Weekly Homework 1 Solution
Dr. Sameti
CE 40242: Signals and Systems
Problem 1.
As long as you don’t mess things up, your definition would be accurate. We talked about
this problem in class, and the subject of discussion was the basic concepts and fundamentals
of the course.
Problem 2.
By definition, we have these equations:
CT signals: Et1 −t2 =
R t2
t1
T →∞ −T
N2
P
DT signals: EN1 −N2 =
RT
|x(t)|2 dt ⇒ E∞ = lim
|x(t)|2 dt
N
P
|x[n]|2 ⇒ E∞ = lim
N →∞ n=−N
n=N1
|x[n]|2
(a)
(b)
So, we can write these equations as well:
R t2
CT signals: Pt1 −t2 =
t1
|x(t)|2 dt
t2 −t1
N
P2
DT signals: PN1 −N2 =
RT
−T
⇒ P∞ = lim
T →∞
N
P
|x[n]|2
n=N1
N2 −N1
|x(t)|2 dt
2T
⇒ P∞ = lim
N →∞
(c)
|x[n]|2
n=−N
2N +1
(d)
Based on the equations above, the answers would be:
(1)
Proof. For a periodic signal, the integration and summation part in the equation (a) and (b)
would be the sum of infinite positive elements. So the energy converges to infinity.
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(2)
Proof. If the energy of signal is a finite constant number, then in equation (c) and (d), P∞
will be a fraction in which the numerator is a constant number and the denominator is
converging to infinity. So the whole fraction(P∞ ) is converging to zero.
(3)
Proof. Take signal x(t) = t for example. If you calculate both average energy and average
power for this signal using equations (a) and (c), you can see that both of them have an
infinite value.
(4)
Proof. What we are trying to prove is :
∞
X
2 ?
|x[n]| =
n−∞
∞
X
2
|xe [n]| +
n−∞
∞
X
|xo [n]|2
n−∞
We begin from the left side of the equation:
∞
P
|x[n]|2 =
n−∞
∞
P
2
∞
P
|xe [n] + xo [n]|2 =
n−∞
|xe [n]xo [n]|
∞
P
|xe [n]2 + xo [n]2 + 2xe [n]xo [n]| =
n−∞
∞
P
n−∞
|xe [n]|2 +
∞
P
|xo [n]|2 +
n−∞
(∗)
n−∞
As you see, we have an extra summation term which needs to be cleared out in order
to prove our initial and main claim. Here we prove that for every signal x[n], xe [n]xo [n] is
an odd signal:
xe [−n] × xo [−n] = xe [n] × −xo [n] = −xe [n] × xo [n] By definition of an odd signal, we know that the sum of all values of an odd signal is
∞
P
zero. So the term
|xe [n]xo [n]| equals to zero and it’s cleared out from the equation
n−∞
(*).
Problem 3.
We just show the answer figure of parts (1) and (2) here:
(1)
2
(2)
(3) Signal x[ 23 n − 3] can have values as long as n = 3k. So we check the values of x[n]
for possible values of n:
3

...








n = 0 → x[ 23 n − 3] = x[−3] = 0








n = 3 → x[ 23 n − 3] = x[−1] = 1





n = 6 → x[ 23 n − 3] = x[1] = 1






n = 9 → x[ 23 n − 3] = x[3] = −2








n = 12 → x[ 23 n − 3] = x[5] = 0







...
Now we’re ready to sketch x[ 23 n − 3]
(4) Again we will determine the values of signal nx[n2 ] for different values of n and corresponding values of x[n]:


 ...


n = −2 → nx[n2 ] = −2x[4] = −4




 n = −1 → nx[n2 ] = −x[1] = −1
n = 0 → nx[n2 ] = 0


n = 1 → nx[n2 ] = x[1] = 1




n = 2 → nx[n2 ] = 2x[4] = 4



...
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(5) If you try the same steps we took in the last two parts (evaluating the desired signal
by possible values of n and corresponding x[n]), you will end up sketching the graph below:
Problem 4.
First we represent x(t) as sum of some unit step functions:
x(t) = 2u(t + 1) − 3u(t − 1) + u(t − 2) + u(t − 3)
By the definition of unit impulse function, we have
d
u(t
dt
− k) = δ(t − k) then:
d
x(t) = 2δ(t + 1) − 3δ(t − 1) + δ(t − 2) + δ(t − 3)
dt
So the graph of x0 (t) would look like something like this:
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Problem 5.
√
2π
respectively. Since
(1) fundamental period of signals cos( 2t) and sin( 34 t) are √
and 8π
3
2
the two values don’t have least common multiple, signal x(t) is not periodic.
(2) |cos(5t)| is periodic with the fundamental period of T0 = π5 .
(3) In the class, we talked about how to figure out if a discrete time signal is periodic
or not, and to find its fundamental period if it is. For more details, you can see problem
1.35 in the text book. For signal sin(5n), ω0 = 5, so it can’t be written in the form of 2π ∗ r
where r is a real number.
3π
3
= 2π ∗ ( 28
) and gcd(3, 28) = 1, so T0 for this signal equals to
(4) For signal ej 14 n , ω0 = 3π
14
28.
π
Also for the signal ej 3 n , ω0 = pi3 = 2π ∗ ( 16 ) and gcd(1, 6) = 1, so T0 for this signal equals to
6.
Then the fundamental period of signal x[n] is lcm(6, 28) = 84
(5) For any complex number c = rejθ , |c| = r and ]c = θ. Hence, signal x[n] denotes
π
the absolute value of complex number ejcos( 4 n) which is equal to 1. So, signal x[n] for all
values of n, has constant value of 1, then it’s a periodic signal with T0 = 1.
Problem 6.
In order to determine if system y(t) = T {x(t)} is time invariant, first step is to calculate
y(t − t0 ) and second step is to calculate system’s response to x(t − t0 ) which we call y2 (t).
In other words, in second step, we calculate y2 (t) = T {x(t − t0 )}. If y(t − t0 ) = y2 (t), the
system is time invariant, else it’s not. Same goes for discrete time signals.
In order to determine if system y(t) = T {x(t)} is stable, we should prove that for any
finite values of x(t), the response y(t) would be a finite number. If for any finite values of
x(t), the response converges to infinity, then system is not stable. Same goes for discrete
time signals.
If system y(t) = T {x(t)} is causal, then for any given time, the response y(t) only depends on the values of x(t) at that given time or times before it ,and not times after it.
Same goes for discrete time signals.
(1)
(1.a)Time Invariance: first we calculate y(t − t0 ):
y(t − t0 ) =
R t−t0
t−t0 −1
x(τ )sin(τ )dτ
Now, we calculate the response to x2 (t) = x(t − t0 ).
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y2 (t) =
Rt
t−1
x2 (τ )sin(τ )dτ =
Rt
t−1
x(τ − t0 )sin(τ )dτ =
R t−t0
t−t0 −1
x(τ )sin(τ + t0 )
Clearly y2 (t) 6= y(t − t0 ), then the system is not time invariant.
(1.b)Stability: For any value of x(t) ≤ L, y(t) =
system is stable.
Rt
t−1
x(τ )sin(τ )dτ < L(why?). Then
(1.c)Causality: Response of the system at time t0 , can be obtained only by the values
of x(t) at times [t0 − 1, t0 ]. So the system is causal.
(2)
(2.a)Time Invariance:
y(t − t0 ) = x(t − t0 )cos(x(t − t0 ))
y2 (t) = T {x(t − t0 )} = x(t − t0 )cos(x(t − t0 )
y2 (t) = y(t − t0 ) ⇒ system is TI.
(2.b)Stability: For any values of x(t) ≤ L, y(t) ≤ x(t) ≤ L. So for a bounded input,
the output is also bounded and system is stable.
(2.c)Causality: The response of system at time t0 can be calculated by only knowing the
values of x(t) at time t0 , the system is also causal.
(3)
(3.a)Time Invariance:
y[n − n0 ] = cos(n − n0 + 3)x[n − n0 ]
y2 [n] = T {x[n − n0 ]} = cos(n + 3)x[n − n0 ]
y2 [n] 6= y[n − n0 ] ⇒ system is not TI.
(3.b)Stability: For any bounded value of x[n], the response y[n] is smaller than x[n], then
the response is also bounded, so the system is stable.
(3.c)Causality: To obtain the value of y[n] at any given time n0 , it’s enough to know the
amount of x[n] at that time, then the system is causal.
(4)
(4.a)Time Invariance:
y[n − n0 ] = sin( (n−n4 0 )π )x[n − n0 ] + (n − n0 )2 x[cos( (n−n2 0 )π )]
y2 [n] = T {x[n − n0 ]} = sin( nπ
)x[n − n0 ] + n2 x[cos( nπ
) − n0 ]
4
2
7
y2 [n] 6= y[n − n0 ] ⇒ system is not TI.
(4.b)Stability: Take constant input x[n] = 1 as an example(It’s obvious that input is
) + n2 . As you see, the
bounded!). For this input, the output would be y[n] = sin( nπ
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output is not bounded and by increasing the value of n to infinity, the output converges to
infinity, while the input is constant.
(4.c)Causality: Let’s calculate the value of y[−2]:
y[−2] = sin(− π2 )x[−2] + 4x[cos(−π)] = −x[−2] + 4x[−1]
As you see, the value of y[-2] depends on the value of x[-1], then the system is not causal.
(5)
It’s easy to see that the system is TI, stable and causal.
Problem 7.
Let y1 (t) be the response of system T to x1 (t) and y2 (t) be the response to x2 (t). T will be
linear if and only if:
1. the response to ax1 (t) is ay1 (t). (homogenity)
2. the response to x1 (t) + x2 (t) is y1 (t) + y2 (t). (additivity)
(1)
T {ax(t)} =
ax(t)ejaπx(t)
2j
,ay(t) =
ax(t)ejπx(t)
2j
,T {ax(t)} =
6 ay(t) ⇒ System is nonlinear.
(2)
This system is additive, because if y1 (t) = T {x1 (t)} = Re{x1 (t)} and y2 (t) = T {x2 (t)} =
Re{x2 (t)}, then T {x1 (t)+x2 (t)} = Re{x1 (t)+x2 (t)} = Re{x1 (t)}+Re{x2 (t)} = y1 (t)+y2 (t)
To see if the system is homogeneous or not, let’s take the constant value a = j. Now
we need to check if jy1 (t) = T {jx1 (t)}. We know that y1 (t) is a real number, so jy1 (t) is a
total complex number, but T {jx1 (t)} = Re{jx1 (t)} is a real number, so jy1 (t) 6= T {jx1 (t)},
then system is not homogeneous, therefore nonlinear.
(3)
T {ax[n]} = 3ax[n] + 5
,ay[n] = 3ax[n] + 5a
,T {ax[n]} =
6 ay[n] ⇒ System is nonlinear.
(4)
n
P
T {ax[n]} =
n
P
y1 [n] =
a
ax[n−i]
i=1
n
=
n
P
n
=a
n
n
P
x1 [n−i]
i=1
,
n
P
x[n−i]
i=1
y2 [n] =
x[n−i]
i=1
n
= ay[n] ⇒ system is homogeneous.
n
P
x2 [n−i]
i=1
n
,
T {x1 [n] + x2 [n]} =
8
x1 [n−i]+x2 [n−i]
i=1
n
n
P
x1 [n−i]+
i=1
n
P
i=1
n
n
P
x2 [n−i]
=
n
P
x1 [n−i]
i=1
n
+
x2 [n−i]
i=1
n
= y1 [n] + y2 [n] ⇒ system is additive.
Then the system is linear.
(5)
pQ
y1 [n] = n ni=1 x1 [n − i],
T {x1 [n] + x2 [n]} =
p
Qn
n
i=1
y2 [n] =
p
Qn
n
i=1
x2 [n − i]
x1 [n − i] + x2 [n − i] 6= y1 [n] + y2 [n] ⇒ system is nonlinear.
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