# Outputs - Signals and Systems - Exam, Exams for Signals and Systems Theory. Biju Patnaik University of Technology, Rourkela

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Main points of this exam paper are: Outputs, System, Input, Rounded Boxes, Continuous-Time, Differential Equation, Two Integrators
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EECS 20. Midterm No. 2 Solutions November 8, 2002.

1. 40 points (10 points each part). Consider the Simulink diagram shown below:

1

Out1

1 s

Integrator1

1 s

Integrator

.9

Gain

1

In1

yx

This shows an LTI system with one input and one output, both of which are continuous-time signals. The input and output are indicated by the rounded boxes, and are labeled x and y. The gain is 0.9, and the integrators both have initial condition equal to 0.

(a) Write a differential equation (with no integrals, just derivatives) that relates the input x and the output y. Answer:

∀ t ∈ Reals+, 0.9x(t) = ÿ(t). (b) Give the [A, b, c, d] representation of this system.

Solution: Let the state be the outputs of the two integrators, as follows:

∀ t ∈ Reals+, z(t) = [ y(t) ẏ(t)

]

Hence,

A =

[ 0 1 0 0

] b =

[ 0 0.9

] cT = [1 0] d = 0.

(c) Find the frequency response H:Reals → Reals of this system. Solution: Let the input be x(t) = eiωt, in which case the output must be H(ω)eiωt. Substituting these into the differential equation from part (a), we get

0.9eiωt = H(ω)(−ω2eiωt), which we can solve for H(ω) to get

H(ω) = 0.92. (d) Find the output of the system if the input x is given by

∀ t ∈ Reals, x(t) = cos(2t). Solution: Write this input as

∀ t ∈ Reals, x(t) = 0.5(ei2t + e−i2t), which using linearity implies that the output must be

∀ t ∈ Reals, y(t) = 0.5(H(2)ei2t +H(2)e−i2t) = 0.5((0.9/4)ei2t + (0.9/4)e−i2t) = 0.2 cos(2t).

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2. 30 points (5 points each part). Consider continuous-time systems with input x:Reals → Reals and output y:Reals → Reals. Each of the following defines such a system. For each, indicate whether it is linear (L), time-invariant (TI), both (LTI), or neither (N). Note that no partial credit will be given for these questions.

(a) ∀ t ∈ Reals, ẏ(t) = x(t) + 0.9y(t) Answer: LTI (b) ∀ t ∈ Reals, y(t) = cos(2πt)x(t) Answer: L (c) ∀ t ∈ Reals, y(t) = x(t− 1) Answer: LTI (d) ∀ t ∈ Reals, y(t) = x(t) + 0.1(x(t))2 Answer: TI (e) ∀ t ∈ Reals, y(t) = x(t) + 0.1(x(t − 1))2 Answer: TI (f) ∀ t ∈ Reals, y(t) = 0 Answer: LTI

3. 40 points (10 points each part). Consider a discrete-time signal x: Integers → Reals defined by

∀ n ∈ Integers, x(n) = 1cos(3πn/4).

Assume this signal is sampled at 8,000 samples/second.

(a) Give the frequency of the cosine term in Hz (cycles/second).

Solution: The frequency is 3π/4 radians/sample. Divide by 2π radians/cycle and mul- tiply by 8000 samples/second to get 3000 Hz.

(b) Give period of x.

Solution: The period is the smallest positive integer p such that 3πp/4 is a multiple of 2π. Thus, p = 8.

(c) Give the fundamental frequency (in any units, but be sure to give the units).

Solution: ω0 = 2π/p = π/4 radians/sample. Alternatives: 1/8 cycles/sample, or 8000/8 cycles/second = 1000 cycles/second.

(d) Give the coefficients A0, A1, A2, · · · , AK and φ1, φ2, · · · , φK of the Fourier series ex- pansion for x,

x(n) = A0 + Kk=1

Ak cos(0n+ φk)

where

K =

{ (p− 1)/2 if p is odd p/2 if p is even

Solution: A0 = 1, A1 = 0, A2 = 0, A3 = 1, A4 = 0, and φ1 = 0, φ2 = 0, φ3 = π, φ4 = 0, although the phases corresponding to zero amplitude can be anything. The φ3 = π accounts for the minus sign in the definition of x.

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