《數字信號處理》Matlab實驗一．離散信號的FFT分析1.用Matlab編程上機練習。已知：N=2^5。這里Q=0.9+j0.3?？梢酝茖С?，首先根據這個式子計算X(k)的理論值，然后計算輸入序列x(n)的32個值，再利用基2時間抽選的FFT算法，計算x(n)的DFTX(k)，與X(k)的理論值比較（要求計算結果最少6位有效數字）。解：函數代碼：>>functionxn()>>formatlong>>q=0.9+0.3*i;>>wn=exp(-2*pi*i/32);>>xk=(1-q^32)./(1-q*wn.^[0:31])>>xn=q.^[0:31]>>xk1=fft(xn,32)>>diff=xk-xk1具體執行情況：>>functionxn()formatlongq=0.9+0.3*i;wn=exp(-2*pi*i/32);xk=(1-q^32)./(1-q*wn.^[0:31])xk=Columns1through20.5698+3.3840i2.8369+8.8597iColumns3through49.3189-9.8673i1.2052-3.5439iColumns5through61.8846-2.0941i0.8299-1.2413iColumns7through80.9214-1.0753i0.3150-0.0644iColumns9through100.9240-0.8060i0.4202-0.2158iColumns11through120.8513-0.6357i0.5040-0.1701iColumns13through140.6217-0.6931i0.2441-0.8978iColumns15through160.9454-0.2800i0.7139-0.3158iColumns17through180.6723-0.6496i0.0263+0.5093iColumns19through200.5671+0.6914i0.3173+0.9841iColumns21through220.8929+0.7792i0.4066+0.8452iColumns23through240.5847+0.9017i0.9129+0.9283iColumns25through260.0573+0.5531i0.4219+0.9562iColumns27through280.3298+0.3143i0.4513+0.2638iColumns29through300.7214+0.1879i0.0933+1.7793iColumns31through320.9483+1.9802i0.4932+2.6347i>>xn=q.^[0:31]xn=Columns1through21.00000.0000+0.0000iColumns3through40.0000+0.0000i0.0000+0.0000iColumns5through60.0000+0.0000i-0.0000+0.0000iColumns7through8-0.0000+0.0000i-0.0000+0.0000iColumns9through10-0.0000+0.0000i-0.0000+0.0000iColumns11through12-0.0000-0.0000i-0.0000-0.0000iColumns13through14-0.2000-0.4000i-0.3600-0.5200iColumns15through16-0.9680-0.1760i0.4816-0.3488iColumns17through180.2381-0.2695i0.2951-0.1711iColumns19through200.1169-0.4655i0.4449-0.9838iColumns21through220.6955+0.2480i0.5516+0.4319iColumns23through240.3669+0.5542i0.9639+0.2088iColumns25through260.3049+0.9771i-0.5187+0.4709iColumns27through28-0.6081+0.2682i-0.1278+0.5589iColumns29through30-0.4827+0.0647i-0.6538+0.5134iColumns31through32-0.8425-0.4341i-0.1280-0.1434i>>xk1=fft(xn,32)xk1=Columns1through20.5698+3.3839i2.8366+8.8599iColumns3through49.3182-9.8692i1.2051-3.5439iColumns5through61.8845-2.0942i0.8298-1.2413iColumns7through80.9213-1.0754i0.3150-0.0645iColumns9through100.9240-0.8060i0.4202-0.2158iColumns11through120.8514-0.6356i0.5040-0.1701iColumns13through140.6217-0.6931i0.2441-0.8977iColumns15through160.9454-0.2800i0.7139-0.3159iColumns17through180.6723-0.6496i0.0263+0.5093iColumns19through200.5671+0.6913i0.3172+0.9840iColumns21through220.8929+0.7792i0.4065+0.8452iColumns23through240.5846+0.9016i0.9129+0.9283iColumns25through260.0572+0.5531i0.4219+0.9563iColumns27through280.3297+0.3144i0.4512+0.2638iColumns29through300.7213+0.1879i0.0932+1.7793iColumns31through320.9480+1.9802i0.4928+2.6347i>>diff=xk-xk1diff=1.0e-013*Columns1through20.4625+0.8501i0.9504-0.4003iColumns3through40.6010+0.4028i0.4752+0.7001iColumns5through60.5502+0.8501i0.4625+0.8501iColumns7through80.7751+0.9250i0+0.3875iColumns9through100.7751-0.4625i0.3126-0.4625iColumns11through12-0.4625-0.3126i0.4625+0.3875iColumns13through14-0.9250+0.6938i0.3875-0.0781iColumns15through160.3875-0.6156i0+0.9641iColumns17through180.9250-0.7598i-0.4625-0.0422iColumns19through200.4625+0.1172i0.4625+0.3094iColumns21through220.9250+0.4625i0.9250+0.2313iColumns23through240.3875+0.1563i0.3875-0.2313iColumns25through260.8501-0.9250-0.4625iColumns27through280.0127-0.7751i0.7001-0.9250iColumns29through300.16260.7814-0.9250iColumns31through320.4816+0.9250i0.7255-0.8501i由以上結果可知，由基2時間抽選的FFT算法所得到的DFT結果與利用公式法所得的理論值稍有偏差，但誤差較小，從結果可以看出大概在小數點第15位才開始出現誤差，故而用計算機FFT處理數據在精度上是可以接受的。產生誤差的主要原因可能是在計算機進行FFT算法函數的過程中幾次蝶形運算所得的中間值會產生舍位上的誤差，故而與理論值產生一定的偏差。2.假設信號x(n)由下述信號組成：請選擇合適的長度N和窗函數，用DFT分析其頻譜，得到清楚的三根譜線。解：函數代碼：N=1000;%LengthofDFTn=[0:1:N-1];xn=0.001*cos(0.45*n*pi)+sin(0.3*n*pi)-cos(0.302*n*pi-pi/4);Xk=fft(xn,N);k=[0:1:N/2];subplot(2,1,2);stem(k,abs(Xk(1:1:(N/2+1))));title('