lti systems (continuous & discrete) - basics systems (continuous & discrete) - basics 1. a...
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LTI Systems (Continuous & Discrete) - Basics
1. A system with an input x(t) and output y(t) is described by the relation:
y(t) = t. x(t). This system is
(a) linear and time-invariant
(b) linear and time-varying
(c) non-linear & time-invariant
(d) non-linear and time-varying
[GATE 2000: 1 Mark]
Soln. Systems that are linear and time invariant are called LTI systems.
First check for linearity i.e. super position applies
Input output equation is
π(π) = π. π(π)
π. ππ(π) = ππ. ππ(π)
π. ππ(π) = ππ. ππ(π)
So, π[ππ(π) + ππ(π)] = π[π ππ(π) + πππ(π)]
So, system is linear.
Check for time variance
If input is delayed then
ππ (π β ππ) = π π(π β ππ)
If output is delayed then
π(π β ππ) = (π β ππ) π(π β ππ)
Both are not equal, so system is time varying.
Alternative:
Since π(π) is multiplied by t, the function of time, so system is time
varying.
Option (b)
2. Let Ξ΄(t) denote the delta function. The value of the integral
β« πΏ(π‘)
β
ββ
πππ (3π‘
2) ππ‘ ππ
(a) 1
(b) -1
(c) 0
(d) Ο/2
[GATE 2001: 1 Mark]
Soln. πΉ(π) is unit impulse function also known as Dirac Delta
Function:
Defined as
πΉ(π) = {π πππ π β πβ πππ π = π
β« πΉ(π)π π = π
β
ββ
πΉ(π) . π(π) = π(π) . πΉ(π)
π―πππ, π(π) = ππ¨π¬ (ππ
π) ππππ π(π) = ππ¨π¬(π) = π
β« π(π)
β
ββ
πΉ(π) π π = β« π(π)
β
ββ
πΉ(π) π π = β« πΉ
β
ββ
(π) π π = π
Option (a)
3. The function x(t) is shown in the figure. Even and odd parts of unit-step
function u(t) are respectively,
x(t)
t
1
-1
(a) 1
2,
1
2π₯(π‘)
(b) β1
2,
1
2π₯(π‘)
(c) 1
2, β
1
2π₯(π‘)
(d) β1
2, β
1
2π₯(π‘)
[GATE 2005: 1 Mark]
Soln. Even and odd parts of signal are given
by
π¬πππ ππππ =πΆ(π)+πΆ(βπ)
π
ππ π ππππ =πΆ(π)βπΆ(βπ)
π
Let, πΆ(π) = π(π)
1/2
- 1/2
1/2
1
Option (a)
4. The Dirac delta function Ξ΄(t) is defined as
(a) πΏ(π‘) = {1 π‘ = 00 ππ‘βπππ€ππ π
(b) πΏ(π‘) = {β π‘ = 00 ππ‘βπππ€ππ π
and β« πΏ(π‘)ππ‘ = 1β
ββ
(c) πΏ(π‘) = {1 π‘ = 00 ππ‘βπππ€ππ π
and β« πΏ(π‘)ππ‘ = 1β
ββ
(d) πΏ(π‘) = {1 π‘ = 00 ππ‘βπππ€ππ π
and β« πΏ(π‘)ππ‘ = 1β
ββ
[GATE 2006: 1 Mark]
Soln. Dirac delta function is also called unit impulse function.
It is defined as
πΉ(π) = {β π = ππ πππππππππ
πππ β« πΉ(π)π π = π
β
ββ
Option (d)
5. The input and output of a continuous time system are respectively
denoted by x(t) and y(t). Which of the following descriptions correspond
to a casual system?
(a) π¦(π‘) = π₯(π‘ β 2) + π₯(π‘ + 4)
(b) π¦(π‘) = (π‘ β 4) π₯(π‘ + 1)
(c) π¦(π‘) = (π‘ + 4) π₯(π‘ β 1)
(d) π¦(π‘) = (π‘ + 5) π₯(π‘ + 5)
[GATE 2008: 1 Mark]
Soln. A casual system has the output that depends only on the present and
past values of the input.
Expressions (a), (b) and (d) involve terms that require future values
like π(π + π), π(π + π) πππ π(π + π)
So, Option (c)
6. A discrete-time signal π₯[π] = sin(π2 π) , π being an integer, is
(a) Periodic with period Ο
(b) Periodic with period Ο2
(c) Periodic with period Ο/2
(d) Not periodic
[GATE 2014: 1 Mark]
Soln. Discrete time signal is given
π[π] = π¬π’π§(π ππ)
πΊπ, ππ = π π
πΊπ, π΅ =ππ
ππ . π (π΅πππ) π» =
ππ
ππ
Where m is the smallest integer that converts ππ
ππ into integer value
πΊπ, π΅ =ππ
π π . π =
π
π π
Thus there is no integer value of m which could make N integer.
So, the system is not periodic
Option (d)
7. Let h(t) be the impulse response of a linear time invariant system. Then
the response of the system for any input u(t) is
(a) β« β(π)π‘
0π’(π‘ β π)ππ
(b) π
ππ‘β« β(π)
π‘
0π’(π‘ β π)ππ
(c) β« |β« β(π)π‘
0π’(π‘ β π)ππ| ππ‘
π‘
0
(d) β« β2(π)π‘
0π’(π‘ β π)ππ
[GATE 1995: 1 Mark]
Soln. For LTI System π(π) is the impulse response find the response for
any input π(π)
π(π) = π(π) β π(π)
= β« π(π)
β
ββ
π(π β π) π π
πππππ π(π) = π πππ π < π (ππππ ππππ)
πΊπ, π(π) = β« π(π)
π
π
π(π β π) π π
Option (a)
8. The unit impulse response of a linear time invariant system is the unit
step function u(t). For t > 0, the response of the system to an excitation
πβππ‘ π’(π‘), π > 0 will be
(a) π πβππ‘
(b) (1/π) (1 β πβππ‘)
(c) π(1 β πβππ‘)
(d) 1 β πβππ‘
[GATE 1998: 1 Mark]
Soln. π(π) = π(π)
π―(π) =π
π
π(π) = πβπππ(π) πππ π > π
π(π) = πΏ(π)π―(π)
πΏ(π) = π[π(π)] =π
(π+π)
π(π) =π
(π+π) .
π
π=
π
π[
π
πβ
π
π+π]
Taking inverse Laplace transform
π(π) =π
π[π β πβππ]
Option (b)
9. Convolution of π₯(π‘ + 5) with impulse function πΏ(π‘ β 7)is equal to
(a) π₯(π‘ β 12)
(b) π₯(π‘ + 12)
(c) π₯(π‘ β 2)
(d) π₯(π‘ + 2)
[GATE 2002: 1 Mark]
Soln. If π(π) β π(π) = π(π)
Then π(π β ππ) β π(π β ππ) = π(π β ππ β ππ)
π(π + π) β πΉ(π β π) = π(π + π β π) = π(π β π)
Option (c)
10. The impulse response β[π] of a linear time-invariant system is given by
β[π] = π’[π + 3] + π’[π β 2] β 2π’[π β 7] where π’[π] is the unit step
sequence. The above system is
(a) stable but not causal
(b) stable and causal
(c) causal but unstable
(d) unstable and not causal
[GATE 2004: 1 Mark]
Soln. LTI System has impulse response
π(π) = π[π + π] + π[π β π] β ππ[π β π]
where π[π] is unit step sequence. The given impulse response can be
written using the summation form
β π(π) = β π
β
π=βπ
(π + π)
β
π=ββ
+ β π(π β π)
β
π=π
β π β π(π β π)
β
π=π
Due to the third term, after π = π the negative impulse will cancel
with first two terms
πΊπ, β π(π)
β
π=ββ
= β π
π
π=βπ
+ β π
π
π=π
= ππ + π = ππ
For the system to be stable
β« π(π)
β
ββ
π π < β, ππππ ππ ππ ππ πππππ ππ ππππππ
Note while summing, include π = π term also
For bounded input, the output is bounded. So the system is stable.
In the impulse response term note that future value of input i.e.
π(π + π)
So system is not causal.
Option (a)
11. The impulse response h(t) of a linear time-invariant continuous time
system is described by β(π‘) = exp(πΌπ‘) π’(π‘) + exp(π½π‘) π’(βπ‘) where u(t)
denotes the unit step function, and πΌ and π½ are real constants. This
system is stable if
(a) πΌ is positive and π½ is positive
(b) πΌ is negative and π½ is negative
(c) πΌ is positive and π½ is negative
(d) πΌ is negative and π½ is positive
[GATE 2008: 1 Mark]
Soln. An LTI continuous time system is stable if and only if its impulse
response in absolutely integrable
π. π. β« |π(π)|
β
ββ
π π < β
Given
π(π) = ππΆππ(π) + ππ·ππ(βπ)
For stability
πΆπ < π πππ π > π
π. π. πΆ < π
& π·π > π πππ π < π
π. π. π· > π
Thus πΆ ππ β ππ πππ π· is +ve
So, π(π) should be of the form shown in the figure.
12. A system is defined by its impulse response β(π) = 2ππ’(π β 2). The
system is
(a) stable and causal
(b) causal but not stable
(c) stable but not causal
(d) unstable and non-causal
[GATE 2011: 1 Mark]
Soln. A system is defined by its impulse response
π(π) = πππ(π β π)
For causal system
π(π) = π πππ π < π
Hence, the given system is causal
β ππ
β
π=π
= β , πΊπ πππππ ππππππ ππ πππ ππππππ
Option (b)
13. Two systems with impulse responses h1(t) and h2(t) are connected in
cascade. Then the overall impulse response of the cascaded system is
given by
(a) product of h1(t) and h2(t)
(b) sum of h1(t) and h2(t)
(c) convolution of h1(t) and h2(t)
(d) subtraction of h2(t) from h1(t)
[GATE 2013: 1 Mark]
Soln. If the two systems with impulse responses ππ(π) and ππ(π) are
connected in cascade configuration as shown is the figure, then
overall response of the system is the convolution of the individual
impulse responses.
14. Which one of following statements is NOT TRUE for a continuous time
causal and stable LTI system?
(a) All the poles of the system must lie on the left side of the jΟ axis.
(b) Zeros of the system can lie anywhere in the s-plane.
(c) All the poles must lie within |π| = 1.
(d) All the roots of the characteristic equation must be located on the
left side of the jΟ axis.
[GATE 2013: 1 Mark]
Soln. For stability of LTI Systems, all poles of system should lie in the left
half of S plane and no repeated pole should be on imaginary axis
Option (c) is not true for the stable system as |π| = π have one pole in
right hand plane also
Option (c)
15. A continuous, linear time has an impulse response h(t) described by
β(π‘) = {3 πππ 0 β€ π‘30 ππ‘βπππ€ππ π
0 β€ π‘ β€ 3
when a constant input of value 5 is applied to this filter, the steady state
output is--------
[GATE 2014: 1 Mark]
Soln. Given
π(π) = {π πππ π β€ π β€ π
π πππππππππ
this can be plotted as shown
0
3
3t
π(π) = π[π(π) β π(π β π)]
π―(π) = π [π
πβ
πβππ
π]
Given π(π) = π πΊπ, πΏ(π) =π
π
We know that
π(π) = π―(π) πΏ(π)
= π [π
πβ
πβππ
π] .
π
π
Steady state output is given by
π₯π’π¦πβπ
π. π(π)
= π₯π’π¦πβπ
π.ππ
π[π β πβππ
π]
= π₯π’π¦πβπ
ππ(π β πβππ)
π
= ππ π₯π’π¦πβπ
ππβππ L Hospitalβs rule
= ππ Γ π = ππ
Answer : 45
16. The input β3πβ2π‘π’(π‘), where u(t) is the unit step function, is applied to a
system with transfer function πβ2
π+3. If the initial value of the output is -2,
the value of the output at steady state is-------
[GATE 2014: 1 Mark]
Soln. Given
π(π) = βππβππ
Find Laplace transform of π(π)
πΏ(π) =βπ
(π + π)
π¨πππ, π―(π) =πβπ
π+π
π(π) = πΏ(π)π―(π)
π(π) =βπ(π β π)
(π + π)(π + π)
Using final value theorem, the steady state value of π(π) is
π(β) = π₯π’π¦πβπ
ππ(π)
= 0
Answer : 0