Signals and Systems (EKT230)
Tutorial III
1. Consider a continuous-time system with input x(t) and output y(t) related by
y (t )  x(sin( t ))
(a) Is the system causal ?
The system is not causal because the output y(t) at some time may depend on future
values of x(t). For instance y(-π) = x(0)
(b) Is the system linear ?
Consider two arbitrary inputs x1(t) and x2(t).
x1  y1(t) = x1(sin(t))
x2  y2(t) = x2(sin(t))
Let x3(t) be a linear combination of x1(t) and x2(t). That is,
x3(t) = ax1(t) + bx2(t)
where a and b are arbitrary scalars. If x3(t) is the input to the given system, then the
corresponding output y3(t) is
y3(t) = x3(sin(t))
= ax1(sin(t)) + bx2(sin(t))
= ay1(t) + by2(t)
The system is linear
2. Determine whether the corresponding system is linear, time invariant or both.
(a) y[n] = x2[n-2]
(i) linear
x1  y1[n] = x12[n-2]
x2  y2[n] = x22[n-2]
Let x3(t) be a linear combination of x1[n] and x2[n]. That is:
x3(t) = ax1[n] + bx2[n]
where a and b are arbitrary scalars. If x3(t) is the input to the given system, then the
corresponding output y3(t) is
y3[n] = x3 [n  2]
2
= (ax1[n  2]  bx2 [n  2]) 2
= a 2 x1 [n  2]  b 2 x 2 [n  2]  2abx1 [n  2]x 2 [n  2]
2
≠ ay1[n] + by2[n]
The system is not linear
2
-2-
Tutorial III
(ii) time invariant
Consider an arbitrary input x1[n]. Let
y1[n] = x1 [n  2]
2
be the corresponding output. Consider a second input x2[n] obtained by shifting x1[n] in
time:
x2[n] = x1 [n  no ]
The output corresponding to this input is
y2 [n]  x2 [n  2]  x1 [n  2  no ]
2
2
Also note that
y1[n  no ]  x1 [n  2  no ]
2
Therefore
y 2 [n]  y1[n  no ]
The system is time-invariant.
(b) y[n] = Od{x(t)}
i) linear
Consider two arbitrary inputs x1(t) and x2(t).
x1  y1(t) = Od{x1(t)}
x2  y2(t) = Od{x2(t)}
Let x3(t) be a linear combination of x1(t) and x2(t). That is,
x3(t) = ax1(t) + bx2(t)
where a and b are arbitrary scalars. If x3(t) is the input to the given system, then the
corresponding output y3(t) is
y3(t) = Od{x3(t)}
= Od{ax1(t) + bx2(t)}
= aOd{x1(t)} + bOd{x2(t)} = ay1(t) + by2(t)
The system is linear
i) time invariant
Consider an arbitrary input x1[n]. Let
-3y1(t)
=
Tutorial III
Od {x1 (t )} 
x1 (t )  x1 ( t )
2
be the corresponding output. Consider a second input x2[n] obtained by shifting x1[n]
in time:
x2(t) = x1 (t  t o )
The output corresponding to this input is
x (t )  x2 (t )
y 2 (t )  Od {x2 (t )}  2
2
x1 (t  t o )  x1 ( t  t o )

2
Also note that
y1 (t  t o ) 
x1 (t  t o )  x1 (t  t o )
 y 2 (t )
2
Therefore, the system is not time-invariant.
3.
(a) A continuous signal x(t) is shown in Figure 1. Sketch and label each of the following
signals.
(i) x(t) u(1-t)
(ii) x(t) [u (t) - u (t -1)]
(iii) x(t) ( t - 3/2)
(b) From the given signal y[n],
e  n
y[n]   n
 e
3 n  0
0n3
Draw and label the waveform for each of the following signals:
(i)
y[n  3]
(ii)
y2n  2
-4-
Tutorial III
(iii) 2 y[1  n]
4. A linear time invariant system has an impulse response, h(t) and input signal, x(t). Use
convolution to find the response, y(t) for following signals:
x(t )  cos(t )u (t  1)  u (t  3) 
h(t )  u (t )
Convolution, y(t) = x(t)*h(t)
x(t )  cos(t )u (t  1)  u (t  3) 
h(t )  u (t )

For t < -1, y(t) = 0.

For -1 =< t<0
-5-
Tutorial III
t
y (t )   (1) cos( )d
1


For -1 =< t<0
sin(  )

sin( t )


t

1
sin( t )


sin(  )

-6-
Tutorial III
3
y (t )   (1) cos( )d
1

sin(  )

3

sin( 3 )
1


sin(  )

0
 sin( t )

, 1  t  1
y (t )   
 0,
otherwise
5. A continuous-time periodic signal x(t) is real valued and has a fundamental period T = 8.
The non-zero Fourier series coefficient for x(t) are
a1  a1  2, a3  a *3  4 j
Express x(t) in the form

x(t )   Ak cos( k t   k )
k 0
x(t )  a1e j ( 2 / T )t  a 1e  j ( 2 / T )t  a3 e j 3( 2 / T )t  a 3 e  j 3( 2 / T )t
 2e j ( 2 / 8)t  2e  j ( 2 / 8)t  4 je j 3( 2 / 8)t  4 je  j 3( 2 / 8)t

6
 4 cos( t )  8 sin(
t)
4
8

3

 4 cos( t )  8 cos( t  )
4
4
2
6.
A discrete-time periodic signal x[n] is real valued and has a fundamental period N=5.
The non-zero Fourier series coefficients for x[n] are
a 0  1, a 2  a  2 *  e j / 4 , a 4  a * 4  2e j / 3
Express x(t) in the form

x[n]  A0   Ak sin(  k n   k )
k 0
-7-
Tutorial III
x[n]  a0  a 2 e j 2( 2 / N ) n  a 2 e  j 2( 2 / N ) n  a 4 e j 4 ( 2 / N ) n  a  4 e  j 4( 2 / N ) n
 1  e j ( / 4) e j 2( 2 / 5) n  e  j ( / 4) e  j 2( 2 / 5) n  2e j ( / 3) e j 4( 2 / 5) n  2e  j ( / 3) e  j 4( 2 / 5) n
4

8

n  )  4 cos( n  )
5
4
5
3
4
3
8
5
 1  2 sin(
n  )  8 sin(
t )
5
4
5
6
 1  2 cos(
7. Consider a discrete-time LTI system with impulse response
h[n] =
1,
-1
0
0≤n≤2
-2 ≤ n ≤ -1
otherwise
Given that the input to this system is
x[n] 

  [ n  4k ]
k  
determine the Fourier series coefficients of the output y[n].
H (e j )  e 2 j  e j  1  e  j  e 2 j
For x[n], N = 4 and ω0 = π/2. The FS coefficients of the input x[n] are
1
ak  ,
for all n
4
Therefore, the FS coefficients of the output are
bk  a k H (e jk0 )
1
 [1  e jk / 2  e  jk / 2 ]
4
8. Consider a causal discrete-time LTI system whose input x[n] and output y[n] are related by
the following difference equation:
y[n] – ay[n-1] = x[n]
(a) Find the Fourier series representation of the output y[n] for the input:
3
x[n]  sin(
n)
4
e j (3 / 4) n  e  j (3 / 4) n
1 j (3 / 4) n 1  j (3 / 4) n
x[n] 

e
 e
2j
2j
2j
1
1
, a 3  
Thus, a3 
2j
2j
N0 
2
0
N0  8

2
8

(3 / 4) 3
-8-
To find the transfer function, take the Fourier Transform:
Y (e j )  ae  j Y (e j )  X (e j )
Y (e j )(1  ae  j )  X (e j )
Y (e j )
1

 H (e j )
j
 j
X (e ) 1  ae
The output y[n] is,
y[n]   ak H (e jk 2 / N )e jk ( 2 / N ) n
y[n] 
 1

1
1
1
(
)e j 3( 2 / 8) n   ((
)e  j 3( 2 / 8) n ) 
 j 3 ( 2 / 8 )
 j 3( 2 / 8 )
2 j 1  ae
 2 j 1  ae

(b) Find the impulse response h[n]
Take the inverse FT of H(ejω),
h[n] = anu[n]
, |a| < 1
Tutorial III
Signals and Systems (EKT230)
Tutorial III
Appendix 1
Lampiran 1
FOURIER TRANSFORM
Signal
Transform
LAPLACE TRANSFORM
Signal
Transform
1
u(t)
s
1
tu(t)
s2
Z-TRANSFORM
Signal
Transform
 [n]
(t)
1
1
2()
u(t)
1
  ( )
j
(t - )
e-s
 n u[n]
e-atu(t)
1
a  j
e-atu(t)
1
sa
n n u[n]
1
( s  a) 2
[cos(1n)]u[n]
1  z 1 cos 1
1  z 1 2 cos 1  z  2
1
u[n]
1
1
1  z 1
1
1  z  1
z 1
(1  z 1 ) 2
te-atu(t)
a  j 2
te-atu(t)
e-a|t|
2a
2
a 2
[cos(1t )]u(t )
s
2
s  12
[sin( 1n)]u[n]
z 1 sin 1
1  z 1 2 cos 1  z  2
[sin( 1t )]u(t )
1
2
s  12
[r n cos(1n)]u[n]
1  z 1 r cos 1
1  z 1 2r cos 1  r 2 z 2
sa
( s  a ) 2  12
[r sin( 1n)]u[n]
z 1 r sin 1
1  z 1 2r cos 1  r 2 z  2
1
2
e t
2
/2
e
 2 / 2
[e
 at
cos(1t )]u(t )
[e  at sin( 1t )]u (t )
1
( s  a) 2  12
n