matlab编程:冲击响应,卷积,dtft与dft的比较
11页1、(1) Using impz to calculate y(n): (N=8,16,32,64,128,256) The results of N=8, 16, 32 and 64 are in figure 1.Figure 1. The results of N=128 and 256 are in figure 2.Figure 2. (2) Using conv to calculate y(n): (N=8,16,32,64,128,256) The results of N=8, 16, 32 and 64 are in figure 3.Figure 3. The results of N=128 and 256 are in figure 4.Figure 4. (3) Using fft to calculate y(n): (N=8,16,32,64,128,256) The results of N=8, 16, 32 and 64 are in figure 5.Figure 5. The results of N=128 and 256 are in figu
2、re 6.Figure 6. Differences between impz, conv and fft:Results of impz and conv are very similar to each other, the only difference between this two is the range of frequency: convs is almost two times wider than impzs. Meanwhile the result of fft is quite different from that of impz and conv. When value of N is larger, the waveforms of fft are more similar to that of impz and conv. (4) Explore zero-padding: N=8; K=N, 1.5*N, 2*N and 2*N-1, shown in figure 7.Figure 7. N=16; K=N, 1.5*N, 2*N and 2*N
3、-1, shown in figure8.Figure 8. N=32; K=N, 1.5*N, 2*N and 2*N-1, shown in figure 9.Figure 9. N=64; K=N, 1.5*N, 2*N and 2*N-1, shown in figure 10.Figure 10.N=128; K=N, 1.5*N, 2*N and 2*N-1, shown in figure 11.Figure 11. N=256; K=N, 1.5*N, 2*N and 2*N-1, shown in figure 12.Figure 12. Comment: From above results, we can find that when the value of K is increasing, the result of fft becomes more same like that of conv; and then when K is equal to 2*N-1, both conv and fft provide the same results. Sin
4、ce when K increases, the number of sampling points in the frequency domain increases, which renders smaller frequency interval. Hence, the spectrum density will augment and the results of fft gradually become the same as that of conv. According to the analysis above, we can find when K=256 impz and convs results are same, and when K=2*N-1, fft and convs results are same, thus, the conclusion is ,when K=256 and K=2*N-1, all three methods produce the same result.Code: 1. Using impz to calculate y(
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