matlab-Lab1.doc

Assignment of Signals and Systems Lab 1 1.4 Properties of Discrete-Time SystemsBasic ProblemsIntroduction: In these problems,we need to tell and demonstrate which property a given system does not satisfy.Use MATLAB vectors to represent the inputs and outputs,and plot figures to construct a reasoned argument that can demonstrate our judgement.(a) The system yn=Sin(/2)xn) is not linear.Use the signals x1n= n and x2n=2 n to demonstrate how the system violates linearity.Solution:MATLAB 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);x=2*x1+3*x2;y=2*y1+3*y2;yn=sin(pi/2)*x);subplot(121);stem(n,y);xlabel(n);ylabel(y);title(y=2*y1+3*y2);subplot(122);stem(n,yn);xlabel(n);ylabel(yn);title(yn=sin(pi/2)*x);Basic problem aWe suppose a=1,b=1,x1nsin(pi/2)*x1n),x2n-sin(pi/2)*xn);Suppose xn=x1n+x2n;from the figure: yn is not equel to y;So it is not linear.(b) The system yn=xn+xn+1 is not causal.Use the signals xn=un to demonstrate this .Define the MATLAB vectors x and y to represent the input on the interval -5n9，and the output on the interval -6n9 ,respectively.Solution:MATLAB code:u = inline(n = 0);n = -5:9;x = u(n);y = u(n) + u(n + 1);stem(n, y);axis(-5 9 -6 9);We know that yn is not only determined by xn, but also xn+1.So the system is not causal.Intermediate ProblemsIntroduction: In these problems,we need to discover an input or pair of input signals that can demonstrate why the system does not satisfy the property.(c)The system yn=log(xn) is not stable.Solution:MATLAB code:x=0:100;y=log(x);stem(x,y);So when the input is infinity,the output should be infinity too.So the system is not stable.(d)The system yn=Sin(/2)xn) is not invertible.Solution:MATLAB code:x1=0:100;x2=4:104;y1=sin(0.5*pi)*x1);y2=sin(0.5*pi)*x2);subplot(2,1,1),stem(y1);subplot(2,1,2),stem(y2);So we can see the outputs are the same even the inputs are different.And we know the system is not invertible.Advanced ProblemsIntroduction: In these problems,we need to state whether or not the system is linear,time-invariant,causal,stable,and invertible.And we need to construct a counter-argument for each property that the system is not satisfied.(e)yn=x3nSolution:n=-30:30;x=n;y=x.3;stem(n,y);This system is not linear,stable,but invertible ,causal and time-invariant.This system is not linear because it does not satisfy the equation “yax1+bx2=ayx1+byx2”.This system is not stable because when the input is infinity,the output should be infinity too.(f) yn=n xnSolution:This system is not invertible and time -varying, not stable ,but linear.(1) time-varyingMATLAB code:u = inline(n = 0);n = -3:3;x1 = u(n);x2 = u(n - 1);y1 = n .* x1;y2 = n .* x2;subplot(2, 1, 1),stem(n,y1);subplot(2, 1, 2),stem(n,y2);We know that y2ny1n-1 when x1n= n and x2n=x1n-1So the system is time-varying.(2) not invertibleMATLAB code:n = -6:6;x = n;y = n .* x; stem(n, y);We can find the same output when inputs are different.So the system is not invertible.(g)yn=x 2nSolution:This system is not causal and not invertible ,but time-invariant ,linear and stable.MATLAB code:x = inline(n);n = -6:6;y1 = x(2 * n);y2 = x(n);subplot(2, 1, 1),stem(n, y1);subplot(2, 1, 2),stem(n, y2);We know the output on the interval n0 is not 0,so the system is not causal.We can also find that we cant get xn from x2n,so the system is not invertible.1.5 Implementing a First-Order Difference EquationIntroduction:This assignment is about implementing a First-Order Difference Equation.Advanced Problemsa) Write a function y=diffeqn(a,x,yn1) which computes the output yn of the causal system determined by Eq.(1.6). The input vector x contains sn for 0=n=N-1 and yn1 supplies the value of y-1. The output vector y contains yn for 0=n=N-1. The first line poof your M-file should readfunction y = diffeqn(a,x,yn1)Hint:Note that y-1 is necessary for computing y0, which is the first step of the autoregression. Use a for loop in your M-file to compute yn for successively larger values of n, starting with n=0MATLAB code:function y = diffeqn(a,x,yn1)y(1)=yn1;for n=0:30 y(n+2)=a*y(n+1)+x(n+1);endb) Assume that a =1,y-1=0,and that we are only interested in the output over the interval 0=n=30. Use your function to compute the response due to x1n=n and x2n=un,the unit impulse and unit step, respectively. Plot each response using stem.MATLAB code:a=1;yn1=0;x1=1 zeros(1,30);y=diffeqn(a,x1,yn1);stem(-1:30,y);c) Assume again that a=1, but that y-1=-1. Use your function to compute yn over 0=n=30 when the inputs are x1n=un and x2n=2un.Define the outputs produced by the two signals to y1n and y2n ,respectively. Use stem to display both outputs. Use stem to plot (2y1n-y2n). Given that Eq.(1.6) is a linear difference equation ,why isnt this difference identically zero?MATLAB code:a=1;yn1=-1;x2=1 ones(1,30);y=diffeqn(a,x2,yn1);stem(-1:30,y);Because xn is not identically zero .The system is not linear.MATLAB code:a=1;yn1=-1;x1=1 ones(1,30);y1=diffeqn(a,x1,yn1);x2=2 2*ones(1,30);y2=diffeqn(a,x2,yn1);subplot(3,1,1),stem(-1:30,y1);subplot(3,1,2),stem(-1:30,y2);subplot(3,1,3),stem(-1:30,2*y1-y2);d) The causal systems described by Eq.(1.6) are BIBO(bounded-input bounded-output) stable whenever |a| 1. A property of these stable systems is that the effect of the initial condition becomes insignificant for sufficiently large n. Assume a = 1/2 and that x contains xn=un for 0=n=30. Assuming both y-1=0 and y-1=1/2, compute the two output signals yn for 0=n=3