EE1101 Signals and Systems Jan-May 2019

Tutorial 2 Solutions

Solution 1 N1 7
k = 1) respectively. Since = is a
N2 3
(a) Let x1 (t) and x2 (t) be periodic with periods rational number, x[n] is periodic.
T1 and T2 respectively. such that The fundamental period would be N =
L.C.M(7, 3) = 21
x1 (t + T1 ) = x1 (t)

x2 (t + T2 ) = x2 (t) Solution 2
1
For x(t) = x1 (t) + x2 (t) to be periodic, To prove δ(at) = δ(t)
|a|
Let g1 (t) and g2 (t) be generalized functions. Then
x(t + T ) = x1 (t + T ) + x2 (t + T )
the equivalence property states that
Z ∞ Z ∞
=⇒ x(t + T ) = x1 (t + mT1 ) + x2 (t + kT2 )
g1 (t) = g2 (t) ⇐⇒ φ(t)g1 (t)dt = φ(t)g2 (t)dt
So, −∞ −∞
mT1 = kT2 = T for all suitably defined R ∞testing functions φ(t).
T1 k Consider the integral −∞ φ(t)δ(at)dt
= = rational number
T2 m Let τ = at. Consider the case when a > 0 ⇒ τ =
In other words, the sum of two periodic at = |a|t ⇒ dτ = |a|dt
signals is periodic only if the ratio of their Z ∞
1
Z ∞
τ
respective periods can be expressed as a φ(t)δ(at)dt = φ( )δ(τ )dτ
−∞ |a| −∞ a
rational number. Then the fundamental
period is the least common multiple of T1 1 τ
= φ( )
and T2 i.e. T = L.C.M {T1 , T2 } |a| a
τ =0
1
= φ(0)
|a|
(b) (a) Let x(t) = x1 (t) + x2 (t) where
x1 (t) = cos(10πt + 1) and x2 (t) =
−2sin(4πt − 1).
Now, consider the case when a < 0 ⇒ τ = at =
x1 (t) and x2 (t) are periodic with peri-
2π 2π −|a|t ⇒ dτ = −|a|dt
ods given by T1 = and T2 =
10π 4π Z ∞
1
Z −∞
τ
respectively. φ(t)δ(at)dt = − φ( )δ(τ )dτ
T1 2 −∞ |a| ∞ a
Since = is a rational number, Z ∞
T2 5 1 τ
x(t) is periodic. = φ( )δ(τ )dτ
|a| −∞ a
The fundamental period is T = L.C.M
2π 2π 1 τ
{ , }=1 = φ( )
10π 4π |a| a
τ =0
1
(b) For a discrete time signal, the same = φ(0)
|a|
concept as in Q1 can be applied.
Each of the complex exponentials are
2πk
periodic with periods N1 = 4π = 7 Thus for any a,
7 Z ∞
2πk 1
(for k = 2) and N2 = 2π = 3 (for φ(t)δ(at)dt = φ(0)
3 −∞ |a|

1
 R∞
But we know φ(0) = −∞ φ(t)δ(t)dt (d)
∞ Z ∞
1
Z
φ(t)δ(at)dt = φ(t)δ(t)dt
Z ∞
−∞ |a| −∞ y= sin(πt)δ(2t − 3)dt
Z ∞ −∞
1
= φ(t) δ(t)dt Let2t − 3 = u =⇒ du = 2dt
−∞ |a|
1 ∞
 
π(u + 3)
Z
Now by the equivalence property mentioned be- y= sin δ(u)du
2 −∞ 2
fore, 1

π(0 + 3)
Z ∞
1 = sin δ(u)du
δ(at) = δ(t) 2 2 −∞
|a|
1
=−
2
Solution 3
For a Dirac Delta function δ(t), the following two
(e)
properties hold true:

x(t)δ(t − t0 ) = x(t0 )δ(t − t0 ) Z ∞

Z ∞ y= e−t δ(t + 3)dt
δ(τ )dτ = 1 −∞
−∞ Let t + 3 = u
Z ∞
(a) y = e3 e−u δ(u)du
−∞
∞
= e3
Z
y(t) = x(τ )δ(t − τ )dτ
Z−∞
∞
= x(t)δ(t − τ )dτ
−∞ (f)
Z ∞
= x(t) δ(t − τ )dτ
Z−∞
∞
Z ∞
= x(t) δ(τ )dτ y= (t3 + 4)δ(1 − t)dt
−∞ −∞
Z ∞
= x(t) 3
= (1 + 4) δ(1 − t)dt
−∞
Z ∞
(b) =5 δ(t)dt
−∞
Z ∞ =5
y(t) = x(t − τ )δ(τ )dτ
−∞
Let t − τ = u
Z ∞
(g)
y(t) = x(u)δ(t − u)du
−∞
Z ∞
Z ∞
= x(t) δ(t − u)du
−∞ y= x(2 − t)δ(3 − t)dt
−∞
= x(t)
2−t=u
Z ∞
(c) y= x(u)δ(1 + u)du
−∞
Z ∞
Z ∞
y= e−jωt δ(t)dt y = x(−1) δ(1 + u)du
−∞
−∞ Z ∞
Z ∞
=e −jω∗0
δ(t)dt y = x(−1) δ(u)du
−∞
−∞
=1 = x(−1)

2
 (h) (b) x(t) can be expressed as follows.
∞  
π
Z
x−1
y= e cos (x − 5) δ(x − 3)dx
−∞ 2 [x(t) = u(t) − δ(t − 1) − δ(t − 2) − δ(t − 3)
x−1 4 x−5
Let e =e e
Z ∞  
Z t Z t Z t
4 x−5 π x(t)dt = u(t)dt − δ(t − 1)dt
y=e e cos (x − 5) δ(x − 3)dx −∞ −∞ −∞
−∞ 2 Z t Z t
x−5=u − δ(t − 3)dt − δ(t − 3)dt
∞   −∞ −∞
π
Z
4 u
y=e e cos (u) δ(u + 2)du = r(t) − u(t − 1)
−∞ 2

π
Z ∞ − u(t − 2) − u(t − 3)
4 −2
y = e ∗ e cos (−2) δ(u + 2)du
2 −∞
Z ∞
y = −e2 δ(u)du Rt
−∞ −∞ x(t)dt
2
= −e
2

Solution 4 1
(a) x(t) can be expressed as follows.
-1 0 1 2 3 4 t
x(t) = u(t) − 2u(t − 1) + u(t − 3) + δ(t − 3)
-1
Z t Z t Z t
x(t)dt = u(t)dt − 2 u(t − 1)dt -2
−∞ −∞ −∞
Z t Z t
+ u(t − 3)dt + δ(t − 3)dt
−∞ −∞
= r(t) − 2r(t − 1) + r(t − 3) + u(t − 3) 

0, if t≤0

t, if 0≤t<1


Z t 
Rt x(t)dt = t − 1, if 1≤t<2
−∞ x(t)dt −∞ 
t − 2, if 2≤t<3





t − 3, if t≥3

2

1 Solution 5
• Suppose y1 (t) is the output of a system to
-1 0 1 2 3 4 t input x1 (t) and y2 (t) is the output to input
x2 (t).The system is linear if the response of
-1
the system to the input ax1 (t) + bx2 (t) is
ay1 (t) + by2 (t) (a and b can be complex in
-2
general).
• Suppose y(t) is the output of a system to
the input x(t). The system is time invari-
Thus, ant if the output of the system to the input
 x(t − t0 ) is y(t − t0 ).
0, if t≤0
• A system is causal if the output of the sys-


Z t 
t, if 0≤t≤1 tem at any time depends only on the present
x(t)dt =
−∞ 

2 − t, if 1≤t<3 and past values of the input and not on the
t≥3 future values of input.

0, if

3
 • A system is said to be stable if all bounded • If a system is invertible then an inverse sys-
inputs to the system result in bounded out- tem exists which when cascaded with the
puts. original system gives an output equal to the
input to the first system.

Qn Linear Time invariant Causal Stable Invertible

dx
(a)y(t) = dt Yes Yes Yes No No
R 3t
(b)y(t) = −∞ x(τ )dτ Yes No No No Yes
(c)y(t) = x(t/2) Yes No No Yes Yes

x(t) − x(t − 100) t ≥ 0
(d)y(t) = Yes Yes Yes Yes No
0 otherwise
P∞
(e)y(t) = n=−∞ x(t)δ(t − nT ) Yes No Yes No No
(f)y(t) = x(2t − 4) Yes No No Yes Yes

The detailed explanations are given below: Consider the following input to the system.

1 −1 ≤ t ≤ 1
(a) x(t) =
dx(t) 0 otherwise
y(t) =
dt The output of the system to the above in-
Suppose y1 (t) is the output of a system to put will go unbounded at -1 and +1. Hence
input x1 (t) and y2 (t) is the output to input the system is unstable.
x2 (t). Let y(t) be the output of the system
to the input ax1 (t) + bx2 (t). Then The system is not invertible since differ-
entiating any constant will give 0 and hence
d
y(t) = (ax1 (t) + bx2 (t)) different inputs give the same output.
dt
d d (b)
=a (x1 (t)) + b (x2 (t))
dt dt Z 3t
y(t) = x(τ )dτ
= ay1 (t) + by2 (t) −∞

Hence the system is linear. Suppose y1 (t) is the output of a system to

input x1 (t) and y2 (t) is the output to input
Suppose y(t) is the output of a system to x2 (t). Let y(t) be the output of the system
input x(t). Let y1 (t) be the output of the to the input ax1 (t) + bx2 (t).
system to the input x(t − t0 ). Then Z 3t
y(t) = ax1 (τ ) + bx2 (τ )dτ
d −∞
y1 (t) = (x(t − t0 ))
dt Z 3t Z 3t
d =a x1 (τ )dτ + b x2 (τ )dτ
= (x(t − t0 )) −∞ −∞
d(t − t0 )
= ay1 (t) + by2 (t)
= y(t − t0 ) (∵ d(t − t0 ) = dt)
Hence the system is linear.
Hence the system is time invariant. Suppose y(t) is the output of a system to
input x(t). Let y1 (t) be the output of the
dx x(t) − x(t − dt) system to the input x(t − t0 ).
y(t) = = lim Z 3t
dt dt→0 dt
y1 (t) = x(τ − t0 )dτ
Since the output of the system depends on −∞
the present and past values of the input τ − t0 = u
alone the system is causal.
hence dτ = du

4
 Z 3t−t0
not causal.
= x(u)du
−∞
But For all bounded values of the input we get
Z 3t−3t0
y(t − t0 ) = x(τ )dτ bounded values at the output. Hence the
−∞ system is stable.
Hence
y(t − t0 ) 6= y1 (t) The system is invertible since the inverse
Hence the system is time variant. of the system exists which is y(t) = x(2t).

For t > 0 the output of the system depends (d)

on the future values of the input hence the 
system is not causal. x(t) − x(t − 100) t≥0
y(t) =
0 otherwise
Consider an input x(t) = 1 for all val- Suppose y1 (t) is the output of a system to
ues of t.The output of the system goes un- input x1 (t) and y2 (t) is the output to input
bounded.Hence the system is unstable. x2 (t). Let y(t) be the output of the system
to the input ax1 (t) + bx2 (t).
If two different inputs give the same output 
to the above system their values should be ax1 (t) + bx2 (t) − ax1 (t − 100) − bx2 (t − 100)
y(t) =
the same at every t since the limit of the in- 0
tegral depends on t. Hence the two inputs 
have to be the same. Hence the system is a(x1 (t) − x1 (t − 100)) + b(x2 (t) − x2 (t − 100))
y(t) =
invertible. 0
y(t) = ay1 (t) + by2 (t)
(c)
y(t) = x(t/2) Hence the system is linear.
Suppose y1 (t) is the output of a system to
input x1 (t) and y2 (t) is the output to input Suppose y(t) is the output of a system to
x2 (t). Let y(t) be the output of the system input x(t). Let y1 (t) be the output of the
to the input ax1 (t) + bx2 (t). system to the input x(t − t0 ).

y(t) = ax1 (t/2) + bx2 (t/2) x(t − t0 ) − x(t − t0 − 100) t≥0
y1 (t) =
0 otherwise
= ay1 (t) + by2 (t)
But
Hence the system is linear. 
Suppose y(t) is the output of a system to x(t − t0 ) − x(t − t0 − 100) t ≥ t0
y(t−t0 ) =
input x(t). Let y1 (t) be the output of the 0 otherwise
system to the input x(t − t0 ). Hence the system is time invariant.
y1 (t) = x(t/2 − t0 )
The output of the system depends on the
But present and past value of the input .Hence
y(t − t0 ) = x((t − t0 )/2) the system is causal.
= x(t/2 − t0 /2)
The system is stable since bounded inputs
Hence
to the system give bounded outputs.
y(t − t0 ) 6= y1 (t)
Hence the system is time variant. Consider the constant input function. The
constant can be of any value. But they all
For negative values of t we find that the give the same zero output at all values of t.
output of the system depends on the future Hence the system is not invertible.
values of the input. Hence the system is

5
(e) (f)
∞
X y(t) = x(2t − 4)
y(t) = x(t)δ(t − nT )
n=−∞ Suppose y1 (t) is the output of a system to
input x1 (t) and y2 (t) is the output to input
Suppose y1 (t) is the output of a system to x2 (t). Let y(t) be the output of the system
input x1 (t) and y2 (t) is the output to input to the input ax1 (t) + bx2 (t).
x2 (t). Let y(t) be the output of the system
to the input ax1 (t) + bx2 (t). y(t) = ax1 (2t − 4) + bx2 (2t − 4)
= ay1 (t) + by2 (t)
Hence the system is linear.
∞
X Suppose y(t) is the output of a system to
y(t) = (ax1 (t) + bx2 (t))δ(t − nT )
n=−∞
input x(t).Let y1 (t) be the output of the
system to the input x(t − t0 ).
∞
y1 (t) = x(2t − 4 − t0 )
X
= ax1 (t)δ(t − nT ) + bx2 (t)δ(t − nT )
n=−∞
y(t − t0 ) = x(2(t − t0 ) − 4)
= ay1 (t) + by2 (t)
Hence y(t − t0 ) 6= y1 (t) Hence the system is
Hence the system is linear. time variant.

Suppose y(t) is the output of a system to For all values of t > 4 the output of the
input x(t).Let y1 (t) be the output of the system depends on the future value of the
system to the input x(t − t0 ). input. Hence the system is not causal.

∞
X The system is stable since bounded inputs
y1 (t) = x(t − t0 )δ(t − nT )
n=−∞
give bounded outputs. The system per-
forms only a time-shifting and time-scaling
But operation
∞
X
y(t − t0 ) = x(t − t0 )δ(t − t0 − nT ) The system is invertible since the inverse
n=−∞ of the system exists which is y(t) = x( 2t +2).

Hence y(t − t0 ) 6= y1 (t) Hence the system is

time variant. Solution 6
We have,
The system represents an ideal sampler. y[n] = x[n]x[n − 2]
The sampler output has non-zero values
only at t = nT which is equal to the value of (a) The system is not memoryless. Since, the
the input x(t) at the same instant. Hence, output at any time instant n depends on the
it is causal. input at the time instant n-2.

(b) When x[n] = Aδ[n]

The system is not stable since output is a
sequence of impulses and impulse function y[n] = Aδ[n]Aδ[n − 2] = 0
takes an unbounded value.
Aδ[n] is non-zero only at n = 0 and Aδ[n−2]
is non-zero only at n = 2. When both are
The system is not invertible since the in- multiplied, we get zero for all n values.
puts can be different but if they have the
same value at nT then they give the same (c) The system is clearly non-invertible.
output. To prove non-invertibility, we need one
counter-example, which is part (b) itself.

6
Solution 7 x[n] − x[−n]
(c) y[n] = Odd{x[n]} = : System
2
(a) y[n] = x2 [n−2] : System is non-linear and is linear and time variant.
time-invariant. Check for time-invariance: Delay the
Check for Linearity:Let y1 [n] be the re- input x[n] by n0 . Then y1 [n] =
x[n − n0 ] − x[−n − n0 ]
sponse of x1 [n], y2 [n] be the response of . Now delay-
x2 [n] and y[n] be the response of the com- 2
ing the output y[n] by n0 to get
bined input x[n] = ax1 [n] + bx2 [n] . x[n − n0 ] − x[−(n − n0 )]
y[n − n0 ] = =
2
y1 [n] = x21 [n − 2] x[n − n0 ] − x[−n + n0 ]
. Hence the system
2
is time variant.
y2 [n] = x22 [n − 2]
Solution 8
y[n] = x2 [n − 2] Let y(t) be the output of the system for input
= (ax1 [n − 2] + bx2 [n − 2])2 x(t) = est . Then,
= a2 x21 [n − 2] + b2 x22 [n − 2] H{est } = y(t)
+ 2abx1 [n − 2]x2 [n − 2]
6= ay1 [n] + by2 [n] Since the system is time-invariant, we have

H{es(t+t0 ) } = y(t + t0 )
The given system is non-linear.
Check for time-invariance: Delay the input for arbitrary real t . Since the system is linear,
0
x[n] by n0 . Then y1 [n] = x2 [n − 2 − n0 ]. we have
Now delaying the output y[n] by n0 to get
y[n − n0 ] = x2 [n − 2 − n0 ] = y1 [n]. Hence H{es(t+t0 ) } = H{est est0 } = est0 H{est } = est0 y(t)
the system is time invariant.
Thus,
(b) y[n] = x[n + 1] − x[n − 1] : System is linear y(t + t0 ) = est0 y(t)
and time invariant.
Setting t = 0, we obtain
Check for Linearity:Let y1 [n] be the re-
sponse of x1 [n], y2 [n] be the response of y(t0 ) = y(0)est0
x2 [n] and y[n] be the response of the com-
Since t0 is arbitrary, by changing t0 to t, we can
bined input x[n] = ax1 [n] + bx2 [n] .
rewrite the above equation as
y1 [n] = x1 [n + 1] − x1 [n − 1]
y(t) = y(0)est = λest

=⇒ H{est } = λest
y2 [n] = x2 [n + 1] − x2 [n − 1]
where λ = y(0)

y[n] = x[n + 1] − x[n − 1]

Solution 9
= ax1 [n + 1] − ax1 [n − 1]
+ bx2 [n + 1] − bx2 [n − 1] The signal x2 (t) can be represented in terms of
= ay1 [n] + by2 [n] the given signal x1 (t) as follows:

x2 (t) = x1 (t) − x1 (t − 2)
The given system is linear.
Check for time-invariance: Delay the input Given that y1 (t) is the response to the input x1 (t),
x[n] by n0 . Then y1 [n] = x[n + 1 − n0 ] − let y2 (t) be the response to the input x2 (t). Since
x[n − 1 − n0 ]. Now delaying the output y[n] the given system is an LTI system, y2 (t) is given
by n0 to get y[n−n0 ] = x[n+1−n0 ]−x[n− as
1 − n0 ] = y1 [n]. Hence the system is time
invariant. y2 (t) = y1 (t) − y1 (t − 2)

7
 y2 (t)
1
2
0.5

y(t)
0

−0.5

0 1 2 3 4 t
−1
−2 −1 0 1
ω = 8π ω = 10π ω = 8π
t→

-2
(b) – Let y1 (t) be the response of the system
to m1 (t) and y2 (t) be the response of
Solution 10 the system to m2 (t).
 Z t 
(a) y1 (t) = A cos ωc t + ω∆ m1 (τ )dτ
−∞
 Z t 
y(t) = A cos ωc t + ω∆ m(τ )dτ  Z t 
−∞
y2 (t) = A cos ωc t + ω∆ m2 (τ )dτ
−∞
Given ωc = 8π, ω∆ = 2π and m(t) =
u(t + 2) − u(t − 1). Consider y(t) be the response of the
 Z t  system to the input am1 (t) + bm2 (t).


y(t) = A cos 8πt + 2π u(τ + 2) − u(τ − 1) dτ 
−∞
y(t) = A cos ωc t
m(t) is sketched below: Z t   
+ ω∆ am1 (τ ) + bm2 (τ ) dτ
m(t) −∞
6= ay1 (t) + by2 (t)
1
Hence the modulation system is non-
-2 -1 0 1 2 t linear.
– The system has memory as the out-
Case 1: t < −2 put at any instant is dependent on the
In this case, the signal m(t) = 0. input at previous time instants.

– The system is causal as output at any

y(t) = A cos (8πt) instant does not depend on the future
values of input.
Case 2: −2 < t < 1

– Let y1 (t) be the response of the system

 Z t 
y(t) = A cos 8πt + 2π 1 · dτ to m(t − t0 ).
−2  Z t 
= A cos (8πt + 2π(t + 2)) y1 (t) = A cos ωc t + ω∆ m(τ − t0 )dτ
= A cos (10πt + 4π) −∞

However, y(t) delayed by t0 may be ex-

Case 3: t > 1 pressed as:
 Z t−t0 
 Z 1 
y(t) = A cos 8πt + 2π 1 · dτ y(t−t0 ) = A cos ωc (t − t0 ) + ω∆ m(τ )dτ
−∞
−2
= A cos (8πt + 2π(3)) Since y(t − t0 ) 6= y1 (t), the system is
= A cos (8πt + 6π) time-variant.