UNIT -3
LINEAR TIME INVARIENTCONTINUOUS TIME SYSTEMS
CONTINUOUS TIME SYSTEM
• A continuous time system (or Analog system) is a physical device that
operates on a continuous time signal called input or excitation,
according to some well defined rule, to produce another continuous
time signal called output or response.
• The input signal x(t) is transformed by the system into a signal y(t) .
• Diagrammatic representation of continuous time system
– Response , y(t) = H{x(t)}
– Where, H denotes the transformation (also called an operator).
x(t)
y(t)
H
Input signal or
Excitation
Continuous time
system
Output signal
or Response
CONTINUOUS TIME SYSTEM
• A continuous time system (or Analog system) is a physical device that
operates on a continuous time signal called input or excitation,
according to some well defined rule, to produce another continuous
time signal called output or response.
• The input signal x(t) is transformed by the system into a signal y(t) .
• Diagrammatic representation of continuous time system
– Response , y(t) = H{x(t)}
– Where, H denotes the transformation (also called an operator).
x(t)
y(t)
H
Input signal or
Excitation
Continuous time
system
Output signal
or Response
LTI SYSTEM
• A continuous time system is linear if it obeys the principle of
superposition and it is time invarient if its input-output relationship
does not change with time.
• When a continuous time system satisfies the properties of linearity
and time invarience then it is called as LTI system.
• IMPULSE RESPONSE:
• Impulse response , h(t) = H{δ(t)}
δ(t)
h(t)
H
Input signal or
Excitation
Continuous time
system
Output signal
or Response
BLOCK DIAGRAM AND SIGNAL
FLOW GRAPH REPRESENTATION
PROBLEM-1
Construct the block diagram and signal flow graph of the system
described by the equation,
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
+
𝑑𝑦(𝑡)
2
𝑑𝑡
+ 3y(t) =
𝑑𝑥(𝑡)
4
𝑑𝑡
+ 5x(t)
Solution:
Case i: Block diagram and signal flow graph using differntiators
Given that
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
therefore y(t) =
+
𝑑𝑦(𝑡)
2
𝑑𝑡
1 𝑑 2 𝑦(𝑡)
3 𝑑𝑡 2
+ 3y(t) =
𝑑𝑥(𝑡)
4
𝑑𝑡
2 𝑑𝑦(𝑡)
3 𝑑𝑡
𝑑𝑥(𝑡)
𝑑𝑡
-
+4
+ 5x(t)
+ 5x(t) ------( 1)
The equation (1) is used to construct the block diagram and signal flow
graph using differentiator
BLOCK DIAGRAM
SIGNAL FLOW GRAPH
RESPONSE OF LTI-CT SYSTEM IN
TIME DOMAIN
• The general equation governing an LTI continuous time system is,
• The solution of the differential equation is the response y(t) of the
LTI system, which consists of two parts.
• In mathematic, the two parts of the solution y(t) are the
homogeneous solution yh(t) and the particular solution yp(t).
•
Response, y(t) = yh(t)+yp(t)
CONT…
• The homogenous solution is the response of the system where there is
no input.
• The particular solution yp(t) is the solution of difference equation for
specific input signal x(t) for t ≥ 0.
• In signal and systems the two parts of the solution y(t) are called
zero-input response yzi(t) and zero-state response yzs(t).
– Response, y(t) = yzi(t)+yzs(t)
• The zero input response is given by the homogenous solution with
constant evaluated using initial values of output (or initial condition).
CONT…
• The zero-input response is mainly due to initial output conditions (or
initial stored energy) in the system.
• Hence the zero-input response is also called free response or natural
response.
• The zero-state response is given by the sum of homogeneous
solution and particular solution with zero initial conditions.
• The zero-state response is the response of the system due to input
signal and with zero initial output condition.
• Hence the zero-state response is also called forced response.
HOMOGENEOUS & PARTICULAR
SOLUTION
• The homogeneous solution is obtained when x(t) =0. Substituting
x(t)=0 in system general equation.
• Particular Solution:
INPUT SIGNAL, x(t)
PARTICULAR SOLUTION, yp(t)
x(t) = A
yp(t) = K
x(t) = Au(t)
yp(t) = Ku(t)
x(t) = Aeαt (if α≠ λ𝑖)
yp(t) = Keαt
x(t) = Aeαt (if α= λ𝑖)
yp(t) = Kteαt
x(t) = A cos Ω0t
= A sin Ω0t
yp(t) = K1cos Ω0t +K2sin Ω0t
Note : λi is one of the
root of characteristic
polynomial.
PROBLEM
1. Determine the natural response of the system described by the
equation,
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
+
𝑑𝑦(𝑡)
6
𝑑𝑡
+ 5y(t) =
𝑑𝑥(𝑡)
𝑑𝑡
+ 4x(t);
y(0) =1;
𝑑𝑦(𝑡)
|t=0 =-2
𝑑𝑡
Solution
• The natural response is response of the system due to initial conditions
and so it is given by zero-input response.
• Zero-input response, yzi(t) = yh(t)| with constants evaluated using initial condition
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
+
𝑑𝑦(𝑡)
6
𝑑𝑡
+ 5y(t) =
𝑑𝑥(𝑡)
𝑑𝑡
+ 4x(t);
• Homogeneous Solution
• The HS is the solution of the system equation when x(t) =0
PROBLEM-1
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
Let y(t) =
+6
Ceλt ;
𝑑𝑦(𝑡)
𝑑𝑡
+ 5y(t) =0
𝑑𝑦(𝑡)
𝑑𝑡
=
------(2)
𝑑 2 𝑦(𝑡)
λt
Cλe ;
𝑑𝑡 2
= Cλ2eλt
on substituting the above terms in equation (2) we get,
Cλ2eλt + 6Cλeλt+5 Ceλt = 0
(λ2+6 λ+5) Ceλt =0
The characteristic polynomial of the above equation is,
λ2+6 λ+5 =0 ≫ (λ+1) (λ+5)=0 ≫ λ= -1,-5
Now the homogeneous solution is given by
yh(t) = C1eλ1t+ C2eλ2t = C1e-t+ C2e-5t ---- Homogeneous solution
CONT…
• Natural Response (or Zero-input Response)
Zero-input response, yzi(t) = yh(t)| with constants evaluated using initial condition
= C1e-t +C2e-5t | with C1 and C2 evaluated using initial condition
𝑑𝑦𝑧𝑖 (𝑡)
𝑑𝑡
= -C1e-t -5 C2e-5t
At t = 0, yzi(0) = C1 +C2
Given that yzi(0) =1,
At t=0,
𝑑𝑦𝑧𝑖 (𝑡)
𝑑𝑡
Given that
C1 +C2=1
-----(3)
= -C1 -5C2
𝑑𝑦𝑧𝑖 (𝑡)
| t=0 =
𝑑𝑡
-2,
-C1-5C2= -2
On adding equation 3 & 4 we get
• -4C2 = -1 >>>> C2 =1/4
-------(4)
CONT…
• From eqn (3) , C1=1-C2
= 1-1/4 = ¾
Natural response yzi(t) = ¾ e-t + ¼ e-5t; t ≥0
PROBLEM -2
• Determine the complete response of the system described by the
equation,
𝑑 2 𝑦(𝑡) 𝑑𝑦(𝑡)
+5
𝑑𝑡 2
𝑑𝑡
+ 4y(t) =
𝑑𝑥(𝑡)
,
𝑑𝑡
y(0) =0;
𝑑𝑦(𝑡)
|
𝑑𝑡 t=0
=1 for the input x(t) = e-2tu(t)
Solution:
𝑑 2 𝑦(𝑡) 𝑑𝑦(𝑡)
+5
𝑑𝑡 2
𝑑𝑡
+ 4y(t) =
𝑑𝑥(𝑡)
𝑑𝑡
--------(1)
Homogeneous Solution
yh(t) = C1e-4t + C2e-t
Particular Solution
The particular solution is the solution of the system equation for specific
input.
CONT…
• Here input is x(t) = e-2t u(t)
x(t) = e-2t; for t≥0
Therefore
𝑑𝑥(𝑡)
𝑑𝑡
= -2 e-2t
Let the particular solution,
yp(t) = Kx(t)
yp(t) = K
e-2t,
𝑑𝑦𝑝 (𝑡)
𝑑𝑡
=-2Ke-2t;
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
= 4Ke-2t
On substituting the above terms and the input in equation (1), we get,
4Ke-2t-10Ke-2t+4Ke-2t = -2e-2t
CONT…
• On dividing throughout by e-2t we get,
 4K-10K+4K = -2
 -2K = -2 >>>> K=1
Therefore Particular Solution yp(t) = e-2t
 Total (or Complete) Response:
Method -1:
Total response y(t) = yh(t)+yp(t)
y(t) = C1e-4t+C2e-t+e-2t
When t=0, y(t) = y(0) = C1e0+C2e0+e0 = C1+C2+1
Given that y(0) =0, C1+C2+1 =0
-------(3)
CONT…
• Here,
• Now,
𝑑𝑦(𝑡)
= -4C1e-4t – C2e-t-2e-2t
𝑑𝑡
𝑑𝑦(𝑡)
| = -4C1 – C2-2
𝑑𝑡 t=0
Given
𝑑𝑦(𝑡)
| =
𝑑𝑡 t=0
1; -4C1 – C2-2 =1 ------ (4)
On adding eqn (3) and (4) we get,
-3C1 -1 = 1 >>>>> -3C1 =2 >>>>> C1 = - 2/3
From Eqn (3), C2=-1 –C1 = -1+2/3 = -1/3
CONT…
• Total Response, y(t) = -2/3 e-4t -1/3 e-t+e-2t ; t≥ 0
» Or y(t) = (e-2t+ -2/3 e-4t-1/3 e-t) u(t)
Method-2:
Total Response, y(t) = yzi(t)+yzs(t)
Zero-input response, yzi(t) = yh(t)| with constant evaluated using initial condition
yh(t) =C1e-4t + C2e-t
∴
𝑑𝑦ℎ (𝑡)
𝑑𝑡
= -4 C1e-4t - C2e-t
At t=0, yh(0) = C1e0+C2e0 = C1+C2
Given that , yh(t) =0 ,
C1+C2=0
--------(5)
CONT…
𝑑𝑦ℎ (𝑡)
= -4 C1e0
𝑑𝑡
𝑑𝑦 (𝑡)
that, ℎ | t=0 =
𝑑𝑡
• At t=0,
- C2e0 = -4C1-C2
• Given
1, -4C1-C2 = 1
• On adding eqn (5) and (6) we get,
• -3C1=1 >>> C1 = -1/3
• From eqn (5) C2 =-C1 = 1/3
∴ yzi(t) = -1/3e-4t +1/3e-t
-------(6)
CONT…
• Zero-state response, yzs(t) = yh(t)+yp(t)| with constant evaluated using zero initial
condition
• yzs(t) =C1e-4t + C2e-t+e-2t
•
𝑑𝑦𝑧𝑠 (𝑡)
𝑑𝑡
= -4C1e-4t –C2e-t -2e-2t
• At t=0, yzs(0) = C1e0 +C2e0+e0 = C1+C2+1
• At t=0,
𝑑𝑦𝑧𝑠 (𝑡)
𝑑𝑡
= -4C1 – C2 -2
• Since initial conditions are zero,
• C1+C2+1 =0
-----(7)
• -4C1-C2-2 =0
------(8)
• On adding eqn (7) and (8) we get,
• -3C1-1 = 0 >>>> C1 = -1/3
• From the eqn (7), C2 = -1 – C1 = -1+1/3 = -2/3
• ∴ yzs(t) = -1/3 e-4t – 2/3e-t + e-2t
• Total response y(t) = yzi(t)+yzs(t)
= -1/3e-4t +1/3e-t-1/3 e-4t – 2/3e-t + e-2t
= -2/3e-4t -1/3e-t+e-2t ; t≥0
y(t) = (e-2t -2/3e-4t-1/3e-t) u(t)
TRANSFER FUNCTION OF LTI CT SYSTEM
• The transfer function of a continuous time system is defined as the
ratio of Laplace transform of output and Laplace transform of input.
• Let, x(t) = Input of a LTI continuous time system
y(t) = Output / Response of the LTI continuous time system for
the input x(t)
h(t) = Impulse response (i.e., response for impulse input)
H(s) =
𝑌(𝑠)
𝑋(𝑠)
---- the transfer function of LTI continuous time system is also given
by Laplace transform of the impulse response.
• Inverse Laplace transform of transfer function is the impulse response
of the system.
∴ Impulse response, h(t) = 𝑳−𝟏 {𝑯(𝒔)} = 𝑳−𝟏
𝒀(𝒔)
𝑿(𝒔)
CONVOLUTION & DECONVOLUTION
• CONVOLUTION:
The convolution operation is performed to find the response y(t) of
LTI continuous time system from the input x(t) and the impulse
response h(t).
∴ Response, y(t) = x(t) * h(t)
On taking Laplace transform of the above equation we get,
L {y(t)} = L{x(t)*h(t)}
∴ Y(s) = X(s) H(s)
∴ Response, y(t) = L-1{Y(s)}
= L-1{X(s) H(s)}
CONT…
• DECONVOLUTION:
The deconvolution is the operation is performed to extract the input x(t) of an LTI
continuous time system from the response y(t) of the system.
We know that,
X(s) =
𝑌(𝑠)
𝐻(𝑠)
On taking inverse Laplace transform of the above equation we get,
𝑌(𝑠)
}
𝐻(𝑠)
Input x(t) = L-1{X(s)} = L-1{
PROBLEM
• Using Laplace transform determine the complete response of the
system described by the equation,
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
+
𝑑𝑦(𝑡)
5
+4y(t)
𝑑𝑡
𝑑𝑥(𝑡)
=
𝑑𝑡
; y(0) = 0;
𝑑𝑦(𝑡)
|
𝑑𝑡 t=0
=1,
for the input, x(t) = e-2tu(t).
Solution
Method-1
input, x(t) = e-2tu(t)
X(s) = L{x(t)} =
L{e-2tu(t)}
=
1
𝑠+2
Initial value of input, x(0) = x(t)|t=0 = e0u(0) = 1
-----(1)
CONT…
• The given system equation is,

𝑑 2 𝑦(𝑡)
𝑑𝑡 2
+
𝑑𝑦(𝑡)
5
+4y(t)
𝑑𝑡
𝑑𝑥(𝑡)
=
𝑑𝑡
On taking laplace transform of the above equation we get,
s2Y(s)-sy(0)-y’(0)+5[sY(s)-y(0)]+4Y(s) = sX(s) – x(0)
On substituting the initial value of output and input in the above
equation we get,
s2Y(s) - sx0 -1+5[sY(s) –0]+4Y(s) = sX(s)-1
s2Y(s)+5sY(s)+4Y(s) =sX(s)
(s2+5s+4)Y(s) = s x
1
𝑠+2
CONT…
𝑠
(𝑠+2)(𝑠2+5𝑠+4)
𝑠
=
(𝑠+2)(𝑠+1)(𝑠+4)
𝑠
=
(𝑠+1)(𝑠+2)(𝑠+4)
 Y(s) =


By partial fraction expansion technique, Y(s) can be expressed as,
Y(s) =
A=
𝑠
(𝑠+1)(𝑠+2)(𝑠+4)
𝑠
(𝑠+1)(𝑠+2)(𝑠+4)
B= 1 , C =-
2
3
=
𝐴
𝑠+1
+
𝐵
𝑠+2
x (s+1)|s=-1 =
+
𝐶
𝑠+4
−1
(−1+2)(−1+4)
=
−1
3
• ∴ Y(s) =-
1 1
3 𝑠+1
+
1 2 1
𝑠+2 3 𝑠+4
• Total response y(t) = L-1{Y(s)}
= L-1 −
L{e-at u(t)} =
+
1
𝑠+2
−
2 1
3 𝑠+4
1
𝑠+𝑎
=
y(t) =
1 1
3 𝑠+1
1 -t
- e u(t)+e-2tu(t)
3
𝟏 −t −2t
− e +e
𝟑
−
𝟐 −4t
e
𝟑
-
2 -4t
e u(t)
3
u(t)
CONT…
•
•
•
•
Method -2:
Total response y(t) = yzi(t) + yzs(t)
Zero-input Response
On substituting x(t)=0 and y(t)=yzi(t) in the system equation we get,
•
𝑑 2 𝑦𝑧𝑖 (𝑡)
𝑑𝑡 2
+
𝑑𝑦𝑧𝑖 (𝑡)
5
+4yzi(t)
𝑑𝑡
=0
• On taking Laplace transform
• s2Yzi(s)-sy(0)-y’(0)+5[sYzi(s)-y(0)]+4Yzi(s) =0
• On substituting the initial values of output in the above equation
• s2Yzi(s) - sx0 -1+5[sYzi(s)-0]+4Yzi(s) = 0
• (s2+5s+4)Yzi(s) = 1
• Yzi(s) =
1
𝑠 2 +5𝑠+4
=
1
(𝑠+1)(𝑠+4)
• By partial fraction expansion technique, Yzi(s)
• Yzi(s) =
1
(𝑠+1)(𝑠+4)
=
𝐷
𝑠+1
+
𝐸
𝑠+4
• D = 1/3 , E = -1/3
• Yzi(s) =
1
3(𝑠+1)
−
1
3 𝑠+4
• On taking inverse Laplace transform of Yzi(s) we get zero-input
response, yzi(t)
CONT…
• Zero-input response, yzi(t) = L-1
1
3(𝑠+1)
1
3
−
1
3(𝑠+4)
1
3
= e-tu(t) - e-4tu(t)
• Zero-state Response:
• Given that, x(t) = e-2tu(t)
∴ x(0) = x(t)|t=0 = e0u(0) = 1
X(s) = L{x(t)}= L{e-2tu(t)} =
1
𝑠+2
On substituting y(t) = yzs(t) in the system equation
𝑑 2 𝑦𝑧𝑠(𝑡)
𝑑𝑡 2
+
𝑑𝑦𝑧𝑠(𝑡)
5
+4yzs(t)
𝑑𝑡
=
𝑑𝑥(𝑡)
𝑑𝑡
CONT…
• (s2+5s+4)Yzs(s) = sX(s) –x(0)
•
•
•
1
=sx
𝑠+2
𝑠−𝑠−2
(s+1)(s+4)Yzs(s) =
𝑠+2
−2
Yzs(s) =
(𝑠+1)(𝑠+2)(𝑠+4)
(s2+5s+4)Yzs(s)
-1
• By partial fraction expansion technique, Yzs(s) can be expressed as,
• Yzs(s) =
−2
(𝑠+1)(𝑠+2)(𝑠+4)
=
𝐹
(𝑠+1)
+
• ∴ F = -2/3 , G = 1, H = -1/3
• ∴ Yzs(s) = -
2
3(𝑠+1)
+
1
(𝑠+2)
-
1
3(𝑠+4)
𝐺
(𝑠+2)
+
𝐻
(𝑠+4)
CONT…
• On taking inverse Laplace transform of Yzs(s) we get zero- state
response, yzs(t)
• ∴ Zero-state response, yzs(t) = L-1{Yzs(s)}
=
L-1
2
{3(𝑠+1)
1
+
(𝑠+2)
2
3
= − e-tu(t) + e-2tu(t)
1
}
3(𝑠+4)
1
- e-4tu(t)
3
Total/Complete Response
Total response y(t) = yzi(t)+yzs(t)
=
1 −t
e u(t)
3
−
1
3
e−4tu(t)
+
2
−
3
e−tu(t) + e−2tu(t)
• Perform convolution of x1(t)=e-2tcos3tu(t) and x2(t)=4sin3tu(t)
using Laplace transform.
• Solution:
• X1(s) = L{x1(t)} =
L{e-2tcos3tu(t)}
=
(𝑠+2)
(𝑠+2)2 +32
• X2(s) = L{x2(t)}= L{4sin3tu(t)} = 4 x
3
𝑠 2 +32
=
=
(𝑠+2)
𝑠 2 +4𝑠+13
12
𝑠 2 +9
• Now by convolution theorem, x1(t)*x2(t) = L-1{X1(s)X2(s)}
• Let X(s) = X1(s)X2(s)
• ∴ X(s) =
(𝑠+2)
𝑠 2 +4𝑠+13
x
12
𝑠 2 +9
=
12(𝑠+2)
(𝑠 2 +4𝑠+13)(𝑠 2 +9)
• By partial fraction expansion, X(s) can be expressed as,
• X(s) =
12(𝑠+2)
(𝑠 2 +4𝑠+13)(𝑠 2 +9)
=
𝐴𝑠+𝐵
𝑠 2 +4𝑠+13
+
𝐶𝑠+𝐷
𝑠 2 +9
• On cross-multiplying the above equation we get,
• 12(𝑠 + 2) = (𝐴𝑠 + 𝐵)(𝑠 2 + 9)+(Cs+D)(𝑠 2 + 4𝑠 + 13)
• 12s+24 = (A+C)s3+(B+4C+D)s2+(9A+13C+4D)s+(9B+13D) ….(2)
•
•
•
•
On equating the coefficient of s3 terms of equation (2) we get,
A+C = 0 >>>>> A = -C
…..(3)
On equating the coefficient of s2 terms of equation (2) we get,
B+4C+D = 0
…..(4)
CONT…
•
•
•
•
•
•
•
•
•
•
On equating the coefficient of s terms of equation (2) we get,
9A+13C+4D =12
….. (5)
On equating constant of equation (2) we get,
9B+13D = 24
…..(6)
On substituting A=-C in equation (5) we get,
9(-C)+13C+4D = 12
4C + 4D = 12 >>>>> C+D = 3 >>>>> C = 3-D
…. (7)
on substituting C=3-D in eqn (4) we get
B+4(3-D)+D =0
B-3D = -12
……(8)
• Eqn (8) x 9 >>> -9B+27D =108
• Eqn (6) x 1 >>> 9B+13D =24
•
40D = 132
From eqn (7), C = 3-D = 3-3.3 = -0.3
From eqn (3), A = -C =0.3
24−13𝐷 24−13∗3.3
=
=-2.1
9
9
𝐴𝑠+𝐵
𝐶𝑠+𝐷
0.3𝑠−2.1
−0.3𝑠+3.3
+
=
+
𝑠 2 +4𝑠+13
𝑠 2 +9
𝑠 2 +4𝑠+13
𝑠 2 +9
From eqn (6) B =
∴ X(s) =
=
D = 3.3
2.1
0.3
+22
0.3 𝑠−
𝑠 2 + 2∗2𝑠
+32
+
−0.3𝑠+3.3
𝑠 2 +9
• =
0.3(𝑠−7)
(𝑠+2)2 +32
• =
0.3(𝑠+2−9)
(𝑠+2)2 +32
• =
0.3 𝑠+2 −0.3∗9
(𝑠+2)2 +32
• = 0.3
+
−0.3𝑠+3.3
𝑠 2 +9
+
−0.3𝑠+3.3
𝑠 2 +32
𝑠+2
(𝑠+2)2 +32
+
−0.3𝑠+3.3
𝑠 2 +32
3
(𝑠+2)2 +32
- 0.9
𝑠
𝑠 2 +32
-0.3
3
𝑠 2 +32
+ 11
• On taking inverse Laplace transform of the above equation we get,
• x(t) = 0.3𝑒 −2𝑡 𝑐𝑜𝑠3𝑡 − 0.9𝑒 −2𝑡 𝑠𝑖𝑛3𝑡 − 0.3𝑐𝑜𝑠3𝑡 + 1.1𝑠𝑖𝑛3𝑡; 𝑡 ≥ 0
Find the transfer function of the systems governed by the following differential equations.
a.
b.
𝑑2 𝑦 (𝑡)
𝑑𝑡 3
𝑑2 𝑦 (𝑡)
𝑑𝑡 3
+
+
𝑑2 𝑦(𝑡)
4 𝑑𝑡 2
𝑑2 𝑦(𝑡)
8 𝑑𝑡 2
𝑑𝑦(𝑡)
𝑑𝑡
𝑑𝑦(𝑡)
6 𝑑𝑡
𝑑𝑥 𝑡
𝑑𝑡
+3
+ 5𝑦 𝑡 − 2
+𝑥 𝑡
+
+ 1𝑦 𝑡 − 𝑥(𝑡 − 2)
Solution
a) Given that,
𝑑2 𝑦 (𝑡)
𝑑𝑡 3
+
𝑑2 𝑦(𝑡)
4 𝑑𝑡 2
+3
𝑑𝑦(𝑡)
𝑑𝑡
+ 5𝑦 𝑡 − 2
𝑑𝑥 𝑡
𝑑𝑡
+𝑥 𝑡
On taking Laplace transform of the above equation with zero initial conditions we get,
𝑠 3 Y s + 4𝑠 2 Y s + 3sY(s)+5Y(s) = 2sX(s) + X(s)
(𝑠 3
+
4𝑠 2
+ 3𝑠 + 5) 𝑌 𝑠 = (2𝑠 + 1) X(s)
Transfer Function,
𝑌 (𝑠)
𝑋(𝑠)
2𝑠+1
= 𝑠3 +4𝑠2+3𝑠+5
If, £{ x(t)} = X(s),
𝑑𝑛 𝑋 (𝑡)
𝑛𝑋
}=𝑠
𝑛
𝑑𝑡
Then, £{
𝑠
With zero initial conditions
Solution
b) Given that,
𝑑 2 𝑦 (𝑡)
𝑑𝑡 3
+
𝑑 2 𝑦(𝑡)
8
𝑑𝑡 2
+6
𝑑𝑦(𝑡)
𝑑𝑡
+ 1𝑦 𝑡 − 𝑥(𝑡 − 2)
On taking Laplace transform of the above equation with zero initial conditions
we get,
𝑠 3 Y s + 8𝑠 2 Y s + 6sY(s) + 11Y(s) = 𝑒 −2𝑠 + X(s)
(𝑠 3 + 8𝑠 2 + 6𝑠 + 11) 𝑌 𝑠 = 𝑒 −2𝑠 + X(s)
𝑒 −2𝑠 + X(s)
Transfer Function,
𝑌 (𝑠)
𝑋(𝑠)
=
𝑒 −2𝑠
𝑠 3 +8𝑠 2 +6𝑠+11
If, £{ x(t)} = X(s),
Then, £{X(t-a)}=𝑒 −2𝑠 𝑋 𝑠
Find the transfer function of the systems governed by the following impulse responses.
a) h(t) = (2+1) 𝑒 −3𝑡 𝑢 𝑡
b) h(t) = 𝑡 2 𝑢 𝑡 − 𝑒 −4𝑡 𝑢 𝑡 + 𝑒 −7𝑡 𝑢(𝑡)
c) h(t) = u(t) + 0.5 𝑒 −5𝑡 𝑢 𝑡 + 0.2 𝑒 −3𝑡 cos 𝑡 𝑢 𝑡
Solution
a) Impulse response, h(t) = (2+1) 𝑒 −3𝑡 𝑢 𝑡
L 𝑒 −𝑠𝑡 𝑢 𝑡
+
1
,𝐿
𝑠+𝑎
𝑡𝑢 𝑡
=
The transfer function is given by Laplace transform of impulse response.
Transfer function, H(s)
= L { h(t)}
= L {(2+t) 𝑒 −3𝑡 𝑢(𝑡)} = L {2 𝑒 −3𝑡 𝑢(𝑡)} + L {t 𝑒 −3𝑡 𝑢(𝑡)}
= 2L { 𝑒 −3𝑡 𝑢(𝑡)} + L {t u(t)}|𝑠=5+3 = 2 𝑋
=
2
𝑠+3
+
1
(𝑠+3)2
=
2 𝑠+3 +1
(𝑠+3)2
=
1
𝑠+3
+
2𝑠+7
𝑠 2 +6𝑠+9
1
𝑠2
|𝑠=5+3
1
𝑠2
b) Impulse response,
h(t) = 𝑡 2 𝑢 𝑡 − 𝑒 −4𝑡 𝑢 𝑡 + 𝑒 −7𝑡 𝑢(𝑡)
The transfer function is given by Laplace transform of impulse response.
L 𝑡𝑛 𝑢 𝑡
=
𝑛1
𝑠𝑛+1
; 𝐿 𝑒 −𝑎𝑡 𝑢 𝑡
Therefore Transfer function,
H(s) = L {h(t)} = L {𝑡 2 𝑢 𝑡 − 𝑒 −4𝑡 𝑢 𝑡 + 𝑒 −7𝑡 𝑢(𝑡)}
= L{𝑡 2 𝑢(𝑡)} - L{𝑒 −4𝑡 𝑢(𝑡)} + L{𝑒 −7𝑡 𝑢(𝑡)}
2
1
1
−
+
𝑠3
𝑠+4
𝑠+7
2 𝑠+4 𝑠+7 − 𝑠 2 𝑠+7 + 𝑠 3 (𝑠+4)
=
𝑠 3 𝑠+4 (𝑠+7)
2 𝑠 2 +7𝑠+4𝑠+28 − 𝑠 4 −7𝑠 3 + 𝑠 4 +4𝑠 3
=
=
=
𝑠 3 𝑠 2 +11 𝑠+28
−3 𝑠 3 +2 𝑠 2 +22𝑠+56
𝑠 5 +11 𝑠 4 +28 𝑠 3
1
= 𝑠+𝑎
The Unit step response of continuous time systems are given below. Determine the
transfer function of the systems.
a) s(t)=u(t)+𝑒 −2𝑡 𝑢 𝑡
𝑏) s(t)=𝑡 2 u(t)+ t 𝑒 −4𝑡 𝑢 𝑡
𝑐) 𝑠 𝑡 = 𝑡 𝑢 𝑡 + sin au(t)
Solution
a) Unit step response, s(t)= s(t)=u(t)+𝑒 −2𝑡 𝑢 𝑡
On taking laplace transform of unit step response we get,
L{s(t)}=S(s) = L{u(t)+𝑒 −2𝑡 𝑢 𝑡 }
∴ S(s) = L{u(t)+𝑒 −2𝑡 𝑢(𝑡)} = L{u(t)} + L{𝑒 −2𝑡 𝑢 𝑡 }
1
𝑠
= +
Let, U(s)=L{u(t)} =
=S(s) x
1
𝑈(𝑠)
1
𝑠+2
=
𝑠+2+𝑠
𝑠 𝑠+2
=
2𝑠+2
𝑠(𝑠+2)
------ (1)
1
𝑠
; where u(t) is a unit step signal ------(2)
=
2𝑠+2
𝑠(𝑠+2)
xs
2𝑠+2
=
𝑠+2
Using equation (1) and (2)
Find the impulse response of continuous time systems governed by the following transfer
functions.
1
a) H(s) =𝑠2(𝑠−2)
b) H(s) = 𝑠
1
𝑠+1 (𝑠−2)
1
c) H(s) = 𝑠2 +𝑠+1
Solution
a) Transfer function , H(s) =
1
𝑠 2 (𝑠−2)
The impulse response is obtained by taking inverse Laplace transform of the transfer function.
1
𝐴
𝐵
𝐶
∴ Impulse response, h(t) = 𝐿−1 {𝑠2 (𝑠−2)} = 𝐿−1 {𝑠2 + 𝑠 + 𝑠−2}
Using partial fraction expansion technique
1
A = 𝑠2 (𝑠−2) x ⃒s=0
𝑑
B = 𝑑𝑠
1
𝑠 2 𝑠−2
1
1
= 0−2 = - 2
× 𝑠 2 ⃒s=0
1
=
𝑑
𝑑𝑠
1
𝑠−2
⃒s=0
1
C= 𝑠2 (𝑠−2) x (s-2)⃒s=2 =22 = −
1 1
11
1
1
1
−1
1
=(0−2)2 = - 4
1
4
1 1
∴ Impulse response , h(t) = 𝐿−1 {− 2 𝑠2 - 4 𝑠 + 4 𝑠−2}
= - 2 t u(t) - 4 𝑒 2𝑡 𝑢(𝑡) = 4 (𝑒 2𝑡 -2t-1) u(t)
−1
=(𝑠−2)2⃒s=0
1
L{u(t)}= 𝑠
b) Transfer function , H(s) = 𝑠
1
𝑠+1 (𝑠−2)
The impulse response is obtained by taking inverse Laplace transform of the transfer
function.
1
𝐴
∴ Impulse response, h(t) = 𝐿−1 {H(s)} =𝐿−1 {𝑠(𝑠+1)(𝑠−2)} = 𝐿−1 { 𝑠 +
1
1
𝐵
𝑠+1
𝐶
+𝑠−2}
1
A = 𝑠(𝑠+1)(𝑠−2) x s ⃒s=0 = (0+1)(0−2) = - 2
1
1
B = 𝑠(𝑠+1)(𝑠−2) x (s+1) ⃒s=- 1
1
= −1×(−1−2) = 3
Using partial fraction expansion
1
C = 𝑠(𝑠+1)(𝑠−2) x (s-2)⃒s=2
technique
1
1 1
1 1
+
}
3 𝑠+1
6 𝑠−2
1 −𝑡
1
1
𝑒 u(t) + 𝑒 2𝑡 u(t) =
3
6
6
∴ Impulse response, h(t) = 𝐿−1 {1
2
= - u(t) +
1
1
= 2×(2+1) = 6
11
2𝑠
+
(𝑒 2𝑡 + 2𝑒 −𝑡 - 3u(t))
L{ u(t)} = 𝑠
L{𝑒 ±𝑎𝑡 𝑢(𝑡)} =
1
𝑠∓𝑎
The input and impulse response of continuous time systems are given below. Find the
output of the continuous time systems.
h(t) =𝑒 −𝑎𝑡 u(t)
a) x(t) = 𝛿(t),
c) x(t) = 𝑒 −3𝑡 u(t),
b) x(t)=𝑒 −2𝑡 u(t), h(t) = u(t)
h(t) = u(t-1)
d) x(t)= cos4t u(t) + cos7t u(t),
h(t) = 𝛿(t-3)
Solution
a) Given that, x(t) = 𝛿(t) and h(t) = 𝑒 −𝑎𝑡 u(t)
1
𝑠+𝑎
………….....(2)
Response / Output, y(t) = x(t) * h(t)
On taking Laplace transform of the above equation we get,
L{y(t)} = L{x(t)*h(t)}
Using convolution theorem of Laplace transform
=X(s) H(s)
=1 x
1
𝑠+𝑎
=
1
𝑠+𝑎
Response / Output is given by inverse Laplace transform of the above equation.
Using equations (1) and (2)
∴ Response, y(t) = 𝐿−1 {
1
}
𝑠+𝑎
1
𝑠+𝑎
……………..(1)
Let, X(s) = L{x(t)} = L{𝛿(t)} = 1
H(s) = L{h(t)}= L{𝑒 −𝑎𝑡 u(t)} =
L{ 𝛿(t)= 1; L{𝑒 −𝑎𝑡 u(t)} =
= 𝑒 −𝑎𝑡 u(t)
Perform convolution of the following causal signals, using Laplace transform
b) 𝑥1 (t) = 𝑒 −2𝑡 u(t), 𝑥2 (t) = 𝑒 −5𝑡 u(t)
a) 𝑥1 (t) = 2 u(t), 𝑥2 (t) = u(t)
c) 𝑥1 (t) = t u(t), 𝑥2 (t) = 𝑒 −5𝑡 u(t)
Solution
d) 𝑥1 (t) = cos t u(t),
a) Given that, 𝑥1 (t) = 2 u(t)
𝑥2 (t) = t u(t)
L{t u(t)} =
1
𝑠2
𝑥2 (t) = u(t)
1
2
Let, 𝑋1 (s) = L{𝑥1 (t)} = L{2 u(t)} = 2 L{u(t)} = 2 x 𝑠 = 𝑠
𝑋2 (s) = L{𝑥2 (t)} = L{u(t)} =
1
𝑠
…………(1)
………….(2)
From convolution theorem of Laplace transform,
L{𝑥1 (t)*𝑥2 (t)} = 𝑋1 (s) 𝑋2 (s)
2 1
2
= 2
𝑠 𝑠
𝑠
2
𝐿−1 {𝑠2 } [[ =
= x
∴ 𝑥1 (t) * 𝑥2 (t) =
Using equations (1) and (2)
1
2 𝐿−1 {𝑠2}
= 2 x t u(t) = 2 t u(t)
1
L{u(t)} = 𝑠
b) Given that, 𝑥1 (t) =𝑒 −2𝑡 u(t)
𝑥2 (t) = 𝑒 −5𝑡 u(t)
Let , 𝑋1 (s) = L{𝑥1 (t)} = L{𝑒 −2𝑡 u(t)} =
1
L{𝑒 −𝑎𝑡 u(t)} = 𝑠+𝑎
1
𝑠+2
1
𝑠+5
…………………..(1)
𝑋2 (𝑠) = L{𝑥2 (t)} = L{𝑒 −5𝑡 u(t)} =
From convolution theorem of Laplace transform,
L{𝑥1 (t) * 𝑥2 (t)} = 𝑋1 (s) 𝑋2 (s)
1
1
1
= 𝑠+2 x 𝑠+5 = 𝑠+2 (𝑠+5)
∴ 𝑥1 (t) * 𝑥2 (t) = 𝐿−1 {
1
}
𝑠+2 (𝑠+5)
𝐴
1
𝑠+2 (𝑠+5)
x (s+5)⃒s= – 5
=
1
1
=
−2+5
3
1
1
=
−5+2
3
1 1
1 1
1
1
1
} = 𝐿−1 { } 3 𝑠+2
3 𝑠+5
3
𝑠+2
3
1 −2𝑡
1 −5𝑡
1
𝑒
u(t) - 𝑒
u(t) = (𝑒 −2𝑡 - 𝑒 −5𝑡 ) u(t)
3
3
3
∴ 𝑥1 (t) * 𝑥2 (t) =𝐿−1 {
=
𝐵
= 𝐿−1 {𝑠+2 + 𝑠+5}
Using partial fraction expansion technique
1
A = 𝑠+2 (𝑠+5) x (s+2)⃒s= - 2
=
B=
…………………(2)
𝐿−1 {
1
}
𝑠+5
Perform deconvolution operation to extract the signal 𝒙𝟏 (t)
a) 𝑥1 (t) * 𝑥2 (t) = 2t u(t)
c) 𝑥1 (t) * 𝑥2 (t) =
1
(𝑒 −5𝑡 +
25
1
3
b) 𝑥1 (t) * 𝑥2 (t) = (𝑒 −2𝑡 - 𝑒 −5𝑡 )u(t)
; 𝑥2 (t) = u(t)
5t -1)u(t) ; 𝑥2 (t) = 𝑒 −5𝑡 u(t)
Solution
a) Given that, 𝑥1 (t) * 𝑥2 (t) = 2t u(t)
Let, 𝑥1 (t) * 𝑥2 (t) = 𝑥3 (t)
∴ 𝑥3 (t) = 2t u(t)
;
𝑥2 (t) = 𝑒 −5𝑡 u(t)
d) 𝑥1 (t) * 𝑥2 (t) = u(t) – cos t u(t) ; 𝑥2 (t) = t u(t)
; 𝑥2 (t) = u(t)
…………………(1)
Let, 𝑋3 (s) = L{𝑥3 (t)} = L{2t u(t)} =
𝑋2 (s) = L{𝑥2 (t)} = L{u(t)} =
1
𝑠
2
𝑠2
………………….(2)
………………….(3)
On taking Laplace transform of equation (1) we get,
L{𝑥1 (t) * 𝑥2 (t)} = L{𝑥3 (t)}
Using convolution property of Laplace transform
∴ 𝑋1 (s) 𝑋2 (s) = 𝑋3 (s)
𝑋1 (s) =
𝑋3 (s)
𝑋2 (s)
= 𝑋3 (s) x
1
𝑋2 (s)
=
2
2
𝑠2
2
xs=𝑠
∴ 𝑥1 (t) = 𝐿−1 {𝑋1 (s)} = 𝐿−1 {𝑠 } = 2 u(t)
Using equations (2) and (3)
d) Given that, 𝑥1 (t) * 𝑥2 (t) = u(t) – cos t u(t) ; 𝑥2 (t) = t u(t)
…………………..(1)
Let, , 𝑥1 (t) * 𝑥2 (t) = 𝑥3 (t)
∴ 𝑥3 (t) = u(t) – cos t u(t)
1
Let, 𝑋3 (s) = L{𝑥3 (t)} = L {u(t) – cos t u(t)} = 𝑠 -
𝑠
𝑠 2 +1
1
……………………(3)
𝑋2 (s) = L{𝑥2 (t)} = L{t u(t)} = 𝑠2
On taking Laplace transform of equation (1) we get,
L{𝑥1 (t) * 𝑥2 (t)} = L{𝑥3 (t)}
Using convolution property of Laplace transform
𝑋1 (s) 𝑋2 (s) = 𝑋3 (s)
𝑋 (s)
∴ 𝑋1 (s) = 3 = 𝑋3 (s) x
𝑋2 (s)
Using equations (2) and (3)
=s-
𝑠3
𝑠 2 +1
=
∴ 𝑥1 (t) = 𝐿−1 {𝑋1 (s)} =
1
𝑋2 (s)
1
= [𝑠 -
𝑠 𝑠 2 +1 −𝑠3
𝑠 2 +1
𝑠
𝐿−1 {𝑠2 +1}
…………………..(2)
=
𝑠
]
𝑠 2 +1
x 𝑠2
𝑠 3 +𝑠 −𝑠 3
𝑠 2 +1
= cos t u(t)
𝑠
= 𝑠2+1
The impulse response of continuous time systems are given below. Determine the unit step response of the
systems using convolution theorem of Laplace transform,
b) h(t) = 𝑒 −5𝑡 u(t)
a) h(t) = 3t u(t)
d) h(t) = u(t-2)
c) h(t) = u(t-2)
e) h(t) = u(t+2) + u(t-2)
Solution
a) Given that, h(t) =3t u(t)
Let, H(s) = L{h(t)} = L{3t u(t)} = 3 L {t u(t)} = 3 x
U(s) = L{u(t)} =
1
𝑠
1
𝑠2
; where u(t) is unit step signal
=
3
𝑠2
……………….(1)
……………….(2)
Unit step response, s(t) = h(t) * u(t)
On taking Laplace transform of the above equation we get,
L{s(t)} = L{h(t) * u(t)}
Using convolution theorem of Laplace transform
= H(s) U(s)
3
1
= 2 x
=
𝑠
𝑠
3
𝑠3
Using equations (1) and (2)
Unit step response is given by inverse Laplace transform of the above equation.
3
𝑠
3
2
∴ Unit step response, s(t) = 𝐿−1 { 3 } = 𝐿−1 { x
3
2
= 𝐿−1 {
2
𝑠 2+1
3
2
} = 𝑡 2 u(t)
2
𝑠 2+1
}
L{𝑡 𝑛 u(t)} =
𝑛!
𝑠 𝑛+1
Block Diagrams
System Realization
b0 s M  b1s M 1  ...  bM
H ( s) 
s N  a1s N 1  ...  a N
• Realization is a synthesis problem, so there is no unique
way of realizing a system.
• A common realization of H(s) is using
• Integrator
• Scalar multiplier
• Adders
Direct Form I Realization
b0 s 3  b1s 2  b2 s  b3
H ( s)  3
s  a1s 2  a2 s  a3
Divide every term by s with
the highest order s3
b1 b2 b3
b0   2  3
s s
s
H (s) 
a1 a2 a3
1
 2  3
s
s
s

b1 b2 b3  
1

H ( s )   b0   2  3  
s s
s   1  a1  a2  a3


s
s 2 s3

X(s)
H1(s)
W(s)
H2(s)
Y(s)






Direct Form I Realization
X(s)
H1(s)
W(s)
H2(s)

b1 b2 b3  
1


H ( s )   b0   2  3 
s s
s   1  a1  a2  a3


s
s 2 s3

Y(s)






Direct Form II Realization


1

H ( s) 
 1  a1  a2  a3

s
s 2 s3

H2(s)
X(s)


b
b
b
  b0  1  22  33 
s s
s 



W(s)
H1(s)
Y(s)
Example
Find the canonic direct form realization of the following transfer
functions:
a)
b)
c)
d)
5
s7
s
s7
s5
s7
4 s  28
s 2  6s  5
Cascade and Parallel Realizations
4 s  28
H ( s)  2
s  6s  5
Cascade Realization
4 s  28
 4 s  28  1 
H (s) 



( s  1)( s  5)  s  1  s  5 
Parallel Realization
H ( s) 
4s  28
6
2


( s  1)( s  5) s  1 s  5
The complex poles in H(s) should be realized as a secondorder system.
FOURIER TRANSFORM
• Analysis of LTI Continuous Time System Using Fourier Transform:
• Transfer Function of LTI CTS in Frequency Domain
The ratio of Fourier transform of output and the Fourier Transform of input is called
transfer function of LTI continuous time system in frequency domain.
Let, x(t) = Input to the continuous time system
y(t) = Output to the continuous time system
∴ X(jΩ) = Fourier transform of x(t)
Y(jΩ) = Fourier transform of y(t)
𝒀(jΩ)
Now, Transfer function =
𝑿(jΩ)
IMPULSE RESPONSE AND TRANSFER FUNCTION
• For a continuous time system H, if the input is impulse signal δ(t) then
the output is called impulse response, which is denoted by h(t).
δ(t)
h(t)
H
• The response for any input to LTI system is given by convolution of
input and impulse response.
• Symbolically, the convolution operation is denoted as,
• y(t) = x(t) * h(t)
“*” is the symbol of convolution.
• Mathematically, the convolution operation is defined as,
• y(t) = x(t) * h(t) =
+∞
𝑥
−∞
𝜏 ℎ 𝑡 − 𝜏 𝑑𝜏
“𝜏” is the dummy variable for integration
• Now, by convolution property of Fourier transform we get,
• F{x(t) * h(t)} = X(jΩ)H(jΩ)
• F{y(t)} = X(jΩ)H(jΩ)
∴Y(jΩ) = X(jΩ)H(jΩ)
Y(jΩ)
∴H(jΩ) = X(jΩ)
• The transfer function in frequency domain is given by Fourier
transform of impulse response.
• Note: only a forced response alone can be computed via frequency
domain.
• Relation Between Fourier and Laplace Transform
• X(jΩ) = X(s)|s=jΩ
PROBLEM
• Determine the convolution of x1(t) = e-2t u(t) and x2(t) = e-6t u(t), using
fourier transform.
• Solution:
• Let X1(jΩ) = Fourier transform of x1(t)
X2(jΩ) = Fourier transform of x2(t)
By convolution property of Fourier transform
F{x1(t) * x2(t)} = X1(jΩ)X2(jΩ)
• Let X(jΩ) = X1(jΩ)X2(jΩ)
= F{e-2t u(t)} x F {e-6t u(t)}
1
=
jΩ +2
x
1
jΩ +6
• By partial fraction expansion technique X(jΩ) can be expressed as,
1
𝐾1
𝐾2
• X(jΩ) =
=
+
(jΩ +2)(jΩ +6) (jΩ +2) (jΩ +6)
1
1
1
• K1 =
x (jΩ + 2)| jΩ=-2 =
= = 0.25
−2+6 4
(jΩ +2)(jΩ +6)
1
1
1
• K2 =
x (jΩ + 6)| jΩ=-6 =
= - =- 0.25
−6+2
4
(jΩ +2)(jΩ +6)
0.25
∴X(jΩ) = jΩ +2 -
0.25
jΩ +6
• On taking inverse Fourier transform of the above equation we get,
• x(t) = 0.25e-2tu(t)-0.25e-6tu(t)
= 0.25(e-2t-e-6t)u(t)
The impulse response of an LTI system is h(t) = 2e-3tu(t). Find the response
of the system for the input x(t) = 2e-5tu(t), using Fourier transform.
• Solution
• Given that, x(t) = 2e-5tu(t)
∴ X(jΩ) = F{x(t)} = F{2e-5tu(t)} =
2
jΩ+5
……(1)
• Given that, h(t) = 2e-3tu(t)
∴ H(jΩ) = F{h(t)} = F{2e-3tu(t)} =
2
jΩ+3
For LTI system, the response, y(t) = x(t) * h(t)
On taking Fourier transform of equation (3) we get,
F{y(t)} = F{x(t)*h(t)}
Let F{y(t)} = Y(jΩ)
Y(jΩ) = F{x(t)*h(t)}
= X(jΩ)H(jΩ)
2
2
4
=
𝑥
=
𝑗Ω+5
𝑗Ω+3
(𝑗Ω+5)(𝑗Ω+3)
.……(2)
……..(3)
• By partial fraction expansion technique X(jΩ) can be expressed as,
1
𝐾1
𝐾2
• Y(jΩ) =
=
+
(jΩ +5)(jΩ +3) (jΩ +5) (jΩ +3)
1
• K1 =
x (jΩ + 5)| jΩ=-5 = -2
(jΩ +5)(jΩ +3)
1
• K2 =
x (jΩ + 3)| jΩ=-3 = 2
(jΩ +5)(jΩ +3)
∴ Y(jΩ) =
2
jΩ +5
+
2
jΩ +3
• On taking inverse Fourier transform of the above equation we get,
• x(t) = -2e-5tu(t)+2e-3tu(t)
= 2(e-3t-e-5t)u(t)