midterm with solutions

8/13/2019 Midterm With Solutions

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Spring 2013

ELEC 310: Digital Signal Processing I

Midterm (50 Minutes)

The exam is closed-book and closed-notes. However, you are allowed to bring oneformula sheet (8.5 11, single-side) to the exam. Hand in your formula sheettogether with the exam paper.

There are four problems. The problems are fully independent. Some of the questionswithin the problems are also independent.

Clearly show the final answer and the major steps for each question.

By my signature below, I agree to the conditions stated above and I agree that I willwork independently on the questions during the exam.

Name:

ID Number:

Email:

Signature:


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Question 1 (4 points)

Determine the minimum number of bits required to represent each quantized value of thefollowing discrete-time signal

x[n] = 6 sin(0.3n),

such that the maximum quantization error is smaller than 0.1.

Answers:

Needq/2


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Consider a discrete-time LTI system described by the following equation

y[n] =x[n]3x[n 1] +x[n 2] +x[n 3].

(a) Determine the impulse response of the system. Is the system causal?

h[n] =[n]3[n 1] +[n 2] +[n 3].

So the system is causal.

(b) Calculate the output of the system when the signal u[n] u[n 2], where u[n] is theunit step signal, is applied at the input.

Since x[n] =[n] +[n 1], y [n] =h[n] +h[n 1] ={ , 0, 1, 2, 2, 2, 1, 0, }.

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The following continuous-time signal is sampled with sampling period T = 0.01,

x(t) = cos(150t).

(a) Determine the corresponding discrete-time signalx[n].

x[n] =x(t)|t=nT= cos(150nT) = cos(1.5n).

(b) The DT signal from part (a) is processed by a discrete-time LTI system with thefollowing frequency response

H() = 2ej0.5, || .

What is output signal of the system?

The frequency ofx[n] is 1.5, which is greater than .Rewritex[n] as cos(1.5n) = cos(0.5n).

H(0.5) = 2ej0.50.5.

y[n] =|H(0.5)| cos(0.5n + H(0.5)) = 2 cos(0.5n 0.25).

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Consider the system given below with T = 0.001. The discrete-time system in the middle

is an ideal lowpass filter with cutoff frequency at 3/5, i.e. has frequency response H()given by

H() = 1, ||< 3/5;0, 3/5 ||< .

(a) Determine the frequency response of the overall continuous-time system for inputsignal that are band limited to 1000. Note that s= 2000.

H() =H()|=T =

1, ||< 600;

0, 600 ||< 1000.

(b) Assume that the following continuous-time signalxc(t) = 0.5 cos(400t)+sin(800t),is processed by the system. Determine the output signal yr(t).

Note thatYr() =Xc()H(). So the frequency component of 800in xc(t) is removed.yr(t) = 0.5 cos(400t).

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(c Bonus question, 3 points) What is yr(t) ifxc(t) = 0.5 cos(400t) + cos(1600t)?

xc(t) is not bandlimited to s/2. Aliasing occurs.Apply H() to the aliased version of the input signal, which is xc(t) = 1.5 cos(400t).

Therefore, we have yr(t) = 1.5 cos(400t).

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