Question 1 (sampling, aliasing, and reconstruction)Consider the continuous time function given by the input signal,

xa(t) = 3 cos(70πt+ 0.6π) + 2 sin(240πt).

a) What is the Nyquist rate for this signal?Solution: The Nyquist rate is

FNyq = 2 maxF1, F2= 2 maxΩ1/(2π),Ω2/(2π)= 2 max70π/(2π), 240π/(2π)= 240.

b) The signal xa(t) is sampled at Fs = 200 samples per unit time. Write anequation for the discrete time sampled signal, x(n).Solution: The discrete time signal is computed by substituting t = n/Fs,

x(n) = xa(t)|t=n/Fs

= 3 cos(70π(n/200) + 0.6π) + 2 sin(240π(n/200))

= 3 cos(0.35πn+ 0.6π) + 2 sin(1.2πn)

= 3 cos(0.35πn+ 0.6π) + 2 sin((1.2− 2)πn)

= 3 cos(0.35πn+ 0.6π) + 2 sin(−0.8πn)

= 3 cos(0.35πn+ 0.6π)− 2 sin(0.8πn).

c) Write MATLAB code to plot the discrete time sampled signal, x(n). Youcan assume that the n vector comprised of time indices has already beendefined. Based on n, you need to compute x(n) and then plot it.Solution: The problem can be solved as follows. (Many students forgotasterisks or wrote π instead of pi.)

x=3*cos(0.35*pi*n+0.6*pi)-2*sin(0.8*pi*n);

plot(n,x);

d) Was xa(t) aliased when sampling at Fs = 200 samples per unit time?Make sure to justify your answer.Solution: Because Fs = 200 < FNyq = 240, aliasing happens. This can beseen in part (b), where sin(1.2πn) becomes − sin(0.8πn).e) The discrete time signal x(n) is passed through an ideal digital to analog(D/A) converter at the same sampling rate of Fs = 200 samples per unittime. Determine the continuous time output, xa(t).Solution: The first component in x(n), 3 cos(0.35πn+0.6π), was not aliased.Therefore, ideal D/A conversion returns the original component 3 cos(70πt+0.6π). This result corresponds to substituting n = 200t. Therefore, thesecond component in the discrete time signal x(n), −2 sin(0.8πn), becomes−2 sin(160πt). In summary,

xa(t) = 3 cos(70πt+ 0.6π)− 2 sin(160πt).

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Question 2 (properties of systems)Consider the discrete time system, y(n) = T [x(n)] = x(n2).

a) Is the system time invariant? If yes, please sketch the system block di-agram using a block diagram (multipliers, adders, z−1 delay blocks, and soon). If not, please justify your statement.Solution: The system is time variant. That is, it is not time invariant. Tosee this, it suffices to provide a counter example. In our counter example,shifting the original input, x1(n), will not result in a shift of y1(n), the outputcorresponding to the original input.Consider x1(n) = δ(n). The input is nonzero only when n = 0. Therefore, theoutput y1(n) is nonzero when n2 = 0, meaning that n = 0. For other indicesn, n2 is nonzero, and y2(n) becomes zero in these cases. To summarize,x1(n) = δ(n) and y1(n) = δ(n).Next, consider a modified input, x2(n) = δ(n − 1). The input is one if andonly if n = 1, else the input is zero. The output, y2(n), is one if and onlyif n2 = 1, meaning that n ∈ −1,+1, else the output is zero. Becausey2(n) = 1 for n ∈ −1,+1, else zero, we can express the output as y2(n) =δ(n− 1) + δ(n+ 1).We see that x2(n) is a shifted version of x1(n). In contrast, y2(n) is not ashifted version of y1(n), because y2 is nonzero at two indices, whereas y1 isnonzero only at one time index. This counter example shows that the systemis time variant.b) Consider a linear time invariant (LTI) system whose transfer function inthe Fourier domain is given by H(ω) = Y (ω)/X(ω) = 1−e−jw. Is this systembounded input bounded output (BIBO) stable? Justify your answer. (Hint:recall that the transfer function in the Fourier domain is closely related tothat in the z domain.)Solution: First, we convert from the Fourier to z-transform domain viaz = ejω,

H(z) = 1− z−1.

The region of convergence (ROC) of H includes the entire complex planeexcept for the origin (the ROC does not include z = 0). Therefore, the ROCis ROCH = z : z 6= 0. We know that H is BIBO stable if its ROC includesthe unit circle. Because the ROC includes the entire complex plane (exceptfor z = 0), it includes the unit circle, and so H is BIBO stable.Another way to see that H is BIBO is to take the inverse z-transform ofH(z), yielding h(n) = δ(n)− δ(n− 1). We can express this as h = 1− 1.Because ‖h‖1 = |1|+ | − 1| = 2 <∞, H is BIBO stable.

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