信号与系统奥本海姆matlab Lab1.docx

Matlab Lab1 锦豪 明哲 1.4 Basic problem Intrduction:y1[n]=f(x1[n]),y2[n]=f(x2[n]),y[n]=f(ax1[n]+bx2[n]),if y[n]=ay1[n]+by2[n], the system is linear. Let x(n)=0(n0),y(n)=0(n0),the system is causal. (a). The system y[n] = sin((π/2)x[n]) is not linear. Use the signals x1[n] =δ[n] and x2[n] = 2δ[n] to demonstrate how the system violates linearity. Code: n=[-10:10]; x1=[zeros(1,10) 1 zeros(1,10)]; x2=[zeros(1,10) 2 zeros(1,10)]; y1=sin((pi/2)*x1); y2=sin((pi/2)*x2); y3=y1+y2; y4=sin((pi/2)*(x1+x2)); subplot(2,1,1),stem(n,y3); title(the sum of y1[n] and y2[n] when input is x1[n] and x2[n]) subplot(2,1,2),stem(n,y4); title(the input is x1[n]+x2[n]) Figure: solution: the pictured above show the sum of y1[n] and y2[n] when the input is x1[n] and x2[n] ,y1[0]+y2[0]=1 ,the picture under show the output when the input is x1[n]+x2[n],y[0]=-1, y1[0]+y2[0]≠y[0].So the system y[n] = sin((π/2)x[n])is not linear (b). The system y[n] = x[n] + x[n + 1] is not causal. Use the signal x[n] = u[n] to demonstrate this. Define the MATLAB vectors x and y to represent the input on the interval -5 n 9, and the output on the interval -6 n 9, respectively. Code: ni=[-5:9]; no=[-6:9]; x1=[zeros(1,5) ones(1,10)]; x2=[0 x1]; x3=[x1,1]; y=x2+x3; subplot(2,1,1),stem(ni,x1); title(the input x[n]=u[n]) subplot(2,1,2),stem(no,y); title(the output y[n]) Figure: solution: we easily see the y[-1]=1,but x[n]=0 when n=-1.The output of causal system only depends on the moment and the moment before the input to the system.So The system y[n] = x[n] + x[n + 1] is not causal Intermediate Problems (c). The system y[n] = log(x[n]) is not stable. Code: n=[-10:10]; x=[zeros(1,10) 1 zeros(1,10)]; y=log(x); subplot(2,1,1),stem(n,x); title(the input x[n]=u[n]) subplot(2,1,2),stem(n,y); title(the output y[n]) Figure: Solution: the input x[n]=u[n],when n=0 ,then x[0]=1,y[0]=0,otherwise x[n]=0,so log(0) tends to minus infinity , y[n] is unbounded. And stable system has the bounded output. So the