数字信号处理米特拉第四版实验五答案

2
Amplitude
Amplitude
T = 0.004 msec
1 0.5
0 -0.5
-1 0
1 0.5
0 -0.5
-1 0
Continuous-time signal xa(t)
Fra Baidu bibliotek
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time, msec
Discrete-time signal x[n]
1
0.5
0
-0.5
-1
0
5 10 15 20 25 30 35 40 45 50
Time index n
Amplitude
Amplitude
3
Amplitude
Amplitude
T = 0.15 msec
Continuous-time signal xa(t) 1
0.5
0
-0.5
-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time, msec Discrete-time signal x[n]
The sampling period in seconds is - T = 0.1 msec.
Q5.3 The effects of the two axis commands are – The first axis command sets the minimum and maximum values for the x-axis and the y-axis in the upper plot. The second axis command does the same thing for the lower plot. In each axis command, the first two parameters are the minimum and maximum values for the x-axis. The third and fourth parameters are the minimum and maximum values for the y-axis.
1
Answers:
Q5.1
The plots of the continuous-time signal and its sampled version generated by running Program
P5_1 are shown below:
Amplitude
Continuous-time signal xa(t) 1
Time index n
Amplitude
5
The plots of the continuous-time sinusoidal signal of frequency 7 kHz and its sampled version
generated by running a modified Program P5_1 are shown below: (Again, the book is in error. For the same reason as above, the frequency here is 7 kHz, not 7 Hz.)
Time index n
Amplitude
Based on these results we make the following observations – In all three cases, we have T = 0.1 msec. Originally, in Q5.1. we had f = 13kHz. This gives a maximum sampling period of 1/26 » 0.0385 msec to satisfy the Nyquist criterion, as we observed above.
1
0.5
0
-0.5
-1
0
1
2
3
4
5
6
Time index n
T = 0.2 msec
Continuous-time signal xa(t) 1 0.5 0 -0.5 -1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time, msec
Discrete-time signal x[n] 1 0.5 0 -0.5 -1
0.5
0
-0.5
-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time, msec Discrete-time signal x[n]
1
0.5
0
-0.5
-1
0
1
2
3
4
5
6
7
8
9 10
Time index n
Amplitude
Q5.2 The frequency of the sinusoidal signal in Hz is – f = 13 kHz. This is because the time scale shown in the upper graph is in msec. Thus, the sinusoid goes through 13 cycles in 1 msec, or 13,000 cycles in one sec.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time index n
Amplitude
Amplitude
4
Based on these results we make the following observations – Since the “analog” waveform goes through 13 cycles in the 1 msec shown in the graph, there will be aliasing unless we get at least two samples per cycle, or 26 samples total on the graph. 26 samples in 1 msec is a sampling rate of 26 kHz, which requires T < 1/26 » 0.0385 msec to avoid
50
100
150
200
250
Time index n
T = 0.02 msec
Continuous-time signal xa(t) 1
0.5
0
-0.5
-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time, msec Discrete-time signal x[n]
Name: SOLUTION (Havlicek) Section:
Laboratory Exercise 5
DIGITAL PROCESSING OF CONTINUOUS-TIME SIGNALS
5.1 THE SAMPLING PROCESS IN THE TIME-DOMAIN Project 5.1 Sampling of a Sinusoidal Signal
aliasing. With T=0.004 msec. there is no aliasing and the discrete-time waveform has an appearance that is very similar to that of the “analog” waveform. With T=0.02 msec, there is still no aliasing, but the sampling rate is much closer to being “critical,” i.e., much closer to the Nyquist rate. Consequently, the appearance of the discrete-time waveform is less similar to the “analog waveform,” although perfect reconstruction is still possible. With T=0.15 msec, there is significant aliasing and the discrete-time waveform has the appearance of an “analog” waveform of much lower frequency. Finally, with T=0.2 msec, there is again severe aliasing which causes the discrete-time waveform to have the appearance of an “analog” waveform of lower frequency.
Amplitude
Continuous-time signal xa(t) 1
0.5
0
-0.5
-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time, msec Discrete-time signal x[n]
1
0.5
0
-0.5
-1
0
1
2
3
4
5
6
7
8
9 10
Amplitude
Continuous-time signal xa(t) 1
0.5
0
-0.5
-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time, msec Discrete-time signal x[n]
1
0.5
0
-0.5
-1
0
1
2
3
4
5
6
7
8
9 10
A copy of Program P5_1 is given below:
% Program P5_1 % Illustration of the Sampling Process % in the Time-Domain clf; t = 0:0.0005:1; f = 13; xa = cos(2*pi*f*t); subplot(2,1,1) plot(t,xa);grid xlabel('Time, msec');ylabel('Amplitude'); title('Continuous-time signal x_{a}(t)'); axis([0 1 -1.2 1.2]) subplot(2,1,2); T = 0.1; n = 0:T:1; xs = cos(2*pi*f*n); k = 0:length(n)-1; stem(k,xs);grid; xlabel('Time index n');ylabel('Amplitude'); title('Discrete-time signal x[n]'); axis([0 (length(n)-1) -1.2 1.2])
Q5.4
The plots of the continuous-time signal and its sampled version generated by running Program
P5_1 for the following four values of the sampling period are shown below:
Q5.5 The plots of the continuous-time sinusoidal signal of frequency 3 kHz and its sampled version
generated by running a modified Program P5_1 are shown below: (NOTE: the book is in error about the frequency. Since the “t” in the program is in units of msec, the frequency here is 3 kHz, not 3 Hz.)
Since 0.1 > 1/26, the Nyquist criterion is not satisfied and there is aliasing in the original discrete-time waveform of Q5.1. In Q5.5 with f = 3 kHz, the maximum sampling period that satisfies the Nyquist criterion is 1/6 » 0.1667 msec, Therefore,