Tutorial I
1.
Sketch the waveforms of the following signals:
(i) x(t) = u(t+1) - 2u(t) + u(t-1)
ans:
(ii) x(t) = -u(t+3) + 2u(t+1) – 2u(t-1) + u(t-3)
ans:
(iii) x(t) = r(t+2) – r(t+1) -r(t-1)+r(t-2)
ans:
Tutorial I
EKT 230 Signals & Systems
Tutorial I
2.
EKT 230 Signals & Systems
Let x[n] and y[n] be given in Figure 1(a) and 1(b), respectively. Carefully sketch the
following signals:
Figure 1(a) and (b)
(i)
x[3n-1]
ans:
(ii)
y[2-2n]
ans:
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Tutorial I
(iii)
x[n-2] + y[n+2]
ans:
(iv)
x[2n]+y[n-4]
ans:
(v)
x[n+2]y[n-2]
ans:
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EKT 230 Signals & Systems
Tutorial I
(vi)
EKT 230 Signals & Systems
x[n]y[-2-n]
ans:
3.
Let x(t) and y(t) be given in Figs. 2(a) and (b), respectively. Carefully sketch the
following signals:
(a) x(t-1)y(-t)
(b) x(t)y(-1-t)
(c) x(2t)y(t/2+1)
(d) x(4-t)y(t)
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Tutorial I
Answer:
(a) x(t-1)y(-t)
(b) x(t)y(-1-t)
(c) x(2t)y(t/2+1)
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EKT 230 Signals & Systems
Tutorial I
(d) x(4-t)y(t)
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EKT 230 Signals & Systems
Tutorial I
4.
EKT 230 Signals & Systems
A discrete-time signal;
1, 0  n  8
x[n]  
0, otherwise
Using u[n], describe x[n] as the superposition of two step functions.
Answer:
x[n]  u[n]  u[n  10]
5.
Given the signal
x[n]  (n  6)u[n]  u[n  6] , sketch and
label the waveform for each of the following signals:
(i)
x[n  2]
(ii)
x[3  n]
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Tutorial I
EKT 230 Signals & Systems
Answer:
Signal
x[n]  (n  6)u[n]  u[n  6] can be described as
below:
(i) x[n  2]
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
time shifting to right side by 2
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(ii)
x[3n  3]
EKT 230 Signals & Systems

time shifting to right side by 3, continued by time
scaling with scale of 3.
(iii)
x[3  n]  x[n  3]
reflect with respect to origin.
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
time shifting to left side by 3, then
Tutorial I
6.
EKT 230 Signals & Systems
Determine whether the system y[n] = nx[n-1]+2 is :
(i) memoryless
No. y[n] depends on x[n-1].
(ii) time invariant
No.
x[n]  y[n] and y[n-no] = (n- no)x[n- no-1]+2
x1 = x[n- no]  y[n] = nx1[n-1]+2 = nx[n- no-1]+2
y[n] ≠ y[n-no]
(iii) linear
No.
x1  y1[n] = nx1[n-1]+2
x2  y2[n] = nx2[n-1]+2
x3 = x1 + x2  y3[n] = nx3[n-1]+2 = nx1[n-1] + nx2[n-1]+2 ≠ y1[n]+y2[n]
(iv) causal
Yes.
y[n] depends on x[n-1]. (Past)
(v) stable
No.
Let | x[n] | A | y[n] || nx[n  1]  2 |
nx[n  1]  2  nA  2 unbounded
Take x[n]  1n | x[n] | 1 also
|y[n]| = n+2  n+2  ∞ as n  ∞ unbounded
Tutorial I
Tutorial I
7.
EKT 230 Signals & Systems
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 ?
Ans :
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 ?
Ans :
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 corresponding output y3(t) is
y3(t) = x3(sin(t))
= ax1(sin(t)) + bx2(sin(t))
= ay1(t) + by2(t)
Hence, the system is linear.
8.
Determine whether the corresponding system is linear, time invariant or both.
(a) y[n] = x2[n-2]
Ans:
(i) linearity
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
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Tutorial I
EKT 230 Signals & Systems
= (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
2
≠ ay1[n] + by2[n]
The system is not linear
(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) linearity
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
Tutorial I
Tutorial I
EKT 230 Signals & Systems
y3(t) = Od{x3(t)}
= Od{ax1(t) + bx2(t)}
= aOd{x1(t)} + bOd{x2(t)} = ay1(t) + by2(t)
Hence, the system is linear
ii) time invariance
Consider an arbitrary input x1[n]. Let
y1(t) = 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
x (t  t o )  x1 ( t  t o )
 1
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.
Tutorial I