430.306: Signals and Systems

Electrical and Computer Engineering, Seoul National Univ.Spring Semester, 2016

Homework #2, Due: In class @ April 7

Note: No late homework will be accepted.

Problem 1) Prob. 3.3 of Oppenheim (p.251)Solution:The given signal is

x(t) = 2 +1

2ej(2π/3)t +

1

2e−j(2π/3)t − 2jej(5π/3)t + 2je−j(5π/3)t

= 2 +1

2ej(2π/6)t +

1

2e−j(2π/6)t − 2jej5(2π/6)t + 2je−j5(2π/6)t

From this, we may conclude that the fundamental frequency of x(t) is 2π/6 = π/3. The non-zeroFourier series coefficients of x(t) are:

a0 = 2, a2 = a−2 = 12 , a5 = a∗−5 = −2j

Problem 2) Prob. 3.11 of Oppenheim (p.252)Solution:Since the Fourier series coefficients repeat every N = 10, we have a1 = a11 = 5. Furthermore,since x[n] is real and even, ak is also real and even. Therefore, a1 = a−1 = 5. We are also giventhat

1

10

9∑n=0

|x[n]|2 = 50

Using Parseval’s relation, ∑k=N

|ak|2 = 50

8∑k=−1

|ak|2 = 50

|a−1|2 + |a1|2 + a20 +

8∑k=2

|ak|2 = 50

a20 +

8∑k=2

|ak|2 = 0 (1)

Therefore, ak = 0 for k = 2, . . . , 8. Now using the synthesis eq.(3.94), we have

x[n] =∑

k=<N>

akej 2πNkn =

8∑k=−1

akej 2π10kn

= 5ej2π10n + 5e−j

2π10n

= 10cos(π

5n)

(2)

Problem 3) Prob. 3.12 of Oppenheim (p.252)Solution:Using the multiplication property, we have

x1[n]x2[n] ←→∑

l=<N>

albk−l =3∑

k=0

albk−l

←→ a0bk + a1bk−1 + a2bk−2 + a3bk−3

←→ bk + 2bk−1 + 2bk−2 + 2bk−3

(3)

Since bk is 1 for all values of k, it is clear that bk + 2bk−1 + 2bk−2 + 2bk−3 will be 6 for all valuesof k. Therefore,

x1[n]x2[n]←→ 6, for all k

Problem 4) Prob. 3.13 of Oppenheim (p.253)Solution:Let us first evaluate the Fourier series coefficients of x(t). Clearly, since x(t) is real and odd, akis purely imaginary and odd. Therefore, a0 = 0. Now,

ak =1

8

∫ 8

0x(t)e−j(2π/8)ktdt

=1

8

∫ 4

0e−j(2π/8)ktdt− 1

8

∫ 8

4e−j(2π)/8)ktdt

=1

jπk

[1− e−jπk

](4)

Clearly, the above expression evaluates to zero for all even values of k. Therefore,

ak =

{0, k = 0,±2,±4, . . .2jπk , k = ±1,±3,±5, . . .

When x(t) is passed through an LTI system with frequency response H(jw), the output y(t) isgiven by

y(t) =

∞∑k=−∞

akH(jkω0)ejkω0

where ω0 = 2πT = π

4 . Since ak is non zero only for odd values of k, we need to evaluate the abovesummation only for odd k. Furthermore, note that

H(jkω0) = H(jk(π/4)) =sin(kπ)

k(π/4)

is always zero for odd values of k. Therefore,

y(t) = 0

.

Problem 5) Prob. 3.23 of Oppenheim (p.256)Solution:(a) First let us consider a signal y(t) with FS coefficients

bk =sin(kπ)/4

kπ.

From Example 3.5, we know that y(t) must be a periodic square wave which over one period is

y(t) =

{1, |t| < 1/2

0, 1/2 < |t| < 2

Now, note that b0 = 1/4. Let us define another signal z(t) = −1/4 whose only nonzero FScoefficient is c0 = −1/4. The signal p(t) = y(t) + z(t) will have FS coefficients

dk = ak + ck =

{0, k = 0sin(kπ/4)

kπ , otherwise

Now, note that ak = dkej(π/2)k. Therefore, the signal x(t) = p(t+ 1).

(d) The FS coefficients ak may be written as the sum of two sets of FS coefficients bk andck, where

bk = 1, for all k

and

ck =

{1, k odd

0, k even

The FS coefficients bk correspond to the signal

y(t) =

∞∑k=−∞

δ(t− 4k)

and the FS coefficients ck correspond to the signal

z(t) =∞∑

k=−∞ej(π/2)tδ(t− 2k).

Therefore,

x(t) = y(t) + p(t) =

∞∑k=−∞

δ(t− 4k) +

∞∑k∞

ej(π/2)tδ(t− 2k)

Problem 6) Prob. 3.26 of Oppenheim (p.257)Solution:(a) If x(t) is real, then x(t) = x∗(t). This implies that for x(t) real ak = a∗−k. Since this is not

true in this case problem, x(t) is not real.

(b) If x(t) is even, then x(t) = x(−t) and ak = a−k. Since this is true for this case, x(t) iseven.

(c) We have

g(t) =dx(t)

dt←→ bk = jk

2π

T0ak.

Therefore,

bk =

{0, k = 0

−k(12

)|k| (2πT0

), otherwise

Problem 7) Prob. 3.37 of Oppenheim (p.261)Solution:The frequency response of the system may be easily shown to be

H(ejw) =1

1− 12e−jw −

1

1− 11−2e−jw

(a) The Fourier series coefficients of x[n] are

ak = 14 for all k.

Also, N=4. Therefore, the Fourier series coefficients of y[n] are

bk = akH(ej2kπ/N ) =1

4

[1

1− 12e−jπk/2 −

1

1− 2e−jπk/2

].

(b) In this case, the Fourier series coefficients of x[n] are

ak = 16

[1 + 2 cos

(kπ3

)], for all k.

Also, N=6. Therefore, the Fourier series coefficients of y[n] are

bk = akH(ej2kπ/N ) =1

6

[1 + 2 cos

(kπ

3

)][1

1− 12e−jπk/3 −

1

1− 2e−jπk/3

].

Problem 8) Prob. 3.51 of Oppenheim (p.266)Solution:We have

ej4(2π/8)nx[n] = ejπnx[n] = (−1)nx[n]←→ ak−4

and therefore,(−1)n+1x[n]←→ −ak−4.

If ak = −ak−4, then x[0] = x[±2] = x[±4] = . . . = 0.Now, note that in the signal p[n] = x[n− 1], p[±1] = p[±3] = . . . = 0.

Now let the signal z[n] = (1 + (−1)n)/2.Clearly, the signal y[n] = z[n]p[n] = p[n] because p[n] is zero whenever z[n] is zero. Therefore,y[n] = x[n− 1]. The FS coefficients of y[n] are ake

−j(2π/8).

Problem 9) Prob. 3.56 of Oppenheim (p.269)Solution:(a) We have

x[n]←→ ak and x∗[n]←→ a∗−k.

Using the multiplication property,

x[n]x∗[n] = |x[n]|2 ←→∑

l=<N>

alal+k ∗ .

(b) From above, it is clear that the answer is yes.