Unit 2
Signal system
1. What is an LSI system?
An LSI system 
is a system that has two essential properties:
- Linearity:
+ 
= a 
+ b 
, that is, it obeys linear superposition.
2. Shift-invariance: this means that if we shift the input in time (or shift the entries in a vector) then the output is shifted by the same amount.
Mathematically, we can say that if 
= 
shift invariance means that
= y(t+τ)
These two properties are independent:
Example
f(x(t)) = x(t)2 is shift-invariant but not linear, and matrix multiplication by an arbitary matrix is linear but (typically) not shift-invariant.
2.What is an impulse response?
If the input to the system is unit impulse response, then the output of the system is known as impulse response of the system denoted by h(n). That is
h(n) = T [ δ(n)]…………………………………………………………..(1)
The arbitrary sequence x(n) can be represented as weighted sum of discrete impulses given by
x(n) = 
δ(n-k) --------------------------------------------(2)
Then the output
y(n) = T [ 
δ(n-k)] -----------------------------------------------------(3)
3. What is step response?
Figure. Impulse response and step response
The step response of a discrete-time LTI system is the convolution of the unit step with the impulse response-
s[n]=u[n]*h[n]. -------------------(1)
Commutative property of convolution,
s[n]=h[n]*u[n]. ----------------------------------------(2)
That means s[n] is the response to the input h[n] of a discrete-time LTI system with unit impulse response u[n].
Figure. Output response
s[n] = 
u[n-k]…………………………………(1)
Since u [n- k]< 0 is for n- k< 0, i.e. k> n and 1 for n -k> 0, i.e. k≤ n.
s[n] = 
------------------------------------------------(2)
That is the step response of the discrete LTI system is the running sum of its impulse response.
s[n-1] = 
s[n] - s[n-1] = 
- = 
s[n] - s[n-1] = 
+ h[n] - = 
h[n] = s[n] – s[n-1]
From here h[n] can be recovered from s[n] , the impulse response of discrete-time LTI system is the first difference of its step response.
For continuous time system
The unit step response is the running integral of its impulse response.
s(t) = 
) dτ
The unit impulse response is the first derivative of the unit step response: -
h(t)= ds(t)/dt = s’(t)
4. Find the step response of the system whose impulse response is given by
u(t+1) – u(t-1)
s(t) = 
) dτ
s(t) = 
(t+1) – u(t-1) dτ
= 
τ - 
τ
= τ |-1 t – τ | 1 t
=(t+1) – (t-1)
=2
e -3t u(t) – e-2t u(t)
s(t) = 
) dτ
s(t) = 
-3t dτ + 
-2t dτ
s(t) = -1/3 [e -3t -1+1/2 [e -2t -1 ]
= -1/3 e -3t +1/3 +1/2 e -2t -1/2
=-1/3 e -3t +1/2 e -2t -1/2
= -1/3 e -3t +1/2 e -2t -1/6 for t
0
x(t) = -1/3 e -3t u(t)+1/2 e -2t u(t)-1/6 u(t)
5. What is convolution? Derive the expression ?
The input signal to the system is x(n) and the output is y(n) .
x(n) y(n) = T[x(n)]
If the input to the system is unit impulse then the output of the system is known as impulse response of the system denoted by h(n).
h(n) = T [ ∂(n) ]
The arbitrary sequence x(n) can be represented aa a weighted sum of discrete impulses given by
x(n) = 
∂(n-k)
Then the output
y(n) = T [ 
∂(n-k)]
Using linear property of the system , interchange the system operator T with the summation x(k) to obtain
y(n) = 
T[ ∂(n-k)]
Now
y(n) = 
h(n,k)
For a time-invariant system
h(n,k) = h(n-k)
Therefore
y(n) = 
h(n-k)
Thus the output of the LTI system is given by the weighted sum of the time-shifted responses.
This sum is termed as convolution sum represented as
y(n) = x(n) * h(n) where * denotes convolution.
6.The impulse response h[n] of a discrete-time LTI system. Determine and sketch the output y[n] of this system to the input x[n]
h[n] = 
[n] + 
[n-1] + 
[n-3] - 
[n-4] - 
[n-5]
x[n] = 
[n-2] - 
[n-4]
x[n] * h[n] = x[n] * {
[n] + 
[n-1] + 
[n-3] - 
[n-4] - 
[n-5]}
=x[n] + x[n-1] + x[n-2] + x[n-3] -x[n-4] – x[n-5]
y[n] = 
[n-2] - 
[n-4] + 
[n-3] - 
[n-5] + 
[n-4] - 
[n-6] + 
[n-5] - 
[n-7] –
[n-6] + 
[n-8] - 
[n-7] + 
[n-9]
= 
[n-2] + 
[n-3] -2 
[n-6] -2 
[n-6] - 
[n-7] + 
[n-8]+ 
[n-9]
y[n] ={ 0,0,1,1,0,0,-2,-2,1,1}
7.Explain the stability of LSI system?
A system is said to be stable if a bounded input sequence produces bounded output sequence. For bounded input sequence x(n) if the output is unbounded the system is said to be unstable system.
Let x(n) be a bounded input sequence satisfying |x(n) | ≤ Mx < ∞. Let h(n) be the impulse response of the system then the output y(n) can be found by convolution sum . That is
y(n) = 
h(n-k) = = 
x(n-k)----------------------(1)
The magnitude of the output is given by
|y(n)| = | 
x(n-k)|-------------------------(2)
We know that the magnitude of the sum of terms is less than or equal to the sum of the magnitudes, hence
|y(n)| = | 
x(n-k)|≤ 
|| x(n-k)| --------------------(3)
Let the bounded value of the input be equal to M then eq(3) can be written as
|y(n) | ≤ M 
|------------------------------(4)
The above condition is satisfied when
| < ∞ ------------------------(5)
Therefore, the necessary and sufficient condition for stability is
| < ∞ -----------------------(6)
Therefore, the LTI system is stable if its impulse response is absolutely summable.
8. Explain the casuality of LSI system?
If x(t) = ∂(t) then according to convolution theorem
y(t) = x(t) * h(t) = h(t) * ∂(t) = h(t)
A casual system is one where the output at the present instant does not anticipate input from future instants.
X(t) , t<to ------- y(t) , t<to
An LTI system is casual if and only if
h(t) = L (δ(t)) =0 , t<0
The impulse response of the system has to be zero for negative time for the system to be casual.
For discrete time system the impulse response of the sequence h[n] of LTI system has to be right -sided sequence
h[n] = L {δ(n)} =0, n<0
9. For the given impulse response h(n) of the system determine if the system are casual or stable.
h(n) = 2 n u(-n)
The given system is non casual since h(n) ≠0 for n<0. For stability the impulse response must be absolutely summable.
< ∞
n u(-n) = 
n = 
(1/2) n = 1+1/2 +………………. = 1/ 1-1/2 = 2
Hence system is stable.
10. An LTI system is specified by equation
d 2 y(t)/ dt 2 + 5 d y(t) /dt + 6 y(t) = dx(t) / dt + 4 x(t)
The input is x(t) = e -t u(t)
Find the natural response for initial conditions :
y(0+) = 3 d y(0+)/dt =0
For forced response and total response.
Given d 2 y(t) / dt 2 + 5 d y(t)/dt + 6 y(t) = d x(t)/dt + 4 x(t)
The natural response can be obtained by equating input terms in differential equation to zero.
d 2 y(t) / dt 2 + 5 d y(t)/dt + 6 y(t)=0
The characteristic equation is
- 2 + 5
+ 6=0
- 1=-2 and
2 = -3
The homogenous solution is
Yh(t) = c1 e -2t + c2 e -3t
y(0+) = c1 + c2
d y(0+)/dt = 2 c1 e -2t + (-3) c2 e -3t |t=0
= 2c1 – 3c2
From initial conditions
y(0+) = 3
Dy(0+)/dt =0
Comparing we get c1 + c2 =3
2c1 + 3c2 =0.
c2 =-6 and c1=9
Yn(t) = 9 e -2t – 6 e -3t
Forced response :
The homogeneous solution is given by
Yn(t) = c1 e -2t + c2 e -3t
For input x(t) = e -t
The particular solution is of the form
Yp(t) = k e -t
From which
Dyp(t)/dt = -k e-t
d2 yp(t)/dt2 = k e-t
Therefore,
d 2 yp(t)/dt2 + 5 d yp(t)/dt + 6 yp(t) = dx(t)/dt + 4 x(t)
k e -t – 5 k e-t + 6 k -t = - e-t + 4 e-t
2k = 3
K=1.5
Yp(t) = 1.5 e -t.
The forced response is the sum of homogeneous solution for particular solution.
Yf(t) = c1 e -2t + c2 e -3t + 1.5 e -t
y(0+) = c1 + c2 +1.5
Dy(0+)/dt = -2c1 e -2t -3 c2 e -3t – 1.5 e -t |t=0
To obtain forced response equate initial conditions to zero
C1 + C2 = -1.5
2C1 + 3C2 = -1.5
Solving for c1 and c2 we get
C2 = 1.5
C1 =-3
y(t) = yn(t) + yf(t)
Substituting yn(t) and yf(t) we get
y(t) = 6 e -2t – 4.5 e -3t + 1.5 e -t
y(0) = c1 + c2 + 1.5 =3
Dy(0+)/dt = -2c1 -3c2 -1.5 =0
c1 =6 and c2 =-4.5
y(t) = 6 e -2t -4.5 e -3t + 1.5 e -t
11. Consider the difference equation given by
y[n] + 0.5 y[n-1] = (-0.8) n u[n]
The solution is composed of natural response and forced response of the system.
y[n]= yh[n] + yp[n]
Where the particular solution satisfies for n≥0 and the homogeneous solution satisfies
Yh[n] + 0.5 yh[n-1] = 0
For n ≥0 yp(n) = y(-0.8) n
Then we get
Y(-0.8) n + 0.5Y(-0.8) n-1 = (-0.8) n
Y[ 1+0.5(-0.8) -1] = 1
Y=8/3
Which yields yp(n) = 8/3 (-0.8) n
To determine yh(n)
Yh[n] = A z n
1+ 0.5 z -1 =0
Z=-0.5
y[n]= yh[n] + yp[n]
= A (-0.5) n + 8/3 (-0.8) n
y[n] = -0.5 y[n-1] + (-0.8 ) n u[n]
n=0
y[0] = -0.5 y[-1] + (-0.8) 0
= 0+1=1
y[0] = 1 = A(-0.5) 0 + 8/3 (-0.8 ) 0
A + 8/3 =1
A= -5/3
y[n] = -5/3 (-0.5) n + 8/3 (-0.8) n
12.Explain the frequency response and its relation to impulse response?
The relationship between the impulse response and the frequency response is one of the foundations of signal processing. A system frequency response is the Fourier transform of its impulse response.
Since h[ ] is the common symbol for the impulse response, H[ ] is used for the frequency response. Systems are described in the time domain by convolution, that is: x[n] ∗ h[n] = y[n].
In the frequency domain, the input spectrum is multiplied by the frequency response, resulting in the output spectrum. As an equation: X[f] × H[f] = Y[f]. In other words, convolution in the time domain corresponds to multiplication in the frequency domain.
Figure. System operation in time and frequency domain
Unit 2
Signal system
1. What is an LSI system?
An LSI system 
is a system that has two essential properties:
- Linearity:
+ 
= a 
+ b 
, that is, it obeys linear superposition.
2. Shift-invariance: this means that if we shift the input in time (or shift the entries in a vector) then the output is shifted by the same amount.
Mathematically, we can say that if 
= 
shift invariance means that
= y(t+τ)
These two properties are independent:
Example
f(x(t)) = x(t)2 is shift-invariant but not linear, and matrix multiplication by an arbitary matrix is linear but (typically) not shift-invariant.
2.What is an impulse response?
If the input to the system is unit impulse response, then the output of the system is known as impulse response of the system denoted by h(n). That is
h(n) = T [ δ(n)]…………………………………………………………..(1)
The arbitrary sequence x(n) can be represented as weighted sum of discrete impulses given by
x(n) = 
δ(n-k) --------------------------------------------(2)
Then the output
y(n) = T [ 
δ(n-k)] -----------------------------------------------------(3)
3. What is step response?
Figure. Impulse response and step response
The step response of a discrete-time LTI system is the convolution of the unit step with the impulse response-
s[n]=u[n]*h[n]. -------------------(1)
Commutative property of convolution,
s[n]=h[n]*u[n]. ----------------------------------------(2)
That means s[n] is the response to the input h[n] of a discrete-time LTI system with unit impulse response u[n].
Figure. Output response
s[n] = 
u[n-k]…………………………………(1)
Since u [n- k]< 0 is for n- k< 0, i.e. k> n and 1 for n -k> 0, i.e. k≤ n.
s[n] = 
------------------------------------------------(2)
That is the step response of the discrete LTI system is the running sum of its impulse response.
s[n-1] = 
s[n] - s[n-1] = 
- = 
s[n] - s[n-1] = 
+ h[n] - = 
h[n] = s[n] – s[n-1]
From here h[n] can be recovered from s[n] , the impulse response of discrete-time LTI system is the first difference of its step response.
For continuous time system
The unit step response is the running integral of its impulse response.
s(t) = 
) dτ
The unit impulse response is the first derivative of the unit step response: -
h(t)= ds(t)/dt = s’(t)
4. Find the step response of the system whose impulse response is given by
u(t+1) – u(t-1)
s(t) = 
) dτ
s(t) = 
(t+1) – u(t-1) dτ
= 
τ - 
τ
= τ |-1 t – τ | 1 t
=(t+1) – (t-1)
=2
e -3t u(t) – e-2t u(t)
s(t) = 
) dτ
s(t) = 
-3t dτ + 
-2t dτ
s(t) = -1/3 [e -3t -1+1/2 [e -2t -1 ]
= -1/3 e -3t +1/3 +1/2 e -2t -1/2
=-1/3 e -3t +1/2 e -2t -1/2
= -1/3 e -3t +1/2 e -2t -1/6 for t
0
x(t) = -1/3 e -3t u(t)+1/2 e -2t u(t)-1/6 u(t)
5. What is convolution? Derive the expression ?
The input signal to the system is x(n) and the output is y(n) .
x(n) y(n) = T[x(n)]
If the input to the system is unit impulse then the output of the system is known as impulse response of the system denoted by h(n).
h(n) = T [ ∂(n) ]
The arbitrary sequence x(n) can be represented aa a weighted sum of discrete impulses given by
x(n) = 
∂(n-k)
Then the output
y(n) = T [ 
∂(n-k)]
Using linear property of the system , interchange the system operator T with the summation x(k) to obtain
y(n) = 
T[ ∂(n-k)]
Now
y(n) = 
h(n,k)
For a time-invariant system
h(n,k) = h(n-k)
Therefore
y(n) = 
h(n-k)
Thus the output of the LTI system is given by the weighted sum of the time-shifted responses.
This sum is termed as convolution sum represented as
y(n) = x(n) * h(n) where * denotes convolution.
6.The impulse response h[n] of a discrete-time LTI system. Determine and sketch the output y[n] of this system to the input x[n]
h[n] = 
[n] + 
[n-1] + 
[n-3] - 
[n-4] - 
[n-5]
x[n] = 
[n-2] - 
[n-4]
x[n] * h[n] = x[n] * {
[n] + 
[n-1] + 
[n-3] - 
[n-4] - 
[n-5]}
=x[n] + x[n-1] + x[n-2] + x[n-3] -x[n-4] – x[n-5]
y[n] = 
[n-2] - 
[n-4] + 
[n-3] - 
[n-5] + 
[n-4] - 
[n-6] + 
[n-5] - 
[n-7] –
[n-6] + 
[n-8] - 
[n-7] + 
[n-9]
= 
[n-2] + 
[n-3] -2 
[n-6] -2 
[n-6] - 
[n-7] + 
[n-8]+ 
[n-9]
y[n] ={ 0,0,1,1,0,0,-2,-2,1,1}
7.Explain the stability of LSI system?
A system is said to be stable if a bounded input sequence produces bounded output sequence. For bounded input sequence x(n) if the output is unbounded the system is said to be unstable system.
Let x(n) be a bounded input sequence satisfying |x(n) | ≤ Mx < ∞. Let h(n) be the impulse response of the system then the output y(n) can be found by convolution sum . That is
y(n) = 
h(n-k) = = 
x(n-k)----------------------(1)
The magnitude of the output is given by
|y(n)| = | 
x(n-k)|-------------------------(2)
We know that the magnitude of the sum of terms is less than or equal to the sum of the magnitudes, hence
|y(n)| = | 
x(n-k)|≤ 
|| x(n-k)| --------------------(3)
Let the bounded value of the input be equal to M then eq(3) can be written as
|y(n) | ≤ M 
|------------------------------(4)
The above condition is satisfied when
| < ∞ ------------------------(5)
Therefore, the necessary and sufficient condition for stability is
| < ∞ -----------------------(6)
Therefore, the LTI system is stable if its impulse response is absolutely summable.
8. Explain the casuality of LSI system?
If x(t) = ∂(t) then according to convolution theorem
y(t) = x(t) * h(t) = h(t) * ∂(t) = h(t)
A casual system is one where the output at the present instant does not anticipate input from future instants.
X(t) , t<to ------- y(t) , t<to
An LTI system is casual if and only if
h(t) = L (δ(t)) =0 , t<0
The impulse response of the system has to be zero for negative time for the system to be casual.
For discrete time system the impulse response of the sequence h[n] of LTI system has to be right -sided sequence
h[n] = L {δ(n)} =0, n<0
9. For the given impulse response h(n) of the system determine if the system are casual or stable.
h(n) = 2 n u(-n)
The given system is non casual since h(n) ≠0 for n<0. For stability the impulse response must be absolutely summable.
< ∞
n u(-n) = 
n = 
(1/2) n = 1+1/2 +………………. = 1/ 1-1/2 = 2
Hence system is stable.
10. An LTI system is specified by equation
d 2 y(t)/ dt 2 + 5 d y(t) /dt + 6 y(t) = dx(t) / dt + 4 x(t)
The input is x(t) = e -t u(t)
Find the natural response for initial conditions :
y(0+) = 3 d y(0+)/dt =0
For forced response and total response.
Given d 2 y(t) / dt 2 + 5 d y(t)/dt + 6 y(t) = d x(t)/dt + 4 x(t)
The natural response can be obtained by equating input terms in differential equation to zero.
d 2 y(t) / dt 2 + 5 d y(t)/dt + 6 y(t)=0
The characteristic equation is
- 2 + 5
+ 6=0
- 1=-2 and
2 = -3
The homogenous solution is
Yh(t) = c1 e -2t + c2 e -3t
y(0+) = c1 + c2
d y(0+)/dt = 2 c1 e -2t + (-3) c2 e -3t |t=0
= 2c1 – 3c2
From initial conditions
y(0+) = 3
Dy(0+)/dt =0
Comparing we get c1 + c2 =3
2c1 + 3c2 =0.
c2 =-6 and c1=9
Yn(t) = 9 e -2t – 6 e -3t
Forced response :
The homogeneous solution is given by
Yn(t) = c1 e -2t + c2 e -3t
For input x(t) = e -t
The particular solution is of the form
Yp(t) = k e -t
From which
Dyp(t)/dt = -k e-t
d2 yp(t)/dt2 = k e-t
Therefore,
d 2 yp(t)/dt2 + 5 d yp(t)/dt + 6 yp(t) = dx(t)/dt + 4 x(t)
k e -t – 5 k e-t + 6 k -t = - e-t + 4 e-t
2k = 3
K=1.5
Yp(t) = 1.5 e -t.
The forced response is the sum of homogeneous solution for particular solution.
Yf(t) = c1 e -2t + c2 e -3t + 1.5 e -t
y(0+) = c1 + c2 +1.5
Dy(0+)/dt = -2c1 e -2t -3 c2 e -3t – 1.5 e -t |t=0
To obtain forced response equate initial conditions to zero
C1 + C2 = -1.5
2C1 + 3C2 = -1.5
Solving for c1 and c2 we get
C2 = 1.5
C1 =-3
y(t) = yn(t) + yf(t)
Substituting yn(t) and yf(t) we get
y(t) = 6 e -2t – 4.5 e -3t + 1.5 e -t
y(0) = c1 + c2 + 1.5 =3
Dy(0+)/dt = -2c1 -3c2 -1.5 =0
c1 =6 and c2 =-4.5
y(t) = 6 e -2t -4.5 e -3t + 1.5 e -t
11. Consider the difference equation given by
y[n] + 0.5 y[n-1] = (-0.8) n u[n]
The solution is composed of natural response and forced response of the system.
y[n]= yh[n] + yp[n]
Where the particular solution satisfies for n≥0 and the homogeneous solution satisfies
Yh[n] + 0.5 yh[n-1] = 0
For n ≥0 yp(n) = y(-0.8) n
Then we get
Y(-0.8) n + 0.5Y(-0.8) n-1 = (-0.8) n
Y[ 1+0.5(-0.8) -1] = 1
Y=8/3
Which yields yp(n) = 8/3 (-0.8) n
To determine yh(n)
Yh[n] = A z n
1+ 0.5 z -1 =0
Z=-0.5
y[n]= yh[n] + yp[n]
= A (-0.5) n + 8/3 (-0.8) n
y[n] = -0.5 y[n-1] + (-0.8 ) n u[n]
n=0
y[0] = -0.5 y[-1] + (-0.8) 0
= 0+1=1
y[0] = 1 = A(-0.5) 0 + 8/3 (-0.8 ) 0
A + 8/3 =1
A= -5/3
y[n] = -5/3 (-0.5) n + 8/3 (-0.8) n
12.Explain the frequency response and its relation to impulse response?
The relationship between the impulse response and the frequency response is one of the foundations of signal processing. A system frequency response is the Fourier transform of its impulse response.
Since h[ ] is the common symbol for the impulse response, H[ ] is used for the frequency response. Systems are described in the time domain by convolution, that is: x[n] ∗ h[n] = y[n].
In the frequency domain, the input spectrum is multiplied by the frequency response, resulting in the output spectrum. As an equation: X[f] × H[f] = Y[f]. In other words, convolution in the time domain corresponds to multiplication in the frequency domain.
Figure. System operation in time and frequency domain
