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ECE 314 Homework 6: Signals and Systems - Probs 3.48-3.51, Special Probs 1 & 4, Exercises of Signals and Systems

The solutions to selected problems from ECE 314 - Signals and Systems homework 6. The problems involve calculating the Fourier transforms of given signals, determining the impulse response of a difference equation, and finding the zero-input and zero-initial-condition responses. The solutions are presented analytically and graphically.

What you will learn

  • How to find the Fourier transform X[k] of the given signal x[n]?
  • What is the impulse response h(n) of the given difference equation?
  • What is the fundamental frequency of the given signal x[n]?

Typology: Exercises

2019/2020

Uploaded on 06/08/2020

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ECE 314 Signals and Systems Spring/2006
Solutions to Homework 6
Problem 3.48
(b) x[n] = 2 sin ¡14π
19 n¢+ cos ¡10π
19 n¢+ 1
Solution: The fundamental frequency of the signal above is ωo= 2π/19.The
signal may be rewritten as
x[n] = jej7ωon+jej7ωon1
2ej5ωon+1
2ej5ωon+ 1.
By inspection, we conclude that
X[k] =
1, k = 0,
1/2, k = 5,14
j, k = 7,
j, k = 12,
0,otherwise (for 0 k < 19)
(d) x[n] as depicted in Figure P3.48 (a).
Solution: From the graph we can see that the period of x[n] is N= 8.
X[k] = 1
N
N1
X
n=0
x[n]ejnok
=1
8
7
X
n=0
x[n]ejn π
4k
=1
8£ej2π
4kej6π
4k¤
=1
8£ej6π
4kej6π
4k¤
=j
4sin µ3π
2k
1
FALL 2007
pf3
pf4
pf5

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Download ECE 314 Homework 6: Signals and Systems - Probs 3.48-3.51, Special Probs 1 & 4 and more Exercises Signals and Systems in PDF only on Docsity!

ECE 314 – Signals and Systems – Spring/

Solutions to Homework 6

Problem 3. (b) x[n] = 2 sin

( 14 π 19 n

  • cos

( 10 π 19 n

Solution: The fundamental frequency of the signal above is ωo = 2π/19.The signal may be rewritten as

x[n] = −jej^7 ωon^ + je−j^7 ωon^

ej^5 ωon^ +

e−j^5 ωon^ + 1.

By inspection, we conclude that

X[k] =

1 , k = 0, 1 / 2 , k = 5, 14 −j, k = 7, j, k = 12, 0 , otherwise (for 0 ≤ k < 19)

(d) x[n] as depicted in Figure P3.48 (a). Solution: From the graph we can see that the period of x[n] is N = 8.

X[k] =

N

N∑ − 1

n=

x[n]e−jnΩok

∑^7

n=

x[n]e−jn^ π 4 k

[

e−j^2 π 4 k − e−j^6 π 4 k]

[

ej^6 π 4 k − e−j^6 π 4 k]

j 4 sin

3 π 2 k

Problem 3. (d) X[k] as depicted in Figure P3.49(a).

Solution: We notice from the graph that X[k] is periodic with period N = 7 (Ωo = 2π/7). From the definition, we may write

x[n] =

N∑ − 1

k=

X[k]ejkΩon

= −je−j^8 π 7 n

  • je−j^6 π 7 n .

Problem 3. (a) x(t) = sin(3πt) + cos(4πt).

Solution: The fundamental frequency of the signal above is ωo = π. We may write

x(t) =

2 j

ej^3 πt^ − e−j^3 πt

ej^4 πt^ + e−j^4 πt

Hence, we have

X[k] =

j/ 2 , k = − 3 , −j/ 2 , k = 3 1 / 2 , k = − 4 , 4 0 , otherwise.

(d) x(t) as depicted in Figure P3.50(a).

Solution: From the graph, we can see that T = 1. We can calculate X[k] using the formula over the period [0, 1]:

X[k] =

0

sin(πt)e−j^2 πktdt

2 j

0

ej(π−^2 πk)t^ − e−j(π+2πk)t

dt

=

π − 2 πk

π + 2πk

  • when n + 2 ∈ [0, 10] ⇒ n ∈ [− 2 , 8],

z(n) = 2

∑^ n+

k=

(n − k + 2))

∑^ n+

j=

j

= 2(n + 3)(n + 2)/ 2 = n^2 + 5n + 6;

  • when n − 16 < 0 and n + 2 > 10 ⇒ n ∈ (8, 16),

z(n) = 2

∑^10

k=

(n − k + 2))

∑^ n+

j=n− 8

j

= 2 · 11(2n − 6)/ 2 = 22 n − 66;

  • when n − 16 ∈ [0, 10] ⇒ n ∈ [16, 26],

z(n) = 2

∑^10

k=n− 16

(n − k + 2))

∑^18

j=n− 8

j

= 2(n + 10)(−n + 27)/ 2 = −n^2 + 17n + 270;

  • and when n − 16 > 10 ⇒ n > 26, z(n) = 0.

(b) What are the values of n for which z(n) 6 = 0? Solution: As we have seen from the previous item, z(n) 6 = 0 when n ∈ [− 2 , 26].

Special Problem 4 Consider the difference equation

y(n) =

y(n − 1) −

y(n − 2) + u(n) +

)n u(n)

with initial conditions y(−2) = 1 and y(−1) = 3.

(a) Find the zero-input response y 0 x(n). Solution: If we consider that the input is zero, the difference equation above becomes the homogeneous equation:

y(n) =

y(n − 1) −

y(n − 2).

Now, let us suppose that the response for the homogeneous equation is y 0 x(n) = rn, where r is a constant. Then, we have

rn^ =

rn−^1 −

rn−^2 ,

and from that expression, we can derive the characteristic equation

r^2 −

r +

The roots of this characteristic equation are: r 1 = 1 and r 2 = 1/3. Since, any linear combination of the responses y 1 (n) = rn 1 and y 2 (n) = r 2 n is also a response to the homogeneous equation, we may write the zero-input response as

y 0 x(n) = c 1 + c 2

)n ,

where c 1 and c 2 are constants. To determine the zero-input response corresponding to the initial condi- tions described in the problem statement, we have to determine the constants c 1 and c 2. This can be done by evaluating y 0 x(n) for n = − 2 , −1: { h(−1) = c 1 + 3c 2 = 3 h(−2) = c 1 + 9c 2 = 1

c 1 = 4 c 2 = −^13

where x(n) is the input signal, given by

x(n) = u(n) +

)n u(n).

Hence,

y 0 ic(n) =

[

u(n) −

)n u(n)

]

[

u(n) +

)n u(n)

]

We can, then, apply the distributive property of the convolution with respect to addition, and use the following identities:

  1. u(n) ∗ u(n) = (n + 1)u(n);
  2. u(n) ∗ αnu(n) = 1 − 1 α−nα+1 u(n);
  3. αnu(n) ∗ αnu(n) = (n + 1)αnu(n). Thus,

y 0 ic(n) =

u(n) ∗ u(n) +

u(n) ∗

)n u(n)

)n u(n) ∗ u(n) −

)n u(n) ∗

)n u(n)

(n + 1)u(n) +

[

1 − (1/3)n+

]

u(n) −

)n (n + 1)u(n)

(d) Find and plot the total response y(n) using the results obtained in (a)- (c). Evaluate y(n) at n = 0,... , 5. Compute y(n) interactively (directly from the difference equation) for n = 0,... , 5, and compare to the values obtained from the total solution. Solution: The total solution is

y(n) = y 0 x(n) + y 0 ic(n)

= 4 −

)n+

(n + 1)u(n) +

[

1 − (1/3)n+

]

u(n)

)n (n + 1)u(n) (1)

The results have been calculated using MATLAB (see file hw6 prob4 d.m). The graph below shows the results.