Periodic Repetition - Signals and Systems - Exam, Exams for Signals and Systems Theory. Biju Patnaik University of Technology, Rourkela
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rajo1 April 2013

Periodic Repetition - 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: Periodic Repetition, Input the Signal, Exponential, Signal Names, Graphically, Discrete-Time System, Not Linear
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EECS 120. Solutions to Midterm No. 1, February 17, 2000.

1. 20 points Find the expression for the frequency response from x to y in terms of H1,H2,H3 for the system depicted in:

(a) Part (a) of Figure 1. Ans Fix ω ∈ Reals and take as input the signal x : t → X(ω)ejωt. Since H1 is LTI, the other signals are of the form w : t → W (ω)ejωt and y :→ Y (ω)ejωt. Moreover,

W (ω) = X(ω) + Y (ω), Y (ω) = H1(ω)W (ω),

from which we obtain the desired frequency response,

H4(ω) = Y (ω) X(ω) =

H1(ω) 1−H1(ω)

(b) Part (b) of Figure 1. Ans Rewrite part (b) of the figure as (c), and introduce the signal names u, v. Take as input the signal x : t → X(ω)ejωt. Since H1,H2,H3 are all LTI, u, v, y are all exponential,

u : t → U(ω)ejωt, v : t → V (ω)ejωt, y : t → Y (ω)ejωt. From the previous part we know V (ω) = H4(ω)U(ω). And then, following the same argument as in the previous part,

Y (ω) X(ω)

= H4(ω)H3(ω)

1 −H2(ω)H3(ω)H4(ω) .

And substituting for H4 from the previous part,

Y (ω) X(ω) =

H4(ω)H3(ω) 1−H2(ω)H3(ω)H4(ω) =

H1(ω)H3(ω) 1−H1(ω)−H1(ω)H2(ω)H3(ω)

2. 20 points Let f, g, x, y be as in Figure 2.

(a) Determine f ∗ g. Ans Since f is periodic with period 2, f ∗ g is periodic with period 2 and it can be obtained as the periodic repetition of f1 ∗ g where f1(t) = f(t), 0 ≤ t ≤ 2 and f1(t) = 0, otherwise. f1 and f1 ∗ g are shown graphically in the figure.

(b) Determine x ∗ y. Ans Arguing in the same way, x ∗ y is the periodic repetition of x1 ∗ y, with period 1.5, and we can use the calculation of f1 ∗ g as shown in the figure.

3. 20 points Give an example of a discrete-time system H that is:

(a) Not linear; Ans Take H(x)(n) = (x(n))2. This is not linear since H(2x) = 4H(x) = 2H(x).

1

(a)

x y

H2ω)

H3(ω)+

(b)

x H1(ω)+ y

H1(ω)

x y

H2ω)

H3(ω)+

(c)

H1(ω)+

w

H4(ω)

u v

H4(ω)

Figure 1: System for Problem 1

1

1 2-1-2

f

1

0.5

g

1.5-1 0.5

1

1

x

y

1

-1.5

20

f1

20

0.5

0.5 1.0

f1*g

0 1.5

y1

1.50

0.5

0.5 1.0

y1*x

Figure 2: Signals for Problem 2

2

(b) Linear and time-varying; Ans Take H(x)(n) = x(2n). This is not time-invariant, since H(D1x)(n) = x(2n − 1) = D1(Hx)(n) = H(x)(n− 1) = x(2n − 2).

(c) LTI but not causal; Ans Take H(x)(n) = x(n+ 1). This is not causal, because its impulse response is n → δ(n + 1), so that h(1) = 1 = 0.

(d) LTI, causal, but not memoryless. Ans Take H(x)(n) = x(n− 1). This is not memoryless, because the output at n depends on the input at n− 1.

4. 20 points Suppose a periodic signal x : Reals → Comps with fundamental frequency ωx has the Fourier series representation:

∀t, x(t) =

k=−∞ Xke

jkωxt.

(a) Let y be the signal ∀t, y(t) = x(t− τ), where τ is a fixed number. What is the Fourier series representation of y? Ans We have ∀t,

y(t) = x(t− τ) =

k=−∞ Xke

jkωx(t−τ) =

k=−∞ Xke

−jωxτejkωxt

=

k=−∞ Yke

jkωyt,

where

Yk = Xke−jωxτ and ωy = ωx.

(b) Let z be the signal ∀t, z(t) = x(2t). What is the fundamental frequency ωz of z in terms of ωx? What is the Fourier series representation of z? Ans We have ∀t,

z(t) = x(2t) =

k=−∞ Xke

jk2ωxt

k=−∞ Zke

jkωzt

where

ωz = 2ωx and Zk = Xk.

(c) Let w be the signal ∀t, w(t) = z(−t). What is the Fourier series representation of w?

3

Ans We have ∀t,

w(t) = z(−t) =

k=−∞ Zke

−jkωzt

=

k=−∞ Wke

jkωwt

where

ωw = ωz = 2ωx and Wk = Z−k = X−k.

4

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