信号与系统奥本海姆-Lab3.docx

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3.5 Synthesizing Signals with the Discrete-Time Fourier Series The discrete-time Fourier series (DTFS) is a frequency-domain representation for periodic discrete-time sequences. The synthesis and analysis equations for the DTFS are given by Eqs. (3.1) and (3.2). This exercise contains three sets of problems to give you practice working with these equations. The Basic Problems allow you to synthesize a very simple periodic discrete-time signal from its DTFS coefficients. In the Intermediate Problems, you will both analyze a set of periodic discrete-time signals to obtain their DTFS coefficients, and construct one of these signals by adding in a few coefficients at a time. For the Advanced Problem, you will write a function to find the DTFS coefficients of an arbitrary periodic discrete-time signal from one period of samples. Basic Problems In these problems, you will synthesize a periodic discrete-time signal with period N = 5 and the following DTFS coefficients (a).Based on the DTFS coefficients,do you expect x[n] to be complex-valued, purely real, or purely imaginary? Why? Code: N=[0:4]; a=[1 2*exp(-i*pi/3) exp(i*pi/4) exp(-i*pi/4) 2*exp(i*pi/3)]; x=length(N)*ifft(a); subplot(2,1,1),plot(N,real(x),'*-') ,xlabel('N') ,ylabel('real(x)') subplot(2,1,2),plot(N,imag(x),'*-') ,xlabel('N') ,ylabel('imag(x)') So the x[n] is purely real. (b). Using the DTFS coefficients given above, determine the values of a0 through a4 and specify a vector a containing these values. Solution: a(0)=1, a(2)=a(-2)*=exp(i*pi/4),so a(-2)=exp(-i*pi/4), and the period N=5 a(3)=a((-2)+5)=a(-2) =exp(-i*pi/4). a(4)=a(-4)*=2*exp(i*pi/3),so a(-4)=2*exp(-i*pi/3) a(1)=a((-4)+5)= a(-4)=2*exp(-i*pi/3); a=[1 2*exp(-i*pi/3) exp(i*pi/4) exp(-i*pi/4) 2*exp(i*pi/3)]; (c). Using the vector a of DTFS coefficients and the synthesis equation, define a new vector x containing one period of the signal x[n] for 0 <= n<= 4. You can either write out the summation explicitly or you may find

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