 2
c) To perform the desired filtering function, we should prevent any aliasing under 6 kHz. Suppose
the sampling period is Ts. 12kHz in CTFT will be mapped to 24πTs in DTFT. 6kHz in CTFT
will be mapped to 12πTs in DTFT. −24πTs + 2π ≥ 12πTs, which will result as Ts ≤ 1 msec. 18
d) 300 Hz in CTFT will be mapped to 2π0.3 = π rad/sample in DTFT. 6 kHz in CTFT will be 18 30
mapped to 2π6 = 2πrad/sample in DTFT. The frequency response of Hd,2(ω) can be plotted 18 3
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