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
( 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
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:
- u(n) ∗ u(n) = (n + 1)u(n);
- u(n) ∗ αnu(n) = 1 − 1 α−nα+1 u(n);
- 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.
